for Intermediate Algebra with Applications, 7e by
Aufmann/Barker/Lockwood
Exclusively from Cengage Learning, Enhanced WebAssign® offers an extensive online program for Intermediate Algebra with Applications, Seventh Edition, by Aufmann/Barker/Lockwood, to encourage the practice that’s so critical for concept mastery. The meticulously crafted pedagogy and exercises in Intermediate Algebra with Applications, Seventh Edition, become even more effective in Enhanced WebAssign, supplemented by multimedia tutorial support and immediate feedback as students complete their assignments.
Read It selections
Watch It videos
Practice It mini-tutorials Master It tutorials
New! Personal Study Plans Available for all new Developmental Mathematics titles and editions, including Intermediate Algebra with Applications, Seventh Edition, by Aufmann/Barker/Lockwood, the Personal Study Plan in Enhanced WebAssign provides chapter and section assessments that show students what material they know and what areas require more work. For items that they answer incorrectly, students can click on links to related study resources such as videos, tutorials, or reading materials. Color-coded progress indicators let students see how well they are doing on different topics. Instructors decide what chapters and sections to include and whether to include the plan as part of the final grade or as a study guide with no scoring involved.
Other Key Features
Online content for each of Cengage Learning’s Developmental Mathematics texts includes some or all of these features: • Read It eBook pages, Watch It videos, Practice It mini-tutorials, Master It tutorials, and Chat About It links • New! Premium eBook with highlighting, note-taking, and search features as well as links to multimedia resources • Algorithmic problems based on up to 2,000 text exercises (and more than 3,500 exercises in selected texts) • Complete solutions to selected algorithmic problems versus answers only (viewable to students at the instructor’s discretion) • New! MathPad makes it easy for students to enter mathematical symbols into their answers • GraphPad enables students to graph lines, segments, parabolas, and circles as they answer questions Screenshots shown here are for illustrative purposes only, and may include content from a variety of mathematics courses.
For more information and sample assignments: www.webassign.net/cengage Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
eCompanion for
Intermediate Algebra with Applications Seventh Edition
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
eCompanion for
Intermediate Algebra with Applications Seventh Edition
Richard N. Aufmann Palomar College
Vernon C. Barker Palomar College
Joanne S. Lockwood Nashua Community College
Australia • Brazil • Japan • Korea • Mexico • Singapore • Spain • United Kingdom • United States
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
This is an electronic version of the print textbook. Due to electronic rights restrictions, some third party content may be suppressed. Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. The publisher reserves the right to remove content from this title at any time if subsequent rights restrictions require it. For valuable information on pricing, previous editions, changes to current editions, and alternate formats, please visit www.cengage.com/highered to search by ISBN#, author, title, or keyword for materials in your areas of interest.
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
eCompanion for Intermediate Algebra with Applications, Seventh Edition Richard N. Aufmann, Vernon C. Barker, Joanne S. Lockwood Acquisitions Editor: Marc Bove Developmental Editor: Erin Brown Assistant Editor: Shaun Williams
© 2011 Brooks/Cole, Cengage Learning ALL RIGHTS RESERVED. No part of this work covered by the copyright herein may be reproduced, transmitted, stored, or used in any form or by any means graphic, electronic, or mechanical, including but not limited to photocopying, recording, scanning, digitizing, taping, Web distribution, information networks, or information storage and retrieval systems, except as permitted under Section 107 or 108 of the 1976 United States Copyright Act, without the prior written permission of the publisher.
Editorial Assistant: Kyle O’Loughlin Media Editor: Heleny Wong Marketing Manager: Gordon Lee Marketing Assistant: Erica O’Connell Marketing Communications Manager: Mary Anne Payumo Content Project Manager: Cheryll Linthicum Art Director: Vernon Boes Print Buyer: Karen Hunt Rights Acquisitions Account Manager, Text: Roberta Broyer Rights Acquisitions Account Manager, Image: Don Schlotman Production Service: Scratchgravel Publishing Services Text Designer: Scratchgravel Publishing Services Photo Researcher: Joshua Brown Copy Editor: Scratchgravel Publishing Services Cover Designer: Irene Morris Cover Image: John Foxx/Getty Images Compositor: Scratchgravel Publishing Services
For product information and technology assistance, contact us at Cengage Learning Customer & Sales Support, 1-800-354-9706. For permission to use material from this text or product, submit all requests online at www.cengage.com/permissions. Further permissions questions can be emailed to
[email protected].
Library of Congress Control Number: 2010926979 ISBN-13: 978-0-8400-5428-9 ISBN-10: 0-8400-5428-9 Brooks/Cole 20 Davis Drive Belmont, CA 94002-3098 USA Cengage Learning is a leading provider of customized learning solutions with office locations around the globe, including Singapore, the United Kingdom, Australia, Mexico, Brazil, and Japan. Locate your local office at: www.cengage.com/global. Cengage Learning products are represented in Canada by Nelson Education, Ltd. To learn more about Brooks/Cole, visit www.cengage.com/brookscole. Purchase any of our products at your local college store or at our preferred online store www.CengageBrain.com.
Printed in the United States of America 1 2 3 4 5 6 7 8 14 13 12 11 10
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Intro The eCompanion for the text Intermediate Algebra with Applications, Seventh Edition, is an abridged, print version of the corresponding eBook. The eCompanion contains selected definitions, examples, and exercises from the eBook. The eCompanion may be used in conjunction with Enhanced WebAssign for Intermediate Algebra with Applications, Seventh Edition, which includes the eBook for Intermediate Algebra with Applications, Seventh Edition. As you work through the contents of the eCompanion, you can reference the eBook for additional instruction and support, as needed. If you have an account with Enhanced WebAssign, you can access the eBook in two ways. If you purchase the eBook, you can access it through the “My eBooks” link in the Student View of Enhanced WebAssign. The eBook as it appears in “My eBooks” contains links to video instruction and tutorials and includes the full text of Intermediate Algebra with Applications, Seventh Edition. If you do not purchase the eBook, but you have an account with Enhanced WebAssign, you can access individual sections of the eBook through the “Read It” buttons that accompany questions. The eBook as it appears through the “Read It” buttons does not contain links to video instruction and tutorials; it is a static version of the textbook. If you would like more practice, then additional problems and answers are available in Enhanced WebAssign. The problems are organized by section, and you can select problems to practice in Enhanced WebAssign that are similar to those found in the corresponding sections of the eCompanion. For more information and access to Enhanced WebAssign, talk to your instructor or visit www.webassign.net/cengage.
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
v
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Cont
Chapter 1 Review of Real Numbers 1
1.1 section 1.2 section 1.3 section 1.4 section
Introduction to Real Numbers 2 Operations on Rational Numbers 5 Variable Expressions 10 Verbal Expressions and Variable Expressions 12
Chapter 2 First-Degree Equations and Inequalities 17
2.1 section 2.2 section 2.3 section 2.4 section 2.5 section 2.6 section
Equations in One Variable 18 Coin, Stamp, and Integer Problems 21 Value Mixture and Motion Problems 23 Applications: Problems Involving Percent 26 Inequalities in One Variable 30 Absolute Value Equations and Inequalities 34
Chapter 3 Linear Functions and Inequalities in Two Variables 37
3.1 section 3.2 section 3.3 section 3.4 section 3.5 section 3.6 section 3.7 section
The Rectangular Coordinate System 38 Introduction to Functions 42 Linear Functions 45 Slope of a Straight Line 48 Finding Equations of Lines 51 Parallel and Perpendicular Lines 53 Inequalities in Two Variables 56
Chapter 4 Systems of Equations and Inequalities 59
4.1 Solving Systems of Linear Equations by Graphing and by the Substitution Method 60 section 4.2 Solving Systems of Linear Equations by the Addition Method 62 section 4.3 Solving Systems of Equations by Using Determinants and by Using Matrices 66 section 4.4 Application Problems 76 section 4.5 Solving Systems of Linear Inequalities 79 section
Chapter 5 Polynomials and Exponents 81
5.1 section 5.2 section 5.3 section 5.4 section 5.5 section 5.6 section 5.7 section
Exponential Expressions 82 Introduction to Polynomials 86 Multiplication of Polynomials 88 Division of Polynomials 91 Factoring Polynomials 95 Special Factoring 99 Solving Equations by Factoring 103
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
vii
Chapter 6 Rational Expressions 105 s 106 ons 108
i
6.3 section 6.4 section 6.5 section 6.6 section
Complex Fractions 111 Rational Equations 112 Proportions and Variation 116 Literal Equations 120
Chapter 7 Rational Exponents and Radicals 121 7.1 section 7.2 section 7.3 section 7.4 section 7.5 section
Rational Exponents and Radical Expressions 122 Operations on Radical Expressions 125 Radical Functions 130 Solving Equations Containing Radical Expressions 132 Complex Numbers 135
Chapter 8 Quadratic Equations and Inequalities 141 8.1 Solving Quadratic Equations by Factoring or by Taking Square Roots 142 8.2 Solving Quadratic Equations by Completing the Square and by Using the Quadratic Formula 144 section 8.3 Equations That Are Reducible to Quadratic Equations 147 section 8.4 Applications of Quadratic Equations 150 section 8.5 Nonlinear Inequalities 152 section 8.6 Properties of Quadratic Functions 154 section 8.7 Applications of Quadratic Functions 156 section section
Chapter 9 Functions and Relations 159 9.1 Translations of Graphs 160 section 9.2 Algebra of Functions 161 section 9.3 One-to-One and Inverse Functions 163 section
Chapter 10 Exponential and Logarithmic Functions 167 10.1 10.2 section 10.3 section 10.4 section 10.5 section section
Exponential Functions 168 Introduction to Logarithms 170 Graphs of Logarithmic Functions 174 Exponential and Logarithmic Equations 175 Applications of Exponential and Logarithmic Functions 178
Chapter 11 Sequences and Series 181 11.1 section 11.2 section 11.3 section 11.4 section
Introduction to Sequences and Series 182 Arithmetic Sequences and Series 184 Geometric Sequences and Series 187 Binomial Expansions 192
Chapter 12 Conic Sections 195 12.1 The Parabola 196 section 12.2 The Circle 198 section
viii Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
section i i
12.3 The Ellipse and Hyperbola 200 ms of Equations 203 d Systems of Inequalities 205
Answers 207
Important Concepts Cards (in back of book)
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
ix
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
� AIM f
Started Welcome to the eCompanion for Intermediate Algebra with Applications, Seventh Edition! Students come to intermediate algebra with varied backgrounds and different experiences in learning math. We are committed to your success in learning mathematics and have developed many tools and resources to support you along the way. Want to excel in this course? Read on to learn the skills you’ll need and how best to use this book to get the results you want.
�
Take Note
Motivation alone won’t lead to success. For example, suppose a person who cannot swim is rowed out to the middle of a lake and thrown overboard. That person has a lot of motivation to swim, but will most likely drown without some help. You’ll need motivation and learning in order to succeed.
Make the Commitment
You’ll nd many real-life problems in this book, relating to sports, money, cars, music, and more. We hope that these topics will help you understand how you will use mathematics in your real life. However, to learn all of the necessary skills and how you can apply them to your life outside this course, you need to stay motivated. THINK ABOUT WHY YOU WANT TO SUCCEED IN THIS COURSE. LIST THE REASONS HERE (NOT IN YOUR HEAD . . . ON THE PAPER!):
We also know that this course may be a requirement for you to graduate or complete your major. That’s okay. If you have a goal for the future, such as becoming a nurse or a teacher, you will need to succeed in mathematics rst. Picture yourself where you want to be, and use this image to stay on track. Stay committed to success! With practice, you will improve your math skills. Skeptical? Think about when you rst learned to ride a bike or drive a car. You probably felt selfconscious and worried that you might fail. But with time and practice, it became second nature to you. You will also need to put in the time and practice to do well in mathematics. Think of us as your “driving” instructors. We’ll lead you along the path to success, but we need you to stay focused and energized along the way.
© Cengage Learning/Photodisc
Motivate Yourself
© Cengage Learning/Photodisc
LIST A SITUATION IN WHICH YOU ACCOMPLISHED YOUR GOAL BY SPENDING TIME PRACTICING AND PERFECTING YOUR SKILLS (SUCH AS LEARNING TO PLAY THE PIANO OR PLAY BASKETBALL):
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
xi
The attendance policy will tell you: • How many classes you can miss without a penalty • What to do if you miss an exam or quiz • Whether you can get the lecture notes from the professor if you miss a class
�
Take Note
When planning your schedule, give some thought to how much time you realistically have available each week. For example, if you work 40 hours a week, take 15 units, spend the recommended study time given at the right, and sleep 8 hours a day, you will use over 80% of the available hours in a week. That leaves less than 20% of the hours in a week for family, friends, eating, recreation, and other activities. Visit http://college. cengage.com/masterstudent/ shared/content/time_chart/ chart.html and use the Interactive Time Chart to see how you’re spending your time—you may be surprised.
Ready, Set, Succeed!
On the rst day of class, your instructor will hand out a syllabus k of this syllabus as your personal roadmap cs you need to learn) and the dates you need to arrive at those destinations (dates by which you need to learn the topics). Learning mathematics is a journey. But, to get the most out of this course, you’ll need to know what the important stops are and what skills you’ll need to learn for your arrival at those stops.
GET THE BASICS
Flip open to the table of contents and look at it next to your syllabus. Identify when your major exams are and what material you’ll need to learn by those dates. For example, if you know you have an exam in the second month of the semester, how many chapters of this text will you need to learn by then? What homework do you have to do during this time? Managing this important information will help keep you on track for success. We know how busy you are outside of school. Do you have a full-time or a part-time job? Do you have children? Visit your family often? Play basketball or write for the school newspaper? It can be stressful to balance all of the important activities and responsibilities in your life. Making a time management plan will help you create a schedule that gives you enough time for everything you need to do. MANAGE YOUR TIME
Let’s get started! Create a weekly schedule.
© Cengage Learning/Photodisc
Take Note
First, list all of your responsibilities that take up certain set hours during the week. Be sure to include: � � �
each class you are taking time you spend at work any other commitments (child care, tutoring, volunteering, etc.)
Then, list all of your responsibilities that are more exible. Remember to make time for:
•
STUDYING You’lleach needother. to study to succeed, but luckily you get to choose Practice teaching what times work We’ve found that best you for canyou. learnKeep in mind: what you know to spend twice as much time studying as they instructors ask students •a lotMostabout when have(3tohours explain it to � 6 hours of study). do you in class of class someone else. • Try studying in chunks. We’ve found it works better to study an hour each day, rather studying for 6 hours on one day. classthan notes. Couldn’t • Compare lastbeve minutes Studyingthecan even more helpful if you’re able to do it right after your •understand of class? Missedwhen classthe because class meets, material is fresh in your mind. wereEating sick? well Chances � you MEALS gives are you energy and stamina for attending classes and someone in your group has the studying. notes for the topics you missed. � ENTERTAINMENT It’s impossible to stay focused on your responsibilities 100% of the test time.questions. Giving yourself a break for entertainment will reduce your • Brainstorm stress and help keep on track. Agree on what topics you’ll talk about and how • Make a plan for youryoumeeting. � EXERCISE to overall health. at your long you’ll beExercise meeting. contributes When you make a plan, you’llYou’ll be surend thatyou’re you make the most productive when you have both a healthy mind and a healthy body. of your meeting. �
© Cengage Learning/Photodisc
�
Take a look at your syllabus to see if your instructor has an attendance policy that is part of your overall grade in the course.
It takes hard work and commitment to succeed, but we know you can do it! Doing well in mathematics is just one step you’ll take along the path to success. We are condent that if you follow our suggestions, you will succeed. Good luck!
xii Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
eCompanion for
Intermediate Algebra with Applications Seventh Edition
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Chapter 1 Image copyright Hnatenko Oleksandr, 2010. Used under license from Shutterstock.com.
Review of Real Numbers
Prep Test For Exercises 1 to 8, add, subtract, multiply, or divide.
1.
5 7 + 12 30
2.
8 7 − 15 20
3.
5 6
4.
4 2 ÷ 15 5
5. 8 + 29.34 + 7.065
6. 92 − 18.37
7. 2.19(3.4)
8. 32.436 ÷ 0.6
9. Which of the following numbers are greater than –8? a. –6 b. –10 c. 0
⋅
4 15
d. 8
10. Match each fraction with its decimal equivalent. 1 2 7 b. 10 3 c. 4 89 d. 100 a.
A. 0.75 B. 0.89 C. 0.5 D. 0.7
1 Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
se c t io n
1.1 Introduction to Real Numbers
Objective
A set is a collection of objects. The objects are called elements of the set. Sets are denoted by placing braces around the elements in the set. It is convenient to use a variable to represent, or stand for, any one of the elements of a set. For instance, the statement “x is an element of the set {0, 2, 4, 6}” means that x can be replaced by 0, 2, 4, or 6. The set {0, 2, 4, 6} is called the domain of the variable. The symbol for “is an element of” is ∈; the symbol for “is not an element of ” is ∉. For example, 2 ∈ {0, 2, 4, 6} and 7 ∉ {0, 2, 4, 6}. Definition of Inequality Symbols If a and b are two real numbers and a is to the left of b on the number line, then a is less than b. This is written a < b. If a and b are two real numbers and a is to the right of b on the number line, then a is greater than b. This is written a > b.
Example 1
Let y = {−7, 0, 6}. For which values of y is the inequality y < 4 a true statement?
Solution y < 4 −7 < 4
True
0 < 4
True
6 < 4
False
Step
• Replace y by each element of the set and determine whether the inequality is true.
The inequality is true for –7 and 0. The absolute value of a number is a measure of its distance from zero on the number line. The symbol for absolute value is
.
Absolute Value The absolute value of a positive number is the number itself. The absolute value of a negative number is the opposite of the negative number. The absolute value of zero is zero.
Example 2
Solution
Let z ∈ {−11, 0, 8}. Evaluate z for each element of the set.
z −11 = 11
0 = 0
8 = 8
Step
• Replace z by each element of the set and evaluate.
2 Chapter 1 Review of Real Numbers Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Exercises Solve. 1. Let x ∈ {−3, 0, ∈ {−4, 0, 4}. For which values of z is x < 5 true? is z > −2 true? 3. Let w ∈ {−2, −1, 0, 1}. For which values of 4. Let p ∈ {−10, −5, 0, 5}. For which values of w is w ≤ −1 true? p is p ≥ 0 true? 5. Let q ∈ {−3, 0, 7}. Evaluate q for each 6. Let m ∈ {−6, −2, 0, 1, 4}. Evaluate − m for each element of the set. element of the set.
Objective 2 Set operations and interval notation
The roster method of writing a set encloses a list of the elements of the set in braces.
Example 3
Solution
Use the roster method to write the set of natural numbers less than 10. {1, 2, 3, 4, 5, 6, 7, 8, 9} A second method of representing a set is set-builder notation. Set-builder notation can be used to describe almost any set, but it is especially useful when writing infinite sets. In set-builder notation, the set of integers greater than –3 is written {x | x > −3, x ∈ integers} and is read “the set of all x such that x is greater than –3 and x is an element of the integers.”
Example 4
Solution
Use set-builder notation to write the set of real numbers less than 7. { x | x < 7, x ∈ real numbers} The union of two sets, written A ∪ B, is the set of all elements that belong to either A or B. In set-builder notation, this is written A ∪ B = {x | x ∈ A or x ∈ B} The intersection of two sets, written A ∩ B, is the set of all elements that are common to both A and B. In set-builder notation, this is written A ∩ B = {x | x ∈ A and x ∈ B}
Example 5
Solution
Find C ∪ D and C ∩ D given C = {1, 5, 9, 13, 17} and D = {3, 5, 7, 9, 11} . C ∪ D = {1, 3, 5, 7, 9, 11, 13, 17} C ∩ D = {5, 9} The graph of {x | x > −2, x ∈ real numbers} is shown below. The set is the real numbers greater than –2. The parenthesis on the graph indicates that –2 is not included in the set. −5 −4 −3 −2 −1
0
1
2
3
4
5
The graph of {x | x ≥ −2, x ∈ real numbers} is shown below. The set is the real numbers greater than or equal to –2. The bracket at –2 indicates that –2 is included in the set. −5 −4 −3 −2 −1
0
1
2
3
4
5
For the remainder of this section, all variables will represent real numbers. Using this convention, the above set is written {x | x ≥ −2}.
Section 1.1 Introduction to Real Numbers 3
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example 6
Graph: {x | x ≤ 3}
Soluti Some sets can also be expressed using interval notation. For example, the interval notation (–3, 2] indicates the interval of all real numbers greater than –3 and less than or equal to 2. An interval is said to be closed if it includes both endpoints; it is open if it does not include either endpoint. An interval is half-open if one endpoint is included and the other is not. In each example given below, –3 and 2 are the endpoints of the interval. In each case, the set notation, the interval notation, and the graph of the set are shown.
Example 7
Solution
{x | −3 < x < 2}
(–3, 2) Open interval
−5 −4 −3 −2 −1
0
1
2
3
4
5
{x | −3 ≤ x ≤ 2}
[–3, 2] Closed interval
−5 −4 −3 −2 −1
0
1
2
3
4
5
{x | −3 ≤ x < 2}
[–3, 2) Half-open interval
−5 −4 −3 −2 −1
0
1
2
3
4
5
{x | −3 < x ≤ 2}
(–3, 2] Half-open interval
−5 −4 −3 −2 −1
0
1
2
3
4
5
Write {x | 0 < x ≤ 5} in interval notation. (0, 5]
Step
• This is the set of real numbers greater than 0 and less than or equal to 5.
To indicate an interval that extends forever in one or both directions using interval notation, we use the infinity symbol ∞ or the negative infinity symbol -∞. {x | x > 1}
(1, ∞)
{x | x ≥ 1}
[1, ∞)
{x | x < 1}
(-∞, 1)
{x | x ≤ 1}
(-∞, 1]
{x | −∞ < x < ∞}
(-∞, ∞)
−5 −4 −3 −2 −1
0
1
2
3
4
5
−5 −4 −3 −2 −1
0
1
2
3
4
5
−5 −4 −3 −2 −1
0
1
2
3
4
5
−5 −4 −3 −2 −1
0
1
2
3
4
5
−5 −4 −3 −2 −1
0
1
2
3
4
5
Note that a parenthesis is used with ∞ or -∞.
Example 8
Solution
Write (-∞, 9] using set-builder notation. { x | x ≤ 9}
Exercises Use the roster method to write the set. 7. The integers between –3 and 5
8. The integers between –4 and 0
Use set-builder notation to write the set. 9. The real numbers greater than or equal to -2
10. The real numbers less than or equal to 2
4 Chapter 1 Review of Real Numbers Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Find A ∪ B and A ∩ B. 11. A = {6, 12, 18
−4, −2, 0, 2, 4}, B = {0, 4, 8}
13. A = {2, 3, 4, 5, 8}, B = {9, 10}
14. A = {2, 4}, B = {0, 1, 2, 3, 4, 5}
Graph. 15. {x | −1 < x < 5}
16. {x | −1 ≤ x ≤ 1}
17. {x | x < 2}
18. {x | x ≥ 1}
19. (–2, 5)
20. [–3, 2]
21. (-∞, -1)
22. [3, ∞)
25. [5, ∞)
26. [-2, ∞)
29. {x | x < 1}
30. {x | x ≥ 3}
Write each interval in set-builder notation. 23. (0, 8)
24. [–5, 7]
Write each set of real numbers in interval notation. 27. {x | −2 < x < 4}
se c t io n
28. {x | −2 ≤ x < 6}
1.2 Operations on Rational Numbers
Objective 1 Operations on integers
Sign Rules for Addition Numbers that have the same sign To add numbers with the same sign, add the absolute values of the numbers. Then attach the sign of the addends. Numbers that have different signs To add numbers with different signs, find the absolute value of each number. Subtract the smaller of these absolute values from the larger. Then attach the sign of the number with the larger absolute value.
Example 1
Add. A. –65 + (–48) B. 27 + (–53)
Solution
A. The signs are the same. Add the absolute values of the numbers. Then attach the sign of the addends.
−65 + ( −48) = −113
B. The signs are different.
27 + ( −53)
Find the absolute value of each number.
27 = 27
Subtract the smaller absolute value from the larger.
53 − 27 = 26
Because −53 > 27 , attach the sign of –53.
27 + ( −53) = −26
−53 = 53
Section 1.2 Operations on Rational Numbers 5
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Subtraction of Real Numbers = a + ( − b).
Example 2
Subtract. A. 48 – (–22) B. –31 – 18
Solution
A. Write the subtraction as the addition of the opposite number.
= 70
Add the numbers.
B. Write the subtraction as the addition of the opposite number.
48 − ( −22) = 48 + 22
−31 − 18 = −31 + ( −18)
= −49
Then add.
Sign Rules for Multiplication Numbers that have the same sign The product of two numbers with the same sign is positive. Numbers that have different signs The product of two numbers with different signs is negative.
Example 3
Solution
Multiply. A. –4(–9) B. 84(–4) A. The product of two numbers with the same sign is positive. B. The product of two numbers with different signs is negative.
−4( −9) = 36 84( −4) = −336
The sign rules for dividing are the same as the sign rules for multiplying.
Example 4
Solution
Divide. A.
−54 B. ( −21) ÷ ( −7 ) 9
−54 = −6 9
A. The quotient of two numbers with different signs is negative. B. The quotient of two numbers with the same sign is positive.
( −21) ÷ ( −7) = 3
Exercises Simplify. 1. –18 + (–12)
2. –18 – 7
3. 5 – 22
4. 16(–60)
5. 3 ⋅ 4( −8)
6. 18 ÷ ( −3)
7. −60 ÷ ( −12)
8. (–9)(–2)(–3)(10)
6 Chapter 1 Review of Real Numbers Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Objective 2 Operations on rational numbers
s written as fractions, first rewrite the fractions as denominator. A common denominator is the product of the denominators of the fractions. The lowest common denominator (LCD) is the least common multiple (LCM) of the denominators.
Example 5
Solution
Add:
5 7 + − 6 8
5 7 5 = + − 6 8 6
⋅4 4
−7 + 8
=
20 −21 + 24 24
=
20 + ( −21) 24
= –
Example 6
Solution
Subtract: − −
3
⋅ 3
Steps
• The LCD is 24. Write each fraction in terms of the common denominator.
• Add the numerators, and place the sum over the common denominator.
1 24
3 2 − − 4 3
3 2 3 3 2 − − = − ⋅ − − 4 3 4 3 3
= −
= −
4
⋅ 4
8 9 − − 12 12
Steps
9 − ( −8) −9 + 8 = 12 12 1 = – 12
• The LCD is 12. Write each fraction in terms of the common denominator. • Subtract the numerators, and place the difference over the common denominator.
The product of two fractions is the product of the numerators over the product of the denominators. a b
Example 7
Solution
Multiply: − −
⋅
c ac = d bd
5 8 ⋅ 12 15
5 8 ⋅ = − 5⋅8 12 15 12 ⋅ 15
Steps
• The product is negative. Multiply the numerators, and multiply the denominators.
• Write the answer in simplest form.
40 180
= −
= −
20 ⋅ 2 20 ⋅ 9
= −
2 9
Section 1.2 Operations on Rational Numbers 7
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
To divide two fractions, multiply the first fraction by the reciprocal of the divisor.
⋅d c
Example 8
Solution
Divide:
3 9 ÷ − 8 16
3 9 3 16 ÷ − = ⋅ − 8 16 8 9
= −
3 ⋅ 16 8⋅9
= −
2 3
Steps
• Multiply by the reciprocal of the divisor.
• Write the answer in simplest form.
Exercises Perform the indicated operation. 9. −
4 2 + 5 3
10. −
7 6 13. − − 8 35
14.
1 5 + − 2 6
5 7 − 6 8
11.
1 3 − 5 4
12. −
5 1 − − 6 4
15.
5 3 ÷ − 8 4
16. −
14 7 ÷ − 12 15
Objective 3 Exponential expressions
An exponent indicates how many times the factor, called the base, occurs in the multiplication. The multiplication 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 is in factored form. The exponential expression 25 is in exponential form. Play close attention to the negative sign in parts A and B in the next example.
Example 9
Solution
( )
4
Simplify: A. −6 B. −64 A. ( −6)4 = ( −6)( −6)( −6)( −6) = 1296 B. −64 = −(6 ⋅ 6 ⋅ 6 ⋅ 6) = –1296
Exercises Simplify. 17. 53
18. −23
19. ( −5)3
20. ( −8)2
21. −34
22. 22 ⋅ 34
23. −22 ⋅ 32
24. ( −2)2 ⋅ 32
8 Chapter 1 Review of Real Numbers Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Objective 4 The Order of Operations Agreement
ment Step 1 Perform operations inside grouping symbols. Grouping symbols include parentheses ( ), brackets [ ], the absolute value symbol , and the fraction bar. Step 2 Simplify exponential expressions. Step 3 Do multiplication and division as they occur from left to right. Step 4 Do addition and subtraction as they occur from left to right.
Example 10
Solution
Simplify: 8 − 8−
17 − 2 3 ⋅2 4+1
17 − 2 3 ⋅2 4+1
= 8−
15 3 ⋅2 5
Steps
• Perform operations above and below the fraction bar.
= 8−
15 ⋅8 5
• Simplify exponential expressions.
= 8 − 3⋅8
• Do multiplication and division as they occur from left to right.
= 8 − 24
= −16
• Do addition and subtraction as they occur from left to right.
Exercises Simplify. 25. 5 − 3(8 ÷ 4)2
4(5 − 2) 28. 2 ÷ 4 4 − 22
26. 42 − (5 − 2)2 ⋅ 3
29.
3+ 11 16
2 3
27. 6 − 32 (5 − 2 ⋅ 3)
8 − 2 1 30. 16 − 4 ÷ 3 − 6 2
Section 1.2 Operations on Rational Numbers 9
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
se c t io n
1.3 Variable Expressions
Objective
The Properties of the Real Numbers describe the way operations on numbers can be performed. Following is a list of some of the Properties of the Real Numbers. The Commutative Property of Addition
The Commutative Property of Multiplication
a+b = b+a
a⋅b = b⋅a
The Associative Property of Addition
The Associative Property of Multiplication
( a + b) + c = a + ( b + c )
( a ⋅ b) ⋅ c = a ⋅ ( b ⋅ c )
The Addition Property of Zero
The Multiplication Property of Zero
a+0 = 0+a = a
a⋅0 = 0⋅a = 0
The Multiplication Property of One
The Inverse Property of Addition
a ⋅1 = 1⋅ a = a
a + ( − a) = ( − a) + a = 0
The Inverse Property of Multiplication
The Distributive Property a( b + c ) = ab + ac
1 1 a ⋅ = ⋅ a = 1, a ≠ 0 a a
Example 1
Complete the statement by using the Inverse Property of Addition.
3x + ? = 0
Solution
3x + (–3x) = 0
Example 2
Complete the statement by using the Commutative Property of Multiplication.
1 x = ? x 4
Solution
1 1 x = x 4 4
Example 3
() ()
Identify the property that justifies the statement.
3( x + 4) = 3x + 12
The Distributive Property
Solution
10 Chapter 1 Review of Real Numbers Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Exercises Use the Properties of Real Numbers to complete the statement.
Pillaged by LibraryPirate
1. Commutative Property of Multiplication 2. The Associative Property of Addition 3 ⋅ 4 = 4 ⋅ ? (3 + 4) + 5 = ? + ( 4 + 5)
3. The Distributive Property 4. The Multiplication Property of One 3( x + 2) = 3x + ? ? ⋅ x = x
Identify the property that justifies the statement. 1 6. ( −12) − = 1 12
5. −8 + 8 = 0
7. 6( x + y ) = 6 x + 6 y
Objective 2 Evaluate variable expressions
Replacing the variable in a variable expression by a numerical value and then simplifying the resulting expression is called evaluating the variable expression.
Example 4
Solution
Evaluate a 2 − ( ab − c ) when a = −2, b = 3, and c = −4. a 2 − ( ab − c ) ( −2)2 − [( −2)(3) − ( −4)]
= ( −2)2 − [ −6 − ( −4)]
= ( −2)2 − [ −2]
= 4 − [ −2]
= 6
Steps
• Replace each variable in the expression with its value. • Use the Order of Operations Agreement to simplify the resulting numerical expression.
Exercises Evaluate the variable expression when a = 2, b = 3, c = −1, and d = −4. 8. b2 − ( d − c )2
9. ( b − 2 a )2 + c
10. − d 3 + 4 ac
Objective 3 Simplify variable expressions
Like terms of a variable expression are terms with the same variable part. Constant terms are like terms.
To combine like terms, use the Distributive Property ba + ca = ( b + c ) a to add the coefficients.
Like terms 4x
–
5
+
7x2
+
3x
–
9
Like terms
3x + 2 x = (3 + 2) x = 5x
Section 1.3 Variable Expressions 11
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example 5
Simplify: 2( x + y ) + 3( y − 3x )
Soluti
Steps
= (2 x − 9 x ) + (2 y + 3 y )
= −7 x + 5 y
Example 6
Solution
• Use the Distributive Property to remove parentheses. • Use the Commutative and Associative Properties of Addition to rearrange and group like terms. • Combine like terms.
Simplify: 4 y − 2[ x − 3( x + y ) − 5 y ] 4 y − 2[ x − 3( x + y ) − 5 y ]
= 4 y − 2[ x − 3x − 3 y − 5 y]
Steps
= 4 y − 2[ −2 x − 8 y ]
= 4 y + 4 x + 16 y
= 4 x + 20 y
• Use the Distributive Property to remove parentheses. • Combine like terms. • Use the Distributive Property to remove brackets. • Combine like terms.
Exercises Simplify. 11. −2 a − 3(3a − 7 )
12. 3x − 2(5x − 7 )
13. 2 x − 3( x − 2 y )
14. −2( x − 3 y ) + 2(3 y − 5x )
15. 5(3a − 2b) − 3( −6 a + 5b)
16. 3x − 2[ y − 2( x + 3[2 x + 3 y ])]
17. 4 − 2(7 x − 2 y ) − 3( −2 x + 3 y )
18. 2 x − 4[ x − 4( y − 2[5 y + 3])]
19. 3x + 8( x − 4) − 3( 2 x − y )
se c t io n
1.4 Verbal Expressions and Variable Expressions
Objective 1 Translate a verbal expression into a variable expression One of the major skills required in applied mathematics is translating a verbal expression into a mathematical expression. Doing so requires recognizing the verbal phrases that translate into mathematical operations. Following is a partial list of the verbal phrases used to indicate the different mathematical operations. Addition
more than
8 more than w
w+8
added to
x added to 9
9+x
the sum of
the sum of z and 9
z+9
the total of
the total of r and s
r+s
increased by
x increased by 7
x+7
12 Chapter 1 Review of Real Numbers Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Subtraction
less than ween
Solution
the difference between x and 1
x–1
minus
z minus 7
z–7
decreased by
17 decreased by a
17 – a
Multiplication
times
negative 2 times c
–2c
the product of
the product of x and y
xy
multiplied by
3 multiplied by n
3n
of
three-fourths of m
3 m 4
twice
twice d
2d
Division
divided by
v divided by 15
v 15
the quotient of
the quotient of y and 3
y 3
ratio
the ratio of x to 7
x 7
Power
the square of or the second power of
the square of x
x2
the cube of or the third power of
the cube of r
r3
the fifth power of
the fifth power of a
a5
Example 1
b – 12
12 less than b
Translate and simplify “the total of five times a number and twice the difference between the number and three.” the unknown number: n
Steps
five times the number: 5n
the difference between the number and three: n – 3
twice the difference between the number and three: 2(n – 3)
5n + 2( n − 3)
= 5n + 2 n − 6
= 7n − 6
Example 2
Solution
• Assign a variable to one of the unknown quantities. • Use the assigned variable to write an expression for any other unknown quantity.
• Use the assigned variable to write the variable expression. • Simplify the variable expression.
Translate and simplify “a number decreased by the difference between eight and twice the number.” the unknown number: n
the difference between 8 and twice the unknown number: 8 – 2n
n − (8 − 2 n)
= n − 8 + 2n
= 3n – 8 Section 1.4 Verbal Expressions and Variable Expressions 13
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Exercises Translate into a variable expr 1. a number minus the sum 2. five times the product of eight and a number 3. the difference between the product of seventeen and a number and twice the number 4. the sum of five times a number and twelve added to the product of fifteen and the number 5. a number decreased by the difference between five and the number 6. the difference between the square of a number and the total of twelve and the square of the number
Objective 2 Application problems
Many of the applications of mathematics require that you identify the unknown quantity, assign a variable to that quantity, and then attempt to express other unknowns in terms of that quantity.
Example 3
Solution
Example 4
Solution
Example 5
Solution
A cyclist is riding at a rate that is twice the speed of a runner. Express the speed of the cyclist in terms of the speed of the runner. the speed of the runner: r the speed of the cyclist: 2r A mixture of candy contains 3 lb more of milk chocolate than of caramel. Express the amount of milk chocolate in the mixture in terms of the amount of caramel in the mixture. the number of pounds of caramel: C There are three more pounds of chocolate than caramel: C + 3 The length of a rectangle is 2 ft more than 3 times the width. Express the length of the rectangle in terms of the width. the width of the rectangle: W The length is 2 more than 3 times the width: 3W + 2
Exercises 7. The distance from Earth to the sun is approximately 390 times the distance from Earth to the moon. Express the distance from Earth to the sun in terms of the distance from Earth to the moon. 8. Heart disease is responsible for seven times as many deaths each year as are accidents. Express the number of deaths caused each year by heart disease in terms of the number of deaths each year caused by accidents.
14 Chapter 1 Review of Real Numbers Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
9. A financial advisor has invested $10,000 in two accounts. If one account contains x dollars, express the amount in the se 10. A fishing line 3 ft long is cut into two pieces, one shorter than the other. Express the length of the shorter piece in terms of the length of the longer piece. 11. The measure of angle A of a triangle is twice the measure of angle B. The measure of angle C is twice the measure of angle A. Write expressions for angle A and angle C in terms of angle B. 12. The length of a rectangle is three more than twice the width. Express the length of the rectangle in terms of the width.
Section 1.4 Verbal Expressions and Variable Expressions 15
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Chapter 2
Image copyright Anyka, 2010. Used under license from Shutterstock.com.
First-Degree Equations and Inequalities
Prep Test For Exercises 1 to 5, add, subtract, multiply, or divide.
1. 8 – 12
3.
5 4 5. − 8 5
−18 −6
2. -9 + 3
3 4 4. − − 4 3
For Exercises 6 to 9, simplify.
6. 3x − 5 + 7 x
7. 6( x − 2) + 3
8. n + ( n + 2) + ( n + 4)
9. 0.08 x + 0.05( 400 − x )
10. Twenty ounces of a snack mixture contains nuts and pretzels. Let n represent the number of ounces of nuts in the mixture. Express the number of ounces of pretzels in the mixture in terms of n.
17 Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
se c t io n
2.1 Equations in One Variable
Objective
and Multiplication Properties of Equations An equation expresses the equality of two mathematical expressions. The expressions can be either numerical or variable expressions.
2 + 8 = 10 x + 8 = 11 Equations x 2 + 2 y = 7
The replacement values of the variable that will make an equation true are called the roots, or solutions, of the equation. The solution of the equation x + 2 = 5 is 3. Solving an equation means finding a solution of the equation. The simplest equation to solve is an equation of the form variable = constant, because the constant is the solution. The Addition Property of Equations can be used to rewrite an equation in this form. The Addition Property of Equations If a, b, and c are algebraic expressions, then the equation a = b has the same solutions as the equation a + c = b + c. The Addition Property of Equations states that the same quantity can be added to each side of an equation without changing the solution of the equation.
Example 1
Solve: x − 3 = 7
Solution
x−3= 7
x −3+3 = 7 +3 x = 10
Steps
• Add the opposite of the constant term –3 to each side of the equation. • Simplify.
The solution is 10. The Multiplication Property of Equations If a, b, and c are algebraic expressions, and c ≠ 0, then the equation a = b has the same solutions as the equation ac = bc. The Multiplication Property of Equations states that we can multiply each side of an equation by the same nonzero number without changing the solution of the equation.
Example 2
Solve: −
Solution
3 x = 12 4
4 3 4 − − x = − (12) 3 4 3
−
3 x = 12 4
x = −16
Steps
• Multiply each side of the equation 4 3 by − , the reciprocal of − . 3 4
• Simplify.
The solution is –16. Because division is defined in terms of multiplication, the Multiplication Property of Equations enables us to divide each side of an equation by the same nonzero number without changing the solution of the equation.
18 Chapter 2 First-Degree Equations and Inequalities Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example 3
Solve: −3x = 12
−3x 12 = −3 −3
Steps
x = −4
• Divide each side of the equation by –3, the coefficient of x. • Simplify.
The solution is –4. When solving an equation, it is often necessary to apply both the Addition and the Multiplication Properties of Equations.
Example 4
Solve: 6 x − 8 = 3x + 7
Solution
6 x − 8 = 3x + 7
6 x − 3x − 8 = 3x − 3x + 7
Steps
3x − 8 = 7
3x − 8 + 8 = 7 + 8
3x = 15 3x 15 = 3 3 x = 5
• Subtract 3x from each side of the equation. • Simplify.
• Add 8 to each side of the equation.
• Simplify.
• Divide each side of the equation by 3.
• Simplify.
The solution is 5.
Exercises Solve. 1. a + 3 = −7
2. −12 = x − 3
3. 8 x = 4
5 4 4. − x = 8 5
5. 3x + 5x = 12
6. 7 x − 9 = 3 − 4 x
7. 2 − 3t = 3t − 4
8. 7 − 5t = 2t − 9
9. 3a − 5a = 8a + 4
Objective 2 Solving equations using the Distributive Property
When an equation contains parentheses, one of the steps in solving the equation requires the use of the Distributive Property.
Example 5
Solution
Solve: 5( 2 x − 7 ) + 2 = 3( 4 − x ) − 12 5( 2 x − 7 ) + 2 = 3( 4 − x ) − 12
10 x − 35 + 2 = 12 − 3x − 12
−33 = −13x
33 = x 13 33 The solution is . 13
10 x − 33 = −3x
Steps
• Use the Distributive Property.
• Simplify.
• Subtract 10x from each side of the equation. • Divide each side of the equation by –13.
Section 2.1 Equations in One Variable 19
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
To solve an equation that contains fractions, first clear denominators by multiplying each ltiple (LCM) of the denominators.
Example 6
Solve:
12
−
9
=
2
3x − 2 x x − = 12 9 2
Solution
Steps
• The LCM of 12, 9, and 2 is 36.
• Multiply each side of the equation by the LCM.
36(3x − 2) 36 x 36 x − = 12 9 2
• Use the Distributive Property.
3(3x − 2) − 4 x = 18 x
• Simplify.
9 x − 6 − 4 x = 18 x
5x − 6 = 18 x
−6 = 13x
3x − 2 x x 36 − = 36 9 12 2
6 = x 13 6 The solution is − . 13
−
Exercises Solve. 10. 2 x + 2( x + 1) = 10
11. 5( 2 − b) = −3( b − 3)
12. 3 − 2( y − 3) = 4 y − 7
13. 3( y − 5) − 5 y = 2 y + 9
14. −4(7 y − 1) + 5 y = −2(3 y + 4) − 3 y
15. 5 − 6[2t − 2( 2t + 3)] = 8 − t
Objective 3 Application problems
Solving application problems is primarily a skill in translating sentences into equations and then solving the equations. An equation states that two mathematical expressions are equal. Therefore, to translate a sentence into an equation requires recognizing the words or phrases that mean equals. These phrases include “is,” “is equal to,” “amounts to,” and “represents.”
Example 7
The charges for a long-distance telephone call are $2.14 for the first three minutes and $.47 for each additional minute or fraction of a minute. If the charges for a longdistance call were $17.65, how many minutes did the phone call last?
Strategy
To find the length of the phone call in minutes, write and solve an equation using n to represent the total number of minutes of the call. Then n - 3 is the number of additional minutes after the first three minutes of the phone call. The fixed charge for the three minutes plus the charge for the additional minutes is the total cost of the phone call.
20 Chapter 2 First-Degree Equations and Inequalities Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Solution 2.14 + 0.47( n − 3) = 17.65
0.47 n = 16.92
n = 36
The phone call lasted 36 min.
Exercises 16. The Fahrenheit temperature is 59°. This is 32° more than Celsius temperature.
9 of the Celsius temperature. Find the 5
17. A local feed store sells a 100-pound bag of feed for $10.90. If a customer buys more than one bag, each additional bag costs $10.50. A customer bought $84.40 worth of feed. How many 100-pound bags of feed did this customer purchase?
se c t io n
2.2 Coin, Stamp, and Integer Problems
Objective 1 Coin and stamp problems
In solving problems dealing with coins and stamps of different values, it is necessary to represent the value of the coins or stamps in the same unit of money. The unit of money is frequently cents. For example,
Example 1
Strategy
The value of five 8¢ stamps is 5 ⋅ 8, or 40, cents.
The value of seven quarters is 7 ⋅ 25, or 175, cents. A coin bank contains $1.80 in nickels and dimes. In all, there are twenty-two coins in the bank. Find the number of nickels and the number of dimes in the bank. c Number of nickels: x
Number of dimes: 22 – x Number
Value
Total value
Nickel
x
5
5x
Dime
22 – x
10
10(22 – x)
c The sum of the total values of each denomination of coin equals the total value of all the coins (180 cents).
Solution 5 x + 10(22 − x ) = 180
5x + 220 − 10 x = 180
−5x + 220 = 180
−5x = −40
x = 8
Steps
• There are 8 nickels in the bank.
• Substitute the value of x into the variable expression for the number of dimes.
22 − x = 22 − 8 = 14
The bank contains 8 nickels and 14 dimes. Section 2.2 Coin, Stamp, and Integer Problems 21
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example 2
Strategy
A collection of stamps contains 3¢, 10¢, and 15¢ stamps. The number of 10¢ stamps stamps. There are three times as many 15¢ alue of the stamps is $1.56. Find the number c The number of 3¢ stamps: x
The number of 10¢ stamps: 2x + 2
The number of 15¢ stamps: 3x Number
Value
Total value
3¢ stamps
x
3
3x
10¢ stamps
2x + 2
10
10(2x + 2)
15¢ stamps
3x
15
45x
c The sum of the total values of the individual types of stamps equals the total value of all the stamps (156 cents).
3x + 10( 2 x + 2) + 45x = 156
Solution
3x + 20 x + 20 + 45x = 156
68 x + 20 = 156
68 x = 136
x = 2
3x = 3( 2) = 6
There are six 15¢ stamps in the collection.
Exercises Solve. 1. A collection of 56 coins has a value of $4.00. The collection contains only nickels and dimes. Find the number of dimes in the collection.
2. A collection of 22 coins has a value of $4.45. The collection contains dimes and quarters. Find the number of quarters in the collection.
3. A stamp collection consists of 3¢, 8¢, and 13¢ stamps. The number of 8¢ stamps is three less than twice the number of 3¢ stamps. The number of 13¢ stamps is twice the number of 8¢ stamps. The total value of all the stamps is $2.53. Find the number of 3¢ stamps in the collection.
Objective 2 Integer problems To solve integer problems, let a variable represent one of the integers. Express each of the other integers in terms of that variable. Remember that for consecutive integer problems, consecutive integers will differ by 1. Consecutive even or consecutive odd integers will differ by 2.
22 Chapter 2 First-Degree Equations and Inequalities Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example 3
The sum of three consecutive even integers is seventy-eight. Find the integers.
Third even integer: n + 4
c The sum of the three even integers is 78.
n + ( n + 2) + ( n + 4) = 78
Solution
3n + 6 = 78
3n = 72
n = 24
n + 2 = 24 + 2 = 26
n + 4 = 24 + 4 = 28
The three consecutive even integers are 24, 26, and 28.
Exercises 4. The sum of two integers is ten. Three times the larger integer is three less than eight times the smaller integer. Find the integers.
5. The sum of three consecutive integers is negative fifty-seven. Find the integers.
6. Find three consecutive odd integers such that three times the middle integer is seven more than the sum of the first and third integers.
se c t io n
2.3 Value Mixture and Motion Problems
Objective 1 Value mixture problems
A solution of a value mixture problem is based on the equation AC = V, where A is the amount of the ingredient, C is the cost per unit of the ingredient, and V is the value of the ingredient. The value of 12 lb of coffee costing $5.25 per pound is
V = AC
V = 12($5.25)
V = $63
Section 2.3 Value Mixture and Motion Problems 23
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example 1
How many ounces of a gold alloy that costs $320 an ounce must be mixed with 100 oz ke a mixture that costs $160 an ounce?
Strate Amount
Cost
Value
$320 alloy
x
320
320x
$100 alloy
100
100
100(100)
x + 100
160
160(x + 100)
Mixture
Solution
c The sum of the values before mixing equals the value after mixing. 320 x + 100(100) = 160( x + 100)
320 x + 10, 000 = 160 x + 16, 000 160 x + 10, 000 = 16, 000
160 x = 6000
x = 37.5
Example 2
Strategy
The mixture must contain 37.5 oz of the $320 gold alloy. A butcher combined hamburger that costs $4.00 per pound with hamburger that costs $2.80 per pound. How many pounds of each were used to make a 75-pound mixture that costs $3.20 per pound? c Pounds of $4.00 hamburger: x
Pounds of $2.80 hamburger: 75 – x Amount
Cost
Value
$4.00 hamburger
x
4.00
4.00x
$2.80 hamburger
75 – x
2.80
2.80(75 – x)
75
3.20
75(3.20)
Mixture
c The sum of the values before mixing equals the value after mixing.
4.00 x + 2.80(75 − x ) = 75(3.20)
Solution
4.00 x + 210 − 2.80 x = 240 1.2 x + 210 = 240
1.2 x = 30
x = 25
75 − x = 75 − 25 = 50
The mixture must contain 25 lb of the $4.00 hamburger and 50 lb of the $2.80 hamburger.
Exercises Solve. 1. A restaurant chef mixes 20 lb of snow peas costing $1.99 a pound with 14 lb of petite onions costing $1.19 a pound to make a vegetable medley for the evening meal. Find the cost per pound of the mixture.
24 Chapter 2 First-Degree Equations and Inequalities Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
2. A coffee merchant combines coffee costing $5.50 per pound with coffee costing $3.00 per pound. How many pounds of ng $4.00 per pound?
3. Find the cost per ounce of salad dressing mixture made from 64 oz of olive oil that costs $8.29 and 20 oz of vinegar that costs $1.99. Round to the nearest cent.
4. A silversmith combined pure silver that costs $5.20 an ounce with 50 oz of a silver alloy that costs $2.80 an ounce. How many ounces of the pure silver were used to make an alloy of silver that costs $4.40 an ounce?
Objective 2 Uniform motion problems
Any object that travels at a constant speed in a straight line is said to be in uniform motion. Uniform motion means that the speed and direction of an object do not change. For instance, a train traveling at a constant speed of 50 mph on a straight track is in uniform motion. The solution of a uniform motion problem is based on the equation d = rt , where d is the distance traveled, r is the rate of travel, and t is the time spent traveling. For instance, suppose a train travels for 2 h at an average speed of 45 mph. Because the time (2 h) and the rate (45 mph) are known, we can find the distance traveled by solving the equation d = rt for d. d = rt = 45(2) = 90
Example 3
Strategy
A long-distance runner started a course running at an average speed of 6 mph. One and one-half hours later, a cyclist traveled the same course at an average speed of 12 mph. How long after the runner started did the cyclist overtake the runner? c Unknown time for the cyclist: t
Rate
Time
Distance
Runner
6
t + 1.5
6(t + 1.5)
Cyclist
12
t
12t
c The runner and cyclist traveled the same distance.
6( t + 1.5) = 12 t
Solution
6t + 9 = 12t
9 = 6t
3 = t 2
t + 1.5 = 1.5 + 1.5
Time for the runner: t + 1.5
Steps
• The cyclist traveled for 1.5 h.
• Substitute the value of t into the variable expression for the runner’s time.
=3
The cyclist overtook the runner 3 h after the runner started.
Section 2.3 Value Mixture and Motion Problems 25
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example 4
Strategy
Two small planes start from the same point and fly in opposite directions. The first nd plane. In 4 h the planes are 1160 mi apart. c Rate of the second plane: r
Rate of the first plane: r + 30 First plane
Rate
Time
Distance
r + 30
4
4(r + 30)
r
4
4r
Second plane
c The total distance traveled by the two planes is 1160 mi.
Solution
4( r + 30) + 4r = 1160
4r + 120 + 4r = 1160
8r + 120 = 1160
8r = 1040
r = 130
r + 30 = 130 + 30 = 160
The first plane is traveling 160 mph.
The second plane is traveling 130 mph.
Exercises 5. It takes a student 30 min to drive from school to work, a distance of 20 mi. At what average rate of speed does the student drive? 6. At 9:00 A.M., Katrina starts down an exercise path and walks at a rate of 4 mph. At 9:30 A.M., Carla leaves from the same place and begins chasing Katrina, catching her at 10:00 A.M. Find Carla’s jogging rate. 7. A bicyclist traveling at 18 mph overtakes an in-line skater who is traveling at 10 mph and had a 0.5-hour head start. How far from the starting point does the bicyclist overtake the in-line skater? 8. Two planes start from the same point and fly in opposite directions. The first plane is flying 50 mph slower than the second plane. In 2.5 h, the planes are 1400 mi apart. Find the rate of each plane.
se c t io n
2.4 Applications: Problems Involving Percent
Objective 1 Investment problems The annual simple interest that an investment earns is given by the equation Pr = I , where P is the principal, or the amount invested, r is the simple interest rate, and I is the simple interest. The solution of an investment problem is based on this equation. The annual interest rate on a $3000 investment is 9%. The annual simple interest earned on the investment is
26 Chapter 2 First-Degree Equations and Inequalities Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
I = Pr
.09)
Example 1
Strategy
An investment of $4000 is made at an annual simple interest rate of 4.9%. How much additional money must be invested at an annual simple interest rate of 7.4% so that the total interest earned is 6.4% of the total investment? c Additional amount to be invested at 7.4%: x Principal, P Amount at 4.9%
4000
Amount at 7.4%
x
Amount at 6.4%
4000 + x
Interest rate, r
=
Interest earned, I
0.049
=
0.049(4000)
0.074
=
0.074x
0.064
=
0.064(4000 + x)
c The sum of the amounts of interest earned by the two investments equals the interest earned by the total investment.
0.049(4000) + 0.074 x = 0.064(4000 + x )
Solution
196 + 0.074 x = 256 + 0.064 x
196 + 0.01x = 256
0.01x = 60
x = 6000
Example 2
Strategy
$6000 must be invested at an annual simple interest rate of 7.4%. You have a total of $8000 invested in two simple interest accounts. On one account, a money market fund, the annual simple interest rate is 11.5%. On the second account, a bond fund, the annual simple interest rate is 9.75%. The total annual interest earned by the two accounts is $823.75. How much do you have invested in each account? c Amount invested at 11.5%: x c Amount invested at 9.75%: 8000 – x Principal, P Amount at 11.5%
x
Amount at 9.75%
8000 – x
c The total interest earned is $823.75.
0.115 x + 0.0975(8000 − x ) = 823.75
Solution
0.115x + 780 − 0.0975x = 823.75
0.0175x + 780 = 823.75
· · ·
Interest rate, r
=
Interest earned, I
0.115
=
0.115x
0.0975
=
0.0975(8000 – x)
0.0175x = 43.75 x = 2500
The amount invested at 9.75% is 8000 – x.
Replace x with 2500 and evaluate.
· · · ·
8000 − x = 8000 − 2500 = 5500
The amount invested at 11.5% is $2500.
The amount invested at 9.75% is $5500. Section 2.4 Applications: Problems Involving Percent 27
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Exercises Solve. 1. Kristi invests $1500 at a ari invests $2000 at a 6.75% annual simple interest rate. Which of the two will earn the greater amount of interest after one year? How much greater? 2. An investment club invested $5000 at an annual simple interest rate of 8.4%. How much additional money must be invested at an annual simple interest rate of 10.5% so that the total interest earned will be 9% of the total investment? 3. An investment club invested $6000 at an annual simple interest rate of 4.0%. How much additional money must be invested at an annual simple interest rate of 6.5% so that the total annual interest earned will be 5% of the total investment? 4. An account executive deposited $42,000 into two simple interest accounts. On the tax-free account the annual simple interest rate is 3.5%, and on the money market fund the annual simple interest rate is 4.5%. How much should be invested in each account so that both accounts earn the same annual interest?
Objective 2 Percent mixture problems
The solution of a percent mixture problem is based on the equation Ar = Q , where A is the amount of solution or alloy, r is the percent of concentration, and Q is the quantity of a substance in the solution or alloy. The number of grams of silver in 50 g of a 40% silver alloy is
Example 3
Strategy
Q = Ar
Q = 50(0.40)
Q = 20 How many grams of pure acid must be added to 60 g of an 8% acid solution to make a 20% acid solution? c Grams of pure acid: x Amount
Percent
Quantity
Pure acid (100%)
x
1.00
x
8%
60
0.08
0.08(60)
20%
x + 60
0.20
0.20(x + 60)
c The sum of the quantities before mixing equals the quantity after mixing.
x + 0.08(60) = 0.20( x + 60)
Solution
x + 4.8 = 0.20 x + 12
0.8 x + 4.8 = 12
0.8 x = 7.2
x = 9
To make the 20% acid solution, 9 g of pure acid must be used.
28 Chapter 2 First-Degree Equations and Inequalities Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example 4
Strategy
A chemist mixes an 11% acid solution with a 4% acid solution. How many milliliters ist use to make a 700-milliliter solution that is 6% c The total amount of solution is 700 ml.
c Amount of 11% solution: x
c Amount of 4% solution: 700 – x Amount of solution, A 11% solution
x
4% solution
700 – x
6% solution
700
· · · ·
Percent of concentration, r
=
Quantity of substance, Q
0.11
=
0.11x
0.04
=
0.04(700 – x)
0.06
=
0.06(700)
c The sum of the amounts of acid in the 11% solution and the 4% solution is equal to the amount of acid in the 6% solution.
0.11 x + 0.04(700 − x ) = 0.06(700)
Solution
0.11x + 28 − 0.04 x = 42
0.07 x + 28 = 42
0.07 x = 14
The amount of 4% solution is 700 – x. Replace x by 200 and evaluate.
x = 200 700 − x = 700 − 200 = 500
The chemist should use 200 ml of the 11% solution and 500 ml of the 4% solution.
Exercises 5. A chemist mixed 100 ml of an 8% saline solution with 60 ml of a 5% saline solution. Find the percent concentration of the resulting mixture.
6. How many pounds of a 12% aluminum alloy must be mixed with 400 lb of a 30% aluminum alloy to make a 20% aluminum alloy?
7. A hospital staff member mixed a 65% disinfectant solution with a 15% disinfectant solution. How many liters of each were used to make 50 L of a 40% disinfectant solution?
8. How many quarts of water must be added to 5 qt of an 80% antifreeze solution to make a 50% antifreeze solution?
Section 2.4 Applications: Problems Involving Percent 29
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
se c t io n
2.5 Inequalities in One Variable
Objective
The solution set of an inequality is a set of numbers, each element of which, when substituted for the variable, results in a true inequality. There are many values of the variable x that will make the inequality x − 1 < 4 true. The solution set of the inequality is any number less than 5. The solution set can be written in set-builder notation as {x | x < 5}. The graph of the solution set of x − 1 < 4 is shown at the right.
−5 −4 −3 −2 −1 0
1
2
3
4
5
In solving an inequality, use the Addition and Multiplication Properties of Inequalities to rewrite the inequality in the form variable < constant or variable > constant. The Addition Property of Inequalities If a > b and c is a real number, then the inequalities a > b and a + c > b + c have the same solution set. If a < b and c is a real number, then the inequalities a < b and a + c < b + c have the same solution set. The Addition Property of Inequalities states that the same number can be added to each side of an inequality without changing the solution set of the inequality. This property is also true for an inequality that contains the symbol ≤ or ≥ .
Example 1
Solve: x + 5 ≤ 2
Solution
x+5−5≤ 2−5
x+5≤ 2 Step
x ≤ −3
• Subtract 5 from each side of the inequality.
The solution is { x | x ≤ –3}. This can also be written in interval notation as (–∞, –3]. The Multiplication Property of Inequalities Rule 1 If a > b and c > 0, then the inequalities a > b and ac > bc have the same solution set. If a < b and c > 0, then the inequalities a < b and ac < bc have the same solution set. Rule 2 If a > b and c < 0, then the inequalities a > b and ac < bc have the same solution set. If a < b and c < 0, then the inequalities a < b and ac > bc have the same solution set. Rule 1 states that when each side of an inequality is multiplied by a positive number, the inequality symbol remains the same. However, Rule 2 states that when each side of an inequality is multiplied by a negative number, the inequality symbol must be reversed.
30 Chapter 2 First-Degree Equations and Inequalities Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Here are some examples of this property. Rule 2 3 > 2
2 < 5 2( −4) > 5( −4)
3( 4) > 2( 4)
2( 4) < 5( 4)
3( −4) < 2( −4)
12 > 8
8 < 20
−12 < −8
−8 > −20
Because division is defined in terms of multiplication, when each side of an inequality is divided by a positive number, the inequality symbol remains the same. When each side of an inequality is divided by a negative number, the inequality symbol must be reversed. The Multiplication Property of Inequalities is also true for the symbol ≤ or ≥ .
Example 2
Solve: −3x > 9. Write the solution set in interval notation.
Solution −3x > 9
−3x 9 < −3 −3
Step
x < −3
• Divide each side of the inequality by –3 and reverse the inequality symbol.
(–∞, –3) The next example uses both the Addition and Multiplication Property of Inequalities.
Example 3
Solve: x + 3 > 4 x + 6. Write the answer using set-builder notation.
Solution x + 3 > 4 x + 6.
−3x + 3 > 6
Steps
• Subtract 4x from each side of the inequality.
• Subtract 3 from each side of the inequality.
x < −1
{ x | x < −1}
• Divide each side of the inequality by –3 and reverse the inequality symbol. • Write the solution set.
−3x > 3
Example 4
Solution
Solve: 5( x − 2) ≥ 9 x − 3( 2 x − 4). Write the solution set in set-builder notation. 5( x − 2) ≥ 9 x − 3( 2 x − 4)
5x − 10 ≥ 9 x − 6 x + 12
Steps
5x − 10 ≥ 3x + 12
2 x − 10 ≥ 12
2 x ≥ 22
• Subtract 3x from each side of the inequality. • Add 10 to each side of the inequality.
x ≥ 11
• Divide each side of the inequality by 2.
• Use the Distributive Property to remove parentheses. • Simplify.
{ x | x ≥ 11}
Exercises Solve. For Exercises 1 to 4, write the answer in set-builder notation. For Exercises 5 to 8, write the answer in interval notation. 1. x − 3 < 2
2. −3x ≤ −9
3. 4 x + 3 ≤ −1
4. 7 − 2 x ≥ 1
5. x + 7 ≥ 4 x − 8
6. 3x + 1 ≤ 7 x − 15
7. 6 − 2( x − 4) ≤ 2 x + 10
8. 4( 2 x − 1) > 3x − 2(3x − 5)
Section 2.5 Inequalities in One Variable 31
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Objective 2 Solve compound inequalities
two inequalities with a connective word such ow are compound inequalities. 2 x < 4 and 3x − 2 > 8 2 x + 3 > 5 or x + 2 < 5 The solution set of a compound inequality with the connective word and is the set of all elements common to the solution sets of both inequalities. Therefore, it is the intersection of the solution sets of the two inequalities.
Example 5
Solution
Solve: 11 − 2 x > −3 and 7 − 3x < 4. Write the answer in set-builder notation. 11 − 2 x > −3
and 7 − 3x < 4
−2 x > −14
−3x < −3
x < 7
x >1
{x | x < 7}
{x | x < 7} ∩ {x | x > 1} = { x | 1 < x < 7}
{ x | 1 < x < 7}
Example 6
Solution
Steps
• Solve each inequality.
• Find the intersection of the solution sets.
{x | x > 1}
Solve: −3 < 2 x + 1 < 5 . Write the answer in set-builder notation. This inequality is the the compound inequality −3 < 2 x + 1 and 2 x + 1 < 5.
−3 < 2 x + 1 and 2 x + 1 < 5
−4 < 2x
2x < 4
−2 < x
x < 2
{ x | x > −2} { x | x < 2}
{x | x > −2} ∩ {x | x < 2} = { x | −2 < x < 2}
{ x | −2 < x < 2} The inequality in the last example is frequently solved as follows.
−3 < 2 x + 1 < 5
Subtract 1 from each of the three parts of the inequality.
−3 − 1 < 2 x + 1 − 1 < 5 − 1
−4 < 2 x < 4
Divide each of the three parts of the inequality by the coefficient 2.
−4 2x 4 < < 2 2 2 −2 < x < 2 { x | −2 < x < 2}
The solution set of a compound inequality with the connective word or is the union of the solution sets of the two inequalities.
Example 7
Solution
Solve: 3 − 4 x > 7 or 4 x + 5 < 9. Write the answer in set-builder notation. 3 − 4 x > 7 or 4 x + 5 < 9
−4 x > 4
• Solve each inequality.
• Find the union of the solution sets.
4x < 4
x < −1
x <1
{x | x < −1}
{x | x < 1}
{x | x < −1} ∪ {x | x < 1} = { x | x < 1}
Steps
32 Chapter 2 First-Degree Equations and Inequalities Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Exercises Solve. For Exercises in set-builder notatio
xercises12 to 14, write the answer
9. 3x < 6 and x + 2 > 1
10. 2 x < 6 or x − 4 > 1
11. 4 x + 1 < 5 and 4 x + 7 > −1
12. 3x + 7 < 10 or 2 x − 1 > 5
13. −5 < 3x + 4 < 16
14. −2 < 3x + 7 < 1
Objective 3 Applications
Example 8
Strategy
To find the number of miles, write and solve an inequality using N to represent the number of miles.
Solution
Cost of Company A car < cost of Company B car
Company A rents cars for $6 a day and 14¢ for every mile driven. Company B rents cars for $12 a day and 8¢ for every mile driven. You want to rent a car for 5 days. How many miles can you drive a Company A car during the 5 days if it is to cost less than a Company B car?
6(5) + 0.14 N < 12(5) + 0.08 N
30 + 0.14 N < 60 + 0.08 N
30 + 0.06 N < 60
0.06 N < 30
N < 500
It is less expensive to rent from Company A if the car is driven less than 500 mi.
Exercises Solve. 15. A cellular phone company offers its customers a rate of $49 for up to 200 min per month of cellular phone time, or a rate of $25 per month plus $.40 for each minute of cellular phone time. For how many minutes per month can a customer who chooses the second option use a cellular phone before the charges exceed those of the first option?
16. AirTouch advertises text messaging for $6.95 per month for up to 400 messages, and $.10 per message thereafter. A competitor advertises text messaging for $3.95 per month for up to 400 messages and $.15 per message thereafter. For what number of messages per month is the AirTouch plan less expensive?
Section 2.5 Inequalities in One Variable 33
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
se c t io n
2.6 Absolute Value Equations
Objective 1 Absolute value equations
An equation containing an absolute value symbol is called an absolute value equation.
Absolute x+2 = 8 value equations 3x − 4 = 5x − 9 x = 3
Absolute Value Equations If a ≥ 0 and x = a, then x = a or x = − a. Given x = 3, then x = 3 or x = −3.
Example 1
Solve: 2 x − 3 = 5 2x − 3 = 5
Solution
2x − 3 = 5
2 x − 3 = −5
2x = 8
2 x = −2
x = 4
x = −1
Steps
• Remove the absolute value symbol and write two equations. • Solve the equations.
The solutions are –1 and 4.
Exercises Solve. 1. x = 7
2. x + 2 = 3
3. 3x = 12
4. 4 x = 6
5. 2 x − 5 = 4
6. 2 − 5x = 2
7. x − 2 − 2 = 3
8. 6 x − 5 − 2 = 4
Objective 2 Absolute value inequalities The solutions of the absolute value inequality x − 1 < 3 are the numbers whose distance from 1 is less than 3. Therefore, the solutions are the numbers greater than –2 and less than 4. The solution set is { x | −2 < x < 4}. Note: In this text, solutions to absolute value inequalities will always be written in set-builder notation. Distance Distance less than 3 less than 3 −5 −4 −3 −2 −1 0
1
2
3
4
5
Absolute Value Inequalities of the Form ax + b < c To solve an absolute value inequality of the form ax + b < c, solve the equivalent compound inequality − c < ax + b < c.
34 Chapter 2 First-Degree Equations and Inequalities Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example 2
Solve: 4 x − 3 < 5
Steps
−5 + 3 < 4 x − 3 + 3 < 5 + 3
−5 < 4 x − 3 < 5
−2 < 4 x < 8 −2 4x 8 < < 4 4 4 1 − < x < 2 2
1 x − < x < 2 2
• Solve the equivalent compound inequality. • Add 3 to each of the three parts of the inequality.
• Divide each of the three parts of the inequality by 4.
• Write the solution set.
The solutions of the absolute value inequality x + 1 > 2 are the numbers whose distance from –1 is greater than 2. Therefore, the solutions are the numbers less than –3 or greater than 1. The solution set is { x | x < −3 or x > 1}. Absolute Value Inequalities of the Form ax + b > c To solve an absolute value inequality of the form ax + b > c, solve the equivalent compound inequality ax + b < − c or ax + b > c.
Example 3
Solution
Solve: 2 x − 1 > 7 2x − 1 > 7 2 x − 1 < −7 or 2 x − 1 > 7
2 x < −6
2x > 8
x < −3
x > 4
{x | x < −3} {x | x > 4}
{x | x < −3} ∪ {x | x > 4}
Steps
• Solve the equivalent compound inequality.
• Find the union of the solution sets.
= { x | x < −3 or x > 4}
Exercises Solve. Write the answer in set-builder notation. 9. x > 3
10. x < 5
11. x − 5 ≤ 1
12. 5x + 2 > 12
13. 2 x + 1 < 5
14. 7 − 2 x > 9
Objective 3 Applications
The tolerance of a component, or part, is the acceptable amount by which the component may vary from a given measurement. For example, the diameter of a piston may vary from the given measurement of 9 cm by 0.001 cm. This is written as 9 cm ± 0.001 cm, which is read “9 centimeters plus or minus 0.001 centimeter.” The maximum diameter,
Section 2.6 Absolute Value Equations and Inequalities 35
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
or upper limit, of the piston is 9 cm + 0.001 cm = 9.001 cm. The minimum diameter, or = 8.999 cm. could also be found by solving the absolute value inequality d − 9 ≤ 0.001, where d is the diameter of the piston. d − 9 ≤ 0.001 −0.001 ≤ d − 9 ≤ 0.001
−0.001 + 9 ≤ d − 9 + 9 ≤ 0.001 + 9
8.999 ≤ d ≤ 9.001 The lower and upper limits of the diameter are 8.999 cm and 9.001 cm.
Example 4
A doctor has prescribed 2 cc of medication for a patient. The tolerance is 0.03 cc. Find the lower and upper limits of the amount of medication to be given.
Strategy
Let p represent the prescribed amount of medication, T the tolerance, and m the given amount of medication. Solve the absolute value inequality m − p ≤ T for m.
Solution
m− p ≤ T
m − 2 ≤ 0.03
Steps
−0.03 ≤ m − 2 ≤ 0.03
1.97 ≤ m ≤ 2.03
• Substitute the values of p and T into the inequality. • Solve the equivalent compound inequality.
The lower and upper limits of the amount of medication to be given to the patient are 1.97 cc and 2.03 cc.
Exercises Solve. 15. The diameter of a bushing is 1.75 in. The bushing has a tolerance of 0.008 in. Find the lower and upper limits of the diameter of the bushing. 16. A doctor has prescribed 2.5 cc of medication for a patient. The tolerance is 0.2 cc. Find the lower and upper limits of the amount of medication to be given.
36 Chapter 2 First-Degree Equations and Inequalities Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Chapter 3
Image copyright George Muresan, 2010. Used under license from Shutterstock.com.
Linear Functions and Inequalities in Two Variables
Prep Test 2. Simplify:
( −6)2 + ( −8)2
1. Simplify: -4(x - 3)
3. Simplify:
3 − ( −5) 2−6
4. Evaluate -2x + 5 for x = -3.
5. Evaluate
2r for r = 5. r −1
6. Evaluate 2p3 - 3p + 4 for p = -1.
x1 + x2 for x1 = 7 8. Given 3x - 4y = 12, find the 2 and x2 = -5. value of x when y = 0.
7. Evaluate
37 Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
se c t io n
3.1 The Rectangular
Objective 1 Points on a rectangular coordinate system
A rectangular coordinate system is formed by two number lines, one horizontal and one vertical, that intersect at the zero point of each line. The point of intersection is called the origin. The two lines are called coordinate axes, or simply axes. Each point in the plane can be identified by a pair of numbers called an ordered pair. The coordinates of a point are the numbers in the ordered pair associated with the point.
y Quadrant II 5 4 3 2 1
Horizontal axis
–5 –4 –3 –2 –1 0 –1 –2 –3 –4 –5
Quadrant I Vertical axis
1 2 3 4 5
x
Origin
The xy-coordinate system is used to graph Quadrant III Quadrant IV equations in two variables. A solution of an equation in two variables is an ordered pair whose coordinates make the equation a true statement.
Example 1
Solution
x x−2 4 y = 4−2 y = 2
The ordered-pair solution is (4, 2).
Example 2
Solution
x that corresponds to x = 4. x−2
Determine the ordered pair solution of y = y =
Step
• Replace x by 4 and solve for y.
Graph the ordered-pair solutions of y = x 2 − 1 when x = -2, -1, 0, 1, and 2. Substitute each value of x into the equation and solve for y. Then graph the resulting ordered pairs. It is convenient to record the ordered-pair solutions in a table similar to the one shown below. x
y = x2 – 1 2
y
-2
y = ( −2) − 1
3
(-2, 3)
-1
y = ( −1)2 − 1
0
(-1, 0)
2
y
(x, y)
4
( −2, 3)
(2, 3)
2
(−1, 0) –4
–2
(1, 0) 0
0
y = 0 −1
–1
(0, -1)
–2
1
y = 12 − 1
0
(1, 0)
–4
2
y = 22 − 1
3
(2, 3)
2
(0, −1)
4
x
Exercises 1. Graph the ordered pairs (0, –1), (2, 0), 2. Graph the ordered pairs (–1, –3), (0, –4), (3, 2), and (−1, 4). (0, 4), and (3, –2) .
3. Draw a line through all points whose x-coordinate is 2. 4. Draw a line through all points whose y-coordinate is –3.
38 Chapter 3 Linear Functions and Inequalities in Two Variables Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
5. Graph the ordered-pair solutions of y = x 2 when x =
6. Graph the ordered-pair solutions of x + 1 when x = −5, -3, 0, 3, and 5.
Objective 2 Find the length and midpoint of a line segment
The distance between two points in an xy-coordinate system can be calculated by using the Pythagorean Theorem. Pythagorean Theorem If a and b are the legs of a right triangle and c is the length of the hypotenuse, then a 2 + b2 = c 2 .
c
b
a
Distance Formula y
If P1 ( x1 , y1 ) and P2 ( x2 , y2 ) are two points in a plane, then the distance d between the two points is given by d =
P2(x 2, y 2) d
( x2 − x1 )2 + ( y2 − y1 )2 . P1(x 1, y 1)
| y2 − y1|
x
Q(x 2, y 1)
|x2 − x1|
The midpoint of a line segment is equidistant from its endpoints. The Midpoint Formula If P1 ( x1 , y1 ) and P2 ( x2 , y2 ) are the endpoints of a line segment, then the coordinates of the midpoint ( xm , ym ) of the line segment are given by xm =
Example 3
Solution
x1 + x2 y + y2 and ym = 1 2 2
Find the distance between the points whose coordinates are (-3, 2) and (4, -1). d =
( x2 − x1 )2 + ( y2 − y1 )2
Steps
• Use the distance formula.
• ( x 1, y 1) = ( −3, 2); ( x 2 , y 2 ) = (4, −1)
=
[4 − ( −3)]2 + [( −1) − 2]2
=
7 2 + ( −3)2 =
=
Example 4
Solution
x1 + x2 2 −5 + ( −3) = 2
xm =
58
Find the coordinates of the midpoint of the line segment with endpoints (–5, 4) and (–3, 7).
49 + 9
= –4
y1 + y2 2 4+7 = 2 11 = 2
ym =
The coordinates of the midpoint are −4,
Steps
• Use the midpoint formula.
• ( x 1, y 1) = ( −5, 4); ( x 2 , y 2 ) = ( −3, 7)
11 . 2
Section 3.1 The Rectangular Coordinate System 39
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Exercises Find the distance between the points.
point of the line segment connecting the
7. P1 (3, 5) and P2 ( 4, 1)
8. P1 ( −2, 3) and P2 (5, −1)
9. P1 (0, 3) and P2 ( −2, 4)
10. P1 (6, −1) and P2 ( −3, −2)
Objective 3 Graph a scatter diagram
The study of the relationship between two variables may begin with a scatter diagram, which is a graph of the ordered pairs of the known data.
Example 5
The scatter diagram below shows the number of computer majors, in thousands, in the United States and Canada in the fall of each year from 1999 to 2003. (Source: Computing Research Association Taulbee Survey)
A. How many students were computer majors in the fall of 2000?
B. In the fall of which year was the number of computer majors less than 19,000?
Number of computer majors (in thousands)
y 24 23 22 21 20 19 18 17 1999 2000 2001 2002 2003
This jagged line means the graph does not begin at zero.
x
Year
Strategy
A. Read the graph. Find 2000 on the horizontal axis. Follow that mark up vertically to determine the number of computer majors in the fall of that year.
B. Find the horizontal line that represents 19,000. Determine the year for which the number of students is below that line.
A. In the fall of 2000, 23,000 students were computer majors.
Solution
B. The number of computer majors was less than 19,000 in 2003.
Exercises Solve.
a. What was the consumer debt in 2001? b. In which year was consumer debt first greater than $1.9 trillion?
c. Looking at the trend of the data, would you expect consumer debt in 2004 to be greater than or less than consumer debt in 2003?
y Consumer debt (in trillions of dollars)
11. The scatter diagram at the right shows how much consumers owed in the years 1999 through 2003. (Source: Federal Reserve)
2.0 1.8 1.6 1.4 '99 '00 '01 '02 '03
x
Year
40 Chapter 3 Linear Functions and Inequalities in Two Variables Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Pillaged by LibraryPirate
Year, x Percent, y
’20
’30
’40
’50
’60
’70
’80
’90
’00
35
41
37
62
78
90
93
95
97
y Percent of households with telephones
12. As shown in the table below, the percent of households with telephones has increased over the years. (Source: U.S. Census Bureau/Federal Communications Commission) Draw a scatter diagram for these data.
90 70 50 30 ’20
’40
’60
Year
’80
’00
x
Objective 4 Average rate of change
Population of California Year
Population (in millions)
1850
0.1
1860
0.4
1870
0.6
1880
0.9
1890
1.2
1900
1.5
1910
2.4
1920
3.4
1930
5.7
1940
6.9
1950
10.6
1960
15.7
1970
20.0
1980
23.7
1990
29.8
2000
33.9
The table at the left shows the population of California each decade from 1850 to 2000. (Source: U.S. Bureau of the Census) One way to describe how the population of California has changed over time is to calculate the average rate of change in population for a given time period. For example, we can find the average rate of change in population from 1990 to 2000. Change in population Change in years
=
population in 2000 − population in 1990 2000 − 1990
=
33.9 − 29.8 4.1 = = 0.41 2000 − 1990 10
The average rate of change was 0.41 million, or 410,000, people per year. This means that on average, from 1990 to 2000, the population of California increased 410,000 people per year.
Average Rate of Change The average rate of change of y with respect to x is
Example 6
Solution
change in y . change in x
Find the average annual rate of change in the population of California from 1980 to 1990. In 1980, the population was 23.7 million: (1980, 23.7) In 1990, the population was 29.8 million: (1990, 29.8)
change in y change in x population in 1990 − population in 1980 = 1990 − 1980 29.8 − 23.7 = 1990 − 1980 6.1 = = 0.61 10 The average rate of change in the population was 0.61 million, or 610,000, people per year. Note from this example that the average rate of change in the population of California is not constant. It varies for different time periods.
Average rate of change =
Section 3.1 The Rectangular Coordinate System 41
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Exercises Solve. 13. On November 5 in mid-s s 34°F. At 2 P.M. on the same day, the temperature was 58°F. Find the average rate of change in temperature per hour.
14. In the year 2000, there were 375,000 applications for trademark registration in the United States. In 2005, there were 324,000 trademark registration applications. What is the average rate of change in trademark registration applications per year?
se c t io n
3.2 Introduction to Functions
Objective 1 Evaluate a function
Definition of Relation A relation is a set of ordered pairs. Definition of Function A function is a relation in which no two ordered pairs have the same first coordinate and different second coordinates. The domain of a function or relation is the set of the first coordinates of all the ordered pairs of the function or relation. The range is the set of the second coordinates of all the ordered pairs of the function or relation.
Example 1
What are the domain and range of the following relation? Is the relation a function?
{(2, 4), (3, 6), (4, 8), (5, 10), (4, 6), (6, 10)}
Domain = {2, 3, 4, 5, 6}
Steps
• The domain is the set of first coordinates.
Range = {4, 6, 8, 10}
• The range is the set of second coordinates.
Because there are two ordered pairs, (4, 8) and (4, 6), with the same first coordinate and different second coordinates, the relation is not a function.
Solution
A function is frequently given by an equation. To emphasize that the equation represents a function, function notation is used. The square function is written in function notation as f ( x ) = x 2 . The symbol f ( x) is read “the value of f at x” or “f of x.” The process of determining f ( x) for a given value of x is called evaluating a function. Just as the variable x is commonly used to represent a number, the letter f is commonly used to name a function. Other letters can be used.
Example 2
Let q( r ) = 2r 3 + 5r 2 − 6. Evaluate q( −3).
Solution q( r ) = 2r 3 + 5r 2 − 6
q( −3) = 2( −3)3 + 5( −3)2 − 6
= 2( −27 ) + 5(9) − 6 = −54 + 45 − 6
Steps
• Replace r by –3.
• Simplify.
q( −3) = −15
42 Chapter 3 Linear Functions and Inequalities in Two Variables Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Evaluating a function produces an element in the range of the function. In Example 2, –3 e range of q. The ordered pair (–3, –15) is an element
Exercises For Exercises 1 and 2, determine the domain and range of the relation. Is the relation a function? 1. {(1, 1), (2, 4), (3, 7), (4, 10), (5, 13)}
2. {(–2, –4), (2, 4), (–1, 1), (1, 1), (–3, 9)}
3. Given f ( x ) = 3x − 2, evaluate f (5).
4. Given g ( x ) = x 2 − x , evaluate g( −3).
5. Evaluate h(t ) = −2t 2 + 3t − 1 when t = 4.
6. Evaluate y ( x ) = x 2 + 2 when x = 0.
7. Find the element in the range of h( x ) = 2 x 2 + 3x − 4 corresponding to x = 2.
Objective 2 Graph a function
The graph of a function is the graph of the ordered pairs that belong to the function. To graph a function, plot the ordered pairs found by evaluating the function for various elements in the domain of the function. Then connect the dots to form the graph.
Example 3
Solution
Graph h( x ) = x 2 − 3 by evaluating the function when x = −3, -2, -1, 0, 1, 2, and 3. Plot the resulting ordered pairs. Then connect the points to form the graph. Evaluate the function for the given values of x. This will produce some ordered pairs of the function. The results can be recorded in a table. y = h( x ) = x 2 – 3
(x, y)
-3
h( −3) = ( −3)2 − 3 = 6
(-3, 6)
-2
h( −2) = ( −2)2 − 3 = 1
(-2, 1)
4
h( −1) = ( −1) − 3 = −2
(-1, -2)
2
0
h(0) = (0)2 − 3 = −3
(0, -3)
1
h(1) = (1)2 − 3 = −2
(1, -2)
2
h( 2) = ( 2)2 − 3 = 1
(2, 1)
3
h(3) = (3)2 − 3 = 6
(3, 6)
x
-1
2
y 6
–4
–2 0
2
4
x
–2
Connect the points to form the graph.
Section 3.2 Introduction to Functions 43
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example 4
Graph f ( x ) = − x + 2 by evaluating the function when x = −3, -2, -1, 0, 1, 2, en connect the points to form the graph. of x. This will produce some ordered pairs of the function. The results can be recorded in a table.
Soluti
x
y = f ( x) = – x + 2
(x, y)
y
-3
f ( −3) = − −3 + 2 = −1
(-3, -1)
4
-2
f ( −2) = − −2 + 2 = 0
(-2, 0)
2
-1
f ( −1) = − −1 + 2 = 1
(-1, 1)
0
f (0) = − 0 + 2 = 2
(0, 2)
1
f (1) = − 1 + 2 = 1
(1, 1)
2
f ( 2) = − 2 + 2 = 0
(2, 0)
3
f (3) = − 3 + 2 = −1
(3, -1)
–4
–2
0
2
4
x
–2 –4
Connect the points to form the graph. It is important to understand that every point on the graph of a function is an ordered pair of the function, and every ordered pair that belongs to the function corresponds to a point on the graph.
Exercises Graph each function by evaluating the function at the given values of x. Plot the resulting ordered pairs. Then connect the points to form the graph. 8. g ( x ) = −2 x + 2 9. h( x ) = x 2 − 4 x 10. f ( x ) = x − 3 x = −2, –1, 0, 1, 2, 3 x = −1, 0, 1, 2, 3, 4, 5 x = −6, –4, –2, 0, 2, 4, 6
Objective 3 Vertical line test Vertical Line Test for the Graph of a Function If every vertical line intersects a graph at most once, then the graph is the graph of a function. This graphical interpretation of a function is often described by saying that each value in the domain of the function is paired with exactly one value in the range of the function. y
y
8
(−1, 3)
0
(−3, − 6)
–4 –8
(5, 7)
4
–4
–8
4 2
4
8
x
(3, −2)
0 –2
For each x, there is exactly one value of y. For instance, when x = −3, y = −6. This is the graph of a function.
(4, 2) 2
4
6
x
(4, −2)
–4
Some values of x can be paired with more than one value of y. For instance, 4 can be paired with -2 and 2. This is not the graph of a function.
44 Chapter 3 Linear Functions and Inequalities in Two Variables Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example 5
Use the vertical line test to determine whether the graph is the graph of a function. B.
y
x
Solution
x
y
A. As shown at the right, there are vertical lines that intersect the graph at more than one point. Therefore, the graph is not the graph of a function.
x
y
B. For the graph at the right, every vertical line intersects the graph at most once. Therefore, the graph is the graph of a function.
x
Exercises Use the vertical line test to determine whether the graph is the graph of a function. 11.
12.
y
–4
–2
8
2
2
4
0
2
4
x
–4
–2
0
2
4
x
–8
–4
–2
–4
y
4
–2
13.
y
4
se c t io n
–4
0
4
8
x
–4
–8
3.3 Linear Functions
Objective 1 Graph a linear function
A function that can be written in the form f ( x ) = mx + b (or y = mx + b) is called a linear function because its graph is a straight line. Examples of linear functions are shown at the right. Note that the exponent on each variable is 1.
f ( x) = 2x + 5
(m = 2, b = 5)
P(t ) = 3t − 2
(m = 3, b = -2)
y = -2x
(m = -2, b = 0)
y = −
2 x + 1 3
g( z) = z − 2
2 m = − 3 , b = 1 (m = 1, b = -2)
Section 3.3 Linear Functions 45
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Because the graph of a linear function is a straight line, and two points determine a straight awn by finding only two of the ordered pairs d that you find at least three ordered pairs to
Example 1
Graph: f ( x ) = −
Solution
3 x−3 2
x
y = f ( x)
0 -2
–3 0
-4
3 y 4
Steps
• Find at least three ordered pairs. When the coefficient of x is a fraction, choose values of x that will simplify the evaluations. The ordered pairs can be displayed in a table.
• Graph the ordered pairs and draw a line through the points.
2 –4
–2
0
2
4
x
–4
Exercises Graph. 1. y = 3x − 4
2. y = −2 x + 3
1 3. y = − x + 2 3
4. y =
3 x −1 5
Objective 2 Graph an equation of the form Ax + By = C
In some cases, a linear equation has the form Ax + By = C. In such a case, it may be convenient to solve the equation for y to get the equation in the form y = mx + b.
Example 2
Solution
Graph: 3x + 2 y = 6 Solve the equation for y.
3x + 2 y = 6
3x − 3x + 2 y = −3x + 6
2 y = −3x + 6
2y −3x 6 = + 2 2 2
y = −
3 x+3 2
Steps
• Subtract 3x from each side of the equation.
• Divide each side of the equation by 2.
• Simplify. The equation is now in the form y = mx + b.
46 Chapter 3 Linear Functions and Inequalities in Two Variables Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
x
y
4
–3 y 4
• Find at least three ordered pairs. We have chosen 0, 2, and 4 for x. However, any value of x could be chosen.
• Graph the ordered pairs and draw a line through the points.
(0, 3)
2
(2, 0) –4
–2
0
2
x
4
–2
(4, −3)
–4
A second method of graphing straight lines uses the intercepts of the graph. The point at which a graph crosses the x-axis is called the x-intercept. The y-coordinate of this point is 0. The point at which a graph crosses the y-axis is called the y-intercept. The x-coordinate of this point is 0.
Example 3
Solution
Graph 4 x − y = 4 by using the x- and y-intercepts. x-intercept: 4 x − y = 4
4x − 0 = 4
4x = 4
x =1
The x-intercept is (1, 0).
y-intercept: 4 x − y = 4
4(0) − y = 4
−y = 4
• To find the x-intercept, let y = 0.
• To find the y-intercept, let x = 0.
• Graph the intercepts (1, 0) and (0, –4). Then draw a line through the points.
y = −4
Steps
The y-intercept is (0, –4). y 4 2
(1, 0) –4
–2
0
2
x
4
–2
–4
(0, – 4)
Exercises For Exercises 5 to 8, solve the equation for y and then graph. 5. 2 x − y = 3
6. 2 x + 5 y = 10
7. 2 x − 3 y = 12
8. 3x − 2 y = 8
For Exercises 9 to 12, graph the equation by first finding the x- and y-intercepts. 9. x − 2 y = −4
10. 3x + y = 3
11. 2 x − 3 y = 9
12. 4 x − 3 y = 6
Section 3.3 Linear Functions 47
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Objective 3 Applications
Example
ing, the daily caloric allowance for a woman can be approximated by the equation C = −7.5 A + 2187.5, where C is the caloric intake and A is the age of the woman.
A. Graph this equation for 25 ≤ A ≤ 75.
B. The point whose coordinates are (45, 1850) is on the graph. Write a sentence that describes the meaning of this ordered pair.
Solution
A.
C 2000
Calories
1800
(45, 1850)
1600 1400 1200
25
35
45
55
65
75
A
Age in years
B. The ordered pair means that the caloric allowance for a 45-year-old woman is 1850 calories per day.
Exercises D Distance traveled (in feet)
13. The Kingda Ka roller coaster at Six Flags Great Adventure in Jackson, N.J., has a maximum speed of 188 ft/s. The equation that describes the total number of feet traveled by the roller coaster in t seconds at this speed is given by D = 188t. Use the coordinate axes at the right to graph this equation for 0 ≤ t ≤ 10. The point (5, 940) is on this graph. Write a sentence that describes the meaning of this ordered pair.
1000
500
5
10
t
Seconds
se c t io n
3.4 Slope of a Straight Line
Objective 1 Find the slope of a line given two points The slope of a line containing two points is the ratio of the change in the y values between the two points to the change in the x values. The line containing the points whose coordinates are (-1, -3) and (5, 2) is shown below. The change in the y values is the difference between the y-coordinates of the two points.
y
The change in the x values is the difference between the x-coordinates of the two points. Change in x is x2 − x1 = 5 − ( −1) = 6 The slope of the line between the two points is the ratio of the change in y to the change in x.
(5, 2)
2
Change in y is y2 − y1 = 2 − ( −3) = 5 –2
(−1, −3)
0
2
4
–2 –4
x Change in y 2 − (−3) = 5
Change in x 5 − (−1) = 6
48 Chapter 3 Linear Functions and Inequalities in Two Variables Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Slope Formula e two points P1 ( x1 , y1 ) and P2 ( x2 , y2 ) is m =
Example 1
Find the slope of the line containing the points (2, -5) and (-4, 2).
Solution
Let ( x1 , y1 ) = ( 2, −5) and ( x2 , y2 ) = ( −4, 2).
7 The slope of the line is − . 6
Example 2
y2 − y1 , x1 ≠ x2 x2 − x1
m =
y2 − y1 2 − ( −5) 7 = = x2 − x1 −4 − 2 −6
Find the slope of the line containing the points (-3, 4) and (5, 4).
Solution
Let ( x1 , y1 ) = ( −3, 4) and ( x2 , y2 ) = (5, 4).
The slope of the line is 0. This is a horizontal line.
Example 3
Solution
m =
Find the slope of the line containing the points (6, -1) and (6, 7). Let ( x1 , y1 ) = (6, −1) and ( x2 , y2 ) = (6, 7 ).
m =
y2 − y1 4−4 0 = = = 0 x2 − x1 5 − ( −3) 8
y2 − y1 7 − ( −1) 8 = = x2 − x1 6−6 0
Division by 0 is undefined.
The line has no slope. This is a vertical line.
Exercises Find the slope of the line between the two points. 1. P1 (1, 3), P2 (3, 1)
2. P1 ( 2, 3), P2 (5, 1)
3. P1 ( −1, 4), P2 ( 2, 5)
4. P1 (3, −2), P2 (1, 4)
5. P1 ( 2, 4), P2 ( 2, −2)
6. P1 (3, 4), P2 (0, 4)
Objective 2 Graph a line given a point and the slope
3 x + 4 is shown at the right. The 4 points (-4, 7) and (4, 1) are on the graph. The slope of the line is The graph of y = −
7 −1 6 3 m = = = − −4 − 4 −8 4
6
(0, 4)
y=– 3 x + 4 4
2
Note that the slope of the line has the same value as the coefficient of x. The y-coordinate of the y-intercept, (0, 4), is 4, the constant term in the equation.
y (− 4, 7)
(4, 1) –4
–2
0
2
4
x
Section 3.4 Slope of a Straight Line 49
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Slope-Intercept Form of a Straight Line ope-intercept form of a straight line. of x. The y-intercept is (0, b).
Example 4
Solution
5 x − 4 by using the slope and y-intercept. 3 5 change in y m = = 3 change in x
Graph y =
The y-intercept is (0, -4).
Beginning at the y-intercept, move right 3 units (change in x) and then up 5 units (change in y). The point whose coordinates are (3, 1) is a second point on the graph. Draw a line through the points (0, -4) and (3, 1).
Example 5
Solution
y 4 2
–4
–2
(3, 1)
0
up 5
–2
(0, –4)
x
4
right 3
4 Graph the line that passes through the point (-2, 3) and has slope − . 3 4 −4 y m = − = 3 3 4 Graph the point (-2, 3). Beginning at that point, move right 3 units (change in x) and then down 4 units (change in y). The point whose coordinates are (1, -1) is a second point on the graph. Draw a line through the points (-2, 3) and (1, -1).
right 3
(–2, 3) –4
–2
2 0 –2
down 4 2
4
x
(1, –1)
–4
Exercises For Exercises 7 to 10, graph by using the slope and the y-intercept. 7. y =
1 x + 2 2
8. y =
2 x − 3 3
9. y =
3 x 4
11. Graph the line that passes through the point (–1, –3) and has slope
4 . 3
12. Graph the line that passes through the point (–2, –3) and has slope
5 . 4
10. y = −
1 x+2 2
13. Graph the line that passes through the point (–3, 0) and has slope –3.
14. Graph the line that passes through the point (2, 0) and has slope –1.
50 Chapter 3 Linear Functions and Inequalities in Two Variables Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
se c t io n
3.5 Finding Equations of Lines
Obj
iven a point and the slope When the slope of a line and a point on the line are known, the equation of the line can be determined. If the particular point is the y-intercept, use the slope-intercept form, y = mx + b, to find the equation.
Example 1
Solution
Find the equation of the line that contains the point (0, 3) and has slope The known point is the y-intercept (0, 3). The slope is
y = mx + b
y =
1 x+3 2
Step
1 . 2
1 . 2
1 , the slope. Replace b by 2 3, the y-coordinate of the y-intercept.
• Replace m by
One method of finding the equation of a line when the slope and any point on the line are known involves using the point-slope formula. Point-Slope Formula Let m be the slope of a line, and let ( x1 , y1 ) be the coordinates of a point on the line. The equation of the line can be found using the point-slope formula y − y1 = m( x − x1 ).
Example 2
Solution
Find the equation of the line that contains the point (-2, 4) and has slope 2. y − y1 = m( x − x1 )
y − 4 = 2[ x − ( −2)]
Steps
• Use the point-slope formula.
• Substitute the slope, 2, and the coordinates of the point (-2, 4) into the formula.
y − 4 = 2[ x + 2] y − 4 = 2 x + 4
y = 2x + 8
The equation of the line is y = 2 x + 8.
Exercises Find the equation of the line that contains the given point and has the given slope. 1. Point (0, 5), m = 2
4. Point (–1, 7), m = −3
2. Point (0, –2), m = −3
1 2
5. Point (5, 1), m = −
4 5
3. Point (2, 3), m =
1 2
6. Point (3, 5), m = −
2 1 3 2
Objective 2 Find the equation of a line given two points
To find the equation of a line given two points, first find the slope of the line between the points. Then use one of the points and the point-slope formula to find the equation of the line.
Section 3.5 Finding Equations of Lines 51
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example 3
Find the equation of the line that contains the points (3, 2) and (-5, 6).
Soluti 2
y − y1 = m( x − x1 )
1 y − 2 = − ( x − 3) 2
y − 2 = −
1 3 x+ 2 2
y = −
1 7 x+ 2 2
Steps
• Find the slope. Let ( x 1, y 1) = (3, 2) and let ( x 2 , y 2 ) = ( −5, 6).
• Use the point-slope formula.
1
The equation of the line is y = −
1 • Substitute the slope, − , and the 2 coordinates of either point into the formula. We used (3, 2).
1 7 x+ . 2 2
Exercises Find the equation of the line that contains the given points. 7. P1 ( −1, 3), P2 ( 2, 4)
8. P1 ( −1, 1), P2 ( 4, 4)
10. P1 ( −3, 3), P2 ( −2, 3)
11. P1 (0, 3), P2 (3, 0)
9. P1 ( 4, 1), P2 (3, −2) 12. P1 (1, −3), P2 ( −2, 4)
Objective 3 Application problems
Linear functions can be used to model a variety of applications in science and business. For each application, data are collected and the independent and dependent variables are selected. Then a linear function is determined that models the data.
Example 4
Strategy
Select the independent and dependent variables. Because we want to determine the population of adults 65 years of age or older, that quantity is the dependent variable, y. The year is the independent variable.
From the data, the ordered pair (1950, 13), where the y-coordinate is in millions of people, is a point on the line. The slope of the line is the rate of increase, 0.5 million per year.
y − y1 = m( x − x1 )
Steps
• Use the point-slope formula.
y − 13 = 0.5[ x − 1950]
y − 13 = 0.5x − 975
• Substitute the slope, 0.5, and the coordinates of the point (1950, 13) into the formula.
In 1950, there were 13 million adults 65 years old or older in the United States. Data from the Census Bureau show that the population of these adults has been increasing at a constant rate of approximately 0.5 million per year. This rate of increase is expected to continue through the year 2010. Find the equation of the line that approximates the population of adults 65 years old or older in terms of the year. Use your equation to approximate the population of these adults in 2005.
Solution
y = 0.5x − 962
The linear function is f ( x ) = 0.5 x – 962.
52 Chapter 3 Linear Functions and Inequalities in Two Variables Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
f ( x ) = 0.5x − 962
= 1002.5 − 962
= 40.5
• Evaluate the function at 2005 to predict the population in 2005.
The equation gives an estimate of 40.5 million people 65 years old or older in 2005.
Exercises 13. The pilot of a Boeing 757 jet takes off from Boston’s Logan Airport, which is at sea level, and climbs to a cruising altitude of 32,000 ft at a constant rate of 1200 ft/min. Write a linear equation for the height of the plane in terms of the time after take-off. Use your equation to find the height of the plane 11 min after take-off. 14. There are approximately 126 calories in a 2-ounce serving of lean hamburger and approximately 189 calories in a 3-ounce serving. Write a linear equation for the number of calories in lean hamburger in terms of the size of the serving. Use your equation to estimate the number of calories in a 5-ounce serving of lean hamburger.
se c t io n
3.6 Parallel and Perpendicular Lines
Objective 1 Find equations of parallel and perpendicular lines
Slopes of Parallel Lines Two nonvertical lines in the plane with slopes of m1 and m2 are parallel if and only if m1 = m2 . Any two vertical lines are parallel.
Example 1
Solution
m1 =
−1 − 1 −2 2 = = −5 − ( −2) −3 3
Steps
• Find the slope of the line through (-2, 1) and (-5, -1).
m2 =
2−0 2 = 4−1 3
• Find the slope of the line through (1, 0) and (4, 2).
The slopes are the same. The lines are parallel.
Example 2
Solution
Is the line that contains the points (-2, 1) and (-5, -1) parallel to the line that contains the points (1, 0) and (4, 2)?
Are the graphs of y = −
3 3 x − 5 is in the form y = mx + b. The slope of the line is − . 2 2 Write 2 x + 3 y = 6 in the form y = mx + b.
2x + 3y = 6
The equation y = −
3 x − 5 and 2 x + 3 y = 6 parallel? 2
3 y = −2 x + 6 y = −
2 x+2 3
2 The slope of this line is − . 3 The slopes of the two lines are not the same. The lines are not parallel. Section 3.6 Parallel and Perpendicular Lines 53
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example 3 Soluti
Find the equation of the line that contains the point (3, -1) and is parallel to the graph the unknown line is the same as the slope of the line 3x − 2 y = 4. To find its slope, solve the equation for y. 3x − 2 y = 4
−2 y = −3x + 4 y =
3 x−2 2
3 . Use this slope and the given point (3, -1) to find the equation of the 2 unknown line.
The slope is
y − y1 = m( x − x1 ) y − ( −1) =
y +1 =
y =
3 ( x − 3) 2 3 9 x− 2 2 3 11 x− 2 2 3 11 x− . 2 2
The equation of the line is y =
Two lines that intersect at right angles are perpendicular lines. Slopes of Perpendicular Lines If m1 and m2 are the slopes of two lines, neither of which is vertical, then the lines are perpendicular if and only if m1 ⋅ m2 = −1. A vertical line is perpendicular to a horizontal line.
Example 4
Solution
Is the line that contains the points (4, 2) and (-2, 5) perpendicular to the line that contains the points (-4, 3) and (-3, 5)? Find the slope of each line. 5−2 3 1 = = − −2 − 4 −6 2
m1 =
1 m1 ⋅ m2 = − ( 2) = −1 2
Example 5
Solution
m2 =
5−3 2 = = 2 −3 − ( −4) 1
Because m1 ⋅ m2 = −1, the lines are perpendicular. Are the graphs of 4 x − y = −2 and x + 4 y = −12 perpendicular? Solve each equation for y. 4 x − y = −2
x + 4 y = −12
− y = −4 x − 2 y = 4 x + 2 m1 = 4
4 y = − x − 12 y = −
1 1 x − 3 m2 = − 4 4
1 m1 ⋅ m2 = 4 − = −1 4
The product of the slopes is -1. The graphs are perpendicular.
54 Chapter 3 Linear Functions and Inequalities in Two Variables Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example 6
Find the equation of the line that contains the point (3, -4) and is perpendicular to the line by solving 2 x − y = −3 for y.
2 x − y = −3
− y = −2 x − 3 y = 2 x + 3 m = 2
1 The slope of the line perpendicular to the given line is − . 2 1 Find the equation of the line with slope − that passes through (3, -4). 2 Steps • Use the point-slope formula. y − y1 = m( x − x1 )
1 y − ( −4) = − ( x − 3) 2
y+4 = −
1 3 x+ 2 2
y = −
1 5 x− 2 2
The equation of the line is y = −
1 • m = − , ( x 1, y 1) = (3, −4) 2
1 5 x − . 2 2
Exercises 1. Are the graphs of 2 x + 3 y = 2 2. Are the graphs of 2 x − 4 y = 3 and and 2 x + 3 y = −4 parallel? 2 x + 4 y = −3 parallel? 3. Are the graphs of x − 4 y = 2 and 4. Are the graphs of y = x − 1 and 4 x + y = 8 perpendicular? 4 x + 3 y = −7 perpendicular? 5. Is the line that contains the points (3, 2) and (1, 6) 6. Is the line that contains the points (4, -3) and (2, 5) parallel to the line that contains the points (-1, 3) parallel to the line that contains the points (-2, -3) and (-1, -1)? and (-4, 1)? 7. Is the line that contains the points (-3, 2) and (4, -1) 8. Is the line that contains the points (3, 5) and (-3, 3) perpendicular to the line that contains the points perpendicular to the line that contains the points (1, 3) and (-2, -4)? (2, -5) and (-4, 4)? 9. Find the equation of the line that contains the point 10. Find the equation of the line that contains the point (-2, -4) and is parallel to the graph of 2 x − 3 y = 2. (3, 2) and is parallel to the graph of 3x + y = −3. 11. Find the equation of the line that contains the point 12. Find the equation of the line that contains the point (4, 1) and is perpendicular to the graph of (-1, 3) and is perpendicular to the graph of y = −3x + 4. 2 x + 4 y = −1.
Section 3.6 Parallel and Perpendicular Lines 55
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
se c t io n
3.7 Inequalities in Two Variables
Objective
ality in two variables y
The graph of the linear equation y = x − 1 separates the plane into three sets: the set of points on the line, the set of points above the line, and the set of points below the line.
4
above y>x−1
The set of points on the line are the solutions of the equation y = x − 1. The set of points above the line are the solutions of the inequality y > x − 1. These points form a half-plane. The set of points below the line are the solutions of the inequality y < x − 1. These points also form a half-plane.
(2, 4)
2
(2, 1) –4
–2
0 –2 –4
y=x−1
2
4
x
(2, –2) below y<x−1
An inequality of the form y > mx + b or Ax + By > C is a linear inequality in two variables. (The inequality symbol could be replaced by ≥, <, or ≤ .)
Example 1
Solution
Graph the solution set of 3x − 4 y < 12 . 3x − 4 y < 12
−4 y < −3x + 12 y >
3 x−3 4
Graph y =
3 x − 3. 4
If the inequality contains ≤ or ≥, the line belongs to the solution set and is shown by a solid line. If the inequality contains < or >, the line is not part of the solution set and is shown by a dashed line. If the inequality contains > or ≥, shade the upper half-plane. If the inequality contains < or ≤, shade the lower half-plane.
Example 2
Solution
y 4 2 –4
–2
0
2
4
x
–2 –4
Graph the solution set of x + 2 y ≤ 4. x + 2y ≤ 4
Steps
• Solve the inequality for y.
• Graph y = −
2 y ≤ −x + 4 y ≤ −
1 x+2 2
y 4
1 x + 2 as a solid 2 line. Shade the lower half-plane.
2 –4 –2 0 –2
2
4
x
–4
56 Chapter 3 Linear Functions and Inequalities in Two Variables Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Exercises Graph the solution se 1. x + 3 y < 4
3. −3x + 2 y > 2
Section 3.7 Inequalities in Two Variables 57
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Chapter 4
Image copyright Sophie Bengtsson, 2010. Used under license from Shutterstock.com.
Systems of Equations and Inequalities
Prep Test 3 1 1. Simplify: 10 x + y 2. Evaluate 3x + 2 y − z for x = −1, 2 5 y = 4, and z = −2.
3. Given 3x − 2 z = 4, find the value of x when z = −2.
4. Solve: 3x + 4( −2 x − 5) = −5
5. Solve: 0.45x + 0.06( − x + 4000) = 630
6. Graph: y =
7. Graph: 3x − 2 y = 6
1 x−4 2
3 8. Graph: y > − x + 1 5
59 Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
se c t io n
4.1 Solving Systems of Linear Equations the Substitution Method
Objective 1 Solve a system of linear equations by graphing
A system of equations is two or more equations considered together. The system at the right is a system of two linear equations in two variables. The graphs of the equations are straight lines.
3x + 4 y = 7 2x − 3y = 6
A solution of a system of equations in two variables is an ordered pair that is a solution of each equation of the system. There are three possibilities for a system of linear equations in two variables. 1. The graphs intersect at one point.
y
The solution of the system of equations is the ordered pair (x, y) whose coordinates are the point of intersection.
x
The system of equations is independent. y
2. The lines are parallel and never intersect. There is no solution of the system of equations.
x
The system of equations is inconsistent. 3. The graphs are the same line, and they intersect at infinitely many points.
y
There are an infinite number of solutions of the system of equations.
x
The system of equations is dependent. Example 1 Solve by graphing: 3x − y = 4 3x + y = 2 y Solution 4
Steps
• Graph each line. • The lines intersect. The system of equations is independent.
Steps
• Graph each line. • The lines are parallel. The system of equations is inconsistent.
2 –4
–2
0
2
4
x
–2
–4
Example 2
The solution is (1, –1). Solve by graphing: 2 x + 3 y = 6 y = −
y
Solution 4 2 –4
–2
0
2
2 x+1 3
x
–2
–4
Because the lines do not intersect, the system of equations has no solution.
60 Chapter 4 Systems of Equations and Inequalities Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example 3
Solve by graphing: 3x − 4 y = 12 x−3
y
Solution
Steps
4 2 –4
–2
0
2
4
• Graph each line. • The lines are the same. The system of equations is dependent.
x
–2
–4
3 The solutions are the ordered pairs x , x − 3 . 4
Exercises Solve by graphing. 1. x − y = −2 x + 2 y = 10
1 2. y = x − 5 3. y = x − 2 4. 2 x − 5 y = 10 2 2x + y = 4 2 x − 2 y = 8 y = x−2 5
Objective 2 Solve a system of linear equations by the substitution method
A graphical solution of a system of equations is based on approximating the coordinates of a point of intersection. An algebraic method called the substitution method can be used to find an exact solution of a system of equations. Example 4 Solve by the substitution method: 3x + y = 5 4x + 5 y = 3 Solution
(1)
3x + y = 5
(2)
4x + 5 y = 3
3x + y = 5
y = –3 x + 5
(3)
4x + 5 y = 3
4 x + 5( −3 x + 5) = 3
4 x − 15x + 25 = 3
−11x = −22
x = 2
y = −3x + 5
y = −3( 2) + 5
y = −6 + 5
y = −1
The solution is (2, –1).
Steps
• Solve equation (1) for y. The result is equation (3). • Use equation (2). Substitute –3x + 5 for y and solve for x.
• Now substitute the value of x into equation (3) and solve for y.
Section 4.1 Solving Systems of Linear Equations by Graphing and by the Substitution Method 61
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example 5
Solve by substitution: 3x − 3 y = 2
Soluti
y = x+2
(2)
3x − 3( x + 2) = 2
3x − 3x − 6 = 2
−6 = 2
Steps
• Equation (2) states that y = x + 2.
• Substitute x + 2 in equation (1).
−6 = 2 is not a true equation. The system of equations is inconsistent. The system does not have a solution.
Example 6 Solve by substitution: 9 x + 3 y = 12 y = −3x + 4
Solution
(1) 9 x + 3 y = 12 y = −3x + 4
(2)
9 x + 3( −3x + 4) = 12
9 x − 9 x + 12 = 12
12 = 12
Steps
• Equation (2) states that y = –3x + 4.
• Substitute –3x + 4 for y in equation (1).
12 = 12 is a true equation. The system of equations is dependent. The solutions are the ordered pairs (x, –3x + 4).
Exercises Solve by the substitution method. 5. 4 x − 3 y = 5 6. x + 3 y = 5 7. 6 x + 2 y = 7 8. 2 x − 4 y = 16 y = 2 x − 3 2 x + 3 y = 4 y = −3x + 2 − x + 2 y = −8
se c t io n
4.2 Solving Systems of Linear Equations by the Addition Method
Objective 1 Solve a system of two linear equations in two variables by the addition method
The addition method is an alternative method for solving a system of equations. This method is based on the Addition Property of Equations. Use the addition method when it is not convenient to solve one equation for one variable in terms of another variable. Note, for the system of equations at the right, the effect of adding equation (2) to equation (1). Because -3y and 3y are additive inverses, adding the equations results in an equation with only one variable. The solution of the resulting equation is the first component of the ordered-pair solution of the system.
(1) 5x − 3 y = 14 (2) 2 x + 3 y = −7
7x + 0 y = 7
7x = 7
7x = 7
x = 1
62 Chapter 4 Systems of Equations and Inequalities Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
The second component is found by substituting the d then solving
(1) 5x − 3 y = 14
5(1) − 3 y = 14
5 − 3 y = 14
−3 y = 9 y = −3
The solution is (1, –3). Sometimes adding the two equations does not eliminate one of the variables. In this case, use the Multiplication Property of Equations to rewrite one or both of the equations so that when the equations are added, one of the variables is eliminated. Example 1 Solve by the addition method: 3x + 4 y = 2 2 x + 5 y = −1 Solution
(1)
3x + 4 y = 2
(2)
2 x + 5 y = −1
2
–3
( 3 x + 4 y ) = 2( 2)
6x + 8 y = 4
−6 x − 15 y = 3
−7 y = 7
(1)
y = −1 3x + 4 y = 2
3x + 4( –1) = 2
3x − 4 = 2
3x = 6
x = 2
• We will number the equations (1) and (2).
• Eliminate x. Multiply equation (1) by 2 and equation (2) by –3. Note how the constants are selected. The negative sign is used so that the coefficients will be additive inverses. • The coefficients of the x-terms are additive inverses.
( 2 x + 5 y ) = −3( −1)
Steps
• Add the equations.
• Solve for y.
• Substitute the value of y into one of the equations and solve for x. Equation (1) is used here.
The solution is (2, –1).
Example 2 Solve by the addition method: 4 x − 8 y = 36 3x − 6 y = 27
Solution
(1)
4 x − 8 y = 36
(2)
3x − 6 y = 27
3( 4 x − 8 y ) = 3(36)
−4(3x − 6 y ) = −4( 27 )
12 x − 24 y = 108
−12 x + 24 y = −108
0 = 0
Steps
• To eliminate x, multiply each side of equation (1) by 3 and each side of equation (2) by –4.
• Add the equations.
0 = 0 is a true equation. The system of equations is dependent. The solutions are 1 9 the ordered pairs x , x − . 2 2
Section 4.2 Solving Systems of Linear Equations by the Addition Method 63
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example 3
Solve the by addition method: 2 x + y = 5 = 6
Soluti
4x + 2 y = 6
Eliminate y.
−2( 2 x + y ) = −2(5)
4x + 2 y = 6
−4 x − 2 y = −10
4x + 2 y = 6
Add the equations.
0 x + 0 y = −4
0 = −4
0 = −4 is not a true equation. The system is inconsistent and therefore has no solution.
Exercises Solve by the addition method. 1. 3x + y = 7 2. 2 x + 5 y = 9 3. 3x + 2 y = 16 x + 2 y = 4 4 x − 7 y = −16 2 x − 3 y = −11
4. 4 x − 6 y = 5 5. 3x − y = 4 6. 2 x − 5 y = 13 2 x − 3 y = 7 6 x − 2 y = 8 5x + 3 y = 17
Objective 2 Solve a system of three linear equations in three variables by the addition method
An equation of the form Ax + By + Cz = D, where A, B, and C are coefficients and D is a constant, is a linear equation in three variables. Examples of these equations are shown below.
2 x + 4 y − 3z = 7
x − 6 y + z = −3
The graph of a linear equation in three variables is a plane. That is, if all the solutions of a linear equation in three variables were plotted in an xyz-coordinate system, the graph would look like a large piece of paper extending infinitely. The graph of x + y + z = 3 is shown at the right.
z 3
3 3
y
x
The solution of a linear system of equations in three variables is an ordered triple that is a solution of each equation in the system of equations. 64 Chapter 4 Systems of Equations and Inequalities Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example 4
Solution
(1) 3x − y + 2 z = 1
(2) 2 x + 3 y + 3z = 4
(3)
Solve: 3x − y + 2 z = 1
3x − y + 2 z = 1
x + y − 4 z = −9
4 x − 2 z = −8
(4)
2 x − z = −4
9x − 3y + 6z = 3
2 x + 3 y + 3z = 4
(5)
(4)
(5)
2 x − z = −4
• Eliminate y. Add equations (1) and (3).
• Multiply each side of the equation by 1 . 2
• Multiply equation (1) by 3 and add to equation (2).
• Solve the system of two equations.
• Multiply equation (4) by 9 and add to equation (5).
• Replace x by –1 in equation (4).
• Replace x by –1 and z by 2 in equation (1), (2), or (3). Equation (3) is used here.
11x + 9 z = 7 18 x − 9 z = −36
11x + 9 z = 7
29 x = −29
x = −1
2 x − z = −4
2( −1) − z = −4
−2 − z = −4
− z = −2
z = 2
x + y − 4 z = −9
−1 + y − 4( 2) = −9
−1 + y − 8 = −9
−9 + y = −9
Steps
11x + 9 z = 7
x + y − 4 z = −9
y = 0
The solution is (–1, 0, 2).
Section 4.2 Solving Systems of Linear Equations by the Addition Method 65
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example 5
Solution
(1)
2x − 3y − z = 1
(2)
x + 4 y + 3z = 2
(3) 4 x − 6 y − 2 z = 5
Solve: 2 x − 3 y − z = 1
2x − 3y − z = 1
−2 x − 8 y − 6 z = −4
−11 y − 7 z = −3
−4 x + 6 y + 2 z = −2
4x − 6 y − 2z = 5
0 = 3
Steps
• Eliminate x from equations (1) and (2). Multiply equation (2) by –2 and add to equation (1).
• Eliminate x from equations (1) and (3). Multiply equation (1) by –2 and add to equation (3). The result is not a true equation.
The system is inconsistent.
Exercises Solve by the addition method. 7. x + 3 y + z = 6 8. x − 2 y + z = 6 9. 2 y + z = 7 10. x − 3 y + 2 z = 1 3x + y − z = −2 x + 3 y + z = 16 2 x − z = 3 x − 2 y + 3z = 5 2 x + 2 y − z = 1 3x - y - z = 12 x − y = 3 2 x − 6 y + 4 z = 3
se c t io n
4.3 Solving Systems of Equations by Using Determinants and by Using Matrices
Objective 1 Evaluate determinants A matrix is a rectangular array of numbers. Each number in the matrix is called an element of the matrix. The matrix at the right, with three rows and four columns, is called a 3 × 4 (“3 by 4”) matrix. A square matrix is one that has the same number of rows as columns. A 2 × 2 matrix and a 3 × 3 matrix are shown at the right. Associated with every square matrix is a number called its determinant.
1 −3 2 4 A = 0 4 −3 2 6 −5 4 −1 4 0 1 5 −3 7 2 1 4
−1 3 5 2
Determinant of a 2 2 Matrix a The determinant of a 2 × 2 matrix 11 a21 this determinant is given by the formula a11
a12
a21
a22
a12 a11 is written a22 a21
a12 a22
. The value of
= a11a22 − a21a12
66 Chapter 4 Systems of Equations and Inequalities Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Note that vertical bars are used to represent the determinant and that brackets are used to
Example 1
Solution
Find the value of the determinant
3 4 . −1 2
3 4 = 3 ⋅ 2 − ( −1)( 4) = 6 − ( −4) = 10 −1 2 The value of the determinant is 10. For a square matrix whose order is 3 × 3 or greater, the value of the determinant of that matrix is found by using 2 × 2 determinants. The minor of an element in a 3 × 3 determinant is the 2 × 2 determinant that is obtained by eliminating the row and column that contain that element.
Example 2
Solution
2 −3 4 Find the minor of –3 for the determinant 0 4 8 . −1 3 6 The minor of –3 is the 2 × 2 determinant created by eliminating the row and column that contain –3.
2 −3 4 Eliminate the row and column as shown: 0 4 8 −1 3 6
The minor of –3 is
0 8 . −1 6
Definition of a Cofactor The cofactor of an element of a determinant is ( −1)i + j times the minor of that element, where i is the row number of the element and j is the column number of the element.
Example 3
Solution
3 −2 1 For the determinant 2 −5 −4 , find the cofactor of –2 and of –5. 0 3 1 Because –2 is in the first row and the second column, i = 1 and j = 2. Therefore, i + j = 1 + 2 = 3, and ( −1)i + j = ( −1)3 = −1. 2 −4 . 0 1
The cofactor of –2 is ( −1)
Because –5 is in the second row and the second column, i = 2 and j = 2. Therefore, i + j = 2 + 2 = 4, and ( −1)i + j = ( −1)4 = 1.
The cofactor of –5 is 1 ⋅
3 1 . 0 1
The value of a 3 × 3 or larger determinant can be found by expanding by cofactors of any row or any column.
Section 4.3 Solving Systems of Equations by Using Determinants and by Using Matrices 67
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
2
Example Solution
−3
2 −1 . 2 2
Expand by cofactors of the first row. 2 −3 2 3 −1 1 −1 1 3 1 3 −1 = 2 − ( −3) +2 −2 2 0 2 0 −2 0 −2 2
= 2(6 − 2) − ( −3)( 2 − 0) + 2( −2 − 0)
= 2( 4) − ( −3)( 2) + 2( −2) = 8 − ( −6) + ( −4)
= 10
Any row or column can be used to find the value of a determinant by expanding by cofactors. Example 5 is a rework of Example 4 but using the second column. Note that the answers are the same.
Example 5
2 −3 2 Find the value of the determinant 1 3 −1 . 0 −2 2 2 −3 2 1 −1 2 2 2 2 1 3 −1 = −3( −1)1+ 2 + 3( −1)2 + 2 + ( −2)( −1)3+ 2 0 2 0 2 1 −1 0 −2 2
Solution
1 −1 2 2 2 2 + 3⋅1 + ( −2)( −1) 0 2 0 2 1 −1
= −3( −1)
= 3( 2 − 0) + 3( 4 − 0) + 2( −2 − 2)
= 3( 2) + 3( 4) + 2( −4) = 6 + 12 + ( −8)
= 10
Exercises Evaluate the determinant. 5 1 1. −1 2
6 −2 2. −3 4
4 1 3 3. 2 −2 1 3 1 2
3 −1 2 4. 0 1 2 3 2 −2
Objective 2 Solve a system of linear equations by using Cramer’s Rule Cramer’s Rule is a method of solving a system of linear equations by using determinants.
68 Chapter 4 Systems of Equations and Inequalities Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Cramer’s Rule a11 x + a12 y = b1
ations y =
Dy D
Example 6
Dx and D
, where D =
a21 x + a22 y = b2
is given by x =
a11
a12
a21
a22
, Dx =
b1
a12
b2
a22
, Dy =
a11
b1
a21
b2
, and D ≠ 0.
Solve by using Cramer’s Rule: 2 x − 3 y = 8 5x + 6 y = 11
2 −3 = 27 5 6
Steps
• Find the value of the coefficient determinant.
Dx =
8 −3 = 81 11 6
• Find the value of each of the numerator determinants.
Dy =
2 8 = −18 5 11
• Use Cramer’s Rule to write the solutions.
Solution D =
Dx 81 = = 3 D 27
x =
Dy
y =
D
=
−18 2 = − 27 3
2 The solution is 3, − . 3 A problem with Cramer’s Rule is that when D = 0 (the determinant of the coefficient matrix is zero), Cramer’s Rule cannot be used. Cramer’s Rule for Three Equations in Three Variables a11 x + a12 y + a13 z = b1 The solution of the system of equations a21 x + a22 y + a23 z = b2 is given a31 x + a32 y + a33 z = b3 by x =
Dy Dx D , y = , and z = z , where D D D a11
a12
a13
b1
a12
a13
a11
b1
a13
D = a21
a22
a23 , Dx = b2
a22
a23 , Dy = a21
b2
a23 ,
a31
a32
a33
a32
a33
b3
a33
b3
a11
a12
b1
Dz = a21
a22
b2 , and D ≠ 0.
a31
a32
b3
a31
Section 4.3 Solving Systems of Equations by Using Determinants and by Using Matrices 69
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example 7
Solution
Solve by using Cramer’s Rule: 3x − y + z = 5 − 2 z = −3 + z = 4 3 −1 1 D = 1 2 −2 = 28 2 3 1
Steps
• Find the value of the coefficient determinant.
5 −1 1 Dx = −3 2 −2 = 28 4 3 1
• Find the value of each of the numerator determinants.
3 5 1 Dy = 1 −3 −2 = 0 2 4 1
3 −1 5 Dz = 1 2 −3 = 56 2 3 4
x =
• Use Cramer’s Rule to write the solution.
Dx 28 = =1 D 28 Dy 0 y = = = 0 D 28
Dz 56 = = 2 D 28
z =
The solution is (1, 0, 2).
Exercises Solve by using Cramer’s Rule 5. 2 x − 5 y = 26 6. 2 x + 3 y = 4 7. 2 x − y + 3z = 9 8. 3x − 2 y + z = 2 5x + 3 y = 3 6 x − 12 y = −5 x + 4 y + 4 z = 5 2 x + 3 y + 2 z = −6 3x + 2 y + 2 z = 5 3x − y + z = 0
Objective 3 Solve a system of linear equations by using matrices By considering only the coefficients and constants for the following system of equations, we can form the corresponding 3 × 4 augmented matrix.
System of Equations
Augmented Matrix
3x − 2 y + z = 2 x − 3z = −2 2x − y + 4z = 5
3 −2 1 2 1 0 −3 −2 2 −1 4 5
Note that when a term is missing from one of the equations of the system, the coefficient of that term is 0, and a 0 is entered in the matrix.
70 Chapter 4 Systems of Equations and Inequalities Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A system of equations can be solved by writing the system in matrix form and then x similar to those performed on the equations of the lementary row operations. Elementary Row Operations 1. Interchange two rows. 2. Multiply all the elements in a row by the same nonzero number. 3. Replace a row by the sum of that row and a multiple of any other row. Each of these elementary row operations has as its basis the operations that can be performed on a system of equations. These operations do not change the solution of the system of equations. Here are some examples of each row operation. 1. Interchange two rows. Original System
This notation means to interchange rows 1 and 2.
2 −3 1 2x − 3y = 1 4 x + 5 y = 13 4 5 13 The solution is (2, 1).
New System 4 5 13 2 −3 1
R1 ↔ R2
4 x + 5 y = 13 2x − 3y = 1 The solution is (2, 1).
2. Multiply all the elements in a row by the same nonzero number.
This notation means to multiply row 2 by 3.
Original System
2x − 3y = 1 4 x + 5 y = 13
2 −3 1 4 5 13
3R2 →
New System 2 −3 1 12 15 39
The solution is (2, 1).
2x − 3y = 1 12 x + 15 y = 39 The solution is (2, 1).
3. Replace a row by the sum of that row and a multiple of any other row.
This notation means to replace row 2 by the sum of that row and -2 times row 1.
Original System
2 −3 1 4 5 13
New System 2 −3 1 0 11 11
2x − 3y = 1 4 x + 5 y = 13
The solution is (2, 1).
Note that we replace the row that follows the addition sign.
−2 R1 + R2 →
2x − 3y = 1 11 y = 11 The solution is (2, 1).
Section 4.3 Solving Systems of Equations by Using Determinants and by Using Matrices 71
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1
Example
3 −4
6 following elementary row operations on A.
A. R1 ↔ R3 B. −2 R3 C. 2 R3 + R1
A. R1 ↔ R3 means to interchange row 1 and row 3.
Solution
−2 −5 3 4 3 2 0 −1 1 3 −4 6
B. −2 R3 means to multiply row 3 by –2.
1 3 −4 6 3 2 0 −1 R1 ↔ R3 −2 −5 3 4
1 3 −4 6 1 3 −4 6 3 2 0 −1 −2 R3 → 3 2 0 −1 −2 −5 4 10 −6 −8 3 4
C. 2 R3 + R1 means to multiply row 3 by 2 and then add the result to row 1.
1 3 −4 6 3 2 0 −1 2 R3 + R1 → −2 −5 3 4
−3 −7 2 14 3 2 0 −1 −2 −5 3 4
Elementary row operations are used to solve a system of equations. The goal is to use the elementary row operations to rewrite the augmented matrix with 1’s down the main diagonal and 0’s to the left of the 1’s in all rows except the first. This is called the row echelon form of the matrix. Here are some examples of the row echelon form.
1 3 −2 0 1 3
1 −2 3 1 1 2.5 −4 0 0 0 1 2
1 −3 2 0 1 3 0 1 4
0 0 1 −2 3
We will follow a very definite procedure to rewrite an augmented matrix in row echelon form. For a 2 × 3 augmented matrix, use elementary row operations to Step 1: Change a11 to a 1. Step 2: Change a21 to a 0. Step 3: Change a22 to a 1.
Example 9 Solution
a12 a13 a22 a23
3 −6 12 Write the matrix in row echelon form. 1 −3 2 Step 1: Change a11 to 1. One way to do this is to multiply row 1 by the reciprocal of a11 .
a11 a21
3 −6 12 1 1 −2 4 R1 → 1 −3 3 1 −3 2 2
Step 2: Change a21 to 0 by multiplying row 1 by the opposite of a21 and then adding to row 2.
1 −2 4 1 −2 4 −2 R1 + R2 → 0 5 −11 1 −3 2
72 Chapter 4 Systems of Equations and Inequalities Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Step 3: Change a22 to 1 by multiplying by the reciprocal of a22 .
0
5 −11 5
2
1 −2 4 0 1 − 11 5
1 −2 4 . A row echelon form of the matrix is 0 1 − 11 5 For a 3 × 4 augmented matrix, use elementary row operations to Step 1: Change a11 to a 1. Step 2: Change a21 and a31 to 0’s.
a11 a21 a 31
a12 a22 a32
a13 a14 a23 a24 a33 a34
Step 3: Change a22 to a 1. Step 4: Change a32 to a 0. Step 5: Change a33 to a 1.
Example 10
Solution
2 1 3 −1 Write 1 3 5 −1 in row echelon form. −3 −1 1 2 Step 1: Change a11 to 1 by interchanging row 1 and row 2. Note: We could have 1 chosen to multiply row 1 by . The sequence of steps to get to row echelon 2 form is not unique.
1 3 5 −1 2 1 3 −1 −3 −1 1 2
−2 R1 + R2 →
1 3 5 −1 0 −5 −7 1 −3 −1 1 2
Change a31 to 0 by multiplying row 1 by the opposite of a31 and then adding to row 3. 1 3 5 −1 0 −5 −7 1 −3 −1 1 2
3R1 + R3 →
1 3 5 −1 0 −5 −7 1 0 8 16 −1
Step 3: Change a22 to 1 by multiplying by the reciprocal of a22 .
R1 ↔ R2
1 3 5 −1 2 1 3 −1 −3 −1 1 2
Step 2: Change a21 to 0 by multiplying row 1 by the opposite of a21 and then adding to row 2.
2 1 3 −1 1 3 5 −1 −3 −1 1 2
1 3 5 −1 0 −5 −7 1 0 8 16 −1
1 − R2 → 5
1 3 5 −1 0 1 7 − 1 5 5 0 8 16 −1
Section 4.3 Solving Systems of Equations by Using Determinants and by Using Matrices 73
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Step 4: Change a32 to 0 by multiplying row 2 by the opposite of a32 and then
−8 R2 + R3 →
Step 5: Change a33 to 1 by multiplying by the reciprocal of a33 .
0 1 7 − 1 5 5 0 8 16 −1
5 −1 1 3 0 1 7 − 1 5 5 24 3 0 0 5 5
1 3 5 −1 0 1 7 − 1 5 5 24 3 0 0 5 5
5 R → 24 3
1 3 5 −1 0 1 7 − 1 5 5 1 0 0 1 8
1 3 5 −1 0 1 7 − 1 A row echelon form of the matrix is 5 5. 1 0 0 1 8 If an augmented matrix is in row echelon form, the corresponding system of equations can be solved by substitution. For instance, consider the following matrix in row echelon form and the corresponding system of equations. 1 −3 4 7 x − 3y + 4z = 7 0 1 3 −6 y + 3z = −6 0 0 1 −1 z = −1 From the last equation of the system above, we have z = −1. Substitute this value into the second equation and solve for y. Thus y = −3.
Substitute y = −3 and z = −1 in the first equation of the system and solve for x. Thus x = 2.
y + 3z = −6
y + 3( −1) = −6
y − 3 = −6
y = −3
x − 3y + 4z = 7
x − 3( −3) + 4( −1) = 7
x+9−4 = 7
x = 2
The solution of the system of equations is (2, –3, –1). The process of solving a system of equations by using elementary row operations is called the Gaussian elimination method.
Example 11
Solve by using the Gaussian elimination method:
Solution
2 x − 5 y = 19 3x + 4 y = −6
Write the augmented matrix and then use elementary row operations to rewrite the matrix in row echelon form. 2 −5 19 Change a11 to 1. 1 3 4 −6 R 2 1
5 19 1 − 2 2 4 −6 3
Steps
• Note that we multiplied R1 by the reciprocal of a11.
74 Chapter 4 Systems of Equations and Inequalities Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
(1)
(2)
Example 12
5 19 1 − 2 2 0 23 − 69 2 2
• Note that we multiplied R1 by the opposite of a21.
5 19 1 − 2 2 1 −3 0
• This is row echelon form.
5 19 Change a to 1. 22 1 − 2 2 2 R 0 23 − 69 23 2 2 2
19
5
x−
5 19 y = 2 2
• Write the system of equations corresponding to the matrix that is in row echelon form.
• Substitute –3 for y in equation (1) and solve for x.
y = −3 x−
5 19 ( −3) = 2 2 x+
15 19 = 2 2 x = 2
The solution is (2, –3). Solve by using the Gaussian elimination method:
x + 2y − z = 9
2 x − y + 2 z = −1
−2 x + 3 y − 2 z = 7
1 2 −1 9 2 −1 2 −1 −2 3 −2 7
Solution
−2 R1 + R2
1 2 −1 9 0 −5 4 −19 −2 3 −2 7
1 2 −1 9 0 −5 4 −19 −2 3 −2 7
Change a31 to 0. 2 R1 + R3
1 2 −1 9 0 −5 4 −19 0 7 −4 25
1 2 −1 9 Change a22 to 1. 0 1 − 4 19 1 5 5 − R2 0 7 −4 25 5
a11 is 1. Change a21 to 0.
1 2 −1 9 0 1 − 4 19 5 5 0 7 −4 25 1 2 −1 9 0 1 − 4 19 5 5 8 − 8 0 0 5 5
Change a32 to 0. −7 R2 + R3
Change a33 to 1. 5 R 8 3
1 2 −1 9 0 −5 4 −19 0 7 −4 25
1 2 −1 9 0 1 − 4 19 5 5 8 − 8 0 0 5 5 1 2 −1 9 0 1 − 4 19 5 5 0 0 1 −1
Steps
• This is row echelon form.
Section 4.3 Solving Systems of Equations by Using Determinants and by Using Matrices 75
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
(1) x + 2 y − z = 9
• Write the system of equations corresponding to the matrix that is in row echelon form.
• Substitute –1 for z in equation (2) and solve for y.
• Substitute –1 for z and 3 for y in equation (1) and solve for x.
z = −1
(3)
y−
4 19 ( −1) = 5 5
4 19 = 5 5 y = 3
x + 2y − z = 9
x + 2( 3) − ( −1) = 9
x+7 = 9
x = 2
y+
The solution is (2, 3, –1).
Exercises Write each matrix in row echelon form. 2 −4 1 9. 3 −7 −1
5 −2 3 10. 3 1 −7
1 4 1 −2 11. 3 11 −1 2 2 3 1 4
−2 6 −1 3 12. 1 −2 2 1 3 −6 7 6
Solve by using the Gaussian elimination method. 13. 3x + y = 6 14. 5x + 2 y = 3 15. x − y − z = 0 16. x − 2 y + 3z = 2 2 x − y = −1 3x + 4 y = 13 3x − y + 5z = −10 2 x + y + 2 z = 5 x + y − 4 z = 12 2 x − 4 y + 6 z = −4
se c t io n
4.4 Application Problems
Objective 1 Rate-of-wind and water-current problems
Example 1
Strategy
Flying with the wind, a plane flew 1000 mi in 5 h. Flying against the wind, the plane could fly only 500 mi in the same amount of time. Find the rate of the plane in calm air and the rate of the wind. c Rate of the plane in still air: p
Rate of the wind: w Rate
Time
Distance
With wind
p+w
5
5( p + w)
Against wind
p–w
5
5( p − w)
c The distance traveled with the wind is 1000 mi.
The distance traveled against the wind is 500 mi.
76 Chapter 4 Systems of Equations and Inequalities Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Solution
5( p + w) = 1000
p + w = 200
p − w = 100
2 p = 300
p = 150
p + w = 200
1 . 5
Steps
• Multiply each side of the equations by
• Add the equations.
• Substitute the value of p into one of the equations.
150 + w = 200
w = 50
The rate of the plane in calm air is 150 mph.
The rate of the wind is 50 mph.
Example 2
Strategy
A rowing team rowing with the current traveled 18 mi in 2 h. Against the current, the team rowed 10 mi in 2 h. Find the rate of the current. c Rate of the rowing team in calm water: t
Rate of the current: c Rate
Time
Distance
With current
t+c
2
2(t + c )
Against current
t-c
2
2(t − c )
c The distance traveled with the current is 18 mi.
The distance traveled against the current is 10 mi.
Solution
2(t + c ) = 18 → t + c = 9
2(t − c ) = 10 → t − c = 5
2t = 14
t = 7
t + c = 9
7+c = 9
c = 2
The rate of the rowing team in calm water is 7 mph.
The rate of the current is 2 mph.
Exercises Solve. 1. Flying with the wind, a small plane flew 320 mi in 2 h. Against the wind, the plane could fly only 280 mi in the same amount of time. Find the rate of the plane in calm air and the rate of the wind. 2. A cabin cruiser traveling with the current went 48 mi in 3 h. Against the current, it took 4 h to travel the same distance. Find the rate of the cabin cruiser in calm water and the rate of the current.
Section 4.4 Application Problems 77
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Objective 2 Application problems
re varieties of those problems solved earlier blems in this section will result in a system of
equations.
Example 3
Strategy
A citrus fruit grower purchased 25 orange trees and 20 grapefruit trees for $290. The next week, at the same prices, the grower bought 20 orange trees and 30 grapefruit trees for $330. Find the cost of an orange tree and the cost of a grapefruit tree. c Cost of an orange tree: x
Cost of a grapefruit tree: y
First purchase: Amount
Unit Cost
Value
25 20
x y
25x 20y
Amount
Unit Cost
Value
20 30
x y
20x 30y
Orange trees Grapefruit trees
Second purchase:
Orange trees Grapefruit trees
c The total of the first purchase was $290.
The total of the second purchase was $330.
25x + 20 y = 290
20 x + 30 y = 330
Solution
4( 25x + 20 y ) = 4 ⋅ 290
−5( 20 x + 30 y ) = −5 ⋅ 330
100 x + 80 y = 1160
−100 x − 150 y = −1650
−70 y = −490
y = 7
25x + 20 y = 290
25x + 20(7) = 290
25x + 140 = 290
25x = 150
x = 6
The cost of an orange tree is $6.
The cost of a grapefruit tree is $7.
78 Chapter 4 Systems of Equations and Inequalities Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Exercises Solve. 3. A merchant mixe he 15-pound mixture cost $40. A second mixture included 12 lb of the cinnamon tea and 8 lb of the spice tea. The 20-pound mixture cost $54. Find the cost per pound of the cinnamon tea and of the spice tea.
4. A contractor buys 20 yd2 of nylon carpet and 28 yd2 of wool carpet for $1360. A second purchase, at the same prices, includes 15 yd2 of nylon carpet and 20 yd2 of wool carpet for $990. Find the cost per square yard of the wool carpet.
se c t io n
4.5 Solving Systems of Linear Inequalities
Objective 1 Graph the solution set of a system of linear inequalities
Two or more inequalities considered together are called a system of inequalities. The solution set of a system of inequalities is the intersection of the solution sets of the individual inequalities. To graph the solution set of a system of inequalities, first graph the solution set of each inequality. The solution set of the system of inequalities is the region of the plane represented by the intersection of the two shaded areas.
Example 1
Graph the solution set. A. y ≥ x − 1 B. 2 x + 3 y > 9 y < −2 x
Solution
y < −
2 x +1 3
A. Shade above the solid line y = x − 1.
y
Shade below the dashed line y = −2 x.
4
The solution set of the system is the intersection of the solution sets of the individual inequalities.
2 –4 – 2 0 –2
2
4
2
4
x
–4
B. Solve 2 x + 3 y > 9 for y.
3 y > −2 x + 9 y > −
2 x+3 3
2 Shade above the dashed line y = − x + 3. 3
Shade below the dashed line y = −
2x + 3y > 9
2 x + 1. 3
y 4 2 –4 –2 0 –2
x
–4
The intersection of the system is the empty set because the solution sets of the two inequalities do not intersect.
Section 4.5 Solving Systems of Linear Inequalities 79
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Exercises Solve. 1. y ≤ x − 3 4 4. 2 x + y ≥ −2 y ≤ − x + 5 y ≥ −2 x + 2 3x − 2 y < 6 6 x + 3 y ≤ 6
80 Chapter 4 Systems of Equations and Inequalities Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Chapter 5 Image copyright Tatiana Grozetskaya, 2010. Used under license from Shutterstock.com.
Polynomials and Exponents
Prep Test
1. Simplify: −4(3 y )
2. Simplify: ( −2)3
3. Simplify: −4 a − 8b + 7 a
4. Simplify: 3x − 2[ y − 4( x + 1) + 5]
5. Simplify: −( x − y ) 6. Write 40 as a product of prime numbers.
7. Find the GCF of 16, 20, and 24. 8. Evaluate x 3 − 2 x 2 + x + 5 for x = −2.
9. Solve: 3x + 1 = 0
81 Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
se c t io n
5.1 Exponential Expressions
Objective
A monomial is a number, a variable, or a product of a number and variables. The examples at the right are monomials. The degree of a monomial is the sum of the exponents of the variables.
x 3x2
degree 1 ( x = x1 ) degree 2
4 x 2 y degree 3 6 x 3 y 4 z 2 degree 9 xn 6
The degree of a nonzero constant term is zero.
degree n degree 0
The expression x4 is an exponential expression. The exponent, 4, indicates the number of times the base, x, occurs as a factor. Rule for Multiplying Exponential Expressions If m and n are positive integers, then x m
Example 1
Solution
⋅ xn
= xm+ n .
Simplify: (5a 2 b4 )( 2 ab5 ) (5a 2 b4 )( 2 ab5 )
= (5 ⋅ 2)( a 2
= 10 a 2 + 1b4 + 5
= 10a 3 b 9
⋅ a)( b4 ⋅ b5 )
Steps
• Use the Commutative and Associative Properties to rearrange and group factors. • Multiply variables with like bases by adding the exponents.
Rule for Simplifying a Power of an Exponential Expression If m and n are positive integers, then ( x m ) n = x mn .
Rule for Simplifying Powers of Products If m, n, and p are positive integers, then ( x m y n ) p = x mp y np .
Example 2
Solution
Simplify: A. ( x 4 )5 B. ( x 2 ) n C. ( 2 a 3b4 )3 D. ( 2 ab)(3a )2 + 5a( 2 a 2 b) A. ( x 4 )5 = x 4⋅5 = x 20
Steps
• Multiply the exponents.
B. ( x 2 ) n = x 2 n
• Multiply the exponents.
C. ( 2 a 3b4 )3 = 21⋅3 a 3⋅3b4⋅3
• Use the Rule for Simplifying Powers of Products.
= 23 a 9 b12
= 8a 9 b12
D. ( 2 ab)(3a )2 + 5a( 2 a 2 b) = ( 2 ab)(32 a 2 ) + 10 a 3b
= ( 2 ab)(9 a 2 ) + 10 a 3b
= 18a 3b + 10 a 3b = 28a 3 b
82 Chapter 5 Polynomials and Exponents Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Exercises Simplify. 1. ( ab3 )( a 3b)
3. (9 xy 2 )( −2 x 2 y 2 )
4. ( −2 ab2 )3
5. ( 2 xy )( −3x 2 yz )( x 2 y 3 z 3 )
7. ( −3ab3 )3 ( −22 a 2 b)2
8. ( 4 ab)2 ( −2 ab2 c3 )3
6. ( 2 a 2 b)3 ( −3ab4 )2
Objective 2 Divide monomials and simplify expressions with negative exponents
Rules for Dividing Exponential Expressions xm = xm− n . xn
If m and n are integers and x ≠ 0, then
Example 3
Solution
Simplify:
a5 b9 a4b
a5 b9 = a5 − 4 b9 − 1 = ab 8 a4b
Step
• Subtract the exponents of the like bases.
Definition of Zero as an Exponent If x ≠ 0, then x 0 = 1. The expression 00 is not defined.
Example 4
Solution
Simplify: (16 z 5 )0 , z ≠ 0 (16 z 5 )0 = 1
Step
• Any nonzero expression to the zero power is 1.
Definition of a Negative Exponent If x ≠ 0 and n is a positive integer, then x−n =
1 1 and − n = x n n x x
Rule for Simplifying Powers of Quotients xm If m, n, and p are integers and y ≠ 0, then n y
p
=
x mp . y np
Following are some examples that show how operations on exponential expressions are performed.
Section 5.1 Exponential Expressions 83
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example
2
−3
−4
3a 2 b−2 c −1 x 2 y −4 C. 27 a −1b2 c −4 x −5 y −2
5
−2
Solution
= 18 x 2 + ( −4 ) y −3+ 5
= 18 x −2 y 2
=
B.
18 y 2 x2
x 2 y −4 = x 2 − ( −5) y −4 − ( −2 ) x −5 y −2
= x 7 y −2
=
−2
a 3b−4 c3 = 9
• Use the Rule for Multiplying Exponential Equations.
• Use the Definition of a Negative Exponent to rewrite the expression without negative exponents. • Use the Rule for Dividing Exponential Expressions.
x7 y2
3a 2 b−2 c −1 C. 27 a −1b2 c −4
Steps
• Use the Definition of a Negative Exponent to rewrite the expression without negative exponents.
• Simplify inside the parentheses by using the Rule for Dividing Exponential Expressions. • Multiply each exponent inside the parentheses by the exponent outside the parentheses. • Use the Definition of a Negative Exponent to rewrite the expression without negative exponents. • Simplify.
−2
=
a −6 b8 c −6 9 −2
=
92 b8 a6 c6
=
81b 8 a6c 6
Exercises Simplify. 9.
x3 x12
10.
13. ( 2 a −1 ) −2 ( 2 a −1 )4
17.
−48ab10 32 a 4 b3
x3 y6 x3 y3
14. (3a ) −3 (9 a −1 ) −2
18.
( −4 x 2 y 3 )2 ( 2 xy 2 )3
11.
a7 b a 2 b4
12.
15. ( x −2 y )2 ( xy ) −2
x −3 y −4 19. −2 x y
−2
2 x −2 y4
16. ( x −1 y 2 ) −3 ( x 2 y −4 ) −3
a −2 b 20. 3 −4 a b
2
Objective 3 Scientific notation To express a number in scientific notation, write the number as the product of a number between 1 and 10 and a power of 10. The form for scientific notation is a × 10 n , where 1 ≤ a < 10 and n is an integer. For numbers greater than 10, move the decimal point to the right of the first digit. The exponent n is positive and equal to the number of places the decimal point has been moved.
965,000
= 9.65 × 105
3,600,000
= 3.6 × 106
92,000,000,000 = 9.2 × 1010
84 Chapter 5 Polynomials and Exponents Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
For numbers less than 1, move the decimal ro digit. bsolute to the number of places the decimal point has been moved.
Example 6
Solution
Example 7
Solution
Example 8
Solution
0.0002
= 2 × 10 −4
0.0000000974
= 9.74 × 10 −8
0.000000000086 = 8.6 × 10 −11
Write 0.000041 in scientific notation. 0.000041 = 4.1 × 10 −5
Step
• The decimal point must be moved 5 places to the right. The exponent is negative.
Step
• Move the decimal point 7 places to the right.
Write 3.3 × 107 in decimal notation. 3.3 × 107 = 33, 000, 000 Simplify:
2, 400, 000, 000 × 0.0000063 0.00009 × 480
2, 400, 000, 000 × 0.0000063 0.00009 × 480
=
2.4 × 109 × 6.3 × 10 −6 9 × 10 −5 × 4.8 × 102
Steps
• Write the numbers in scientific notation.
=
( 2.4)(6.3) × 109 + ( −6 ) − ( −5) − 2 (9)( 4.8)
• Simplify.
= 0.35 × 106
= 3.5 × 105
• Write in scientific notation.
Exercises Write in scientific notation. 21. 0.00000467
22. 0.00000000017
23. 4,300,000
24. 9,800,000,000
26. 6.2 × 10 −12
27. 8.2 × 1015
28. 6.34 × 105
Write in decimal notation. 25. 1.23 × 10 −7
Simplify. Write the answer in scientific notation. 29. (8.9 × 10 −5 )(3.2 × 10 −6 )
0.00056 0.000000000004
32.
30. (0.0000065)(3,200,000,000,000)
33.
31.
9 × 10 −3 6 × 105
(3.2 × 10 −11 )( 2.9 × 1015 ) 8.1 × 10 −3
Section 5.1 Exponential Expressions 85
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Objective 4 Application problems
Example Strategy
the answer in scientific notation.
y? The speed of light is 186,000 mi/s. Write
To find the distance traveled:
c Write the speed of light in scientific notation.
c Write the number of seconds in 1 day in scientific notation.
c Use the equation d = rt , where r is the speed of light and t is the number of seconds in 1 day.
186, 000 = 1.86 × 105
Solution
24 ⋅ 60 ⋅ 60 = 86, 400 = 8.64 × 104
d = rt
d = (1.86 × 105 )(8.64 × 104 )
d = 1.86 × 8.64 × 109
d = 16.0704 × 109
d = 1.60704 × 1010
Light travels 1.60704 × 1010 mi in one day.
Exercises Solve. Write the answer in scientific notation. 34. A computer can do an arithmetic operation in 2 × 10 −9 s. How many arithmetic operations can the computer perform in 1 min? 35. The mass of an electron is 9.109 × 10 −31 kg. The mass of a proton is 1.673 × 10 −27 kg. How many times heavier is a proton than an electron? 36. The mass of Earth is 5.9 × 1024 kg. The mass of the sun is 2 × 1030 kg. How many times heavier is the sun than Earth?
se c t io n
5.2 Introduction to Polynomials
Objective 1 Evaluate polynomial functions A polynomial is a variable expression in which the terms are monomials. A polynomial of one term is a monomial.
5x
A polynomial of two terms is a binomial.
5x2 y + 6x
A polynomial of three terms is a trinomial.
3 x 2 + 9 xy − 5 y
To evaluate a polynomial function, replace the variable by its value and simplify.
86 Chapter 5 Polynomials and Exponents Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example 1
Given P( x ) = x 3 − 3x 2 + 4, evaluate P( −3).
P( −3) = ( −3)3 − 3( −3)2 + 4
P( −3) = −27 − 27 + 4
P( −3) = −50
Step
• Substitute –3 for x and simplify.
The graph of a linear function is a straight line and can be found by plotting just two points. The graph of a polynomial function of degree greater than 1 is a curve. Consequently, many points may have to be found before an accurate graph can be drawn. Evaluating the quadratic function given by the equation f ( x ) = x 2 − x − 6 when x = −3, –2, –1, 0, 1, 2, 3, and 4 gives the points shown in Figure 1. For instance, f ( −3) = 6, so (–3, 6) is graphed; f ( 2) = −4, so (2, –4) is graphed; and f ( 4) = 6, so (4, 6) is graphed. 3 5 Evaluating the function when x is not an integer, such as x = − and x = , produces 2 2 more points to graph, as shown in Figure 2. Connecting the points with a smooth curve results in Figure 3, which is the graph of f. y
−4
−2
y
y
8
8
8
4
4
4
0
2
4
x
–4
–2
0
2
4
x
–4
–2
0
2
4
x
–4 −8
–8
Figure 1
Example 2
Solution
–8
Figure 2
Figure 3
Graph: f ( x ) = x 2 − 2 x
y = f ( x)
-3
7
4
-2
2
2
-1
-1
0
-2
1
-1
2
2
3
7
y
–4
–2
0
2
4
x
–2 –4
Exercises 1. Given P( x ) = −3x 2 − 5x + 8, evaluate P( −5).
2. Given R( x ) = 2 x 3 − 3x 2 + 4 x − 2, evaluate R( 2).
Graph. 3. P( x ) = x 2 − 1
4. R( x ) = x 3 + 2
5. f ( x ) = x 3 − 2 x
Objective 2 Add or subtract polynomials Polynomials can be added by combining like terms.
Section 5.2 Introduction to Polynomials 87
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example 3
Solution
Add: (3x 2 − 2 x + 5) + ( 2 − 3x + 4 x 3 ) + ( x − 3x 2 − x 3 − 4)
+5 Pillaged3xby− 2xLibraryPirate 2
4x
3
Steps
• Arrange the terms of each polynomial in descending order with like terms in the same column.
• Combine like terms in each column.
− 3x + 2
3
2
− x − 3x + x − 4 3x3
− 4x + 3
The additive inverse of the polynomial x 2 + 5x − 4 is –( x 2 + 5 x – 4). To simplify the additive inverse of a polynomial, change the sign of every term inside the parentheses.
− ( x 2 + 5x − 4) = − x 2 − 5 x + 4
To subtract two polynomials, add the additive inverse of the second polynomial to the first.
Example 4
Solution
Subtract: (3x 2 − 2 x + 4) − (7 x 2 + 3x − 12) (3x 2 − 2 x + 4) − (7 x 2 + 3x − 12) (3x 2 − 2 x + 4) + ( −7 x 2 − 3x + 12) 3x 2 − 2 x + 4
Steps
2
−7 x − 3x + 12 −4 x 2 − 5 x + 16
• Rewrite the subtraction as the addition of the additive inverse. • Arrange the terms of each polynomial in descending order in a vertical format. • Combine like terms in each column.
Exercises Simplify. 6. (5x 2 + 2 x − 7 ) + ( x 2 − 8 x + 12)
7. (3x 2 − 2 x + 7 ) + ( −3x 2 + 2 x − 12)
8. ( x 2 + 8 − 3x ) − ( 2 x 2 − 3x + 7 )
9. ( 2 x 2 + 3x − 7 ) − (5x 2 − 1 − 8 x )
se c t io n
5.3 Multiplication of Polynomials
Objective 1 Multiply a polynomial by a monomial
To multiply a polynomial by a monomial, use the Distributive Property and the Rule for Multiplying Exponential Expressions.
Example 1
Multiply.
A. −5x ( x 2 − 2 x + 3) B. x 2 − x[3 − x ( x − 2) + 3]
A. −5x ( x 2 − 2 x + 3)
Solution
= −5x ( x 2 ) − ( −5x )( 2 x ) + ( −5x )(3)
Steps
• Use the Distributive Property.
= −5 x 3 + 10 x 2 − 15 x
• Use the Rule for Multiplying Exponential Expressions.
88 Chapter 5 Polynomials and Exponents Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
B. x 2 − x[3 − x ( x − 2) + 3]
]
= x 2 − x[6 − x 2 + 2 x ]
= x2 − 6x + x3 − 2x2
= x3 − x2 − 6x
• Use the Distributive Property to remove the inner grouping symbols. • Combine like terms. • Use the Distributive Property to remove the brackets. • Combine like terms and write the polynomial in descending order.
Exercises Simplify. 1. 2 x ( x − 3)
2. −4 y 2 ( 4 y − 6 y 2 )
3. 4b(3b3 − 12b2 − 6)
4. x − 2 x ( x − 2)
5. 2 y 2 − y[3 − 2( y − 4) − y ]
6. 3x 2 − x[ x − 2(3x − 4)]
Objective 2 Multiply two polynomials
The product of two polynomials is the polynomial obtained by multiplying each term of one polynomial by each term of the other polynomial and then combining like terms. A convenient method of multiplying two polynomials is to use a vertical format similar to that used for multiplication of whole numbers. Like terms are written in the same column. Combine like terms.
Example 2
Multiply: ( 4 a 3 − 3a + 7 )( a − 5) 4 a 3 − 3a + 7 a − 5
Solution
2x2 − 2x + 1 3x + 2 2 4 x − 4 x + 2 = 2(2 x 2 − 2 x + 1) 3 = 3 x (2 x 2 − 2 x + 1) 6 x − 6 x 2 + 3x 3 2 6x − 2x − x + 2
− 20 a 3 4a
4 4
+ 15a − 35 2
− 3a + 7 a 3
4a − 20a − 3a 2 + 22a − 35 It is frequently necessary to find the product of two binomials. The product can be found by using a method called FOIL, which is based on the Distributive Property. The letters FOIL stand for First, Outer, Inner, and Last. Multiply: (3x − 2)( 2 x + 5) Multiply the First terms.
( 3 x − 2)( 2 x + 5)
3x ⋅ 2 x = 6 x 2
Multiply the Outer terms.
( 3 x − 2)( 2x + 5)
3x ⋅ 5 = 15x
Multiply the Inner terms.
(3x − 2)( 2x + 5)
−2 ⋅ 2x = −4x
Multiply the Last terms.
(3x − 2)( 2 x + 5)
−2 ⋅ 5 = −10
Add the products. Combine like terms.
(3x − 2)( 2 x + 5)
=
F O I L 2 6 x + 15 x − 4 x − 10
= 6 x 2 + 11 x − 10 Section 5.3 Multiplication of Polynomials 89
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example 3
Multiply: (6 x − 5)(3x − 4) ( −5)(3x ) + ( −5)( −4)
Soluti
= 18 x 2 − 24 x − 15x + 20
= 18 x 2 − 39 x + 20
Exercises Simplify. 7. (5x − 7 )(3x − 8) 10. ( x − 2)( x 2 − 3x + 7 )
8. ( 2 x − 3 y )( 2 x + 5 y )
9. (7 x − 3 y )( 2 x − 9 y )
11. ( x + 3)( x 2 + 5x − 8)
12. ( x + 5)( x 3 − 3x + 4)
Objective 3 Multiply polynomials that have special products
Using FOIL, a pattern can be found for the product of the sum and difference of two terms and for the square of a binomial. ( a + b)( a − b) = a 2 − ab + ab − b2
The Sum and Difference of Two Terms
= a 2 − b2
Square of the first term
Square of the second term
The Square of a Binomial
( a + b)2 = ( a + b)( a + b) = a 2 + ab + ab + b2 = a 2 + 2 ab + b2
Example 4
Solution
Square of the first term
Twice the product of the two terms
Square of the second term Simplify. A. ( 4 x + 3)( 4 x − 3) B. ( 2 x − 3 y )2 A. ( 4 x + 3)( 4 x − 3) = ( 4 x )2 − 32
• This is the sum and difference of two terms.
• This is the square of a binomial.
= 16 x 2 − 9
Steps
B. ( 2 x − 3 y )2
= ( 2 x )2 + 2( 2 x )( −3 y ) + ( −3 y )2
= 4 x 2 − 12 xy + 9 y 2
Exercises Simplify. 13. ( a − 4)( a + 4)
14. (3x − 2)(3x + 2)
15. ( b − 11)( b + 11)
16. (3a + 5b)2
17. (5x − 4 y )2
18. (3a − 4b)2
90 Chapter 5 Polynomials and Exponents Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Objective 4 Application problems
Ex
+ 3) ft. The width is ( x − 5) ft. Find the area of the rectangle in terms of the variable x. Strategy
To find the area, replace the variables L and W in the equation A = LW by the given values, and solve for A.
x–5 2x + 3
A = LW
Solution
A = ( 2 x + 3)( x − 5)
A = 2 x 2 − 10 x + 3x − 15
A = 2 x 2 − 7 x − 15
The area is (2 x 2 − 7 x − 15) ft2.
Exercises 19. The length of a rectangle is (3x + 3) ft. The width is (x – 4) ft. Find the area of the rectangle in terms of the variable x. 20. The base of a triangle is (x + 2) ft. The height is (2x – 3) ft. Find the area of the triangle in terms of the variable x.
se c t io n
5.4 Division of Polynomials
Objective 1 Divide a polynomial by a monomial
To divide a polynomial by a monomial, divide each term of the numerator by the denominator, and write the sum of the quotients. Divide each term of the polynomial 8 x 3 − 4 x 2 + 6 x by the monomial 2x.
8x 3 − 4 x 2 + 6 x 8x 3 4x2 6x = − + 2x 2x 2x 2x = 4x2 − 2x + 3
Simplify.
Example 1
Solution
6 p 4 − 9 p 3 − 12 p 2 3 p2 4 3 6 p − 9 p − 12 p 2 6 p4 9 p 3 12 p 2 = − − 2 2 3p 3p 3 p2 3 p2
Divide:
= 2 p2 − 3 p − 4
Example 2
Solution
Divide:
• Divide each term of the polynomial by 3p 2 . • Simplify.
18a 2 b + 27 a 2 b2 − 9 ab2 9a 2 b
18a 2 b + 27 a 2 b2 − 9 ab2 9a 2 b 18a 2 b 27 a 2 b2 9 ab2 + − 9a 2 b 9a 2 b 9a 2 b
=
= 2 + 3b −
Steps
b a
Steps
• Divide each term of the polynomial by 9a 2 b.
• Simplify.
Section 5.4 Division of Polynomials 91
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Exercises Simplify. 1.
4a − 8 4
2.
6w + 4w 2w
3.
8z 3 − 6 z 2 −2 z
4.
3t 3 − 9t 2 + 12t 3t
5.
12 y 2 − 9 y + 6 −3 y
6.
12 x 2 y 2 − 16 x 2 y + 20 xy 2 4x2 y
Objective 2 Divide polynomials The division method illustrated in Objective 1 is appropriate only when the divisor is a monomial. To divide two polynomials in which the divisor is not a monomial, use a method similar to that used for division of whole numbers. Divide: ( x 2 + 5x − 7 ) ÷ ( x + 3) Step 1
)
)
Think: x x 2 =
x
x + 3 x 2 + 5x − 7
Step 2
)
Multiply: x ( x + 3) = x 2 + 3 x Subtract: ( x 2 + 5x ) − ( x 2 + 3x ) = 2x Bring down –7.
x2 + 3x 2x − 7 x+2
x + 3 x + 5x − 7
x2 = x x
2
)
Think: x 2 x =
x 2 + 3x 2x − 7 2x + 6 −13
2x = 2 x
Multiply: 2( x + 3) = 2 x + 6 Subtract: ( 2 x − 7 ) − ( 2 x + 6) = −13 The remainder is –13.
( x 2 + 5x − 7 ) ÷ ( x + 3) = x + 2 –
13 x+3
When dividing polynomials, make sure the terms are in descending order. Here is an example. Divide:
6 − 6x2 + 4x3 2x + 3
Arrange the terms in descending order. Note that there is no term of x in 4 x 3 − 6 x 2 + 6. Insert a zero as 0x for the missing term so that like terms will be in the same columns.
)
2x2 − 6x + 9
2x + 3 4x3 − 4x3 +
6 x 2 + 0x + 6 6x2 −12 x 2 + 0 x −12 x 2 − 18 x 18 x + 6 18 x + 27 −21
4x3 − 6x2 + 6 21 = 2x2 − 6x + 9 − 2x + 3 2x + 3 92 Chapter 5 Polynomials and Exponents Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example 3
Divide. A.
12 x 2 − 11x + 10
B.
x3 + 1 x+1
)
4 x − 5 12 x 2 − 11x + 10
12 x 2 − 15x 4 x + 10 4x − 5 15
12 x 2 − 11x + 10 15 = 3x + 1 + 4x − 5 4x − 5
B.
)
x2 − x + 1
x + 1 x3 + 0x2 + 0x + 1
x3 + x2 − x2 + 0x − x2 − x x+1 x+1 0
x3 + 1 = x2 – x + 1 x +1
Exercises Simplify. 7. ( x 2 + 3x − 40) ÷ ( x − 5)
8. (12 x 2 + 13x − 14) ÷ (3x − 2)
9. ( x 3 − 3x 2 + 2) ÷ ( x − 3)
11.
10. ( x 3 + 4 x 2 − 8) ÷ ( x + 4)
4 − 7 x + 5x 2 − x 3 x−3
12.
4 + 6 x − 3x 2 + 2 x 3 2x + 1
Objective 3 Synthetic division Synthetic division is a shorter method of dividing a polynomial by a binomial of the form x - a. This method of dividing uses only the coefficients of the variable terms and the constant term. Following is an example.
Section 5.4 Division of Polynomials 93
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
)
(
Divide: 3x 2 − 10 x + 7 ÷ ( x − 2) x − a = x − 2; a = 2
Value of a Coefficients of the dividend 2
3
–10
7
–10 6 –4
7
–10 6 –4
7 –8 –1
3
Bring down the 3. Multiply 2 ⋅ 3 and add the product (6) to –10. The result is –4.
2
Multiply 2(–4) and add the product (–8) to 7. The result is –1.
2
3 3 3 3
Coefficients of the quotient
( 3 x 2 − 10 x + 7 ) ÷ ( x − 2) = 3 x − 4 −
The degree of the first term of the quotient is one degree less than the degree of the first term of the dividend.
Example 4
1 x −2
Check: (3x − 4)( x − 2) − 1
= 3x 2 − 6 x − 4 x + 8 − 1
= 3x 2 − 10 x + 7
Divide: A. (5x 2 − 3x + 7 ) ÷ ( x − 1) B. (3x 4 − 8 x 2 + 2 x + 1) ÷ ( x + 2)
Solution
Remainder
A. 1
5
5
= 5x + 2 +
B. –2 3
7 2 9
Steps
• x – a = x – 1; a = 1
• Insert a zero for the missing term.
• x – a = x + 2; a = –2
(5x 2 − 3x + 7 ) ÷ ( x − 1)
–3 5 2
3
0 –6 –6
9 x−1 –8 12 4
2 –8 –6
1 12 13
(3x 4 − 8 x 2 + 2 x + 1) ÷ ( x + 2)
= 3x3 − 6x2 + 4x − 6 +
13 x+2
Exercises Divide by using synthetic division. 13. ( 2 x 2 − 6 x − 8) ÷ ( x + 1)
14. (3x 2 + 19 x + 20) ÷ ( x + 5)
15. ( 2 x 3 − x 2 + 6 x + 9) ÷ ( x + 1)
16. ( x 3 − 6 x 2 + 11x − 6) ÷ ( x − 3)
17. ( x 3 + x − 2) ÷ ( x + 1)
18. (6 x − 3x 2 + x 3 − 9) ÷ ( x + 2)
94 Chapter 5 Polynomials and Exponents Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Objective 4 Evaluate a polynomial using synthetic division
ing synthetic division and the following theorem. Remainder Theorem If the polynomial P( x ) is divided by x − a, the remainder is P( a ). Evaluate P( x ) = x 4 − 3x 2 + 4 x − 5 when x = −2 by using the Remainder Theorem. The value at which the polynomial is evaluated –2 1
0 –2 –2
1
–3 4 1
4 –2 2
–5 –4 –9
A 0 is inserted for the x3 term. The remainder
P( −2) = −9
Example 5
Solution
Use the Remainder Theorem to evaluate P( −2) when P( x ) = x 3 − 3x 2 + x + 3. Use synthetic division with a = −2. –2 1 1
–3 –2 –5
1 10 11
3 –22 –19
By the Remainder Theorem, P( −2) = −19.
Exercises Use the Remainder Theorem to evaluate the polynomial. 19. P( x ) = 2 x 2 − 3x − 1; P(3)
20. Q( x ) = 3x 2 − 5x − 1; Q( 2)
21. P( z ) = 2 z 3 − 4 z 2 + 3z − 1; P( −2)
22. R(t ) = 3t 3 + t 2 − 4t + 2; R( −3)
23. F ( x ) = 2 x 4 − x 3 − 2 x − 5; F( −3)
24. R(t ) = 4t 4 − 3t 2 + 5; R( −3)
se c t io n
5.5 Factoring Polynomials
Objective 1 Factor a monomial from a polynomial To factor a polynomial means to write the polynomial as a product of other polynomials. In the example at the right, 3x is the GCF of the terms 3x2 and 6x. 3x is a monomial factor of the terms of the binomial. x + 2 is a binomial factor of 3x 2 + 6 x.
Multiply
Polynomial 3x2 + 6x
=
Factors 3x (x + 2)
Factor
Section 5.5 Factoring Polynomials 95
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example 1
Factor: 4 x 3 y 2 + 12 x 3 y + 20 xy 2 y2 .
Soluti
⋅ x3 ⋅ y2 12 x 3 y = 22 ⋅ 3 ⋅ x 3 ⋅ y 20 xy 2 = 22 ⋅ 5 ⋅ x ⋅ y 2 GCF = 22 ⋅ x ⋅ y = 4xy
Rewrite each term of the polynomial as a product with the GCF as one of the factors.
4 x 3 y 2 + 12 x 3 y + 20 xy 2
4 x 3 y 2 = 22
= 4 xy ( x 2 y ) + 4 xy (3x 2 ) + 4 xy (5 y )
= 4 xy ( x 2 y + 3 x 2 + 5 y )
Exercises Factor. 1. 6 a 2 − 15a
2. 32b2 + 12b
3. 9 x 2 + 14 y 2
4. 16 x 2 − 12 x + 24
5. 2 x 5 + 3x 4 − 4 x 2
6. 5b2 − 10b3 + 25b4
Objective 2 Factor by grouping In the examples at the right, the binomials in parentheses are binomial factors.
4 x 4 ( 2 x − 3) −2r 2 s(5r + 2s)
The Distributive Property is used to factor a common binomial factor from an expression. Factor: 4 a( 2b + 3) − 5( 2b + 3) The common binomial factor is ( 2b + 3). Use the Distributive Property to write the expression as a product of factors.
4 a( 2b + 3) − 5( 2b + 3) = (2b + 3)(4a − 5)
Consider the binomial y − x. Factoring –1 from this binomial gives y − x = −( x − y )
This equation is sometimes used to factor a common binomial from an expression. Some polynomials can be factored by grouping terms so that a common binomial factor is found. Factor: 8 y 2 + 4 y − 6 ay − 3a 8 y 2 + 4 y − 6 ay − 3a
= (8 y 2 + 4 y ) − (6 ay + 3a )
Factor the GCF from each group.
= 4 y ( 2 y + 1) − 3a( 2 y + 1)
Write the expression as a product of factors.
= (2 y + 1)(4 y − 3a )
Group the first two terms and the last two terms. Note that
−6 ay − 3a = −(6 ay + 3a ).
96 Chapter 5 Polynomials and Exponents Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example 2
Factor. A. 3x ( y − 4) − 2( 4 − y ) B. xy − 4 x − 2 y + 8
4)
Steps
= ( y − 4)(3 x + 2)
• Write the expression as a sum of terms that have a common factor. Note that 4 – y = –(y – 4). • Write the expression as a product of factors.
B. xy − 4 x − 2 y + 8
= ( xy − 4 x ) − ( 2 y − 8)
= x ( y − 4) − 2( y − 4)
= ( y − 4)( x − 2)
• Group the first two terms and the last two terms. • Factor the GCF from each group. • Write the expression as the product of factors.
Exercises Factor by grouping. 7. x ( a + 2) − 2( a + 2)
9. x 2 + 3x + 2 x + 6
8. a( x − 2) − b( 2 − x )
10. x 2 − 5x + 4 x − 20
11. xy + 4 y − 2 x − 8
12. ab + 7 b − 3a − 21
Objective 3 Factor trinomials of the form x2 + bx + c
A quadratic trinomial is a trinomial of the form ax 2 + bx + c, where a and b are nonzero coefficients and c is a nonzero constant. The degree of a quadratic trinomial is 2. To factor a quadratic trinomial means to express the trinomial as the product of two binomials. For example,
Trinomial
Factored Form
2x2 − x − 1
=
( 2 x + 1)( x − 1)
y2 − 3y + 2
=
( y − 1)( y − 2)
In this objective, trinomials of the form x 2 + bx + c ( a = 1) will be factored. The next objective deals with trinomials for which a ≠ 1. The method by which factors of a trinomial are found is based on FOIL. For example, to factor x 2 − 5x − 24, find two numbers whose product is -24 and whose sum is -5. The numbers are -8 and 3. x 2 − 5x − 24 = ( x − 8)( x + 3)
Example 3
Solution
Factor: A. x 2 + 8 x + 12 B. x 2 + 5x − 84 C. x 2 + 3x + 4 A. Find two integers whose product is 12 and whose sum is 8. The numbers are 2 and 6; 2 ⋅ 6 = 12, 2 + 6 = 8.
B. Find two integers whose product is –84 and whose sum is 5. The numbers are -7 and 12; −7 ⋅ 12 = −84, −7 + 12 = 5.
x 2 + 8 x + 12 = ( x + 2)( x + 6)
x 2 + 5x − 84 = ( x + 12)( x – 7)
C. There are no two integers whose product is 4 and whose sum is 3. x 2 + 3 x + 4 is nonfactorable over the integers. Section 5.5 Factoring Polynomials 97
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Exercises Factor. 13. x 2 − 8 x + 15
14. x + 12 x + 20
15. a 2 + 12 a + 11
16. a 2 + a − 72
17. b2 + 2b − 35
18. x 2 + 4 x + 5
Objective 4 Factor trinomials of the form ax2 + bx + c
There are various methods of factoring trinomials of the form ax 2 + bx + c, where a ≠ 1. Factoring by using trial factors and factoring by grouping will be discussed in this objective. Factoring by using trial factors is illustrated first. To use the trial factor method, use the factors of a and the factors of c to write all the possible binomial factors of the trinomial. Then use FOIL to determine the correct factorization. To reduce the number of trial factors that must be considered, remember that if the terms of the trinomial do not have a common factor, then the terms in either one of the binomial factors will not have a common factor.
Example 4
Solution
Factor: A. 2 x 2 − 21x + 10 B. 6 x 2 + 17 x − 10 A. 2 x 2 − 21x + 10 Factors of 2
Factors of 10
1, 2
-1, -10
Steps
• Use negative factors of 10.
• Write trial factors. Use FOIL to check the middle term.
• Find the factors of a (6) and the factors of c (–10).
• Write trial factors. Use FOIL to check the middle term.
-2, -5
Trial Factors
Middle Term
( x − 2)( 2 x − 5)
−5x − 4 x = −9 x
( 2 x − 1)( x − 10)
−20 x − x = −21x
2 x 2 − 21x + 10 = (2 x − 1)( x − 10)
B. 6 x 2 + 17 x − 10 Factors of 6
Factors of –10
1, 6
1, -10
2, 3
-1, 10 2, -5 -2, 5
Trial Factors
Middle Term
( x + 2)(6 x − 5)
−5x + 12 x = 7 x
( x − 2)(6 x + 5)
5x − 12 x = −7 x
( 2 x + 1)(3x − 10)
−20 x + 3x = −17 x
( 2 x − 1)(3x + 10)
20 x − 3x = 17 x
6 x 2 + 17 x − 10 = (2 x − 1)(3 x + 10)
98 Chapter 5 Polynomials and Exponents Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Trinomials of the form ax 2 + bx + c can also be factored by grouping. This method is an Objective 2 of this section. e factors of a ⋅ c whose sum is b. Use the two factors to rewrite the middle term of the trinomial as the sum of two terms. Then factor by grouping to write the factorization of the trinomial. Factor: 4 z 2 − 17 z − 21 Find two factors of –84 [ a ⋅ c = 4( −21)] whose sum is –17, the coefficient of z.
Factors of –84
Sum
1, –84 –1, 84 2, –42 –2, 42 3, –28 –3, 28 4, –21
–83 83 –40 40 –25 25 –17
When the required sum is found, the remaining factors need not be checked.
Use the factors of –84 whose sum is –17 to write –17z as 4 z − 21z. Factor by grouping. Recall that −21z − 21 = −( 21z + 21).
Example 5
Solution
4 z 2 − 17 z − 21 = 4 z 2 + 4 z − 21z − 21
= ( 4 z 2 + 4 z ) − ( 21z + 21)
= 4 z ( z + 1) − 21( z + 1)
= ( z + 1)(4 z − 21)
Factor: 6 x 2 + 7 x − 20 a ⋅ c = −120
Factors of –120
Sum
–1, 120 –2, 60 –3, 40 –4, 30 –5, 24 –6, 20 –8, 15
119 58 37 26 19 14 7
6 x 2 + 7 x − 20
= 6 x 2 + 15x − 8 x − 20
= (6 x 2 + 15x ) − (8 x + 20)
= 3x ( 2 x + 5) − 4( 2 x + 5)
= (2 x + 5)(3 x − 4)
Exercises Factor. Use either the method of trial factors or factoring by grouping. 19. 2 x 2 + 7 x + 3
20. 2 x 2 − 11x − 40
21. 6 y 2 + 5 y − 6
22. 4 y 2 − 15 y + 9
23. 6b2 − b − 35
24. 2 a 2 + 13a + 6
se c t io n
5.6 Special Factoring
Objective 1 Factor the difference of two perfect squares and factor perfect-square trinomials
The factors of the difference of two perfect squares are the sum and difference of the square roots of the perfect squares.
Section 5.6 Special Factoring 99
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Factors of the Difference of Two Perfect Squares ( a + b)( a − b) The sum of two perfect squares, a 2 + b2 , is nonfactorable over the integers.
Example 1
Solution
Factor: 25x 2 − 1 25x 2 − 1 = (5x )2 − 12
Steps
= (5 x + 1)(5 x − 1)
• Write the binomial as the difference of two perfect squares. • The factors are the sum and difference of the square roots of the perfect squares.
A perfect-square trinomial is the square of a binomial. Factors of a Perfect-Square Trinomial a 2 + 2 ab + b2 = ( a + b)2 a 2 − 2 ab + b2 = ( a − b)2
Example 2
Solution
Factor: 4 x 2 − 12 x + 9 Because 4 x 2 is a perfect square and 9 is a perfect square, try factoring 4 x 2 − 12 x + 9 as the square of a binomial. ?
4 x 2 − 12 x + 9 = ( 2 x − 3)2
Check:
( 2 x − 3)2 = ( 2 x − 3)( 2 x − 3) = 4 x 2 − 6 x − 6 x + 9 = 4 x 2 − 12 x + 9
The factorization checks. Thus 4 x 2 − 12 x + 9 = (2 x − 3)2 .
Exercises Factor. 1. x 2 − 16
2. 4 x 2 − 1
3. 81x 2 − 4
4. y 2 − 6 y + 9
5. b2 − 18b + 81
6. 16 x 2 − 40 x + 25
Objective 2 Factor the sum or the difference of two cubes The following rules are used to factor the sum or difference of two perfect cubes. Factoring the Sum or Difference of Two Perfect Cubes a 3 + b3 = ( a + b)( a 2 − ab + b2 ) a 3 − b3 = ( a − b)( a 2 + ab + b2 )
100 Chapter 5 Polynomials and Exponents Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
To factor 8 x 3 − 27: 8 x 3 − 27 = ( 2 x )3 − 33 The terms of the binomial factor are the cube roots of the perfect cubes. The sign of the binomial factor is the same sign as in the given binomial. The trinomial factor is obtained from the binomial factor.
Example 3
Solution
3
= ( 2 x − 3)( 4 x 2 + 6 x + 9)
Square of the first term Opposite of the product of the two terms Square of the last term
3
Factor: x y − 8 x 3 y 3 − 8 = ( xy )3 − 23 = ( xy – 2)( x 2 y 2 + 2 xy + 4)
Steps
• Write as the difference of cubes.
• Use the rule for factoring the difference of cubes.
Exercises Factor. 7. x 3 − 27
8. y 3 + 125
10. 64 a 3 + 27
11. 27 x 3 − 8 y 3
9. 8 x 3 − 1
12. 64 x 3 + 27 y 3
Objective 3 Factor trinomials that are quadratic in form
A trinomial is quadratic in form if it can be written as au 2 + bu + c.
Example 4
Solution
= ( x 2 )2 + 5( x 2 ) + 6
The trinomial is quadratic in form.
= u 2 + 5u + 6
= 2( xy )2 + 3( xy ) − 9
The trinomial is quadratic in form.
= 2u 2 + 3u − 9
Factor: A. 6 x 2 y 2 − xy − 12 B. 2 x 4 + 5x 2 − 12 A. 6 x 2 y 2 − xy − 12
Steps
• The trinomial is quadratic in form when xy = u.
• The trinomial is quadratic in form when x 2 = u.
= (3 xy + 4)(2 xy − 3)
B. 2 x 4 + 5x 2 − 12
2 x 2 y 2 + 3xy − 9
If we let xy = u, the trinomial 2 x 2 y 2 + 3xy − 9 can be written as shown at the right.
x 4 + 5x 2 + 6
If we let x 2 = u, the trinomial x 4 + 5x 2 + 6 can be written as shown at the right.
= ( x 2 + 4)(2 x 2 − 3)
Section 5.6 Special Factoring 101
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Exercises Factor. 13. x 2 y 2 − 8 xy + 15
15. x 4 − 9 x 2 + 18
16. b4 − 13b2 − 90
17. a 4 + 14 a 2 + 45
18. y 4 − 6 y 2 − 16
Objective 4 Factor completely
When factoring a polynomial completely, ask the following questions about the polynomial. 1. Is there a common factor? If so, factor out the GCF. 2. If the polynomial is a binomial, is it the difference of two perfect squares, the sum of two perfect cubes, or the difference of two perfect cubes? If so, factor. 3. If the polynomial is a trinomial, is it a perfect-square trinomial or the product of two binomials? If so, factor. 4. If the polynomial has four terms, can it be factored by grouping? If so, factor. 5. Is each factor nonfactorable over the integers? If not, factor.
Example 5
Solution
Factor: A. 2 x 3 − 2 x 2 − 12 x B. x 4 − 81 C. 2 x 3 + 4 x 2 − 3x − 6 A. 2 x 3 − 2 x 2 − 12 x = 2 x ( x 2 − x − 6) = 2 x ( x + 2)( x − 3)
B. x 4 − 81 = ( x 2 − 9)( x 2 + 9)
= ( x + 3)( x − 3)( x 2 + 9)
C. 2 x 3 + 4 x 2 − 3x − 6 = ( 2 x 3 + 4 x 2 ) − (3x + 6)
= 2 x ( x + 2) − 3( x + 2)
= ( x + 2)(2 x − 3)
Exercises Factor. 19. 12 x 2 − 36 x + 27
20. 7 x 2 − 28
21. y 5 + 6 y 4 − 55 y 3
22. x 4 − 16
23. 3x 3 − 3x 2 + 4 x − 4
24. 3x 4 − 81x
102 Chapter 5 Polynomials and Exponents Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
se c t io n
5.7 Solving Equations by Factoring
Obj
Principle of Zero Products If the product of two factors is zero, then at least one of the factors must be zero. If ab = 0, then a = 0 or b = 0. The Principle of Zero Products is used to solve equations.
Example 1
Solution
Solve: ( x − 4)( x + 2) = 0 By the Principle of Zero Products, if ( x − 4)( x + 2) = 0, then x − 4 = 0 or x + 2 = 0.
x − 4 = 0
x = 4
Example 2
Solution
x+2 = 0 x = −2
The solutions are -2 and 4. Solve: A. 3x 2 + 5x = 2 B. ( x + 4)( x − 3) = 8 A.
3x 2 + 5x = 2
3x 2 + 5x − 2 = 0
Steps
• Write the equation in standard form.
(3x − 1)( x + 2) = 0
• Factor the trinomial.
3x − 1 = 0
3x = 1
• Let each factor equal zero (the Principle of Zero Products). • Solve each equation for x.
• Write the solutions.
x =
x+2 = 0 x = −2
1 3
The solutions are
1 and -2. 3
B. ( x + 4)( x − 3) = 8
x 2 + x − 12 = 8
x 2 + x − 20 = 0
( x + 5)( x − 4) = 0
• Write the equation in standard form by first multiplying the binomials. • Subtract 8 from each side of the equation. • Factor the trinomial.
x + 5 = 0
x−4 = 0
• Let each factor equal zero.
x = 4
• Solve each equation for x.
• Write the solutions.
x = −5
The solutions are -5 and 4.
Recall that a zero of a function f is a number c for which f ( c ) = 0 . The zeros of some quadratic functions can be found by using the Principle of Zero Products.
Example 3
Solution
Find the zeros of f ( x ) = x 2 − 2 x − 8. We must find the values of c for which f ( c ) = 0.
f (c) = 0
c 2 − 2c − 8 = 0
( c + 2)( c − 4) = 0
Steps
• f ( x ) = x 2 − 2x − 8. Thus f (c ) = c 2 − 2c − 8. • Factor and use the Principle of Zero Products.
Section 5.7 Solving Equations by Factoring 103
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
c + 2 = 0
c−4 = 0
Exercises Solve. 1. ( y + 4)( y + 6) = 0
2. ( 2 x + 3)( x − 7 ) = 0
3. y 2 + 4 y − 5 = 0
4. b2 − 4b = 32
5. z ( z − 1) = 20
6. ( x + 2)( x − 6) = 20
8. g ( x ) = x 2 + 2 x − 15
9. g ( x ) = 3x 2 − 7 x − 6
Find the zeros of the function. 7. f ( x ) = x 2 + 3x − 4
Objective 2 Application problems
Example 4
Strategy
Solution
An architect wants to design a fountain to be placed on the front lawn of a new art museum. The base of the fountain is to be 30 ft by 40 ft with a uniform brick walkway around the fountain. The total area of the fountain and the walkway is 2576 ft 2 . How wide is the walkway? Let x represent the width of the brick walkway. Then the width of the fountain and walkway is 30 + 2 x, and the length of the fountain and walkway is 40 + 2 x. Use the formula A = LW , where A = 2576, to write an equation. A = LW
2576 = ( 40 + 2 x )(30 + 2 x )
Steps
• Replace A, L, and W by their values.
2576 = 1200 + 140 x + 4 x 2
• Multiply.
2
0 = 4 x + 140 x − 1376
0 = x 2 + 35x − 344
• Divide each side by 4.
0 = ( x − 8)( x + 43)
• Factor and use the Principle of Zero Products.
x − 8 = 0
x = 8
x + 43 = 0 x = −43
A walkway –43 ft wide would not make sense.
The walkway must be 8 ft wide.
Exercises 10. The length of a rectangle is 5 in. more than twice the width. The area is 168 in2. Find the width and length of the rectangle. 11. An object is thrown downward, with an initial speed of 16 ft/s, from the top of a building 480 ft high. How many seconds later will the object hit the ground? Use the equation d = vt + 16t 2 , where d is the distance in feet, v is the initial speed in feet per second, and t is the time in seconds.
104 Chapter 5 Polynomials and Exponents Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Chapter 6
Image copyright Elenamiv, 2010. Used under license from Shutterstock.com.
Rational Expressions
Prep Test
1. Find the LCM of 10 and 25.
For Exercises 2 to 5, add, subtract, multiply, or divide.
2. −
3 4 ⋅ 8 9
3. −
4 8 ÷ 5 15
4. −
5 7 + 6 8
5. −
3 7 − − 8 12
2 1 − 6. Simplify: 3 4 1 −2 8
7. Evaluate
8. Solve: 4( 2 x + 1) = 3( x − 2)
t t 9. Solve: 10 + = 10(1) 2 5
2x − 3 for x = 2. x − x+1 2
10. Two planes start from the same point and fly in opposite directions. The first plane is flying 20 mph slower than the second plane. In 2 h, the planes are 480 mi apart. Find the rate of each plane.
105 Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
se c t io n
6.1 Introduction to
Objective 1 Find the domain of a rational function
An expression in which the numerator and denominator are polynomials is called a rational expression. Examples of rational expressions are shown at the right.
9 z
3x + 4 2x2 + 1
x3 − x + 1 x 2 − 3x − 5
A function that is written in terms of a rational expression is a rational function. Each of the following equations represents a rational function. f ( x) =
x2 + 3 2x − 1
g (t ) =
3 2 t −4
R( z ) =
z 2 + 3z − 1 z 2 + z − 12
To evaluate a rational function, replace the variable by its value. Then simplify.
Example 1
Given f ( x ) =
Solution f ( x ) =
3x − 4 , find f ( −3). x2 − 2x + 1
3x − 4 x2 − 2x + 1
f ( −3) =
3( −3) − 4 ( −3)2 − 2( −3) + 1
f ( −3) =
−9 − 4 9+6+1
f ( −3) =
−13 16
f ( −3) = −
Step
• Substitute –3 for x.
13 16
Because division by zero is not defined, the domain of a rational function must exclude those numbers for which the value of the polynomial in the denominator is zero.
Example 2
Solution
Find the domain of f ( x ) =
2x − 6 . x 2 − 3x − 4
The domain must exclude values of x for which x 2 − 3x − 4 = 0. Solve this equation for x. Because x 2 − 3x − 4 factors over the integers, solve the equation by using the Principle of Zero Products.
x 2 − 3x − 4 = 0
( x + 1)( x − 4) = 0
x +1 = 0
x = −1
x−4 = 0 x = 4
When x = −1 and x = 4, the value of the denominator is zero. Therefore, these values must be excluded from the domain of f.
The domain is { x | x ≠ −1, 4}.
106 Chapter 6 Rational Expressions Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Exercises 1. Given f ( x ) =
3. Given f ( x ) =
f ( x) =
x−2 , find f (3). 2 x + 3x + 8
4. Given f ( x ) =
2
x−3 , find f (3). 2x − 1
x2 , find f (4). 3x − 3x + 5 2
Find the domain of the function. 5. H ( x ) =
4 x−3
6. R( x ) =
5x 3x + 9
8. f ( x ) =
2x − 1 x + x−6
9. G ( x ) =
3 − 4x x + 4x − 5
2
7. h( x ) =
2
10. k ( x ) =
2x + 1 ( x + 1)( x + 5)
x+1 x2 + 1
Objective 2 Simplify rational expressions
A rational expression is in simplest form when the numerator and denominator have no common factors other than 1. The Multiplication Property of One is used to write a rational expression in simplest form.
Example 3
Solution
Simplify: A.
x 2 − 16 12 + 5x − 2 x 2 B. x 2 + 11x + 28 2 x 2 − 3x − 20
x 2 − 16 ( x + 4)( x − 4) = 2 ( x + 4)( x + 7 ) x + 11x + 28
A.
Steps
• Factor the numerator and the denominator.
• Divide by the common factors.
• Write the answer in simplest form.
• Factor the numerator and the denominator.
• Divide by the common factors. Remember that
1
=
( x + 4) ( x − 4) ( x + 4) ( x + 7 ) 1
B.
=
x −4 x+7
12 + 5x − 2 x 2 ( 4 − x )(3 + 2 x ) = 2 ( x − 4)( 2 x + 5) 2 x − 3x − 20 −1
=
( 4 − x ) (3 + 2 x ) ( x − 4) ( 2 x + 5) 1
= −
2x + 3 2x + 5
4 − x = −( x − 4). Therefore, 4−x −( x − 4) −1 = = = −1. x −4 x −4 1
• Write the answer in simplest form.
Section 6.1 Introduction to Rational Functions 107
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Exercises Simplify. 11.
6 x 3 − 15x 2 12 x 2 − 30 x
12.
14.
x 2 − x − 20 x 2 − 2 x − 15
15.
se c t io n
18a 3 + 24 a 2
6 − x − x2 3x 2 − 10 x + 8
13.
x 2 − 7 x + 12 x 2 − 9 x + 20
16.
3x 2 + 10 x − 8 8 − 14 x + 3x 2
6.2 Operations on Rational Expressions
Objective 1 Multiply and divide rational expressions
The product of two fractions is a fraction whose numerator is the product of the numerators of the two fractions and whose denominator is the product of the denominators of the two fractions. a b
Example 1
⋅
c ac = d bd
Multiply.
A.
2x2 − 6x 3x − 6
⋅
6 x − 12 6x2 + x − 2 B. 8 x 3 − 12 x 2 6x2 + 7 x + 2
A.
2x2 − 6x 3x − 6
⋅
6 x − 12 2 x ( x − 3) = 3( x − 2) 8 x 3 − 12 x 2
Solution
⋅
2
+ 9x + 4 4 − 7 x − 2x2
⋅ 2x
6( x − 2) 4 x 2 ( 2 x − 3) 1
=
12 x ( x − 3) ( x − 2) 2
12 x ( x − 2) ( 2 x − 3)
=
x −3 x (2 x − 3)
1
B.
2
6x + x − 2 6x2 + 7 x + 2
=
=
+ 9x + 4 4 − 7 x − 2x2
⋅ 2x
( 2 x − 1)(3x + 2) (3x + 2)( 2 x + 1) −1
2
⋅ (2 x + 1)( x + 4)
(1 − 2 x )( 4 + x )
1
1
1
( 2 x − 1) (3x + 2) ( 2 x + 1) ( x + 4) (3x + 2) ( 2 x + 1) (1 − 2 x ) ( 4 + x ) 1
1
1
= −1
1
The reciprocal of a rational expression is the rational expression with the numerator and denominator interchanged. Rational expression a 2
a b − 2y 4
b a Reciprocal 4 a 2 − 2 y
To divide two rational expressions, multiply by the reciprocal of the divisor.
108 Chapter 6 Rational Expressions Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example 2
Divide. 8x 2 y
B.
3 y 2 − 10 y + 8 2 y2 − 7 y + 6 ÷ 3 y 2 + 8 y − 16 2 y 2 + 5 y − 12
A.
12 x 2 y 2 − 24 xy 2 4 x 3 y − 8x 2 y ÷ 5z 2 3z 4
Solution
=
12 x 2 y 2 − 24 xy 2 5z 2
=
12 xy 2 ( x − 2) 5z 2
⋅
3z 4 4 x 3 y − 8x 2 y
3z 4 4 x y ( x − 2)
⋅
2
1
=
36 xy 2 z 4 ( x − 2) 20 x 2 yz 2 ( x − 2)
9 yz 2 5x
=
1
B.
2
3 y − 10 y + 8 2 y2 − 7 y + 6 ÷ 3 y 2 + 8 y − 16 2 y 2 + 5 y − 12
=
3 y 2 − 10 y + 8 3 y 2 + 8 y − 16
=
( y − 2)(3 y − 4) (3 y − 4)( y + 4) 1
=
2
⋅ 2 y 2 + 5y
1
− 12 2y − 7y + 6
⋅ ( y + 4)( 2 y
− 3) ( y − 2)( 2 y − 3) 1
1
( y − 2) (3 y − 4) ( y + 4) ( 2 y − 3) (3 y − 4) ( y + 4) ( y − 2) ( 2 y − 3) 1
1
1
= 1
1
Exercises Multiply or divide. 2.
x 2 − 3x + 2 x 2 + x − 12 ⋅ x 2 − 8 x + 15 8 − 2 x − x 2
+ x − 21 4x2 − 9
4.
2x − 6 4 x 2 − 12 x ÷ 6 x 2 − 15x 18 x 3 − 45x 2
2 x 2 − 5x − 3 2 x 2 − 3x − 20 ÷ 2x2 + 7 x + 3 2 x 2 − x − 15
6.
2 x 2 − 13x + 15 6 x 2 + x − 12 ÷ 2 x 2 − 3x − 35 6 x 2 + 13x − 28
2
1.
3x − 15 4x2 − 2x
− 10 x 15x − 75
3.
12 + x − 6 x 2 6 x 2 + 29 x + 28
5.
⋅ 20 x
⋅ 2x
2
Objective 2 Add or subtract rational expressions
Before two rational expressions with different denominators can be added or subtracted, both rational expressions must be expressed in terms of a common denominator. This common denominator is the LCM of the denominators of the rational expressions.
Section 6.2 Operations on Rational Expressions 109
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example 3
3x
Add:
Soluti
2x − 3
3x + 6
+
2
Steps
2x2 + x − 6
=
3x 3x + 6 ⋅x+2+ 2 x − 3 x + 2 ( 2 x − 3)( x + 2)
=
3x 2 + 6 x 3x + 6 + ( 2 x − 3)( x + 2) ( 2 x − 3)( x + 2)
=
(3x 2 + 6 x ) + (3x + 6) ( 2 x − 3)( x + 2)
=
3x 2 + 9 x + 6 ( 2 x − 3)( x + 2)
=
3( x 2 + 3x + 2) ( 2 x − 3)( x + 2)
=
3( x + 2)( x + 1) ( 2 x − 3)( x + 2)
• The LCM of 2x − 3 and 2x 2 + x − 6 is (2x − 3)( x + 2). • Rewrite each fraction in terms of the LCM of the denominators.
• Add the fractions.
• Factor the numerator to determine whether there are common factors in the numerator and denominator.
1
=
3( x + 2)( x + 1)
=
( 2 x − 3) ( x + 2)
3( x + 1) 2x − 3
1
Example 4
Solution
Subtract:
x 4− x − 2 2x − 4 x − 2x
x 4− x − 2 2x − 4 x − 2x
⋅x
4− x x ( x − 2)
=
x 2( x − 2)
=
x2 8 − 2x − 2 x ( x − 2) 2 x ( x − 2)
=
x2 + 2x − 8 2 x ( x − 2)
=
( x + 4)( x − 2) 2 x ( x − 2)
x
−
⋅2 2
Steps
• Write each fraction in terms of the LCM. The LCM is 2x(x – 2).
• Subtract the fractions.
• Divide by the common factor.
1
=
( x + 4)( x − 2) 2 x ( x − 2)
=
x+4 2x
1
Exercises Add or subtract. 7.
2x 3x − x−3 x−5
8.
3a 5a − a −2 a+1
9.
3 2x + x2 − 1 x2 + 2x + 1
110 Chapter 6 Rational Expressions Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
10.
x−2 − x +1
3 − 12 x x2
se c t io n
3x − 4
3x + 6
12.
x +1 x+2 − 2 x + x − 6 x + 4x + 3 2
6.3 Complex Fractions
Objective 1 Simplify complex fractions
A complex fraction is a fraction whose numerator or denominator contains one or more fractions. Examples of complex fractions are shown below. 1 y 1 1 5− 2+ y 2
Example 1
Solution
5+
5
1 x+2 1 x−2+ x+2 x+4+
11 15 7 + 2 2x − 1 + x x+4 x B. Simplify. A. 5 12 17 3− − 2 3x − 8 + x+4 x x 2−
11 15 11 15 2− + 2 + 2 x x x x = A. 5 12 5 12 3− − 2 3− − 2 x x x x 2−
2
⋅ x2 x
11 2 15 2 ⋅x + 2 ⋅x x x = 5 12 2 2 3 ⋅ x − ⋅ x − 2 ⋅ x2 x x
=
2 x 2 − 11x + 15 3x 2 − 5x − 12
=
( 2 x − 5)( x − 3) 2x − 5 = (3x + 4)( x − 3) 3x + 4
• The LCM of the denominators is x 2.
• The LCM of the denominators is x + 4.
2 ⋅ x2 −
7 7 2x − 1 + x + 4 x + 4 B. = 17 17 3x − 8 + 3x − 8 + x+4 x+4 2x − 1 +
⋅x+4
x+4
7 ( x + 4) x + 4 = 17 (3x − 8)( x + 4) + ( x + 4) x+4
=
2x2 + 7 x − 4 + 7 3x 2 + 4 x − 32 + 17
=
2x2 + 7 x + 3 3x 2 + 4 x − 15
=
( 2 x + 1)( x + 3) 2x + 1 = (3x − 5)( x + 3) 3x − 5
Steps
( 2 x − 1)( x + 4) +
Section 6.3 Complex Fractions 111
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Exercises Multiply or divide. 1 x 1. 1 1− 2 x
a 2. a 1 1 + a a2
4 x+2 3. 10 5− x+2
1 6 − 2 x x 4. 4 3 1− + 2 x x
3 10 − 2 x x 5. 11 18 1+ + 2 x x
2 x − 4 6. 6 x +3+ x−4
1+
1−
−
2
1−
se c t io n
2−
x −1+
6.4 Rational Equations
Objective 1 Solve fractional equations
To solve an equation containing fractions, clear denominators by multiplying each side of the equation by the LCM of the denominators. Then solve for the variable.
Example 1
Solve:
Solution
3x 5 = 5− x−5 x−5 3x 5 = 5− x−5 x−5
3x 5 ( x − 5) = ( x − 5) 5 − x − 5 x − 5
5 3x = ( x − 5)5 − ( x − 5) x − 5
3x = 5x − 25 − 5
3x = 5x − 30
−2 x = −30
x = 15
Steps
• Multiply each side of the equation by the LCM of the denominators.
• Simplify.
• Solve the equation for x.
15 checks as a solution. The solution is 15. Occasionally, a value of the variable that appears to be a solution will make one of the denominators zero. In this case, the equation has no solution for that value of the variable.
Example 2
Solve:
Solution
3x 9 = 2+ x−3 x−3 3x 9 = 2+ x−3 x−3
3x 9 ( x − 3) = ( x − 3) 2 + x − 3 x − 3
Steps
• Multiply each side of the equation by the LCM of the denominators.
112 Chapter 6 Rational Expressions Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
9 3x = ( x − 3)2 + ( x − 3) x − 3
3x = 2 x + 3
Example 3
3x 9 = 2+ x−3 x−3
3(3) 9 = 2+ 3−3 3−3
9 9 = 2+ 0 0
Solve:
• Substituting 3 into the equation results in division by zero. Because division by zero is not defined, the equation has no solution.
Step
• Multiply each side of the equation by the LCM of the denominators.
1 5 = 4 x+5 1 5 = 4 x+5
Solution
• Use the Distributive Property.
x = 3
4( x + 5)
1 5 = 4( x + 5) 4 x+5
x + 5 = 4(5)
x + 5 = 20
x = 15
15 checks as a solution.
The solution is 15.
Exercises Solve. 1.
x 5 x + = 2 6 3
2.
4.
3 5 +2 = x−4 x−4
5. 5 +
4 2 = x−4 x−2
8 4a = a−2 a−2
3.
x x +1 = 3 7
6. −
5 4 +1= x+7 x+7
Objective 2 Work problems In solving a work problem, the goal is to determine the time it takes to complete a task. The basic equation that is used to solve work problems is Rate of work × Time worked Part of task completed For example, if a pipe can fill a tank in 5 h, then in 2 h the pipe will fill tank. In t hours, the pipe will fill
1 t of the tank. ×t = 5 5
1 2 of the ×2 = 5 5
Section 6.4 Rational Equations 113
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example 4
Strategy
An electrician requires 12 h to wire a house. The electrician’s apprentice can wire a ne job for 4 h, the electrician quits, and the oes it take the apprentice to finish wiring the c Time required for the apprentice to finish wiring the house: t Rate
Time
Part
Electrician
1 12
4
4 12
Apprentice
1 16
t
t 16
c The sum of the part of the task completed by the electrician and the part of the task completed by the apprentice is 1.
Solution
4 t + =1 12 16
1 t + =1 3 16
1 t 48 + = 48(1) 3 16
16 + 3t = 48
3t = 32
Example 5
Strategy
2 h to finish wiring the house. 3
Two water pipes can fill a tank with water in 6 h. The larger pipe working alone can fill the tank in 9 h. How long would it take the smaller pipe, working alone, to fill the tank? c Time required for the small pipe to fill the tank: x Rate
Time
Part
Large pipe
1 9
6
6 9
Small pipe
1 x
6
6 x
c The sum of the part of the task completed by the large pipe and the part of the task completed by the small pipe is 1.
6 6 + =1 9 x 2 6 + =1 3 x
2 6 3x + = 3x ⋅ 1 3 x
32 3
It takes the apprentice 10
Solution
t =
2 x + 18 = 3x 18 = x
The small pipe working alone will fill the tank in 18 h.
114 Chapter 6 Rational Expressions Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Exercises 7. A large biotech computer can pr it take to process the data if both computers were used?
of its research studies. One to do the same job. How long would
8. Two college students have started their own business building computers from kits. Working alone, one student can build a computer in 20 h. When the second student helps, the students can build a computer in 7.5 h. How long would it take the second student, working alone, to build the computer? 9. One solar heating panel can raise the temperature of water 1 degree in 30 min. A second solar heating panel can raise the temperature 1 degree in 45 min. How long would it take to raise the temperature of the water 1 degree with both solar panels operating?
Objective 3 Uniform motion problems
The basic equation used to solve uniform motion problems is Distance = Rate × Time
An alternative form of this equation can be written by solving the equation for time. This form of the equation is used to solve the following problem. Distance = Time Rate
Example 6
Strategy
A marketing executive traveled 810 mi on a corporate jet in the same amount of time it took to travel an additional 162 mi by helicopter. The rate of the jet was 360 mph greater than the rate of the helicopter. Find the rate of the jet. c Rate of the helicopter: r
Distance
Rate
Time
Jet
810
r + 360
810 r + 360
Helicopter
162
r
162 r
c The time traveled by jet is equal to the time traveled by helicopter.
Solution
Rate of the jet: r + 360
810 162 = r + 360 r
810 162 r ( r + 360) = r ( r + 360) r + 360 r
810r = ( r + 360)162
810r = 162r + 58, 320
648r = 58, 320
r = 90
r + 360 = 90 + 360 = 450
The rate of the jet was 450 mph.
Steps
• The rate of the helicopter was 90 mph.
• Substitute the value of r into the variable expression for the rate of the jet.
Section 6.4 Rational Equations 115
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example 7
Strategy
A plane can fly at a rate of 150 mph in calm air. Traveling with the wind, the plane flew to fly 500 mi against the wind. Find the rate c Rate of the wind: r Distance
Rate
Time
With wind
700
150 + r
700 150 + r
Against wind
500
150 – r
500 150 − r
c The time flying with the wind equals the time flying against the wind.
Solution
700 500 = 150 + r 150 − r
700 500 (150 + r )(150 − r ) = (150 + r )(150 − r ) 150 + r 150 − r
(150 − r )700 = (150 + r )500
105, 000 − 700r = 75, 000 + 500r
30, 000 = 1200r
25 = r
The rate of the wind is 25 mph.
Exercises 10. Two skaters take off for an afternoon of rollerblading in Central Park. The first skater can cover 15 mi in the same time it takes the second skater, traveling 3 mph slower than the first skater, to cover 12 mi. Find the rate of each rollerblader.
11. A commercial jet travels 1620 mi in the same amount of time it takes a corporate jet to travel 1260 mi. The rate of the commercial jet is 120 mph greater than the rate of the corporate jet. Find the rate of each jet.
12. A passenger train travels 295 mi in the same amount of time it takes a freight train to travel 225 mi. The rate of the passenger train is 14 mph greater than the rate of the freight train. Find the rate of each train.
se c t io n
6.5 Proportions and Variation
Objective 1 Proportions A proportion is an equation that states that two ratios or rates are equal. For example, 90 km 45 km 3 x+2 and are proportions. = = 4L 2L 4 16 Note that a proportion is a special kind of fractional equation. Many application problems can be solved by using proportions.
116 Chapter 6 Rational Expressions Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example 1
A stock investment of 50 shares pays a dividend of $106. At this rate, how many rn a dividend of $424? hares that are required, write and solve a proportion using x to represent the additional number of shares. Then 50 + x is the total number of shares of stock.
Solution
106 424 = 50 50 + x
53 424 = 25 50 + x
25(50 + x )
53 424 = 25(50 + x ) 25 50 + x
(50 + x )53 = ( 25)424
2650 + 53x = 10, 600
53x = 7950
x = 150
An additional 150 shares of stock are required.
Example 2
Two pounds of cashews cost $5.80. At this rate, how much would 15 lb of cashews cost?
Strategy
Solution
To find the cost, write and solve a proportion using x to represent the cost.
x (5.80)
2 15 = 5.80 x 2 15 = x (5.80) 5.80 x
2 x = 15(5.80)
2 x = 87
x = 43.5
The cost of 15 lb of cashews is $43.50.
Exercises 1. In a wildlife preserve, 60 ducks are captured, tagged, and then released. Later, 200 ducks are examined, and three of the 200 ducks are found to have tags. Estimate the number of ducks in the preserve.
2. Six ounces of an insecticide are mixed with 15 gal of water to make a spray for spraying an orange grove. At the same rate, how much additional insecticide is required to be mixed with 100 gal of water?
Objective 2 Variation problems A direct variation is a special function that can be expressed as the equation y = kx , where k is a constant. The equation y = kx is read “y varies directly as x” or “y is proportional to x.” The constant is called the constant of variation or the constant of proportionality.
Section 6.5 Proportions and Variation 117
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example 3
Strategy
The amount (A) of medication prescribed for a person varies directly with the person’s kg, 2 ml of medication are prescribed. How d for a person who weighs 75 kg? To find the required amount of medication:
c Write the basic direct variation equation, replace the variables by the given values, and solve for k.
c Write the direct variation equation, replacing k by its value. Substitute 75 for W, and solve for A.
Solution A = kW
2 = k ⋅ 50 1 = k 25 1 A = W 25
A =
1 ⋅ 75 = 3 25
Steps
• This is the direct variation equation.
• Replace W by 75.
The required amount of medication is 3 ml. k An inverse variation is a function that can be expressed as the equation y = , where x k is read “y varies inversely as x” or “y is inversely k is a constant. The equation y = x proportional to x.” k , where n is a positive xn k number. For example, the equation y = 2 is read “y varies inversely as the square of x.” x In general, an inverse variation equation can be written y =
Example 4
Strategy
A company that produces personal computers has determined that the number of computers it can sell (s) is inversely proportional to the price (P) of the computer. Two thousand computers can be sold when the price is $2500. How many computers can be sold when the price of a computer is $2000? To find the number of computers:
c Write the basic inverse variation equation, replace the variables by the given values, and solve for k.
c Write the inverse variation equation, replacing k by its value. Substitute 2000 for P, and solve for s.
Solution
s =
2000 =
k P k 2500
5, 000, 000 = k
s =
5, 000, 000 P
Steps
• This is the inverse variation equation.
s =
5, 000, 000 = 2500 2000
• Replace P by 2000.
At a price of $2000, 2500 computers can be sold.
118 Chapter 6 Rational Expressions Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example 5
Strategy
A combined variation is a variation in which two or more types of variation occur at the y, the volume of a gas varies directly as the temperature kT . . This combined variation is written V = P The pressure (P) of a gas varies directly as the temperature (T) and inversely as the volume (V). When T = 50° and V = 275 in 3 , P = 20 lb/in 2 . Find the pressure of a gas when T = 60° and V = 250 in 3 . To find the pressure:
c Write the basic combined variation equation, replace the variables by the given values, and solve for k.
c Write the combined variation equation, replacing k by its value. Substitute 60 for T and 250 for V, and solve for P.
Solution P =
20 =
kT V k(50) 275
110 = k P =
110T V
Steps
• This is the combined variation equation.
P =
110( 60) = 26.4 250
• Replace T by 60 and V by 250.
The pressure is 26.4 lb/in2.
Exercises 3. The pressure (p) on a diver in the water varies directly as the diver’s depth (d). If the pressure is 3.6 lb/in2 when the depth is 8 ft, what is the pressure when the depth is 30 ft?
4. Leonardo da Vinci observed that the length of a person’s face varies directly as the length of the person’s chin. If a person whose face length is 9 in. has a chin length of 1.5 in., what is the length of a person’s face whose chin length is 1.7 in.?
5. The speed (v) of a gear varies inversely as the number of teeth (t). If a gear that has 48 teeth makes 20 revolutions per minute, how many revolutions per minute will a gear that has 30 teeth make?
6. The current (I) in a wire varies directly as the voltage (v) and inversely as the resistance (r). If the current is 27.5 amps when the voltage is 110 volts and the resistance is 4 ohms, find the current when the voltage is 195 volts and the resistance is 12 ohms.
Section 6.5 Proportions and Variation 119
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
se c t io n
6.6 Literal Equations
Objective
A literal equation is an equation that contains more than one variable. Examples of literal equations are shown at the right.
3x − 2 y = 4 v 2 = v02 + 2 as
Formulas are used to express relationships among physical quantities. A formula is a literal equation that states a rule about measurement. Examples of formulas are shown below.
Example 1
Solution
s = vt − 16t 2
(Physics)
c 2 = a 2 + b2
(Geometry)
A = P(1 + r )t (Business) S = R for C. S −C
A. Solve A = P + Prt for P. B. Solve A.
A = P + Prt
A = (1 + rt ) P
A (1 + rt ) P = 1 + rt 1 + rt
A = P 1 + rt
• Factor P from P + Prt.
• Divide each side of the equation by 1 + rt.
• Multiply each side of the equation by S – C.
• Add CR to each side of the equation. • Subtract S from each side of the equation. • Divide each side of the equation by R.
S = R S −C
B.
Steps
(S − C )
S = (S − C )R S −C S = SR − CR
CR + S = SR CR = SR − S C =
SR − S R
Exercises Solve the formula for the variable given. 9 C + 32; C (Temperature conversion) 5
1. P = 2 L + 2W ; W (Geometry)
2. F =
3. S = C − rC ; C (Business)
4. A = P + Prt; t (Business)
5. I =
E ; R (Physics) R+r
6. P =
R−C ; R (Business) n
Solve the equation for y. 7. 2 x + 3 y = 6
8. x − 2 y = 4
9. 3x − 2 y = 12
10. 3x + 4 y = 12
120 Chapter 6 Rational Expressions Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Chapter 7
Image copyright Ken Knoll, 2010. Used under license from Shutterstock.com.
Rational Exponents and Radicals
Prep Test
1. Complete: 48 = ? ⋅ 3 For Exercises 2 to 7, simplify. 3 3. 6 2
2. 25
4.
1 2 1 − + 2 3 4
5. (3 − 7 x ) − ( 4 − 2 x )
6.
3x 5 y 6 12 x 4 y
7. (3x − 2)2
For Exercises 8 and 9, multiply.
8. ( 2 + 4 x )(5 − 3x )
9. (6 x − 1)(6 x + 1)
10. Solve: x 2 − 14 x − 5 = 10
121 Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
se c t io n
7.1 Rational Exponents
ions
Objective 1 Simplify expressions with rational exponents
( ) 1
n
Because a n
1
= a, the number a n is the number whose nth power is a. 1
252 = 5 because ( 5)2 = 25.
8 3 = 2 because ( 2)3 = 8.
1
1
1
In the expression a n , if a is a negative number and n is a positive even integer, then a n is not a real number. 1
( −4) 2 is not a real number, because there is no real number whose second power is –4. When n is a positive odd integer, a can be a positive or a negative number. 1
( −27 ) 3 = −3 because ( −3)3 = −27.
m
Definition of a n
1
If m and n are positive integers and a n is a real number, then m
Example 1
Solution
( ) 1
m
a n = an
2
−
2
3
Simplify. A. 27 3 B. 32 5 C. ( −49) 2 2
2
A. 27 3 = ( 33 ) 3 2 3
= 3 3
= 32
= 9
B. 32
−
2 5
= ( 25 )
−
2 5
Steps
• Rewrite 27 as 33.
• Use the Rule for Simplifying a Power of an Exponential Expression.
• Simplify.
• Rewrite 32 as 25.
= 2 −2
• Use the Rule for Simplifying a Power of an Exponential Expression.
=
1 22
• Use the Rule of Negative Exponents.
=
1 4
• Simplify.
• The base of the exponential expression is a negative number, while the denominator of the exponent is a positive even number.
3
C. ( −49) 2
3
( −49) 2 is not a real number.
122 Chapter 7 Rational Exponents and Radicals Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example 2
1
2
⋅ b3 ⋅ b
Simplify. A. b 2
−
1 4
3
B. ( x 6 y 4 ) 2
6
= b12
= b 12
+
Steps
• Use the Rule for Multiplying Exponential Expressions.
• Use the Rule for Simplifying Powers of Products.
8 3 − 12 12
11
3
B. ( x 6 y 4 ) 2 = x
3 6 2
y
3 4 2
= x9 y6
Exercises Simplify. 1
3
1. 8 3
−
2. 9 2
1 3
25 5. 49
4. 64
7. y
−
1 6
2 3
⋅y ⋅y
1 2
8.
3 2
2
1
3
b4
4 3
9.
b
−
2 3
⋅x
6. x 5
3
10. ( x 8 y 2 ) 2
−
−
b3 b
1
3. 27
−
4 5
3 2
12. ( x −3 y 6 )
11. ( x 4 y 2 z 6 ) 2
−
1 3
Objective 2 Write exponential expressions as radical expressions and radical expressions as exponential expressions Alternative Notation for the nth Root of a If a is a real number and n is a positive integer, then
1
n
a = an .
In the expression n a , the symbol is called a radical, n is the index of the radical, and a is the radicand. When n = 2, the radical expression represents a square root and the index 2 is usually not written. An exponential expression with a rational exponent can be written as a radical expression. Definition of the nth Root of am 1
m
If a n is a real number, then a n = a
m⋅
1 n
1
= (am ) n =
n
am .
Section 7.1 Rational Exponents and Radical Expressions 123
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example 3
Rewrite the exponential expression as a radical expression. 2
2
Pillaged by LibraryPirate
A. (3x ) 3 B. −2 x 3
A. (3x ) 3 =
3
(3x )2
=
3
9x2
2
Solution
• The –2 is not raised to the power.
= −2 3 x 2
Example 4
• The denominator of the rational exponent is the index of the radical. The numerator is the power of the radicand.
1
2
B. −2 x 3 = −2( x 2 ) 3
Steps
Rewrite the radical expression as an exponential expression.
A.
7
x 5 B.
3
Solution
A.
7
x5 = ( x5 ) 7 = x 7
B.
3
a 3 + b3 = ( a 3 + b 3 ) 3
a 3 + b3 5
1
Steps
• The index of the radical is the denominator of the rational exponent. The power of the radicand is the numerator of the rational exponent.
• Note that (a 3 + b 3 ) 3 ≠ a + b.
1
1
Exercises Rewrite the exponential expression as a radical expression. 1
13. 52
4
3
14. b 3
15. ( 2t ) 2
2
16. −3a 5
Rewrite the radical expression as an exponential expression. 17. 14
18.
3
x
19.
5
b3
20. 2 y x 3
Objective 3 Simplify radical expressions that are roots of perfect powers Every positive number has two square roots, one a positive and one a negative number. For example, because ( 5)2 = 25 and ( −5)2 = 25, there are two square roots of 25: 5 and –5. 25 = 5
The symbol is used to indicate the positive or principal square root. To indicate the negative square root of a number, a negative sign is placed in front of the radical.
a 2 = a and − a 2 = − a . If a is a positive real number, then
For any real number a, a 2 = a and
( a)
2
− 25 = −5
= a.
The cube root of a positive number is positive.
3
8 = 2, because 23 = 8.
The cube root of a negative number is negative.
3
−8 = −2, because ( −2)3 = −8.
For any real number a,
3
a 3 = a.
124 Chapter 7 Rational Exponents and Radicals Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A variable expression is a perfect power if the exponents on the factors are evenly divisible ect powers. Knowledge of these roots is very helpful
in simplifying radical expressions. Square Roots
Example 5
Cube Roots
1 =1
36 = 6
4 = 2
49 = 7
9 = 3
64 = 8
16 = 4
81 = 9
25 = 5
100 = 10
3
3
3
Fourth Roots 4
1 =1
5
1 =1
8 = 2
4
16 = 2
3
27 = 3
4
81 = 3
3
64 = 4
4
256 = 4
125 = 5
4
625 = 5
5
5
1 =1
32 = 2
243 = 3
Simplify. 3
−125a 6 b9 C. − 4 16 a 4 b8
A.
49 x 2 y12 B.
Solution
A.
49 x 2 y12 = 7 xy 6
Steps
B.
−125a 6 b9 = −5a 2 b 3
C. − 4 16 a 4 b8 = −2ab 2
Fifth Roots
3
• The radicand is a perfect square. Divide each exponent by 2. • The radicand is a perfect cube. Divide each exponent by 3. • The radicand is a perfect fourth power. Divide each exponent by 4.
Exercises Simplify. 21.
25.
3
x 2 y10
22. 121 y12
−8 x 3
26.
se c t io n
4
81a 20
23.
3
a 6 b12
24. 3 125b15
27.
5
x 20 y10
28.
5
−32 x15 y 20
7.2 Operations on Radical Expressions
Objective 1 Simplify radical expressions If a number is not a perfect power, its root can only be approximated; examples include 5 and 3 3. These numbers are irrational numbers. Their decimal representations never terminate or repeat. 5 = 2.2360679 …
3
3 = 1.4422495…
A radical expression is not in simplest form when the radicand contains a factor greater than 1 that is a perfect power. The Product Property of Radicals is used to simplify radical expressions whose radicands are not perfect powers.
Section 7.2 Operations on Radical Expressions 125
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Product Property of Radicals then n
Simplify:
ab =
n
a
⋅nb
and
n
a
⋅nb
=
Solution
16 ⋅ 3
16 3
=
Simplify
= 4 3
16 .
3
x7 3
Write the radicand as the product of a perfect cube and a factor that does not contain a perfect cube.
48 =
Use the Product Property of Radicals to write the expression as a product.
Simplify:
Example 1
ab
48
Write the radicand as the product of a perfect square and a factor that does not contain a perfect square.
n
⋅x
x7 =
3
x6
3
x6 3 x
Use the Product Property of Radicals to write the expression as a product.
=
Simplify.
= x2 3 x
4
32 x 7
32 x 7 =
4
16 x 4 ( 2 x 3 )
Steps
=
4
16 x 4 4 2 x 3
= 2x 4 2x3
Simplify: 4
• Write the radicand as the product of a perfect fourth power and factors that do not contain a perfect fourth power. • Use the Product Property of Radicals to write the expression as a product. • Simplify.
Exercises Simplify. 1.
98
2. 128
5.
45x 2 y 3 z 5
6.
3
−125x 2 y 4
3. 3 128
7.
3
−216 x 5 y 6
x4 y3z5
4.
8.
4
64 x 8 y10
Objective 2 Add or subtract radical expressions The Distributive Property is used to simplify the sum or difference of radical expressions that have the same radicand and the same index. For example,
3 5 + 8 5 = ( 3 + 8) 5 = 11 5
2 3 3x − 9 3 3x = ( 2 − 9) 3 3x = −7 3 3 x
126 Chapter 7 Rational Exponents and Radicals Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Simplify: 3 32 x 2 − 2 x 2 + 128 x 2 ine like terms by using the Distributive Property. 3 32 x 2 − 2 x 2 + 128 x 2 = 3 16 x 2 2 − 2 x 2 +
Example 2
64 x 2 2
= 3 ⋅ 4 x 2 − 2 x 2 + 8x 2
= 12 x 2 − 2 x 2 + 8 x 2
= 18 x 2 Add or subtract.
A. 3xy 3 81x 5 y − 3 192 x 8 y 4 B. 4 a 3 54 a 7 b9 + a 2 b 3 128a 4 b6
A. 3xy 3 81x 5 y − 3 192 x 8 y 4
Solution
= 3xy 3 27 x 3 3 3x 2 y −
= 3xy ⋅ 3x 3 3x 2 y − 4 x 2 y 3 3x 2 y
= 9 x 2 y 3 3x 2 y − 4 x 2 y 3 3x 2 y
= 5x2 y 3 3x2 y
3
64 x 6 y 3 3 3x 2 y
B. 4 a 3 54 a 7 b9 + a 2 b 3 128a 4 b6
= 4 a 3 27 a 6 b9 3 2 a + a 2 b 3 64 a 3b6 3 2 a
= 4 a ⋅ 3a 2 b3 3 2 a + a 2 b ⋅ 4 ab2 3 2 a
= 12 a 3b3 3 2 a + 4 a 3b3 3 2 a
= 16a 3 b 3 3 2a
Exercises Simplify. 9.
12.
128 x −
3
128 +
3
98 x
250
10.
18b +
75b
13. 2 3 3a 4 − 3a 3 81a
11. 2 2 x 3 + 4 x 8 x
14. 2b 3 16b2 + 3 128b5
Section 7.2 Operations on Radical Expressions 127
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Objective 3 Multiply radical expressions
multiply radical expressions with the same Multiply:
3
2 a5 b 3 16 a 2 b2 3
Use the Product Property of Radicals to multiply the radicands. Then simplify.
Example 3
Solution
Example 4
Solution
Multiply: 3x
(
3x
(
27 x 2 −
3x
)
27 x 2 −
)
81x 3 −
3x =
81x 2 x −
=
= 9x x − 3x
(
(
A. 2 3 + 4
)(
3−5
)
)(
)
= 2 9 − 10 3 + 4 3 − 20
= 6 − 6 3 − 20
= –14 – 6 3
(
) =(
x +3
)(
2
x +3
32 a 7 b3
=
3
8a 6 b3 3 4 a
= 2a 2 b 3 4a
9x2
3 − 5 B.
B.
3
9x2
Multiply: A. 2 3 + 4
2 a5 b 3 16 a 2 b2 =
(
)
x +3
2
Step
• Use the FOIL method.
)
x +3
x2 + 3 x + 3 x + 9
=
= x+6 x +9
Exercises Simplify. 15.
2 x 2 y 32 xy
18.
3
21.
(
(
27 −
a −2
)(
16.
)
3
3
19.
)
a −3
4 a 2 b3 3 8ab5
3a
(
27 a 2 −
(
)
17.
)
a
2
22. 4 5 + 2
4
12 ab3 4 4 a5 b2
)(
20.
(
5−5 2 5+2
23.
(
x −3
)
)
2
Objective 4 Divide radical expressions The Quotient Property of Radicals is used to divide radical expressions with the same index.
128 Chapter 7 Rational Exponents and Radicals Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Quotient Property of Radicals and b ≠ 0, then
n
Simplify:
a = b
n n
a b
and
n n
a b
=
n
a b
81x 5 y6
3
Use the Quotient Property of Radicals. Then simplify each radical expression. 3
81x 5 = y6
=
=
3
81x 5 3
3
y6
27 x 3 3 3x 2 3
y6
3x 3 3x2 y2
Simplest Form of a Radical Expression A radical expression is in simplest form when all of the following conditions are met. 1. The radicand contains no factor greater than 1 that is a perfect power of the index. 2. There is no fraction under the radical sign. 3. No radical remains in the denominator of the radical expression. The procedure used to remove a radical from a denominator is called rationalizing the denominator. This is illustrated in the next example.
Example 5
Solution
5
Simplify: 5 5x
=
5x 5 5x
⋅
5x
Step
5x
• Multiply the expression by
5x . 5x
5 5x
=
=
5 5x 5x
=
5x x
25x 2
The expressions a + b and a − b, which are the sum and difference of two terms, are called conjugates of each other. The product of conjugates of the form ( a + b)( a − b) is a 2 − b2 .
(
)(
x −3
) ( x)
x +3 =
2
− 32 = x – 9
To simplify a fraction that has a square-root radical expression with two terms in the denominator, multiply the numerator and denominator by the conjugate of the denominator. Then simplify.
Section 7.2 Operations on Radical Expressions 129
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
2−
Example 6
Solution
3+
5
y y
A.
2−
5
3+
2
=
2−
5
3+
2
⋅3− 3−
B.
3+
y
3−
y
=
=
3+
y
3−
y
9 +
3y +
⋅
2 2
=
6 − 2 2 − 3 5 + 10 9−2
=
6 − 2 2 − 3 5 + 10 7
3+
y
3+
y
3y +
( 3) − ( y ) 2
y2
=
2
3 + 2 3y + y 3− y
Exercises Simplify. 24.
27.
30.
32 x 2 2x
9 3a
25.
28.
4−
2
2−
3
31.
se c t io n
42 a 3b5 2
14 a b
40 x 3 y 2 2
80 x y
3−
x
3+
x
3
26.
29.
32.
1 5
2 5+2
2 −
3
2 +
3
7.3 Radical Functions
Objective 1 Find the domain of a radical function A radical function is one that contains a fractional exponent or a variable underneath a radical. The domain of a radical function is a set of real numbers for which the radical expression is a real number. For example, -9 is one number that would be excluded from the domain of f ( x ) =
x + 5 because f ( −9) =
−9 + 5 =
−4 , which is not a real number.
If the index of a radical expression is an even number, the radicand must be greater than or equal to zero to ensure that the value of the expression will be a real number. If the index of a radical expression is an odd number, the radicand may be a positive or a negative number. 130 Chapter 7 Rational Exponents and Radicals Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example 1
State the domain of each function in set-builder notation. x) =
Solution
5
x+4
A. 6 − 4 x ≥ 0
−4 x ≥ −6
x ≤
Steps
• V contains an even root. Therefore, the radicand must be greater than or equal to zero.
• Because R contains an odd root, the radicand may be positive or negative.
3 2
3 The domain is x | x ≤ . 2
B. The domain is { x | x ∈ real numbers}.
Exercises State the domain of each function in set-builder notation. 1. g ( x ) = −2 x + 1
1
2. y ( x ) = −3 3 1 + x
3. F ( x ) = 4(3x − 6) 2
State the domain of each function in interval notation. 1
4. f ( x ) = −2( 4 x − 12) 2
5. Y ( x ) = −6 +
2
6− x
6. g ( x ) = 2( 2 x − 10) 3
Objective 2 Graph a radical function
The graph of a radical function is produced in the same manner as the graph of any other function. The function is evaluated at several values in the domain of the function, and the resulting ordered pairs are graphed. Ordered pairs must be graphed until an accurate graph can be made.
Example 2
Solution
Graph: f ( x ) =
Because f contains an even root, the radicand must be positive. To determine the domain of f, solve the inequality x + 2 ≥ 0. The solution is x ≥ −2, so the domain is {x | x ≥ −2}. Now determine ordered pairs of the function by choosing values of x from the domain. Some possible choices are shown in the following table. x
x+2
f ( x) =
x+2
y
–2
f ( −2) =
−2 + 2 =
0 = 0
0
–1
f ( −1) =
−1 + 2 =
1 =1
1
2
f ( 2) =
2+2 =
4 = 2
2
7
f (7 ) =
7+2 =
9 = 3
3
y 4 2 –2 0 –2
(7, 3)
(2, 2)
2
4
6
8
x
–4
Section 7.3 Radical Functions 131
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example 3
3
Graph: H ( x ) =
x
domain of H is all real numbers. Choose valuate the function for those values. Some possible choices are given in the following table.
Soluti
x
H ( x) =
–8
H ( −8) =
3
H ( −1) =
3
–1 0
x
y
−8 = −2
–2
3
0
H (1) =
3
1 =1
1
H (8) =
3
8 = 2
2
1 8
2 (0, 0) –8 –4
0 = 0
3
4
–1
−1 = −1
H (0) =
y
(8, 2) (1, 1) 4
8
x
–2 –4
Exercises Graph. 7. F ( x ) =
x
8. h( x ) = − 3 x
se c t io n
9. K ( x ) =
3
x + 1
10. g ( x ) =
x −1
7.4 Solving Equations Containing Radical Expressions
Objective 1 Solve equations containing one or more radical expressions
Property of Raising Each Side of an Equation to a Power If two numbers are equal, then the same powers of the numbers are equal. If a = b, then a n = bn .
Example 1
Solution
Solve. A. A.
3
3x − 1 = −4
3x − 2 − 8 = −3 3x − 2 = 5
3x − 2 − 8 = −3 B.
(
Steps
= 52
• Rewrite the equation so that the radical is alone on one side of the equation. • Square each side of the equation.
3x − 2 = 25
• Solve the resulting equation.
3x − 2
)
2
3x = 27
x = 9
132 Chapter 7 Rational Exponents and Radicals Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Check:
3 ⋅ 9 − 2 − 8
-3
27 − 2 − 8
-3
25 − 8
-3
5 − 8
-3
3
(
3
3x − 1
)
3
= ( −4)3
3x − 1 = −64 3x = −63
x = −21 Check: 3
3
• Solve the resulting equation.
• Check the solution.
3x − 1 = −4
3( −21) − 1 3
-4
−63 − 1 -4 3
−64
-4
−4 = −4
• Cube each side of the equation.
3x − 1 = −4
The solution is 9.
B.
• Check the solution.
-3 = -3
The solution is -21.
When you raise both sides of an equation to an even power, the resulting equation may have a solution that is not a solution of the original equation. Therefore, it is necessary to check the solution of a radical equation.
Example 2
Solution
Solve. A. x + 2 x − 1 = 9 B.
x + 1
A. x + 2 x − 1 = 9 2 x −1 = 9− x
(2
x −1
)
2
= (9 − x ) 2
4( x − 1) = 81 − 18 x + x 2
4 x − 4 = 81 − 18 x + x 2
Steps
• Rewrite the equation so that the radical is alone on one side of the equation. • Square each side of the equation.
0 = x 2 − 22 x + 85
0 = ( x − 5)( x − 17 )
• Write the quadratic equation in standard form. • Factor.
• Use the Principle of Zero Products.
x+7 =
x − 5 = 0 x − 17 = 0 x = 5
x = 17
Section 7.4 Solving Equations Containing Radical Expressions 133
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Check: x + 2 x − 1 = 9
x +2 x −1 = 9 + 2 17 − 1 9
5+ 2 4
9
17 + 2 16 9
5 + 2 ⋅ 2
9
17 + 2 ⋅ 4 9
5 + 4 9
9 = 9
9
25 ≠ 9
17 does not check as a solution. It is called an extraneous solution of the equation. The solution is 5.
B.
17 + 8
x+7 =
(
x+7
x +1
) =(
)
2
x +1
2
• A radical appears on each side of the equation.
• Square each side of the equation.
x+7 = x+2 x +1
6 = 2 x
• Simplify the resulting equation.
3=
x
• The equation contains a radical.
32 =
( x)
• Square each side of the equation.
• Check the solution.
9 = x
Check:
x+7 =
x +1
9+7
9 +1
16
3+1
4 = 4
2
The solution is 9.
Exercises Solve. 1.
3x − 2 = 5
4.
2x + 4 = 3 −
2.
2x
5.
3
5x + 2 = 3
x+1 = 2−
3.
x
2x − 3 − 2 = 1
6. 3 x − 2 + 2 = x
Objective 2 Application problems Pythagorean Theorem The square of the hypotenuse, c, of a right triangle is equal to the sum of the squares of the two legs, a and b.
c 2 = a 2 + b2
134 Chapter 7 Rational Exponents and Radicals Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example 3
Strategy
A ladder 20 ft long is leaning against a building. he ladder reach s 8 ft from the nth. To find the distance, use the Pythagorean Theorem. The hypotenuse is the length of the ladder. One leg is the distance from the bottom of the ladder to the base of the building. The distance along the building from the ground to the top of the ladder is the unknown leg.
Solution
c 2 = a 2 + b2
202 = 82 + b2
400 = 64 + b2
336 = b2 1
20 ft
8 ft
1
(336) 2 = ( b2 ) 2
336 = b
18.3 ≈ b The distance is 18.3 ft.
Exercises 7. A 12-foot ladder is leaning against a building. How high on the building will the ladder reach when the bottom of the ladder is 4 ft from the building? Round to the nearest tenth.
8. The sustained wind velocity, v (in meters per second), in a hurricane is given by v = 6.3 1013 − p , where p is the air pressure in millibars (mb). If the velocity of the wind in a hurricane is 64 m/s , what is the air pressure? Round to the nearest tenth. What happens to wind speed in a hurricane as air pressure decreases?
9. The time, T (in days), that it takes a planet to revolve around the sun can be approximated by the equation T = 0.407 d 3 , where d is the mean distance of the planet from the sun in millions of miles. It takes Venus approximately 226 days to complete one revolution of the sun. To the nearest million miles, what is the mean distance of Venus from the sun?
se c t io n
7.5 Complex Numbers
Objective 1 Simplify complex numbers The radical expression −4 is not a real number because there is no real number whose square is -4. However, the solution of an algebraic equation is sometimes the square root of a negative number.
Section 7.5 Complex Numbers 135
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
During the late 17th century, a new number, called an imaginary number, was defined so root. The letter i was chosen to represent the i 2 = −1 Definition of
–a
If a is a positive real number, then the principal square root of negative a is the imaginary number i a. −a = i a Here are some examples:
−16 = i 16 = 4i
−12 = i 12 = 2i 3
−21 = i 21
−1 = i 1 = i Definition of a Complex Number A complex number is a number of the form a + bi, where a and b are real numbers and i = imaginary part.
Example 1
Solution
Example 2
Solution
Simplify:
−1. The number a is the real part of a + bi, and b is the
−80
−80 = i 80 = i 16 ⋅ 5 = 4i 5 Simplify: 25 +
25 + −40 =
−40 25 + i 40 =
25 + i 4 ⋅ 10
= 5 + 2i 10
Exercises Simplify. 1.
−16
2.
−64
4.
−72
5.
12 −
−18
3.
−98
6.
60 −
−48
136 Chapter 7 Rational Exponents and Radicals Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Objective 2 Add or subtract complex numbers
omplex Numbers To add two complex numbers, add the real parts and add the imaginary parts. To subtract two complex numbers, subtract the real parts and subtract the imaginary parts.
Example 3
Solution
( a + bi ) + ( c + di ) = ( a + c ) + ( b + d )i
( a + bi ) − ( c + di ) = ( a − c ) + ( b − d )i
Add: (3 + 2i ) + (6 − 5i ) (3 + 2i ) + (6 − 5i )
= (3 + 6) + ( 2 − 5)i
= 9 − 3i
Example 4
Solution
Step
• Add the real parts and add the imaginary parts.
Subtract: ( −4 + 2i ) − (6 − 8i ) ( −4 + 2i ) − (6 − 8i ) = ( −4 − 6) + ( 2 + 8)i = –10 + 10i
Exercises Add or subtract. 7. ( 2 + 4i ) + (6 − 5i )
8. (6 − 9i ) + ( 4 + 2i )
10. (3 − 5i ) − (8 − 2i )
11. (5 − 3i ) + 2i
9. ( −2 − 4i ) − (6 − 8i ) 12. (9 + 4i ) + 6
Objective 3 Multiply complex numbers When we multiply complex numbers, the term i 2 is frequently a part of the product. Recall that i 2 = −1. Multiply: 2i ⋅ 3i Multiply the imaginary numbers.
2i ⋅ 3i = 6i 2
Replace i 2 by –1. Then simplify.
= 6( −1) = −6
Multiply: 4i(3 − 2i ) Use the Distributive Property to remove parentheses.
4i(3 − 2i ) = 12i − 8i 2
Replace i 2 by –1.
= 12i − 8( −1)
Write the answer in the form a + bi.
= 8 + 12i
The product of two complex numbers can be found by using the FOIL method. For example,
( 2 + 4i )(3 − 5i ) = 6 − 10i + 12i − 20i 2
= 6 + 2i − 20i 2
= 6 + 2i − 20( −1)
= 26 + 2i Section 7.5 Complex Numbers 137
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example 5
Multiply.
i
C. ( 4 + 5i )( 4 − 5i )
A. (3 − 4i )( 2 + 5i )
Solution
1−
1 i 3
D. (6 + i )2
= 6 + 15i − 8i − 20i 2
Steps
• Use the FOIL method.
= 6 + 7 i − 20i 2
• Combine like terms.
= 6 + 7 i − 20( −1)
• Replace i 2 by –1.
= 26 + 7i
• Write the answer in the form a + bi.
9 3 1 B. + i 1 − i 3 10 10
=
9 3 3 1 2 − i+ i− i 10 10 10 10
• Use the FOIL method.
=
9 1 2 − i 10 10
• Combine like terms.
=
9 1 − ( −1) 10 10
• Replace i 2 by –1.
=
9 1 + = 1 10 10
• Simplify.
• The product of conjugates of the form (a + bi )(a − bi ) is a 2 + b 2 .
• (6 + i )2 = (6 + i )(6 + i )
C. ( 4 + 5i )( 4 − 5i ) = 42 + 52
= 16 + 25
= 41
D. (6 + i )2 = 36 + 12i + i 2
= 36 + 12i + ( −1)
= 35 + 12i
Exercises Multiply. 13. (7i)(–9i)
14. ( −6i)( −4i )
15. 2i(6 + 2i)
16. −3i( 4 − 5i )
17. (5 − 2i )(3 + i )
18. (6 + 5i )(3 + 2i )
2 1 19. ( 2 − i ) 5 + 5 i
20. (6 + i )2
21. (5 − 2i )2
Objective 4 Divide complex numbers A fraction containing one or more complex numbers is in simplest form when no imaginary number remains in the denominator.
138 Chapter 7 Rational Exponents and Radicals Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example 6
Solution
Simplify:
4 − 5i
2i
i
2i
Steps
• Multiply the expression by 1 in the form
i . i
2
=
4i − 5i 2i 2
=
4i − 5( −1) 2( −1)
• Replace i 2 by –1.
=
5 + 4i −2
• Simplify.
= −
• Write the answer in the form a + bi.
5 − 2i 2
To simplify a fraction that has a complex number in the denominator, multiply the numerator and denominator by the conjugate of the complex number.
Example 7
Solution
Simplify:
5 − 3i 4 + 2i =
20 − 10i − 12i + 6i 2 42 + 22
=
20 − 22i + 6( −1) 20
=
14 − 22i 20
=
7 − 11i 10
=
7 11 − i 10 10
5 − 3i (5 − 3i ) = 4 + 2i ( 4 + 2i )
⋅ ( 4 − 2i ) ( 4 − 2i )
Exercises Simplify. 22.
3 i
23.
4 5i
24.
2 − 3i −4i
25.
16 + 5i −3i
26.
2 2−i
27.
1 − 3i 3+ i
28.
2 + 12i 5+i
29.
2 − 3i 3+ i
Section 7.5 Complex Numbers 139
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Chapter 8
Image copyright Aleksey Fursov, 2010. Used under license from Shutterstock.com.
Quadratic Equations and Inequalities
Prep Test
1. Simplify:
18
2. Simplify:
−9
3x − 2 4. Evaluate b2 − 4 ac when − 1 x −1 a = 2, b = −4, and c = 1.
3. Simplify:
5. Is 4 x 2 + 28 x + 49 a perfect-square trinomial?
7. Factor: 9 x 2 − 4
9. Solve: x ( x − 1) = x + 15
6. Factor: 4 x 2 − 4 x + 1
8. Graph: {x | x < −1} ∩ {x | x < 4}
10. Solve:
4 16 = x−3 x
141 Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
se c t io n
8.1 Solving Quadratic Equations king Square Roots
Objective 1 Solve quadratic equations by factoring
A quadratic equation is an equation of the form ax 2 + bx + c = 0, where a and b are coefficients, c is a constant, and a ≠ 0.
3x 2 − x + 2 = 0,
a = 3,
b = −1,
c = 2
− x 2 + 4 = 0,
a = −1,
b = 0,
c = 4
6 x 2 − 5x = 0,
a = 6,
b = −5,
c = 0
A quadratic equation is in standard form when the polynomial is in descending order and equal to zero. Because the degree of the polynomial ax 2 + bx + c is 2, a quadratic equation is also called a second-degree equation. In the chapter entitled “Polynomials and Exponents,” the Principle of Zero Products was used to solve some quadratic equations. That procedure is reviewed here. Principle of Zero Products If the product of two factors is zero, then at least one of the factors must be zero. If ab = 0, then a = 0 or b = 0.
Example 1
Solve by factoring: 2 x 2 − 3x = 2
Solution
2 x 2 − 3x = 2
2 x 2 − 3x − 2 = 0
Steps
• Write in standard form.
( 2 x + 1)( x − 2) = 0
• Factor.
2x + 1 = 0
x−2 = 0
• Use the Principle of Zero Products.
x = 2
• Solve for x.
2 x = −1
x = −
1 2
The solutions are −
1 and 2. 2
Exercises Solve. 1. s 2 − s − 6 = 0
2. 2 x 2 − 9 x − 18 = 0
3. v 2 + 10 = 7 v
4. 2r 2 + r = 6
5. 6 x 2 = 23x + 18
6. 6 x 2 = 7 x − 2
142 Chapter 8 Quadratic Equations and Inequalities Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Objective 2 Write a quadratic equation given its solutions
2
Example 2
) = 0 and the fact that r1 and r2 are solutions of this adratic equation given its solutions.
Write a quadratic equation that has solutions 4 and –5.
Solution ( x − r1 )( x − r2 ) = 0
( x − 4)[ x − ( −5)] = 0
Example 3
( x − 4)( x + 5) = 0 2
x + x − 20 = 0
Steps
• Replace r1 by 4 and r2 by –5.
• Simplify.
• Multiply.
Write a quadratic equation that has integer coefficients and has solutions
Solution ( x − r1 )( x − r2 ) = 0
2 1 x − x − = 0 3 2
x2 −
7 1 x+ = 0 6 3
7 1 6 x 2 − x + = 6 ⋅ 0 6 3
2 1 and . 3 2
2 1 and r2 by . 3 2
Steps
• Replace r1 by
• Multiply.
• Multiply each side of the equation by the LCM of the denominators.
6x2 − 7x + 2 = 0
Exercises Write a quadratic equation that has integer coefficients and has as solutions the given pair of numbers. 7. 2 and 5
10. −
8. –2 and –4
5 and –2 3
11.
9. 6 and –1
1 6 and − 2 5
12.
1 3 and − 10 5
Objective 3 Solve quadratic equations by taking square roots
Recall that if x is a real number, then equations by taking square roots.
Example 4
x 2 = x . This fact is used to solve some quadratic
Solve by taking square roots: 3x 2 = 54
Solution 3x 2 = 54 Steps
• Solve for x 2 .
18
x = 3 2
• Take the square root of each side of the equation. • Simplify.
x = ±3 2
x 2 = 18 x2 =
The solutions are −3 2 and 3 2.
Section 8.1 Solving Quadratic Equations by Factoring or by Taking Square Roots 143
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
This method can also be used when the solutions are complex numbers.
Example
Solution
= 0 2 x 2 + 18 = 0
2 x 2 = −18
x 2 = −9
x
2
=
−9
x = 3i
x = ±3i
Steps
• Solve for x 2 .
• Take the square root of each side of the equation. • Simplify.
The solutions are –3i and 3i.
Example 6
Solution
Solve by taking square roots: 3( x − 2)2 + 12 = 0 3( x − 2)2 + 12 = 0
3( x − 2)2 = −12
( x − 2)2 = −4
( x − 2) 2 =
x − 2 = 2i
x − 2 = ±2i
x = 2 + 2i
• Solve for (x – 2)2.
• Take the square root of each side of the equation. Then simplify.
• Solve for x.
−4
x − 2 = 2i
Steps
x − 2 = −2i x = 2 − 2i
The solutions are 2 + 2i and 2 – 2i.
Exercises Solve by taking square roots. 13. y 2 = 49
14. z 2 = −4
15. 4 x 2 − 81 = 0
16. z 2 + 18 = 0
17. 3( y + 3)2 = 27
18. ( s − 2)2 − 24 = 0
19. ( z + 1)2 + 12 = 0
20. ( v − 3)2 − 18 = 0
se c t io n 8.2
Solving Quadratic Equations by Completing the Square and by Using the Quadratic Formula
Objective 1 Solve quadratic equations by completing the square Not all quadratic equations can be solved by factoring, but any quadratic equation can be solved by completing the square. Add to each side of the equation the term that completes the square. Rewrite the equation in the form ( x + a )2 = b. Then take the square root of each side of the equation.
144 Chapter 8 Quadratic Equations and Inequalities Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example 1
Solve by completing the square: x 2 − 4 x − 14 = 0
Solution x − 4 x − 14 = 0 Pillaged by LibraryPirate 2
x 2 − 4 x = 14
Steps
• Add 14 to each side of the equation.
x 2 − 4 x + 4 = 14 + 4
• Add the constant term that completes the square on x 2 − 4x to each side of the equation. 2 1 ( − ) 4 = 4 2
( x − 2)2 = 18
• Factor the perfect-square trinomial.
( x − 2) 2 =
18
x−2 = 3 2
• Take the square root of each side of the equation. • Simplify.
x − 2 = ±3 2
• Solve for x.
x − 2 = 3 2
x − 2 = −3 2
x = 2+3 2
x = 2−3 2
The solutions are 2 + 3 2 and 2 – 3 2 .
Example 2
Solve by completing the square.
A. 4 x 2 − 8 x + 1 = 0 B. x 2 + 4 x + 5 = 0
A. 4 x 2 − 8 x + 1 = 0
Solution
4 x 2 − 8x −1 = 4 4 1 2 x − 2x = − 4 1 x2 − 2x + 1 = − + 1 4 3 ( x − 1)2 = 4
Steps
• Subtract 1 from each side of the equation.
• The coefficient of the x 2 term must be 1. Divide each side of the equation by 4. 2
1 • Complete the square: ( −2) = 1 2
• Factor the perfect-square trinomial.
( x − 1)2 =
3 4
• Take the square root of each side of the equation.
x −1 =
3 2
• Simplify.
x −1 = ±
x −1 =
3 2
x = 1+
x =
4 x 2 − 8 x = −1
3 2 x −1 = −
3 2
2+ 3 2
The solutions are
3 2
x = 1− x =
• Solve for x.
3 2
2− 3 2
2+ 3 2− 3 and . 2 2
Section 8.2 Solving Quadratic Equations by Completing the Square and by Using the Quadratic Formula 145
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
B. x 2 + 4 x + 5 = 0 x + 4 x = −5 Pillaged by LibraryPirate 2
x 2 + 4 x + 4 = −5 + 4
( x + 2)2 = −1
( x + 2) 2 =
−1
x+2 = i
x + 2 = ±i x + 2 = i
x = −2 + i
x + 2 = −i
• Subtract 5 from each side of the equation. • Complete the square. • Factor the perfect-square trinomial. • Take the square root of each side of the equation. Simplify.
• Solve for x.
x = −2 − i
The solutions are -2 + i and -2 - i.
Exercises Solve by completing the square. 1. x 2 − 4 x − 5 = 0
2. r 2 + 4r − 7 = 0
3. v 2 = 4v − 13
4. x 2 = 2 x − 17
5. 2 y 2 + 3 y + 1 = 0
6. 4u 2 − 20u = −9
Objective 2 Solve quadratic equations by using the quadratic formula
A general formula known as the quadratic formula can be derived by applying the method of completing the square to the standard form of a quadratic equation. This formula can be used to solve any quadratic equation. Quadratic Formula The solutions of ax 2 + bx + c = 0, a ≠ 0, are −b +
b2 − 4 ac − b − b2 − 4 ac and 2a 2a
The quadratic formula is frequently written in the form x =
Example 3
Solution
−b ±
b2 − 4 ac 2a
Solve by using the quadratic formula: 2 x 2 + 5x + 3 = 0 x = = =
−b ± −(5) ± −5 ±
b2 − 4 ac 2a (5)2 − 4( 2)(3) 2( 2) 25 − 24 4
Step
• The equation 2x 2 + 5x + 3 = 0 is in standard form. a = 2, b = 5, c = 3. Replace a, b, and c in the quadratic formula with these values.
146 Chapter 8 Quadratic Equations and Inequalities Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
−5 ± 1
x =
−5 ± 1
=
4
4
= −1
x =
−5 − 1 −6 3 = = − 4 4 2
3 The solutions are –1 and − . 2
Example 4
Solve by using the quadratic formula: 4 x 2 = 8 x − 13 4 x 2 = 8 x − 13
Solution
4 x 2 − 8 x + 13 = 0
Steps
• Write the equation in standard form.
• Replace a, b, and c in the quadratic formula by their values. • Simplify.
a = 4, b = −8, c = 13 x = = = =
−b ±
b2 − 4 ac 2a
−( −8) ± 8± 8±
( −8)2 − 4 ⋅ 4 ⋅ 13 2⋅4
64 − 208 8
−144 8
8 ± 12i 2 ± 3i 3 = = 1± i 8 2 2 3 3 The solutions are 1 + i and 1 − i . 2 2 =
• Write the answer in the form a + bi.
Exercises Solve by using the quadratic formula. 7. x 2 − 3x − 10 = 0 10. y 2 − 2 y + 5 = 0
se c t io n 8.3
8. v 2 − 2v − 7 = 0 11. 2r 2 = 4r − 11
9. 2 z 2 − 2 z − 1 = 0 12. 3v 2 + 6v + 1 = 0
Equations That Are Reducible to Quadratic Equations
Objective 1 Equations that are quadratic in form Certain equations that are not quadratic equations can be expressed in quadratic form by making suitable substitutions. An equation is quadratic in form if it can be written as au 2 + bu + c = 0. The key to recognizing equations that are quadratic in form is that when the equation is 1 written in standard form, the exponent on one variable term is the exponent on the other 2 variable term.
Section 8.3 Equations That Are Reducible to Quadratic Equations 147
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example 1
1
Solve: z + 7 z 2 − 18 = 0 1
Soluti
(z ) 1 2
1
+ 7 z 2 − 18 = 0 u 2 + 7u − 18 = 0
• To solve this equation, let z 2 = u .
(u − 2)(u + 9) = 0
• Solve for u by factoring.
• Replace u by z 2 .
• Solve for z by squaring each side of the equation.
1 2
z
( z ) 1 2
u+9 = 0
u = 2
Example 2
1
• The equation z + 7z 2 − 18 = 0 is quadratic in form.
( )
2
u − 2 = 0
Steps
u = −9 1
1
= 2
z 2 = −9
(z )
2
1 2
2
= 2 z = 4
2
= ( −9)2 z = 81
Because each side of an equation was squared, you must check the solutions. 4 checks as a solution. 81 does not check as a solution. The solution is 4. Solve: x 4 + x 2 − 12 = 0
Solution
x 4 + x 2 − 12 = 0
Steps
• The equation is quadratic in form.
( x 2 )2 + ( x 2 ) − 12 = 0
u 2 + u − 12 = 0
• Let x 2 = u .
(u − 3)(u + 4) = 0
• Solve for u by factoring.
u − 3 = 0
u+4 = 0
u = 3
u = −4
x2 = 3
x 2 = −4
• Replace u by x 2 .
x2 =
x2 =
• Solve for x by taking square roots.
3
x = ± 3
The solutions are
−4
x = ±2i 3, − 3, 2i, and -2i.
Exercises Solve. 1. x 4 − 13x 2 + 36 = 0 1
4. p − 3 p 2 + 2 = 0
2. y 4 − 5 y 2 + 4 = 0 1
5. v − 7 v 2 + 12 = 0
3. x 4 + 12 x 2 − 64 = 0 1
6. w − 2 w 2 − 15 = 0
Objective 2 Radical equations Certain equations containing a radical can be solved by first solving the equation for the radical expression and then squaring each side of the equation.
148 Chapter 8 Quadratic Equations and Inequalities Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Remember that when each side of an equation has been squared, the resulting equation herefore, the solutions of a radical equation must be
Example 3
Solution
7y − 3 + 3 = 2y 7y − 3 = 2y − 3
(
7y − 3
Solution
= ( 2 y − 3)2
• Square each side of the equation.
0 = 4 y 2 − 19 y + 12
0 = ( 4 y − 3)( y − 4)
• Write the equation in standard form. • Solve for y by factoring.
• 4 checks as a solution.
•
Steps
• Solve for one of the radical expressions.
• Square each side of the equation.
• Solve for the radical expression.
• Square each side of the equation.
• Write the equation in standard form.
• 0 and 4 check as solutions.
4y − 3 = 0
y−4 = 0
4y = 3
y = 4
y =
Example 4
2
• Solve for the radical expression.
)
Steps
7 y − 3 = 4 y 2 − 12 y + 9
7y − 3 + 3 = 2y
Solve:
3 4
3 does not check as a solution. 4
The solution is 4. 2y + 1 −
Solve:
2y + 1 −
y =1
2y + 1 =
y =1
(
2y + 1
y +1
) =(
)
2
y +1
2
2y + 1 = y + 2 y + 1
y = 2 y
y2 = 2 y
y2 = 4 y
( )
y2 − 4 y = 0
y ( y − 4) = 0 y = 0
2
y−4 = 0 y = 4
The solutions are 0 and 4.
Exercises Solve. 7.
x + 1 + x = 5
10.
x −1−
x = −1
8.
11.
x − 4 + x = 6
y +1 =
y + 5
9.
12.
p + 11 = 1 − p
x+6 +
x+2 = 2
Section 8.3 Equations That Are Reducible to Quadratic Equations 149
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Objective 3 Fractional equations
n multiplied by the LCM of the denominators, ratic equation. The solutions to the resulting equation must be checked, because multiplying each side of an equation by a variable expression may produce an equation that has a solution that is not a solution of the original equation.
Example 5
Solve:
1 1 3 + = r r +1 2 1 1 3 + = r r +1 2
Solution
1 1 3 2 r ( r + 1) + = 2 r ( r + 1) ⋅ 2 r r + 1
2( r + 1) + 2r = r ( r + 1) ⋅ 3
2r + 2 + 2r = ( r 2 + r ) ⋅ 3
Steps
• Multiply each side of the equation by the LCM of the denominators.
• Write the equation in standard form. • Solve for r by factoring.
4r + 2 = 3r 2 + 3r
0 = 3r 2 − r − 2
0 = (3r + 2)( r − 1)
3r + 2 = 0
r −1 = 0
3r = −2
r = −
r =1
2 3
2 and 1 check as solutions. 3 2 The solutions are − and 1. 3 −
Exercises Solve. 10 x−9
14.
y −1 + y = 1 y+2
15.
3r + 2 − 2r = 1 r +2
3 2 − = 1 s 2s − 1
17.
16 16 + = 6 z −2 z+2
18.
2 1 + =1 y +1 y −1
13. x =
16.
se c t io n
8.4 Applications of Quadratic Equations
Objective 1 Application problems The application problems in this section are similar to those problems that were solved earlier in the text. Each of the strategies for the problems in this section will result in a quadratic equation.
150 Chapter 8 Quadratic Equations and Inequalities Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example 1
Strategy
A small pipe takes 16 min longer to empty a tank than does a larger pipe. Working e tank in 6 min. How long would it take each pipe, c The problem is a work problem. c The unknown time of the larger pipe: t
The unknown time of the smaller pipe: t + 16 Part of task completed
Rate of work
⋅
Time worked
=
Larger pipe
1 t
⋅
6
=
6 t
Smaller pipe
1 t + 16
⋅
6
=
6 t + 16
c The sum of the parts of the task completed must equal 1.
Solution
6 6 + =1 t t + 16
6 6 t (t + 16) + = t (t + 16) ⋅ 1 t + 16 t
(t + 16)6 + 6t = t 2 + 16t
6t + 96 + 6t = t 2 + 16t
0 = t 2 + 4t − 96
0 = (t + 12)(t − 8)
t + 12 = 0
t −8 = 0
t = −12
t = 8
The solution t = −12 is not possible because time cannot be a negative number. The time for the smaller pipe is t + 16. Replace t by 8 and evaluate. t + 16 = 8 + 16 = 24 The larger pipe requires 8 min to empty the tank. The smaller pipe requires 24 min to empty the tank.
Example 2
Strategy
In 8 h, two campers rowed 15 mi down a river and then rowed back to their campsite. The rate of the river’s current was 1 mph. Find the rate at which the campers row in calm water. c This is a uniform motion problem. c Unknown rowing rate of the campers: r Distance
÷
Rate
=
Time
Down river
15
÷
r+1
=
15 r +1
Up river
15
÷
r–1
=
15 r −1
c The total time of the trip was 8 h. Section 8.4 Applications of Quadratic Equations 151
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
15
Solution
+
15
= 8
( r + 1)( r − 1) + = ( r + 1)( r − 1)8 r + 1 r − 1
( r − 1)15 + ( r + 1)15 = ( r 2 − 1)8
15r − 15 + 15r + 15 = 8r 2 − 8
30r = 8r 2 − 8
0 = 8r 2 − 30r − 8
0 = 2( 4r 2 − 15r − 4)
0 = 2( 4r + 1)( r − 4)
4r + 1 = 0
4r = −1
r = −
r −4 = 0 r = 4
1 4
1 is not possible because the rate cannot be a negative number. 4 The rowing rate is 4 mph. The solution r = −
Exercises Solve. 1. A small pipe can fill an oil tank in 6 min more time than it takes a larger pipe to fill the same tank. Working together, both pipes can fill the tank in 4 min. How long would it take each pipe, working alone, to fill the tank? 2. A small heating unit takes 8 h longer to melt a piece of iron than does a larger unit. Working together, the heating units can melt the iron in 3 h. How long would it take each heating unit, working alone, to melt the iron? 3. A cyclist traveled 60 mi at a constant rate before reducing the speed by 2 mph. Another 40 mi was traveled at the reduced speed. The total time for the 100-mile trip was 9 h. Find the rate during the first 60 mi. 4. The rate of a single-engine plane in calm air is 100 mph. Flying with the wind, the plane can fly 240 mi in 1 h less time than is required to make the return trip of 240 mi. Find the rate of the wind.
se c t io n
8.5 Nonlinear Inequalities
Objective 1 Solve nonlinear inequalities A quadratic inequality in one variable is one that can be written in the form ax 2 + bx + c < 0 or ax 2 + bx + c > 0, where a ≠ 0. The symbols ≤ and ≥ can also be used.
152 Chapter 8 Quadratic Equations and Inequalities Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Quadratic inequalities can be solved by algebraic means. However, it is often easier to use equalities.
Ex
of x 2 − x − 6 < 0. Solution
x2 − x − 6 < 0
( x − 3)( x + 2) < 0
−5 − 4 −3 −2 −1
0
1
2
3
4
Steps
• Factor the trinomial.
• On a number line, draw vertical lines at the numbers that make each factor equal to zero.
5
x −3 = 0
x − 3 −−−−−−−− −−−−−−−−−−−
+++++
x + 2 −−−−−−−−
+++++
+++++++++++
−5 −4 −3 −2 −1
0
1
2
3
4
x =3
{ x | −2 < x < 3}
The graph of the solution set of the inequality x 2 − x − 6 < 0 is shown at the right.
Example 2
−5 −4 −3 −2 −1
0
1
2
3
4
5
Solve and graph the solution set of 2 x 2 − x − 3 ≥ 0.
Solution
x = −2
• For each factor, place plus signs above the number line for those regions where the factor is positive and place negative signs where the factor is negative. x – 3 is positive for x > 3, and x + 2 is positive for x > −2. Because x 2 − x − 6 < 0, the solution set will be the regions where one factor is positive and the other factor is negative. • Write the solution set.
5
x +2=0
2x2 − x − 3 ≥ 0
( 2 x − 3)( x + 1) ≥ 0 2x − 3 − − − − − − − − − − − − − − − + + + + + + + + + + x + 1 −−−−−−−−−− +++++ ++++++++++
−5 −4 −3 −2 −1
3 x | x ≤ −1 or x ≥ 2
−5 − 4 −3 −2 −1 0
0
1
1
2
3
2
4
3
4
5
5
This method can also be used to solve rational inequalites.
Example 3
Solution
Solve and graph the solution set of x+4 ≥ 0 x−3
x+4 ≥ 0. x−3
x + 4 −−
++++++++++++++++
+++++
x − 3 −−
−−−−−−−−−−−−−−−−
+++++
−5 −4 −3 −2 −1
0
1
2
3
4
5
{ x | x > 3 or x ≤ −4} − 5 − 4 − 3 −2 −1
0
1
2
3
4
5
Section 8.5 Nonlinear Inequalities 153
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Exercises Solve and graph the solution set. Write the answer in set-builder notation.
Pillaged by LibraryPirate
1. x 2 − 3x + 2 ≥ 0
2. x 2 + 5x + 6 > 0
4. x 2 + x − 20 < 0
5.
se c t io n
x−4 > 0 x+2
3. x 2 − x − 12 < 0
6.
x−3 ≤ 0 x+1
8.6 Properties of Quadratic Functions
Objective 1 Graph a quadratic function
A quadratic function is one that can be expressed by the equation f ( x ) = ax 2 + bx + c, a ≠ 0. The graph of this function is called a parabola. In general, the graph of f ( x ) = ax 2 + bx + c, a ≠ 0, resembles a “cup” shape. The parabola opens up when a > 0 and opens down when a < 0. The vertex of the parabola is the point with the smallest y-coordinate when a > 0 and is the point with the largest y-coordinate when a < 0. y
y Axis of Symmetry
Vertex
x
x Axis of Symmetry Vertex a<0
a>0
The axis of symmetry is a line that passes through the vertex of the parabola and is parallel to the y-axis. To understand the axis of symmetry, think of folding the graph along that line. The two portions of the graph will match up. Vertex and Axis of Symmetry of a Parabola Let f ( x ) = ax 2 + bx + c be the equation of a parabola. The coordinates of the b vertex are − , 2a
Example 1
Solution
b b f − . The equation of the axis of symmetry is x = − . 2a 2a
Graph y = x 2 + 2 x − 3 using the vertex and axis of symmetry. x = −
b 2 = − = −1 2a 2(1)
y = x2 + 2x − 3
Steps
y = ( −1)2 + 2( −1) − 3
• Find the x-coordinate of the vertex. a = 1 and b = 2. • Find the y-coordinate of the vertex by replacing x with –1 and solving for y.
y = 1− 2−3 y = −4
• The vertex is (–1, –4). • The axis of symmetry is the line x = –1.
154 Chapter 8 Quadratic Equations and Inequalities Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
x
y
1 2
0 5
Axis of Symmetry
• Find some ordered-pair solutions of the equation and record these in a table. Because the graph is symmetric to the line x = –1, choose values of x greater than –1.
• Graph the ordered-pair solutions on a rectangular coordinate system. Use symmetry to locate points of the graph on the other side of the axis of symmetry. Remember that corresponding points on the graph are the same distance from the axis of symmetry. • Draw a parabola through the points.
y
(– 4, 5)
(2, 5) 4
3 units
3 units
2
(–3, 0)
(1, 0)
−4
0
(–2, –3)
2
x
4
(0, –3)
(–1, – 4)
Exercises Find the vertex and axis of symmetry of the parabola given by each equation. Then sketch its graph. 1. y = −2 x 2 + 4 x
2. y = x 2 − x − 2
3. y = x 2 − 3x + 2
Objective 2 Find the x-intercepts of a parabola
Recall that a point at which a graph crosses the x- or y-axis is called an intercept of the graph. The x-intercepts of the graph of an equation occur when y = 0; the y-intercepts occur when x = 0.
x-intercepts y 2
(−4, 0) −6
2
The graph of y = x + 3x − 4 is shown at the right. The points whose coordinates are (–4, 0) and (1, 0) are x-intercepts of the graph.
−2 0 −2
(1, 0) x 2
− 4 (0, −4) −6
y-intercept
Example 2
Find the x-intercepts of the parabola given by each equation.
A. y = x 2 + 2 x − 2 B. y = 4 x 2 + 4 x + 1
A. y = x 2 + 2 x − 2
Solution
0 = x2 + 2x − 2
x =
b2 − 4 ac 2a
− ( 2) ±
=
=
=
−2 ± 12 2
=
−2 ± 2 3 = −1 ± 2
−2 ±
• Let y = 0.
• The equation is nonfactorable over the integers. Use the quadratic formula to solve for x.
( 2)2 − 4(1)( −2) 2 ⋅1
−b ±
Steps
4+8 2
(
3
The x-intercepts are −1 +
)
(
3 , 0 and −1 −
)
3, 0 .
Section 8.6 Properties of Quadratic Functions 155
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
B. y = 4 x 2 + 4 x + 1
2x + 1 = 0
2 x = −1
x = −
• Let y = 0.
• Solve for x by factoring.
• The equation has a double root.
2x + 1 = 0
1 2
2 x = −1 x = −
1 2
1 The x-intercept is − , 0 . 2
Exercises Find the x-intercepts of the parabola given by each equation. 4. y = x 2 − 4
5. y = x 2 − x − 2
6. y = 2 x 2 − 5x − 3
7. y = x 2 + 4 x − 3
8. y = x 2 + 6 x + 10
9. y = − x 2 + 4 x + 1
se c t io n
8.7 Applications of Quadratic Functions
Objective 1 Minimum and maximum problems
The graph of f ( x ) = x 2 − 2 x + 3 is shown at the right. Because a is positive, the parabola opens up. The vertex of the parabola is the lowest point on the parabola. It is the point that has the minimum y-coordinate. Therefore, the value of the function at this point is a minimum.
The graph of f ( x ) = − x 2 + 2 x + 1 is shown at the right. Because a is negative, the parabola opens down. The vertex of the parabola is the highest point on the parabola. It is the point that has the maximum y-coordinate. Therefore, the value of the function at this point is a maximum.
y 6 4 2
Minimum y-coordinate
0
2
Vertex (1, 2)
4
6
x
y 4
Vertex (1, 2)
2 0
2
Maximum y-coordinate x 4
6
−2 −4
To find the minimum or maximum value of a quadratic function, first find the x-coordinate of the vertex. Then evaluate the function at that value.
Example 1
Solution
Find the maximum or minimum value of the function f ( x ) = −2 x 2 + 4 x + 3. x = −
b 4 = − = 1 2a 2( −2)
f ( x ) = −2 x 2 + 4 x + 3
Steps
• Find the x-coordinate of the vertex. a = –2, b = 4.
• Evaluate the function at x = 1.
156 Chapter 8 Quadratic Equations and Inequalities Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
f (1) = −2(1)2 + 4(1) + 3 pens down. Therefore, the function has a maximum
value.
The maximum value of the function is 5.
Exercises Find the minimum or maximum of the quadratic function. 1. f ( x ) = x 2 − 2 x + 3
2. f ( x ) = −2 x 2 + 4 x − 3
3. f ( x ) = 2 x 2 + 3x − 8
4. f ( x ) = − x 2 − x + 2
Objective 2 Applications of minimum and maximum
Example 2
A mining company has determined that the cost in dollars (c) per ton of mining a mineral is given by c( x ) = 0.2 x 2 − 2 x + 12 where x is the number of tons of the mineral that is mined. Find the number of tons of the mineral that should be mined to minimize the cost. What is the minimum cost?
Strategy
c To find the number of tons that will minimize the cost, find the x-coordinate of the vertex.
c To find the minimum cost, evaluate the function at the x-coordinate of the vertex.
x = −
Solution
b −2 = − = 5 2a 2(0.2)
To minimize the cost, 5 tons should be mined. c( x ) = 0.2 x 2 − 2 x + 12 c(5) = 0.2(5)2 − 2(5) + 12 = 5 − 10 + 12 = 7 The minimum cost per ton is $7.
Example 3
The height in feet (s) of a ball thrown straight up is given by s(t ) = −16t 2 + 64t , where t is the time in seconds. Find the time it takes the ball to reach its maximum height. What is the maximum height?
Strategy
c To find the time it takes the ball to reach its maximum height, find the t-coordinate of the vertex.
c To find the maximum height, evaluate the function at the t-coordinate of the vertex.
t = −
Solution
b 64 = − = 2 2a 2( −16)
The ball reaches its maximum height in 2 s. s(t ) = −16t 2 + 64t s( 2) = −16( 2)2 + 64( 2) = −64 + 128 = 64 The maximum height is 64 ft.
Section 8.7 Applications of Quadratic Functions 157
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Exercises Solve. 5. The height s, in feet, of a rom a cliff 50 ft above an 2 ocean beach is given by the function s(t ) = −16t + 64t + 50, where t is the time in seconds. Find the maximum height above the beach that the rock will attain. 6. A manufacturer of microwave ovens believes that the revenue R, in dollars, that the company receives is related to the price P, in dollars, of an oven by the function R( P ) = 125 P − 0.25 P 2 . What price will give the maximum revenue? 7. A pool is treated with a chemical to reduce the amount of algae. The amount of algae in the pool t days after the treatment can be approximated by the function A(t ) = 40t 2 − 400t + 500. How many days after treatment will the pool have the least amount of algae?
158 Chapter 8 Quadratic Equations and Inequalities Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Chapter 9
Image copyright Adrian Zenz, 2010. Used under license from Shutterstock.com.
Functions and Relations
Prep Test
1. Evaluate x + 4 for x = −6.
2. Given y = − x 2 + 2 x + 1, find the value of y when x = −2.
3. Given f ( x ) = x 2 − 3x + 2, find f ( −4).
4. Evaluate p( r ) = r 2 − 5 when r = 2 + h.
5. Solve x = 2 y + 4 for y.
6. Find the domain and range of the relation {(–2, 4), (3, 5), (4, 6), (6, 5)}. Is the relation a function?
7. What value is excluded from the domain of f ( x ) =
8. Graph: f ( x ) = x
3 ? x−8
9. Graph: f ( x ) = x 2 − 2
159 Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
se c t io n
9.1 Translations of Graphs
Objective
Vertical Translation If f is a function and c is a positive constant, then
Example 1
y = f ( x ) + c is the graph of y = f ( x ) shifted up c units.
y = f ( x ) − c is the graph of y = f ( x ) shifted down c units.
Given the graph of the function f ( x) =
3
y
x shown at the right in 3
gray, graph g ( x ) = x + 3 by using a vertical translation.
Solution
The graph of g ( x ) =
3
x + 3 is the
g
4
f
2 −4
−2
2
0
x
4
−2
3
graph of f ( x ) = x shifted 3 units up. This graph is shown in black at the right.
−4
Horizontal Translation If f is a function and c is a positive constant, then y = f ( x − c ) is the graph of y = f ( x ) shifted to the right c units. y = f ( x + c ) is the graph of y = f ( x ) shifted to the left c units.
Example 2
Solution
Given the graph of the function F ( x ) = x 3 shown at the right in gray, graph G ( x ) = ( x + 2)3 by using a horizontal translation. The graph of G (shown in black) is the graph of F shifted horizontally 2 units to the left. Note that if
y F
G4 2 −4
−2
0
2
x
4
−2 −4
( x + 2)3 = 0, then x = −2.
The result, –2, is one confirmation that the graph should be shifted 2 units to the left. It is possible for a graph to involve both a horizontal and a vertical translation.
Example 3
Solution
Given the graph of f ( x ) = x shown at the right in gray, graph A( x ) = x + 1 − 3 using both a horizontal and a vertical translation. The graph of A includes a horizontal shift of 1 unit to the left and a vertical shift of 3 units down. The graph of A is shown in black at the right.
y
−4
−2
4
f
2
A
0
2
4
x
−2 −4
160 Chapter 9 Functions and Relations Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Exercises Graph. 1. Given the graph
y
h g ( x ) = x + 4. 4 2
2. Given the graph of f ( x ) = x shown in Figure 1 at the right, graph g ( x ) = x − 3. −4
3. Given the graph of f ( x ) = x shown in Figure 1 at the right, graph g ( x ) = x − 1 .
−2
f(x) = | x|
0
2
4
x
−2 −4
Figure 1
4. Given the graph of f ( x ) = x shown in Figure 1 at the right, graph g ( x ) = x + 4 . 5. Given the graph of f ( x ) = x shown in Figure 1 at the right, graph g ( x ) = x + 2 − 1. 6. Given the graph of f ( x ) = x shown in Figure 1 at the right, graph g ( x ) = x − 1 + 2.
se c t io n
9.2 Algebra of Functions
Objective 1 Perform operations on functions
The operations of addition, subtraction, multiplication, and division of functions are defined as follows: Operations on Functions If f and g are functions and x is an element of the domain of each function, then
Example 1
Solution
( f + g )( x ) = f ( x ) + g ( x )
( f − g )( x ) = f ( x ) − g ( x )
(f
f f ( x) g ( x ) = g ( x ) , g ( x ) ≠ 0
⋅ g )( x )
= f ( x) ⋅ g( x)
Given f ( x ) = x 2 − x + 1 and g ( x ) = x 3 − 4, find ( f − g )(3). ( f − g )(3) = f (3) − g (3)
= (32 − 3 + 1) − (33 − 4)
= 7 − 23
= −16
( f − g )(3) = −16
Section 9.2 Algebra of Functions 161
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example 2
Given f ( x ) = x 2 + 2 and g ( x ) = 2 x + 3, find ( f
⋅ g )( −2).
Soluti
= [( −2)2 + 2] [2( −2) + 3]
= 6 ⋅ [ −1]
= −6
(f
Example 3
Solution
⋅ g )( −2)
= −6
f Given f ( x ) = x 2 + 4 x + 4 and g ( x ) = x 3 − 2, find (3). g f f (3) g (3) = g (3)
=
32 + 4(3) + 4 33 − 2
=
25 25
=1
f g (3) = 1
Exercises Given f ( x ) = 2 x 2 − 3 and g ( x ) = −2 x + 4, find: 1. ( f − g )( 2)
2. ( f + g )(0)
3. ( f
g 4. ( −3) f
⋅ g )( −1)
Objective 2 Find the composition of two functions
Composition of functions is another way in which functions can be combined. This method of combining functions uses the output of one function as the input for a second function. Definition of the Composition of Two Functions Let f and g be two functions such that g ( x ) is in the domain of f for all x in the domain of g. Then the composition of the two functions, denoted by f g , is the function whose value at x is given by ( f g )( x ) = f [ g ( x )].
Example 4
Given f ( x ) = x 2 − 1 and g ( x ) = 3x + 4, evaluate each composite function.
A. f [ g (0)] B. g[ f ( x )]
A.
g ( x ) = 3x + 4
g(0) = 3(0) + 4 = 4
f ( x) = x2 − 1
Solution
f [ g (0)] = f ( 4) = 42 − 1 = 15
f [ g (0)] = 15
Steps
• To evaluate f [ g(0)], first evaluate g(0).
• Substitute the value of g(0) for x in f(x); g(0) = 4.
162 Chapter 9 Functions and Relations Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
B. g[ f ( x )] = g ( x 2 − 1)
Example 5
= 3x 2 − 3 + 4
= 3x2 + 1
• f ( x ) = x 2 − 1
• Substitute x 2 − 1 for x in the function g(x); g ( x ) = 3x + 4.
Given g ( x ) = 3x − 2 and h( x ) = x 2 + 1, evaluate each composite function.
A. g[ h(0)] B. h[ g ( x )]
A.
h( x ) = x 2 + 1
h(0) = 0 + 1 = 1
g ( x ) = 3x − 2
g(1) = 3(1) − 2 = 1
Solution
g[ h(0)] = 1
B. h[ g ( x )] = h(3x − 2)
= (3x − 2)2 + 1
= 9 x 2 − 12 x + 4 + 1
= 9 x 2 − 12 x + 5
Exercises Given g ( x ) = x 2 + 3 and h( x ) = x − 2, evaluate the composite function. 5. g[ h(0)]
6. h[ g (0)]
7. g[ h( 4)]
8. h[ g ( −2)]
9. g[ h( x )]
10. h[ g ( x )]
se c t io n
9.3 One-to-One and Inverse Functions
Objective 1 Determine whether a function is one-to-one Recall that a function is a set of ordered pairs in which no two ordered pairs that have the same first component have different second components. This means that given any x, there is only one y that can be paired with that x. A one-to-one function satisfies the additional condition that given any y, there is only one x that can be paired with the given y. One-to-one functions are commonly expressed by writing 1-1. The function given by the equation y = x is not a 1-1 function since, given y = 2, there are two possible values of x, 2 and -2, that can be paired with the given y-value. Just as the vertical line test can be used to determine whether a graph represents a function, a horizontal line test can be used to determine whether the graph of a function represents a 1–1 function.
Section 9.3 One-to-One and Inverse Functions 163
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Horizontal Line Test function if any horizontal line intersects
Example 1
Since any vertical line will intersect the graph at the right at no more than one point, the graph is the graph of a function. Since any horizontal line will intersect the graph at no more than one point, the graph is the graph of a 1–1 function.
y
The graph of a quadratic function is shown at the right. Note that a horizontal line can intersect the graph at more than one point. Therefore, this graph is not the graph of a 1–1 function. In general, f ( x ) = ax 2 + bx + c, a ≠ 0, is not a 1–1 function.
y
x
x
Determine whether or not the graph represents the graph of a 1–1 function. y
A.
B.
y
x
Solution
A.
x
y
Steps
• A vertical line can intersect the graph at more than one point. The graph does not represent a function.
• A horizontal line can intersect the curve at more than one point.
x
This is not the graph of a 1-1 function.
B.
y
x
This is not the graph of a 1-1 function.
164 Chapter 9 Functions and Relations Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Exercises Determine whether t 1.
3.
y
–8
–4
y
8
8
8
4
4
4
0
4
8
x
–8
0
–4
4
8
x
–8
–4
0
–4
–4
–4
–8
–8
–8
4
8
x
Objective 2 Find the inverse of a function
The inverse of a function is the set of ordered pairs formed by reversing the coordinates of each ordered pair of the function. The domain of the inverse function is the range of the function, and the range of the inverse function is the domain of the function. Condition for an Inverse Function A function f has an inverse function if and only if f is a 1–1 function. The symbol f −1 is used to denote the inverse of a 1–1 function f. The symbol f −1 ( x ) is read “f inverse of x.” f −1 ( x ) is not the reciprocal of f ( x ) but is the notation for the inverse of a 1–1 function. To find the inverse of a function, interchange x and y. Then solve for y.
Example 2
Solution
Find the inverse of the function defined by f ( x ) = 2 x − 4. f ( x) = 2x − 4
y = 2x − 4
Steps
x = 2y − 4
• Think of the function as the equation y = 2x – 4. • Interchange x and y.
• Solve for y.
2y = x + 4
y =
1 x+2 2
f −1 ( x ) =
1 x+2 2
The composition of a function and its inverse have a special property. Composition of Inverse Functions Property f −1[ f ( x )] = x and f [ f −1 ( x )] = x
Section 9.3 One-to-One and Inverse Functions 165
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example 3
Solution
Are the functions defined by the equations f ( x ) = −2 x + 3 and g ( x ) = − 1 3 f [ g ( x )] = f − x + 2 2
1 3 = −2 − x + + 3 2 2
= x −3+3
= x
1 3 x+ 2 2
Steps
• Use the property that, for inverses, f [f −1( x )] = f −1[f ( x )] = x .
• f [g ( x )] = x
• g [f ( x )] = x
g[ f ( x )] = g ( −2 x + 3)
1 3 = − ( −2 x + 3) + 2 2
= x−
= x
3 3 + 2 2
The functions are inverses of each other.
Exercises Find the inverse of the function. 4. f ( x ) = 4 x − 8
5. f ( x ) =
1 x − 1 2
6. f ( x ) = −3x − 9
7. f ( x ) =
2 x+4 3
Use the Composition of Inverse Functions Property to determine whether the functions are inverses of each other. 8. g ( x ) = 3x + 2; f ( x ) =
1 2 x− 3 3
10. h( x ) = x + 2; g ( x ) = 2 − x
9. h( x ) = 4 x − 1; f ( x ) =
11. g ( x ) = −
1 1 x+ 4 4
1 1 x − ; h( x ) = −2 x + 1 2 2
166 Chapter 9 Functions and Relations Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Chapter 10 Image copyright Alexander Dashewsky, 2010. Used under license from Shutterstock.com.
Exponential and Logarithmic Functions
Prep Test 1 2. Simplify: 2
−4
1. Simplify: 3−2
3. Complete:
4. Evaluate f ( x ) = x 4 + x 3 for x = −1 and x = 3.
5. Solve: 3x + 7 = x − 5
6. Solve: 16 = x 2 − 6 x
7. Evaluate A(1 + i ) n for A = 5000, i = 0.04, and n = 6. Round to the nearest hundredth.
8. Graph: f ( x ) = x 2 − 1
1 = 2? 8
167 Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
sec t io n
10.1 Exponential Functions
Objective
Definition of an Exponential Function The exponential function with base b is defined by f ( x ) = bx where b > 0, b ≠ 1, and x is any real number. x
1 Evaluate f ( x ) = at x = 2 and x = −3. 2
Example 1 Solution
1 f ( x) = 2
x
1 f ( 2) = 2
2
Example 2
Solution
=
1 f ( −3) = 2
1 4
−3
= 23 = 8
Evaluate f ( x ) = 23x − 1 at x = 1 and x = -1. f ( x ) = 23 x − 1 f (1) = 23(1) − 1 = 22 = 4
f ( −1) = 23( −1) − 1 = 2 −4 =
1 1 = 4 16 2
A frequently used base in applications of exponential functions is an irrational number designated by e. The number e is approximately 2.71828183. It is an irrational number, so it has a nonterminating, nonrepeating decimal representation. Natural Exponential Function The function defined by f ( x ) = e x is called the natural exponential function. y
The ex key on a calculator can be used to evaluate the natural exponential function. The graph of y = e x is shown at the right.
8 4 −8
Example 3
Solution
−4
0
4
x
Evaluate f ( x ) = e x at x = 2, x = -3, and x = π . Round to the nearest ten-thousandth.
f ( x ) = ex f ( 2) = e2
f ( −3) = e−3
≈ 7.3891
≈ 0.0498
f (π ) = eπ ≈ 23.1407
Exercises 2. Given g ( x ) = 2 x + 1 , evaluate:
1. Given f ( x ) = 3x , evaluate: a. f ( 2)
b. f (0)
c. f ( −2)
a. g(3)
b. g(1)
c. g( −3)
168 Chapter 10 Exponential and Logarithmic Functions Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1 3. Given P( x ) =
2x
x
G ( x ) = e 2 , evaluate the following. Round to rest ten-thousandth.
3 b. P 2
a. P(0)
c. P( −2)
a. G( 4)
b. G( −2)
1 c. G 2
5. Given H ( r ) = e− r + 3 , evaluate the following. 6. Given f ( x ) = e2 x , evaluate the following. Round to Round to the nearest ten-thousandth. the nearest ten-thousandth. a. H ( −1)
b. H (3)
c. H (5)
a. f ( −2)
2 b. f − 3
c. f ( 2)
Objective 2 Graph an exponential function
f ( x) = y
x
To graph f ( x ) = 2 x , think of the function as the equation y = 2 x .
–1
Graph the ordered pairs on a rectangular coordinate system. Connect the points with a smooth curve.
1 4 1 = 2
2 −2 =
–2
Choose values of x, and find the corresponding values of y. The results can be recorded in a table.
y
2
−1
0
20 = 1
1
21 = 2
2
22 = 4
3
23 = 8
6 4 2 −2
0
2
x
Note that a vertical line would intersect the graph at only one point. Therefore, by the vertical line test, f ( x ) = 2 x is the graph of a function. Also note that a horizontal line would intersect the graph at only one point. Therefore, f ( x ) = 2 x is the graph of a 1–1 function. Exponential Functions Are 1–1 The exponential function defined by f ( x ) = bx , b > 0, b ≠ 1, is a 1–1 function.
Example 4 Solution
1
Graph. A. f ( x ) = 32 A.
x
2 4
3
0
B. f ( x ) = 2 x − 1
y 1 9 1 3 1
–2
x −1
y 6
–6
0
6
x
–6
Section 10.1 Exponential Functions 169
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
B. f ( x ) = 2 x − 1
x
y
y 6
0
1 2 0
1
1
2
3
3
7
−
–1
–6
0
6
x
–6
Exercises Graph. 8. f ( x ) = 3− x
7. f ( x ) = 3x
9. f ( x ) = 2 x + 1
x
1 11. f ( x ) = 3
10. f ( x ) = 2 x − 1
sec t io n
2 12. f ( x ) = 3
x
10.2 Introduction to Logarithms
Objective 1 Write equivalent exponential and logarithmic equations
Definition of Logarithm If x > 0 and b is a positive constant not equal to 1, then y = log b x is equivalent to b y = x. Read log b x as “the logarithm base b of x” or “the log base b of x.” The following table shows equivalent statements written in both exponential and logarithmic form.
Example 1
Solution
Exponential Form
24 = 16
2 3
10 −1 = 0.1
log10 (0.1) = −1
ba = x
log b x = a
2
=
Logarithmic Form
4 9
log 2 16 = 4 4 log 2 = 2 9 3
Write 45 = 1024 in logarithmic form. 45 = 1024 is equivalent to log 4 1024 = 5.
170 Chapter 10 Exponential and Logarithmic Functions Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example 2
Write log 7 343 = 3 in exponential form. = 343.
The 1–1 property of exponential functions can be used to evaluate some logarithms. Equality of Exponents Property For b > 0, b ≠ 1, if bu = bv , then u = v.
Example 3
Solution
1 Evaluate: log 3 9 1 log 3 = x 9
Steps
• Write an equation. • Write the equation in its equivalent exponential form. 1 • Write in exponential form using 3 9 as the base. • Solve for x using the Equality of Exponents Property.
1 = 3x 9
3−2 = 3x
−2 = x
1 log 3 = −2 9
Example 4
Solution
Solve log 6 x = 2 for x. log 6 x = 2
62 = x
36 = x
Step
• Write log6 x = 2 in its equivalent exponential form.
The solution is 36. Logarithms base 10 are called common logarithms. We usually omit the base, 10, when writing the common logarithm of a number. Therefore, log10 x is written log x. To find the common logarithm of most numbers, a calculator is necessary. A calculator was used to find the value of log 384, shown below. log 384 ≈ 2.584331224 When e (the base of the natural exponential function) is used as a base of a logarithm, the logarithm is referred to as the natural logarithm and is abbreviated ln x. This is read “el en x.” Using a calculator, we find that ln 23 ≈ 3.135494216
Example 5
Solve ln x = -1 for x. Round to the nearest ten-thousandth.
Solution
ln x = -1
e−1 = x
Steps
• ln x is the abbreviation for loge x .
• Write the equation in its equivalent exponential form. Evaluate.
0.3679 ≈ x The solution is 0.3679.
Section 10.2 Introduction to Logarithms 171
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Exercises Write the exponential equatio 1. 52 = 25
3. 4 −2 =
2. 103 = 1000
16
4. e y = x
Write the logarithmic equation in exponential form. 5. log 3 9 = 2
6. log 0.01 = −2
7. log5
1 = −1 5
8. log x = y
Evaluate. 9. log 3 81
10. log5 125
11. log 0.001
12. ln e2
14. log 4 x = 3
15. log 7 x = −1
16. log 6 x = 0
19. ln x = 2
20. ln x = −1.7
Solve for x. 13. log 3 x = 2
Solve for x. Round to the nearest hundredth. 17. log x = 2.5
18. log x = −1.75
Objective 2 The properties of logarithms
Product Property of Logarithms For any positive real numbers x, y, and b, b ≠ 1, log b ( xy ) = log b x + log b y . Quotient Property of Logarithms For any positive real numbers x, y, and b, b ≠ 1, log b
x = log b x − log b y. y
Power Property of Logarithms For any positive real numbers x and b, b ≠ 1, and for any real number r, log b x r = r log b x.
Example 6
Solution
Write the logarithm in expanded form. A. log b
x2 xy B. ln 3 z y
A. log b
xy = log b ( xy ) − log b z z
= log b x + log b y − log b z
Steps
• Use the Quotient Property of Logarithms. • Use the Product Property of Logarithms.
172 Chapter 10 Exponential and Logarithmic Functions Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
B. ln
x2
= ln x 2 − ln y 3
• Use the Quotient Property of Logarithms. • Use the Power Property of Logarithms.
Example 7
Express as a single logarithm with a coefficient of 1.
A. 3 log5 x + log5 y − 2 log5 z B.
A. 3 log5 x + log5 y − 2 log5 z
Solution
1 (log 3 x − 3 log 3 y + log 3 z ) 2
= log5 x 3 + log5 y − log5 z 2
Steps
= log5 x 3 y − log5 z 2
= log 5
B.
x3 y z2
1 (log 3 x − 3 log 3 y + log 3 z ) 2 1 = (log 3 x − log 3 y 3 + log 3 z ) 2
• Use the Power Property of Logarithms. • Use the Product Property of Logarithms.
• Use the Quotient Property of Logarithms.
• Use the Power Property of Logarithms.
=
1 x log + log 3 z 2 3 y 3
• Use the Quotient Property of Logarithms.
=
1 xz log 2 3 y 3
• Use the Product Property of Logarithms.
xz 2 xz = log 3 3 = log 3 y y3
• Use the Power Property of Logarithms. Write the exponential expression as a radical expression.
1
Although only common logarithms and natural logarithms are programmed into a calculator, the logarithms for other positive bases can be found. Change-of-Base Formula log a N =
Example 8
Solution
log b N log b a
Evaluate log 7 32. Round to the nearest ten-thousandth. log 7 32 =
ln 32 ≈ 1.7810 ln 7
Step
• Use the Change-of-Base Formula. N = 32, a = 7, b = e
Exercises Write the logarithm in expanded form. 21. log8 ( xz )
22. log 3 x 5
r 23. ln s
24. log 9 x 2 yz
xy 2 25. ln 4 z
x2 26. log8 2 yz
Section 10.2 Introduction to Logarithms 173
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Express as a single logarithm with a coefficient of 1. 27. 3ln x
29. 2 log 3 x − log 3 y + 2 log 3 z
30. log b x − ( 2 log b y + log b z )
31.
1 (log8 x − log8 y ) 3
32. 2(log 4 s − 2 log 4 t + log 4 r )
Evaluate. Round to the nearest ten-thousandth. 33. log8 6
34. log5 30
sec t io n
35. log5 (0.6)
10.3 Graphs of Logarithmic Functions
Objective 1 Graph logarithmic functions
The graph of a logarithmic function can be drawn by using the relationship between the exponential and logarithmic functions. To graph f ( x ) = log 2 x , think of the function as the equation y = log 2 x . Write the equation.
equivalent
exponential
f ( x ) = log 2 x y = log 2 x
x = 2y
Because the equation is solved for x in terms of y, it is easier to choose values of y and find the corresponding values of x. The results can be recorded in a table.
x 1 4 1 2 1
Graph the ordered pairs on a rectangular coordinate system. Connect the points with a smooth curve.
y –2
y 4
–1 0
2
1
4
2
2 0
2
4
6
x
–2
Applying the vertical line and horizontal line tests reveals that f ( x ) = log 2 x is the graph of a 1–1 function.
Example 1
Solution
Graph. A. f ( x ) = log 3 x B.
f ( x ) = 2 log 3 x
A. f ( x ) = log 3 x
y = log 3 x
Steps
• Substitute y for f(x).
x = 3
• Write the equivalent exponential equation. • Choose values of y, and find the corresponding values of x. Graph the ordered pairs on a rectangular coordinate system. Connect the points with a smooth curve.
x 1 9 1 3 1
3
y
y
y
-2 -1 0
6
–6
0 –6
6
x
1
174 Chapter 10 Exponential and Logarithmic Functions Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
B. f ( x ) = 2 log 3 x
y = log 3 x 2
x = 32
• Substitute y for f(x).
• Solve the equation for log3 x.
• Write the equivalent exponential equation. • Choose values of y, and find the corresponding values of x. Graph the ordered pairs on a rectangular coordinate system. Connect the points with a smooth curve.
y
x
y
1 9 1 3 1
-2
6
-4 –6
0
6
x
–6
0
3
y
2
Exercises Graph. 1. f ( x ) = log 4 x
se c t io n
2. f ( x ) = log 2 ( x + 1)
3. f ( x ) = 3 log 2 x
4. f ( x ) = − log 2 x
10.4 Exponential and Logarithmic Equations
Objective 1 Solve exponential equations
An exponential equation is one in which a variable occurs in the exponent. The examples at the right are exponential equations.
6 2 x + 1 = 63 x − 2 4x = 3 2x +1 = 7
An exponential equation in which each side of the equation can be expressed in terms of the same base can be solved by using the Equality of Exponents Property. Recall that the Equality of Exponents Property states that If bu = bv , then u = v.
In Example 1 below, this property is used to solve an exponential equation.
Example 1
Solve and check: 9 x + 1 = 27 x − 1
Solution
9 x + 1 = 27 x − 1
(32 ) x + 1 = (33 ) x − 1 2x + 2
3
• Rewrite each side of the equation using the same base.
• Use the Equality of Exponents Property to equate the exponents. • Solve the resulting equation.
= 3
2 x + 2 = 3x − 3
Steps
3x − 3
2 = x−3
5 = x
Section 10.4 Exponential and Logarithmic Equations 175
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
9 x + 1 = 27 x − 1
Check:
96
(32 )6
27 4 (33 )4
312 = 312
The solution is 5. When both sides of an exponential equation cannot easily be expressed in terms of the same base, logarithms are used to solve the exponential equation.
Example 2
Solve for x. Round to the nearest ten-thousandth.
A. 4 x = 7 B. 32 x = 4
A.
Solution
4x = 7
log 4 x = log 7
Steps
x log 4 = log 7
log 7 log 4
• Solve for x.
• Take the common logarithm of each side of the equation. • Rewrite using the properties of logarithms.
x =
x ≈ 1.4037
The solution is 1.4037. 32 x = 4
B.
log 32 x = log 4
2 x log 3 = log 4
2x =
log 4 log 3
x =
log 4 2 log 3
x ≈ 0.6309
• Take the common logarithm of each side of the equation. • Rewrite using the properties of logarithms.
• Solve for x.
The solution is 0.6309.
Exercises Solve for x. Round to the nearest ten-thousandth. 1. 54 x − 1 = 5x + 2
2. 93x = 81x − 4
3. 5x = 6
4. (1.5) x = 2
5. 4 − x + 2 = 12
6. 8x + 2 = 16 x
7. 43x + 8 = 12
8. 3− x + 2 = 18
Objective 2 Solve logarithmic equations A logarithmic equation can be solved by using the properties of logarithms.
176 Chapter 10 Exponential and Logarithmic Functions Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Solve: log 9 x + log 9 ( x − 8) = 1 f f
the equation.
log 9 x + log 9 ( x − 8) = 1 log 9 x ( x − 8) = 1
Write the equation in exponential form.
91 = x ( x − 8)
Simplify and solve for x.
9 = x 2 − 8x 0 = x 2 − 8x − 9 0 = ( x − 9)( x + 1) x − 9 = 0
x +1 = 0
x = 9
x = −1
When x is replaced by 9 in the original equation, 9 checks as a solution. When x is replaced by –1, the original equation contains the expression log 9 ( −1). Because the logarithm of a negative number is not a real number, –1 does not check as a solution. The solution of the equation is 9. Some logarithmic equations can be solved by using the 1–1 Property of Logarithms, which states that for x > 0, y > 0, b > 0, b ≠ 1, If log b x = log b y , then x = y. This property is used in Example 3B.
Example 3
Solve for x. A. log 3 ( 2 x − 1) = 2 B. log 2 x − log 2 ( x − 1) = log 2 2
Solution
A. log 3 ( 2 x − 1) = 2
Steps
• Rewrite in exponential form.
• Solve for x.
B. log 2 x − log 2 ( x − 1) = log 2 2
x log 2 = log 2 2 x − 1
• Use the Quotient Property of Logarithms.
• Use the 1–1 Property of Logarithms.
• Solve for x.
32 = 2 x − 1
9 = 2x − 1
10 = 2 x
5 = x
The solution is 5.
x = 2 x −1 x ( x − 1) = ( x − 1)2 x − 1
− x = −2
x = 2
x = 2x − 2
The solution is 2.
Section 10.4 Exponential and Logarithmic Equations 177
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Exercises Solve for x. 3x =1 x +1
9. log 3 ( x + 1) = 2
10. log 2 ( 2 x − 3) = 3
11. log 6
12. log( x − 2) − log x = 3
13. log 4 x + log 4 ( x − 2) = log 4 15
14. log x − 2 = log( x − 4)
sec t io n
10.5 Applications of Exponential and Logarithmic Functions
Objective 1 Application problems
An equation that can be written in the form A = A0 bkt , where A is the size at time t, A0 is the initial size, b > 1, and k is a positive real number, is an exponential growth equation. Compound interest is interest that is computed not only on the original principal but also on the interest already earned. The compound interest formula is an exponential growth equation. The compound interest formula is P = A(1 + i ) n , where A is the original value of an investment, i is the interest rate per compounding period, n is the total number of compounding periods, and P is the value of the investment after n periods.
Example 1
Solution
An investment broker deposits $1000 into an account that earns 12% annual interest compounded quarterly. What is the value of the investment after 2 years? Round to the nearest dollar. i =
12% 0.12 = = 0.03 4 4
Steps
n = 4⋅2 = 8
P = A(1 + i ) n
• Find i, the interest rate per quarter. The quarterly rate is the annual rate divided by 4, the number of quarters in 1 year. • Find n, the number of compounding periods. The investment is compounded quarterly, 4 times a year, for 2 years. • Use the compound interest formula.
P = 1000(1 + 0.03)8
• Replace A, i, and n by their values.
P ≈ 1267
• Solve for P.
The value of the investment after 2 years is $1267. Exponential decay can also be modeled by an exponential equation. One of the most common illustrations of exponential decay is the decay of a radioactive substance. A method by which an archeologist can measure the age of a bone is called carbon dating. Carbon dating is based on a radioactive isotope of carbon called carbon-14, which has a half-life of approximately 5570 years. The exponential decay equation is given by t
1 5570 A = A0 , where A0 is the original amount of carbon-14 present in the bone, t is the 2 age of the bone, and A is the amount of carbon-14 present after t years.
178 Chapter 10 Exponential and Logarithmic Functions Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example 2
A bone that originally contained 100 mg of carbon-14 now has 70 mg of carbon-14. e bone? Round to the nearest year.
Solution
1 5570 A = A0 2
Steps
• Use the exponential decay equation.
• Replace A0 and A by their given values, and solve for t.
• Divide each side of the equation by 100.
• Take the common logarithm of each side of the equation. Then simplify.
t
1 5570 70 = 100 2
70 = 100(0.5) 5570
0.7 = (0.5) 5570
log 0.7 = log(0.5) 5570
log 0.7 =
t
t
t
t log 0.5 5570
5570 log 0.7 = t log 0.5
2866 ≈ t
The age of the bone is approximately 2866 years. A chemist measures the acidity or alkalinity of a solution by the formula pH = − log( H + ), where H+ is the concentration of hydrogen ions in the solution. A neutral solution such as distilled water has a pH of 7, acids have a pH less than 7, and alkaline solutions (also called basic solutions) have a pH greater than 7.
Example 3
Strategy
Solution
pH = − log( H + )
3.6 = − log( H + )
Find the hydrogen ion concentration, H+, of orange juice that has a pH of 3.6. To find the hydrogen ion concentration, replace pH by 3.6 in the equation pH = − log( H + ) and solve for H+.
−3.6 = log( H + )
10 −3.6 = H + 0.00025 ≈ H + The hydrogen ion concentration is approximately 0.00025.
Exercises Solve. 1. An investment broker deposits $1000 into an account that earns 8% annual interest compounded quarterly. What is the value of the investment after 2 years? Round to the nearest dollar.
Section 10.5 Applications of Exponential and Logarithmic Functions 179
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
2. An isotope of technetium is used to prepare images of internal body organs. This isotope has a half-life of approximately 6 h. A p
a. What will be the amo
b. How long (in hours) will it take for the amount of technetium in the patient to reach 20 mg?
3. Find the pH of milk, for which the hydrogen ion concentration is 3.97 × 10 −7. 4. The number of decibels, D, of a sound can be given by the equation D = 10(log I + 16), where I is the power of a sound measured in watts. The loudest sound made by any animal is made by the blue whale and can be heard over 500 mi away. The power of the sound is 630 watts. Find the number of decibels of sound emitted by the blue whale.
180 Chapter 10 Exponential and Logarithmic Functions Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Chapter 11
Image copyright Luciano Mortula, 2010. Used under license from Shutterstock.com.
Sequences and Series
Prep Test
1. Simplify: [3(1) − 2] + [3( 2) − 2] + [3(3) − 2]
2. Evaluate f ( n) =
3. Evaluate a1 + ( n − 1) d for a1 = 2, n = 5, and d = 4.
4. Evaluate a1r n − 1 for a1 = −3, r = −2, and n = 6.
5. Evaluate
6. Simplify:
n for n = 6. n+2
a1 (1 − r n ) for a1 = −2, r = −4, and n = 5. 1− r 4 10 1−
1 10
7. Simplify: ( x + y )2
8. Simplify: ( x + y )3
181 Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
sec t io n
11.1 Introduction to
s
Objective 1 Write the terms of a sequence
A sequence is an ordered list of numbers. Each of the numbers of a sequence is called a term of the sequence. 1, 1, 2, 3, 5, 8 1, 2, 3, 4, 5, 6, 7, 8 1, −1, 1, −1
Examples of other sequences are shown at the right. These sequences are separated into two groups. A finite sequence contains a finite number of terms. An infinite sequence contains an infinite number of terms.
1, 3, 5, 6, … 1 1 1 1, , , , … 2 4 8 1, 1, 2, 3, 5, 8, …
Finite sequences Infinite sequences
A general sequence is shown below. a1 , a2 , a3 , . . . an , . . . The first term is a1 , the second term is a2 , the third term is a3 , and the nth term, also called the general term of the sequence, is an .
Example 1
Solution
Write the first three terms of the sequence whose nth term is given by the formula an = 2 n − 1. an = 2 n − 1 a1 = 2(1) − 1 = 1
Steps
• Replace n by 1.
a2 = 2( 2) − 1 = 3
• Replace n by 2.
a3 = 2( 3) − 1 = 5
• Replace n by 3.
The first term is 1, the second term is 3, and the third term is 5.
Example 2
Solution
Find the eighth and tenth terms of the sequence whose nth term is given by the formula n an = . n+1 an =
n n+1
a8 =
8 8 = 8+1 9
Steps
• Replace n by 8.
a10 =
10 10 = 10 + 1 11
• Replace n by 10.
The eighth term is
8 10 , and the tenth term is . 9 11
Exercises Write the first four terms of the sequence whose nth term is given by the formula. 1. an = n + 1
2. an = 2 − 2 n
3. an = 2 n
182 Chapter 11 Sequences and Series Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
4. an = n −
1 n
5. a = ( −1) n + 1 n
6. an =
( −1) n + 1 n+1
9. an =
n ; a n + 1 12
Find the indicated term of the sequence whose nth term is given by the formula. 7. an = 3n + 4; a12
8. an = n( n − 1); a11
Objective 2 Evaluate a series
The indicated sum of the terms of a sequence is called a series. Given the sequence 10, 11, 12.10, 13.31, the series 10 + 11 + 12.10 + 13.31 can be written. S n is used to indicate the sum of the first n terms of a sequence. For the preceding example, the sums of the series S1 , S2 , S3 , and S4 are shown at the right.
S1 = 10
= 10
S2 = 10 + 11
= 21
S3 = 10 + 11 + 12.10
= 33.10
For the general sequence a1 , a2 , a3 , . . . , an , the series S1 , S2 , S3 , and S n are shown at the right.
S4 = 10 + 11 + 12.10 + 13.31 = 46.41 S1 = a1 S2 = a1 + a2 S3 = a1 + a2 + a3
It is convenient to represent a S n = a1 + a2 + a3 + + an series in a compact form called summation notation, or sigma notation. The Greek letter sigma, Σ, is used to indicate a sum. The first four terms of the sequence whose nth term is given by the formula an = 2 n are 2, 4, 6, 8. The corresponding series is shown at the right written in summation notation and is read “the summation from 1 to 4 of 2n.” The letter n is called the index of the summation.
Example 3
4
∑ 2n n =1
4
To write the terms of the series, replace n by the consecutive integers from 1 to 4.
∑ 2n
= 2(1) + 2( 2) + 2(3) + 2( 4)
The series is 2 + 4 + 6 + 8.
= 2+4+6+8
The sum of the series is 20.
= 20
Evaluate the series. A.
3
∑ (2i − 1)
n =1
B.
i =1
Solution
A.
6
1
∑2n n=3
3
∑ (2i − 1) i =1
= [2(1) − 1] + [2( 2) − 1] + [2( 3) − 1]
Steps
• Replace i by 1, 2, and 3.
= 1+ 3+5
• Write the series.
= 9
• Find the sum of the series.
Section 11.1 Introduction to Sequences and Series 183
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
B.
6
∑
1
n
=
=
= 9
2
( 3) +
2
( 4) +
2
( 5) +
2
( 6)
3 5 +2+ +3 2 2
• Replace n by 3, 4, 5, and 6.
• Write the series.
• Find the sum of the series.
Exercises Evaluate the series. 10.
5
∑ (2n + 3)
11.
7
∑n
14.
n =1
∑ (i + 2 )
12.
i =1
n =1
13.
7
∑ 2i i =1
6
∑ i2
15.
i =1
sec t io n
4
5
∑ (i
2
+ 1)
i =1
11.2 Arithmetic Sequences and Series
Objective 1 Find the nth term of an arithmetic sequence
An arithmetic sequence, or arithmetic progression, is one in which the difference between any two consecutive terms is constant. The difference between consecutive terms is called the common difference of the sequence. Each sequence shown below is an arithmetic sequence. To find the common difference of an arithmetic sequence, subtract the first term from the second term.
2, 7, 12, 17, 22, . . .
Common difference: 5
3, 1, -1, -3, -5, . . .
Common difference: -2
1,
3 5 7 , 2, , 3, 2 2 2
Common difference:
1 2
Formula for the nth Term of an Arithmetic Sequence The nth term of an arithmetic sequence with a common difference of d is given by an = a1 + ( n − 1) d .
Example 1
Solution
Find the 27th term of the arithmetic sequence –4, –1, 2, 5, 8, . . . . d = a2 − a1 = −1 − ( −4) = 3
Steps
• Find the common difference.
an = a1 + ( n − 1) d
• Use the Formula for the nth Term of an Arithmetic Sequence to find the 27th term. n = 27, a1 = -4, d = 3
a27 = −4 + ( 27 − 1)3
= −4 + ( 26)3
= −4 + 78
= 74
184 Chapter 11 Sequences and Series Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example 2
Find the formula for the nth term of the arithmetic sequence -5, -1, 3, 7, . . . .
Solution d = a − a = −1 − ( −5) = 4 Pillaged by LibraryPirate
2
1
an = a1 ( n − 1) d
Steps
• Find the common difference.
• Use the Formula for the nth Term of an Arithmetic Sequence. a1 = -5, d = 4
an = −5 + ( n − 1)4 an = −5 + 4 n − 4 an = 4 n − 9
Example 3
Solution
Find the number of terms in the finite arithmetic sequence 7, 10, 13, . . . , 55. d = a2 − a1 = 10 − 7 = 3
Steps
• Find the common difference.
an = a1 + ( n − 1) d
• Use the Formula for the nth Term of an Arithmetic Sequence. an = 55, a1 = 7, d = 3
• Solve for n.
55 = 7 + ( n − 1)3 55 = 7 + 3n − 3 55 = 3n + 4 51 = 3n 17 = n There are 17 terms in the sequence.
Exercises Find the indicated term of the arithmetic sequence. 1. 1, 11, 21, . . . ; a15
2. –6, –2, 2, . . . ; a15
3. 3, 7, 11, . . . ; a18
Find the formula for the nth term of the arithmetic sequence. 4. 1, 4, 7, . . .
5. 6, 2, –2, . . .
6. 17, 30, 43, . . .
Find the number of terms in the finite arithmetic sequence. 7. –2, 1, 4, . . . , 73
8. 7, 11, 15, . . . , 171
9. 1, 5, 9, . . . , 81
Objective 2 Evaluate an arithmetic series Formula for the Sum of n Terms of an Arithmetic Series Let a1 be the first term of a finite arithmetic sequence, let n be the number of terms, and let an be the last term of the sequence. Then the sum of the series S n is given n by S n = ( a1 + an ). 2
Section 11.2 Arithmetic Sequences and Series 185
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example 4
Find the sum of the first 10 terms of the arithmetic sequence 2, 4, 6, 8, . . . .
Soluti
Steps
• Find the common difference.
• Use the Formula for the nth Term of an Arithmetic Sequence to find the 10th term.
• Use the Formula for the Sum of n Terms of an Arithmetic Series. n = 10, a1 = 2, an = 20
a1 = 3(1) + 1 = 4
Steps
• Find the first term.
a25 = 3( 25) + 1 = 76
• Find the 25th term.
• Use the Formula for the Sum of n Terms of an Arithmetic Series. n = 25, a1 = 4, an = 76
an = a1 + ( n − 1) d a10 = 2 + (10 − 1)2
= 2 + (9)2
= 2 + 18 = 20 n ( a + an ) 2 1 10 = ( 2 + 20) = 5( 22) = 110 2
Sn = S10
Example 5
Evaluate the arithmetic series
25
∑ (3n + 1). n =1
Solution an = 3n + 1
Sn =
n ( a + an ) 2 1
25 ( 4 + 76) 2 25 = (80) 2
S25 =
= 1000
Exercises Find the sum of the indicated number of terms of the arithmetic sequence. 10. 1, 3, 5, . . . ; n = 50
11. 2, 4, 6, . . . ; n = 25
12. 20, 18, 16, . . . ; n = 40
Evaluate the arithmetic series. 13.
15
∑ (3i − 1)
14.
i =1
15
∑ (3i + 4)
15.
i =1
10
∑ (1 − 4n) n =1
Objective 3 Application problems
Example 6
Strategy
The distance a ball rolls down a ramp each second is given by an arithmetic sequence. The distance in feet traveled by the ball during the nth second is given by 2n – 1. Find the distance the ball will travel during the first 10 s. To find the distance:
c Find the first and tenth terms of the sequence.
c Use the Formula for the Sum of n Terms of an Arithmetic Series to find the sum of the first ten terms.
186 Chapter 11 Sequences and Series Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
an = 2 n − 1
Solution
a10 = 2(10) − 1 = 19 Sn =
n ( a + an ) 2 1
S10 =
10 (1 + 19) = 5( 20) = 100 2
The ball will roll 100 ft during the first 10 s.
Exercises Solve. 16. An exercise program calls for walking 12 min each day for a week. Each week thereafter, the amount of time spent walking increases by 6 min per day. In how many weeks will a person be walking 60 min each day?
17. A display of cans in a grocery store consists of 20 cans in the bottom row, 18 cans in the next row, and so on in an arithmetic sequence. The top row has 4 cans. Find the total number of cans in the display.
18. A theater in the round has 52 seats in the first row, 58 seats in the second row, 64 seats in the third row, and so on in an arithmetic sequence. Find the total number of seats in the theater if there are 20 rows of seats.
se c t io n
11.3 Geometric Sequences and Series
Objective 1 Find the nth term of a geometric sequence A geometric sequence, or geometric progression, is one in which each successive term of the sequence is the same nonzero constant multiple of the preceding term. The common multiple is called the common ratio of the sequence. Each of the sequences shown below is a geometric sequence. To find the common ratio of a geometric sequence, divide the second term of the sequence by the first term.
3, 6, 12, 24, 48, . . .
Common ratio: 2
4, -12, 36, -108, 324, . . .
Common ratio: -3
6, 4,
8 16 16 , , , . . . 3 9 9
Common ratio:
2 3
Formula for the nth Term of a Geometric Sequence The nth term of a geometric sequence with first term a1 and common ratio r is given by an = a1r n − 1 .
Section 11.3 Geometric Sequences and Series 187
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example 1
Find the 6th term of the geometric sequence 3, 6, 12, … .
Soluti
Steps
• Find the common ratio.
• Use the Formula for the nth Term of a Geometric Sequence to find the 6th term. n = 6, a1 = 3, r = 2
1
an = a1r n − 1 a6 = 3( 2)6 − 1
Example 2
= 3( 2)5
= 3(32)
= 96
Find a3 for the geometric sequence 8, a2 , a3 , 27, . . . .
Solution an = a1r n − 1
Steps
a4 = a1r 4 − 1 27 = 8r 4 − 1
• Use the Formula for the nth Term of a Geometric Sequence to find the common ratio. The 1st and 4th terms are known. Substitute 4 for n. • a4 = 27, a1 = 8
27 = r3 8
3 = r 2
3 a3 = 8 2
• Take the cube root of each side of the equation.
• Use the Formula for the nth Term of a Geometric Sequence to find a3. 3 n = 3, a1 = 8, r = 2
3−1
2
3 = 8 2
9 = 8 = 18 4
Exercises Find the indicated term of the geometric sequence. 1. 2, 8, 32, . . . ; a9
2. 4, 3,
9 , . . . ; a8 4
3. 6, –4,
8 , . . . ; a7 3
Find a2 and a3 for the geometric sequence. 4. 9, a2 , a3 ,
8 , . . . 3
Objective 2 Finite geometric series
5. 8, a2 , a3 ,
27 , . . . 8
8 6. 3, a2 , a3 , − , . . . 9
Formula for the Sum of n Terms of a Finite Geometric Series Let a1 be the first term of a finite geometric sequence, let n be the number of terms, and let r be the common ratio, r ≠ 1. Then the sum of the series S n is given by a (1 − r n ) . Sn = 1 1− r 188 Chapter 11 Sequences and Series Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example 3
Find the sum of the terms of the geometric sequence 2, 8, 32, 128, 512.
Steps
• Find the common ratio.
• Use the Formula for the Sum of n Terms of a Finite Geometric Series. n = 5, a1 = 2, r = 4
1
Example 4
Solution
Sn =
a1 (1 − r n ) 1− r
S5 =
2(1 − 45 ) 1− 4
=
2(1 − 1024) −3
=
2( −1023) −3
=
−2046 −3
= 682 n
2 Evaluate the geometric series ∑ 5 3 . n =1 6
2 an = 5 3
n
1
2 10 a1 = 5 = 3 3 2 , the base of the exponential expression 3
r = Sn =
a1 (1 − r n ) 1− r
• n = 6, a1 =
Step
2 10 ,r = 3 3
6 2 10 1 − 3 3 S6 = 2 1− 3
10 665 10 64 1− 729 3 729 3 = = 1 1 3 3
=
6650 729
Exercises Find the sum of the indicated number of terms of the geometric sequence. 7. 2, 6, 18, . . . ; n = 7
8. –4, 12, –36, . . . ; n = 7
9. 12, 9,
27 , ...;n=5 4,
Find the sum of the geometric series. 10.
5
∑ ( 2) i
11.
i =1
n
2 ∑ n =1 3 6
12.
1 3 ∑ n =1 2 7
n
Section 11.3 Geometric Sequences and Series 189
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Objective 3 Infinite geometric series
ite geometric sequence is called an infinite An example of an infinite geometric series is shown at the right. The first term is 1. The 1 common ration is . 3
1+
1 1 1 1 1 + + + + + 3 9 27 81 243
Formula for the Sum of an Infinite Geometric Series The sum of an infinite geometric series in which r < 1 and a1 is the first term is a1 given by S = . 1− r When r ≥ 1, the infinite geometric series does not have a finite sum. For example, the sum of the infinite geometric series 1 + 2 + 4 + 8 + increases without limit.
Example 5
Solution
1 1 1 Find the sum of the terms of the infinite geometric sequence 1, − , , − , … . 2 4 8 1 a1 1 Step • The common ratio is − . S = = 2 1− r 1 1 − − 1 − < 1. Use the Formula for 2 2
1 2 = = 3 3 2
the Sum of an Infinite Geometric Series.
The sum of an infinite geometric series can be used to find a fraction that is equivalent to a nonterminating, repeating decimal.
Example 6
Solution
Find an equivalent fraction for 0.12. Write the decimal as an infinite geometric series. The geometric series does not begin 2 1 with the first term. The series begins with . The common ratio is . 100 10 0.12 = 0.1 + 0.02 + 0.002 + 0.0002 +
=
1 2 2 2 + + + + 10 100 1000 10, 000
a1 S = = 1− r
0.12 =
2 2 100 = 100 = 2 1 9 90 1− 10 10
1 2 11 + = 10 90 90
An equivalent fraction is
11 . 90
Steps
• Use the Formula for the Sum of an Infinite Geometric Series.
• Add
1 to the sum of the 10 series.
190 Chapter 11 Sequences and Series Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Exercises Find the sum of the i 13. 3 + 2 +
4 + 3
14. 2 −
4
+
32
+
15.
7 7 7 + + + 10 100 1000
Find an equivalent fraction for the repeating decimal. 16. 0.888
17. 0.4545
18. 0.1666
Objective 4 Application problems
Example 7
Strategy
Solution
On the first swing, the length of the arc through which a pendulum swings is 16 in. The 7 length of each successive swing is of the preceding swing. Find the length of the arc 8 on the fifth swing. Round to the nearest tenth. To find the length of the arc on the fifth swing, use the Formula for the nth Term of a 7 Geometric Sequence. n = 5, a1 = 16, r = 8 an = a1r n − 1 7 a5 = 16 8
5 −1
7 = 16 8
4
2401 38, 416 = ≈ 9.4 = 16 4096 4096
The length of the arc on the fifth swing is 9.4 in.
Exercises Solve. 19. On the first swing, the length of the arc through which a pendulum swings is 18 in. The length of each 3 successive swing is of the preceding swing. What is the total distance the pendulum has traveled 4 during the first five swings? Round to the nearest tenth.
20. To test the bounce of a tennis ball, the ball is dropped from a height of 8 ft. The ball bounces to 80% of its previous height with each bounce. How high does the ball bounce on the fifth bounce? Round to the nearest tenth.
21. Assume the average value of a home increases 5% per year. How much would a house costing $100,000 be worth in 30 years?
Section 11.3 Geometric Sequences and Series 191
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
sec t io n
11.4 Binomial Expansions
Objective
n Factorial n! (which is read “n factorial”) is the product of the first n consecutive natural numbers. 0! is defined to be 1. n! = n( n − 1)( n − 2) ⋅ . . . ⋅ 3 ⋅ 2 ⋅ 1
Example 1
Solution
1!, 5!, and 7! are shown at the right.
1! = 1
5! = 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 120
7 ! = 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 5040 Evaluate:
7! 4 !3!
7! 7 ⋅ 6 ⋅5⋅ 4 ⋅ 3⋅ 2 ⋅1 = 4 !3! ( 4 ⋅ 3 ⋅ 2 ⋅ 1)(3 ⋅ 2 ⋅ 1) = 35
Steps
• Write each factorial as a product.
• Simplify.
n A binomial coefficient is a quotient of factorials. The symbol is used to express this r quotient of factorials. n n! r = ( n − r )! r !
Example 2
Solution
8 Evaluate: 5 8 8! 5 = (8 − 5)!5!
=
8! 8 ⋅ 7 ⋅ 6 ⋅5⋅ 4 ⋅ 3⋅ 2 ⋅1 = = 566 3!5! (3 ⋅ 2 ⋅ 1)(5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1)
Steps
• Write the quotient of the factorials.
• Simplify.
Binomial Expansion Formula n n n n n ( a + b) n = a n + a n − 1b + a n − 2 b2 + + a n − r br + + bn 0 1 2 r n The Binomial Expansion Formula is used below to expand ( a + b)7 . 7 7 7 7 7 7 7 7 ( a + b)7 = a 7 + a 6 b + a 5 b2 + a 4 b3 + a 3b4 + a 2 b5 + ab6 + b7 0 1 2 3 4 5 7 6 = a 7 + 7a 6 b + 21a 5 b 2 + 35a 4 b 3 + 35a 3 b 4 + 21a 2 b 5 + 7ab 6 + b 7
Example 3
Solution
Write ( 4 x + 3 y )3 in expanded form. ( 4 x + 3 y )3
3 3 3 3 = ( 4 x )3 + ( 4 x )2 (3 y ) + ( 4 x )(3 y )2 + (3 y )3 0 1 2 3
192 Chapter 11 Sequences and Series Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
= 1(64 x 3 ) + 3(16 x 2 )(3 y ) + 3( 4 x )(9 y 2 ) + 1( 27 y 3 ) 2
+ 27 y 3
The Binomial Theorem can also be used to write any term of a binomial expansion. Formula for the rth Term in a Binomial Expansion n n − r +1 r −1 The rth term of ( a + b) n is a b . r − 1
Example 4
Solution
Find the fourth term in the expansion of ( x + 3)7 . n n − r +1 r −1 b r − 1 a
Step
7 7 − 4 +1 4 −1 7 = x 4 (3)3 (3) 4 − 1 x 3
= 35x 4 ( 27 )
= 945 x 4
• Use the Formula for the r th Term in a Binomial Expansion. r = 4, n = 7, a = x, b = 3
Exercises Evaluate. 1. 3!
2. 8!
3. 9!
7 4. 2
9 5. 6
9 6. 0
8. ( x − y )5
9. ( 2r − 3)5
Write in expanded form. 7. ( x + y )4 Find the indicated term in the expansion. 10. ( 2 x − 1)7 ; 4th term
11. ( x + 4)5 ; 3rd term
12. ( x − 2)8 ; 8th term
Section 11.4 Binomial Expansions 193
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Chapter 12
Image copyright Kat Mack, 2010. Used under license from Shutterstock.com.
Conic Sections
Prep Test
1. Find the distance between the points whose coordinates are (–2, 3) and (4, –1). Round to the nearest hundredth.
2. Complete the square on x 2 − 8 x. Write the resulting perfect-square trinomial as the square of a binomial.
3. Solve
x2 y2 + = 1 for x when y = 3 and when y = 0. 16 9
4. Solve by the substitution method: 7 x + 4 y = 3 y = x−2 5. Solve by the addition method: 4 x − y = 9 2 x + 3 y = −13
6. Find the axis of symmetry and the vertex of the graph of y = x 2 − 4 x + 2.
7. Graph: f ( x ) = −2 x 2 + 4 x 8. Graph the solution set of 5x − 2 y > 10. 9. Graph the solution set: x + 2 y ≤ 4 x− y ≤ 2
195 Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
sec t io n
12.1 The Parabola
Objective
y
x
Vertex Axis of symmetry
A parabola is a conic section formed by the intersection of a right circular cone and a plane parallel to the side of the cone. Every parabola has an axis of symmetry and a vertex that is on the axis of symmetry. To understand the axis of symmetry, think of folding the paper along that axis. The two halves of the curve will match up. The graph of the equation y = ax 2 + bx + c, a ≠ 0, is a parabola with the axis of symmetry parallel to the y-axis. The parabola opens up when a > 0 and opens down when a < 0. When the parabola opens up, the vertex is the lowest point on the parabola. When the parabola opens down, the vertex is the highest point on the parabola. b . The y-coordinate of the vertex can then be 2a determined by substituting this value of x into y = ax 2 + bx + c and solving for y.
The x-coordinate of the vertex is −
Because the axis of symmetry is parallel to the y-axis and passes through the vertex, the b equation of the axis of symmetry is x = − . 2a
Example 1
Solution
Find the vertex and the axis of symmetry of the parabola given by the equation y = x 2 + 2 x − 3. Then sketch the graph of the parabola. −
b 2 = − = –1 2a 2(1)
y = x2 + 2x − 3
b . 2a
Steps
• The x-coordinate of the vertex is −
• Find the y-coordinate of the vertex by replacing x by –1 and solving for y.
• The axis of symmetry is the line b x = − . 2a
• Because a is positive, the parabola opens up. Use the vertex and axis of symmetry to sketch the graph.
y = ( −1)2 + 2( −1) − 3 y = −4 The vertex is (–1, –4). The axis of symmetry is the line x = –1. y 4 2 –4 –2 0 –2
2
4
x
–4
196 Chapter 12 Conic Sections Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
The graph of an equation of the form 2 parabola. the right left when a is negative.
y Axis of symmetry
Vertex
x
For a parabola of this form, the y-coordinate b of the vertex is − The axis of symmetry 2 a. b is the line y = − 2 a.
Example 2
Solution
Find the vertex and the axis of symmetry of the parabola whose equation is x = 2 y 2 − 8 y + 5. Then sketch its graph. −
b −8 = − = 2 2a 2( 2)
x = 2 y2 − 8 y + 5
Steps
• Find the y-coordinate of the vertex. a = 2, b = -8
• Find the x-coordinate of the vertex by replacing y by 2 and solving for x.
• The axis of symmentry is the line b y = − . 2a • Because a is positive, the parabola opens to the right. Use the vertex and axis of symmetry to sketch the graph.
x = 2( 2)2 − 8( 2) + 5 x = −3 The vertex is (-3, 2). The axis of symmetry is the line y = 2.
y 4 2 –4 –2 0 –2
2
4
x
–4
Exercises Find the vertex and the axis of symmetry of the parabola given by the equation. Then sketch its graph. 1. y = x 2 − 2 x − 4
2. y = − x 2 + 2 x − 3
3. x = y 2 + 6 y + 5
4. x = y 2 − y − 6
5. y = 2 x 2 − 4 x + 1
6. x = −
1 2 y + 2y − 3 2
Section 12.1 The Parabola 197
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
sec t io n
12.2 The Circle
Objective
en graph the circle A circle is a conic section formed by the intersection of a cone and a plane that is parallel to the base of the cone. y
Radius Center
r ( h, k )
x
A circle can be defined as all the points (x, y) in the plane that are a fixed distance from a given point (h, k) called the center. The fixed distance is the radius of the circle. Standard Form of the Equation of a Circle Let r be the radius of a circle and let (h, k) be the coordinates of the center of the circle. Then the equation of the circle is given by ( x − h) 2 + ( y − k ) 2 = r 2
Example 1
Find the equation of the circle with radius 5 and center C(–1, 3). Then sketch its graph.
Solution ( x − h)2 + ( y − k )2 = r 2
y
[ x − ( −1)]2 + ( y − 3)2 = 52
6
( x + 1)2 + ( y − 3)2 = 25
–6
Center: (-1, 3) Radius:
Example 2
Solution
0
6
x
–6
25 = 5
Find the equation of the circle for which a diameter has endpoints P1(–4, –1) and P2(2, 3). xm =
x1 + x2 2
ym =
y1 + y2 2
xm =
−4 + 2 2
ym =
−1 + 3 2
xm = −1
Steps
• Let ( x 1, y 1) = (–4, –1) and ( x 2 , y 2 ) = (2, 3). Find the center of the circle by finding the midpoint of the diameter.
• Find the radius of the circle. Use either point on the circle and the coordinates of the center of the circle. P1 is used here.
• Write the equation of the circle with center C(-1, 1) and radius 13.
ym = 1
Center: ( xm , ym ) = ( −1, 1) r =
( x1 − xm )2 + ( y1 − ym )2
r =
[ −4 − ( −1)]2 + ( −1 − 1)2
r =
9+4
r =
13
( x + 1)2 + ( y − 1)2 = 13
198 Chapter 12 Conic Sections Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Exercises 1. Find the equatio and center C(2,
e equation of the circle with radius 3 ter C(–1, –2). Then sketch its graph.
3. Find the equation of the circle for which a 4. Find the equation of the circle for which a diameter has endpoints P1(–1, 4) and P2(–5, 8). diameter has endpoints P1(2, 3) and P2(5, –2).
Objective 2 Write the equation of a circle in standard form
and then graph the circle
The equation of a circle can also be expressed in general form as x 2 + y 2 + ax + by + c = 0 To rewrite this equation in standard form, it is necessary to complete the square on the x and y terms.
Example 3
Write the equation of the circle x 2 + y 2 + 4 x + 2 y + 1 = 0 in standard form.
Solution
x2 + y2 + 4x + 2 y + 1 = 0
x 2 + y 2 + 4 x + 2 y = −1
( x 2 + 4 x ) + ( y 2 + 2 y ) = −1
( x 2 + 4 x + 4) + ( y 2 + 2 y + 1) = −1 + 4 + 1
Steps
( x 2 + 4 x + 4) + ( y 2 + 2 y + 1) = 4
( x + 2)2 + ( y + 1)2 = 4
• Subtract the constant term from each side of the equation. • Rewrite the equation by grouping terms involving x and terms involving y. • Complete the square on x 2 + 4x and y 2 + 2 y . • Factor each trinomial.
Exercises Write the equation of the circle in standard form. Then sketch its graph. 5. x 2 + y 2 − 2 x + 4 y − 20 = 0
7. x 2 + y 2 − x + 4 y +
13 = 0 4
6. x 2 + y 2 + 6 x + 8 y + 9 = 0
8. x 2 + y 2 − 6 x + 4 y + 4 = 0
Section 12.2 The Circle 199
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
sec t io n
12.3 The Ellipse and Hyperbola
Objective
origin The orbits of the planets around the sun are “oval” shaped. This oval shape can be described as an ellipse, which is another of the conic sections. y
x
There are two axes of symmetry for an ellipse. The intersection of these two axes is the center of the ellipse. Standard Form of the Equation of an Ellipse with Center at the Origin x2 y2 + = 1. The a2 b2 x-intercepts are (a, 0) and (–a, 0). The y-intercepts are (0, b) and (0, –b).
The equation of an ellipse with center at the origin is
Example 1
Sketch a graph of the ellipse given by the equation.
A.
x2 y2 x2 y2 + = 1 B. + =1 9 4 16 12
A.
x2 y2 + =1 9 4
Steps
• a 2 = 9, b 2 = 4
x-intercepts: (3, 0) and (-3, 0)
• The x-intercepts are (a, 0) and (-a, 0).
y-intercepts: (0, 2) and (0, -2)
• The y-intercepts are (0, b) and (0, -b). • Use the intercepts and symmetry to sketch the graph of the ellipse.
Solution
y 4 2 –4 –2 0 –2
2
4
x
–4
B.
x2 y2 + =1 16 12
• a 2 = 16, b 2 = 12
x-intercepts: (4, 0) and (-4, 0)
• The x-intercepts are (a, 0) and (-a, 0).
y-intercepts: 0, 2 3 and 0, −2 3
)
(
4
• The y-intercepts are (0, b) and (0, -b). • Use the intercepts and symmetry to sketch the graph of the ellipse. 2 3 ≈ 3.5
2
)
y
–4 –2 0 –2
(
2
4
x
–4
200 Chapter 12 Conic Sections Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Exercises Sketch the graph of t 1.
x2 y2 + = 1 4 9
2.
4.
x2 y2 + = 1 16 9
5.
= 1
3.
x2 y2 + =1 25 9
x2 y2 + = 1 9 25
6.
x2 y2 + =1 16 36
25
+
16
Objective 2 Graph a hyperbola with center at the origin A hyperbola is a conic section that is formed by the intersection of a right circular cone and a plane perpendicular to the base of the cone. y
x
The hyperbola has two vertices and an axis of symmetry that passes through the vertices. The center of a hyperbola is the point halfway between the vertices. The graphs below show two graphs of a hyperbola with center at the origin. In the first graph, the axis of symmetry that contains the vertices is the x-axis.
y
y
x
In the second graph, the axis of symmetry that contains the vertices is the y-axis.
x
Standard Form of the Equation of a Hyperbola with Center at the Origin The equation of a hyperbola for which the axis of symmetry that contains the x2 y2 vertices is the x-axis is 2 − 2 = 1. The vertices are (a, 0) and (–a, 0). a b The equation of a hyperbola for which the axis of symmetry that contains the y2 x2 vertices is the y-axis is 2 − 2 = 1. The vertices are (0, b) and (0, –b). b a To sketch a hyperbola, it is helpful to draw two lines that are “approached” by the hyperbola. These two lines are called asymptotes. As a point on the hyperbola gets farther from the origin, the hyperbola “gets closer to” the asymptotes.
y
Because the asymptotes are straight lines, their equations are linear equations. The equations of the asymptotes for b a hyperbola with center at the origin are y = x and a b y = − x. a
x
Section 12.3 The Ellipse and Hyperbola 201
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example 2
Sketch a graph of the hyperbola given by the equation.
Solution
A.
x2 y2 − =1 16 4
Axis of symmetry: x-axis
Steps
• a 2 = 16, b 2 = 4
Vertices: (4, 0) and (–4, 0)
• The vertices are (a, 0) and (–a, 0).
Asymptotes: 1 1 y = x and y = − x 2 2
• The asymptotes are y =
y
y = −
• Sketch the asymptotes. Use symmetry and the fact that the hyperbola will approach the asymptotes to sketch its graph.
• a 2 = 25, b 2 = 16
6
–6
0
b x. a
b x and a
x
6
–6
B.
y2 x2 − =1 16 25
Axis of symmetry: y-axis
Vertices: (0, 4) and (0, –4)
• The vertices are (0, b) and (0, –b).
Asymptotes: 4 4 y = x and y = − x 5 5
• The asymptotes are y =
y
6
–6
0
y = −
b x. a
b x and a
• Sketch the asymptotes. Use symmetry and the fact that the hyperbola will approach the asymptotes to sketch its graph.
x
6
–6
Exercises Sketch a graph of the hyperbola given by the equation. 7.
x2 y2 − = 1 9 16
8.
x2 y2 − =1 25 4
9.
y2 x2 − = 1 16 9
10.
y2 x2 − =1 4 9
202 Chapter 12 Conic Sections Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
11.
x2 y2 − = 1 4 25
se c t io n
12.
y2
−
x2 =1 9
12.4 Solving Nonlinear Systems of Equations
Objective 1 Solve nonlinear systems of equations
A nonlinear system of equations is one in which one or more equations of the system are not linear equations. Some examples of nonlinear systems and their graphs are shown here. y
x2 + y2 = 4
4
y = x2 + 2
The graphs intersect at one point. 0
–4
The system of equations has one solution.
4
x
–4
y = x2
y 4
y = −x + 2
2
The graphs intersect at two points. The system of equations has two solutions.
–4
–2
0
2
4
2
4
x
–2 –4
( x + 2) 2 + ( y − 2) 2 = 4
y 4
x = y2
2
The graphs do not intersect. The system of equations has no solutions.
–4
–2
0
x
–2 –4
Nonlinear systems of equations can be solved by using either a substitution method or an addition method.
Example 1
Solve: A. y = 2 x 2 − 3x − 1 B. 3x 2 − 2 y 2 = 26 y = x2 − 2x + 5
Solution
A. (1)
y = 2 x 2 − 3x − 1
y = x2 − 2x + 5
2 x 2 − 3x − 1 = x 2 − 2 x + 5
x2 − x − 6 = 0
( x − 3)( x + 2) = 0
x − 3 = 0
(2)
x = 3
x2 − y2 = 5
Steps
• Use the substitution method.
x+2 = 0 x = −2 Section 12.4 Solving Nonlinear Systems of Equations 203
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
y = 2 x 2 − 3x − 1
y = 18 − 9 − 1
y = 8
y = 2 x 2 − 3x − 1
y = 2( −2)2 − 3( −2) − 1
y = 8+6−1
y = 13
The solutions are (3, 8) and (-2, 13).
B. (1)
(2)
• Use the addition method. Multiply equation (2) by -2.
• Substitute the values of x into equation (2).
x2 − y2 = 5 3x 2 − 2 y 2 = 26
−2 x 2 + 2 y 2 = −10
x 2 = 16
x = ±4
x2 − y2 = 5
x2 − y2 = 5
42 − y 2 = 5
( −4)2 − y 2 = 5
16 − y 2 = 5
16 − y 2 = 5
− y 2 = −11
− y 2 = −11
y 2 = 11
y 2 = 11
• Substitute the values of x into equation (1).
3x 2 − 2 y 2 = 26
y = ± 11
y = ± 11
(
The solutions are 4,
) ( 4, − 11 ) , ( −4,
11 ,
)
(
)
11 , and −4, − 11 .
Exercises Solve. 1. y = x 2 − x − 1 2. y2 = − x + 3 3. x 2 + 2 y 2 = 12 y = 2 x + 9 x − y = 1 2 x − y = 2
4. 2 x 2 − y 2 = 7 5. 2 x 2 + 3 y 2 = 21 6. 2 x 2 + 3 y 2 = 12 2 x − y = 5 x 2 + 2 y 2 = 12 x 2 − y 2 = 25
204 Chapter 12 Conic Sections Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
se c t io n
12.5 Quadratic Inequalities equalities
Objective 1 Graph the solution set of a quadratic inequality in two variables
The graph of a quadratic inequality in two variables is a region of the plane that is bounded by one of the conic sections (parabola, circle, ellipse, or hyperbola).
Example 1
Graph the solution set.
A. y ≤ x 2 + 2 x + 2 B.
A. y ≤ x 2 + 2 x + 2
Solution
y2 x2 − ≥1 9 4
y = x2 + 2x + 2
Steps
y
6
–6
0
6
x
–6
B.
y2 x2 − ≥1 9 4
y2 x2 − =1 9 4
y
6
–6
0
6
x
–6
• Change the inequality to an equality. This is the equation of a parabola that opens up. The vertex is (–1, 1). The axis of symmetry is the line x = –1. • Because the inequality is ≤, the graph is drawn as a solid curve. • Substitute the point (0, 0) into the inequality. Because 0 < 02 + 2(0) + 2 is true, the point (0, 0) should be in the shaded region.
• Change the inequality to an equality. This is the equation of a hyperbola. The vertices are (0, –3) and (0, 3). The equations of the 3 3 asymptotes are y = x and y = − x . 2 2 • Because the inequality is ≥, the graph is drawn as a solid curve. • Substitute the point (0, 0) into the 02 02 − ≥ 1 is not 9 4 true, the point (0, 0) should not be in the shaded region. inequality. Because
Exercises Graph the solution set. 1. y ≤ x 2 − 4 x + 3
4.
y2 x2 − > 1 25 36
2. ( x + 2)2 + ( y − 3)2 > 4
3.
x2 y2 + <1 16 25
5. x ≤ y 2 + 2 y + 1
6.
x2 y2 − ≤1 4 16
Section 12.5 Quadratic Inequalities and Systems of Inequalities 205
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Objective 2 Graph the solution set of a nonlinear system of inequalities
nequalities is the intersection of the solution the solution set of a system of inequalities, first graph the solution set for each inequality. The graph of the solution set of the system of inequalities is the region of the plane represented by the intersection of the two shaded regions.
Example 2
Graph the solution set. A. y > x 2 B. y < x + 2
Solution
A.
Steps
y 8
4 –8 –4 0 –4
4
8
x
–8
y
• Graph the solution set of each inequality.
•
• x 2 + y 2 = 4 is the equation of a circle. Use a solid curve. Shade inside the circle.
4 2 –4 –2 0 –2
• Graph the solution set of each inequality. • y = x 2 is the equation of a parabola. Use a dashed curve. Shade inside the parabola. • y = x + 2 is the equation of a line. Use a dashed line. Shade below the line.
The solution set is the region of the plane represented by the intersection of the solution sets of each inequality.
B.
x2 y2 − ≥1 9 16 x2 + y2 ≤ 4
2
4
x
–4
x2 y2 − = 1 is the equation of a 9 16 hyperbola. Use a solid curve. The point (0, 0) should not be in the shaded region.
The solution sets of the two inequalities do not intersect. This system of inequalities has no real number solution.
Exercises Graph the solution set. 7. y ≤ x2 − 4x + 4 8. x 2 + y 2 < 1 9. x 2 + y 2 < 16 y + x > 4 x + y ≥ 4 y > x+1
x2 y2 − <1 9 4 x2 y2 y < x − 2 y ≥ 2 x − 2 + <1 25 9 10. y > x 2 − 4
11. x ≥ y 2 − 3 y + 2
12.
206 Chapter 12 Conic Sections Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Answers Chapter 1
16.
Prep Test
17.
13 1. 20
18.
11 2. 60 3. 4. 5. 6. 7. 8. 9. 10.
19.
2 9
20. 21.
2 3 44.405 73.63 7.446 54.06 a, c, d a and C; b and D; c and A; d and B.
22. 23. 24. 25. 26. 27. 28. 29. 30.
Section 1.1 1. –3, 0 2. 0, 4 3. –2, –1 4. 0, 5 5. 3, 0, 7 6. –6, –2, 0, 1, 4 7. {–2, –1, 0, 1, 2, 3, 4} 8. {–3, –2, –1} 9. {x | x ≥ −2} 10. {x | x ≤ 2} A ∩ B = {6} 12. A ∪ B = {−4, −2, 0, 2, 4, 8} A ∩ B = {0, 4} 13. A ∪ B = {2, 3, 4, 5, 8, 9, 10} A∩ B = ∅ 14. A ∪ B = {0, 1, 2, 3, 4, 5} A ∩ B = {2, 4} –5 –4 –3 –2 –1
0
1
2
0
1
2
3
4
5
–5 –4 –3 –2 –1
0
1
2
3
4
5
–5 –4 –3 –2 –1
0
1
2
3
4
5
3
4
5
–5 –4 –3 –2 –1
0
1
2
3
4
5
–5 –4 –3 –2 –1
0
1
2
3
4
5
–5 –4 –3 –2 –1
0
1
2
3
4
5
–5 –4 –3 –2 –1
0
1
2
3
4
5
{x | 0 < x < 8} {x | −5 ≤ x ≤ 7} {x | x ≥ 5} {x | x ≥ −2} (–2, 4) [–2, 6) (-∞, 1) [3, ∞)
2 15 4 10. − 3 11 11. − 20 7 12. − 12 9. −
13.
35 48 5 15. − 6 14. −
17. 18. 19. 20. 21.
5 8 125 –8 –125 64 –81
22. 23. 24. 25. 26. 27.
324 –36 36 –7 –11 15
16.
Section 1.2 1. –30 2. –25 3. –17 4. –960 5. –96 6. –6 7. 5 8. –540
11. A ∪ B = {3, 6, 9, 12, 18}
15.
–5 –4 –3 –2 –1
3 20
1 4 16 29. 3 30. 32 28.
Section 1.3 1. 3 2. 3 3. 6 4. 1 5. Inverse Property of Addition 6. Inverse Property of Multiplication 7. Distributive Property 8. 0 9. 0 10. 56 11. −11a + 21 12. −7 x + 14 13. − x + 6 y 14. −12 x + 12 y Answers 207
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
15. 16. 17. 18. 19.
33a − 25b 31x + 34 y 4 − 8x − 5 y −2 x − 144 y − 96 5x + 3 y − 32
Section 1.4 1. n − ( n + 2); –2 2. 5(8n); 40n 3. 17 n − 2 n; 15n 4. (5n + 12) + 15n; 20 n + 12 5. n − (5 − n); 2 n − 5 6. n2 − (12 + n2 ); –12 7. Distance from Earth to the moon: d; 390d 8. Number of deaths each year caused by accidents: d; 7d 9. 10, 000 − x 10. Length of longer piece: L; 3 – L 11. Measure of angle B: x; measure of angle A: 2x; Measure of angle C: 4x 12. Width of the rectangle: W; 2W + 3
Chapter 2 Prep Test 1. –4 2. –6 3. 3 4. 1 1 2 10x – 5 6x – 9 3n + 6 0.03x + 20 20 − n
5. − 6. 7. 8. 9. 10.
Section 2.1 1. –10 2. –9 3.
1 2
4. −
32 25
5.
3
11 7. 1 16 7 2 9. − 5 10. 2 8.
1 2 8 12. 3 13. –6 11.
6 7 15. –33 16. 15°C 17. 8 bags 14.
Section 2.2 1. 24 dimes 2. 15 quarters 3. 5 stamps 4. 3 and 7 5. –20, –19, and –18 6. 5, 7, and 9 Section 2.3 1. $1.66 2. 16 lb of $5.50 coffee; 24 lb of $3.00 coffee 3. $.12 4. 100 oz 5. 40 mph 6. 8 mph 7. 11.25 mi 8. 1st plane: 255 mph; 2nd plane: 305 mph Section 2.4 1. Kari; $26.25 2. $2000 3. $4000 4. $18,375 at 4.5%; 23,625 at 3.5% 5. 6.875%
6. 500 lb 7. 25 L of 65% solution; 25 L of 15% solution 8. 3 qt Section 2.5 1. {x | x < 2. {x | x ≥ 3. {x | x ≤ 4. {x | x ≤ 5. (-∞, 5] 6. [4, ∞) 7. [1, ∞) 14 8. , 11 9. 10. 11. 12. 13. 14. 15. 16.
5} 3} −1} 3}
∞
(–1, 2) ( −∞, 3) ∪ (5, ∞) (–2, 1) {x | x < 1 or x > 3} {x | −3 < x < 4} {x | −3 < x < −2} 60 min more than 460 messages
Section 2.6 1. –7, 7 2. –5, 1 3. –4, 4 3 3 4. − , 2 2 1 9 5. , 2 2 4 6. 0, 5 7. –3, 7 1 11 8. − , 6 6 9. {x | x > 3 or x < −3} 10. {x | −5 < x < 5} 11. {x | 4 ≤ x ≤ 6} 14 12. x | x > 2 or x < − 5 13. {x | −3 < x < 2} 14. {x | x < −1 or x > 8} 15. 1.742 in., 1.758 in. 16. 2.3 cc, 2.7 cc
208 Answers Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
11. a. $1.8 trillion, b. 2003,
Chapter 3
Percent of household with telephones
Prep Test 1. –4x + 12 2. 10 3. –2 4. 11 5. 2.5 6. 5 7. 1 8. 4
0
x
4
−4
2.
y 4
−4
0
x
4
−4
3.
y 4
−4
0
x
4
4.
0
x
4
’80
’00
0
4
x
−4
4.
y 4
−4
4
x
−4
5.
y 4
−4
0
4
x
−4
6.
y 4
−4
0
4
x
−4
7.
y 4
0
−4
x
8
0
8.
x
−4
x
y 4
0
x
8
−4
0
4
x
−4
9. (–4, 0); (0, 2)
y
y
8
0
4
x
8
−8
4
4
−4
−8
10. −8
4
11. Function 12. Not a function 13. Function
−4
0
4
x
−4
10. (1, 0); (0, 3) y
4
7.
7 17; , 2
3
8.
3 65; , 2
1
10.
0
y
−8
y
−4
’60
Year
−8
−4
6.
’40
−4
x
8
y
y
30
9.
x
4
4
−4
x
4
’20
−8
0
5.
9.
3.
50
−4
4
8
4
70
0 −4
Section 3.2 1. Domain: {1, 2, 3, 4, 5}; Range: {1, 4, 7, 10, 13}; It is a function. 2. Domain: {–3, –2, –1, 1, 2}; Range: {–4, 1, 4, 9}; It is a function. 3. 13 4. 12 5. –21 6. 2 7. 10 y 8.
y
−4
90
−4
y
−4
13. 3°F per hour 14. –10,200 trademark registration applications per year
Section 3.1 y 1. −4
2.
7 5; −1, 2 3 3 82; , − 2 2
Section 3.3 y 1.
−4
0 −4
4
x
−4
4
−4
0
4
x
9 11. , 2
0 ; (0, –3) y 4
−4
0
4
x
−4
Answers 209
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
3 12. , 2
12.
0 ; (0, –2)
10. y = −3x + 11
y
1 1 x− 3 3 12. y = 2 x + 5 11. y =
y 4
−4
0
4
−4
13.
x
y
Section 3.7 y 1.
4
13. Distance traveled (in feet)
D
−4
1000
14.
10
t
Seconds The roller coaster travels 940 ft in 5 s.
Section 3.4 1. –1
1 3 4. –3 5. Undefined 6. 0 y 7. 3.
4
4
x
−4
8.
y 4
−4
0
4
x
9.
y 4
−4
0
4
x
−4
10.
y 4
−4
0
4
x
−4
11.
y 4
−4
0
0
4
x
0
4
x
−4
2. 4
y
x
4
−4
−4
0
4
x
−4
Section 3.5 1. y = 2 x + 5 2. y = −3x − 2 1 3. y = x + 2 2 4. y = −3x + 4
2 3
4
y
−4
0
x
4
500
−4
4
−4
(5, 940)
5
2. −
0
4 5. y = − x + 5 5 2 6. y = − x + 7 3 1 10 7. y = x + 3 3 3 8 8. y = x + 5 5 9. y = 3x − 11 10. y = 3 11. y = − x + 3 7 2 12. y = − x − 3 3 13. y = 1200 x; 13,200 ft 14. y = 63x; 315 calories Section 3.6 1. Yes 2. No 3. Yes 4. No 5. No 6. No 7. Yes 8. No 2 8 9. y = x − 3 3
3.
y 4
−4
4
x
−4
Chapter 4 Prep Test 1. 6 x + 5 y 2. 7 3. 0 4. –3 5. 1000 y 6. 4
−4
0
4
x
−4
7.
y 4
−4
0
4
x
−4
8.
y 4
−4
0
4
x
−4
Section 4.1 1. (2, 4) 2. (3, –2) 3. No solution 2 4. x , x − 2 5
210 Answers Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
5. (2, 1) 6. (–1, 2) 7. No solution
13. (1, 3)
1 8. x , x − 4 2
16. No solution
3. −18 x 3 y 4
Section 4.2 1. (2, 1) 1 2. − , 2 3. 4. 5. 6. 7. 8. 9. 10.
2. 6 a 3b8
2
(2, 5) No solution (x, 3x – 4) (4, –1) (–1, 2, 1) (6, 2, 4) (4, 1, 5) No solution
Section 4.4 1. Plane: 150 mph; wind: 10 mph 2. Cabin cruiser: 14 mph; current: 2 mph 3. Cinnamon tea: $2.50/lb; spice tea: $3.00/lb 4. $30
6. 72 a8 b11
Section 4.5 y 1.
11.
−4
0
2.
0
3.
8. −128a5 b8 c9 9.
10. y 3 a5 b3 2 12. 2 4 x y
0
4
x
16.
4
x
y
−4
4
1 2 5 2
Chapter 5
2 1 − 5 10. 1 0
3 5 26
Prep Test 1. –12y 2. –8
1 4 1 0 1 4 0 0 1
−2 −8 32 − 19
1 2 3 1 2 0 1
−3
−
3 2 5 2 3
3. 4. 5. 6. 7. 8.
x
−4
y6 x3
17. −
3b7 2a3
18. 2x 19. x 2 y10
4.
4 a2
1 2187a 1 15. 6 x
4
−4
1 x9
14.
y
1 −2 9. 0 1
0
7. −432 a 7 b11
13.
−4
8. (–1, –2, 1)
0
x
y
−4
7. (1, –1, 2)
12.
4
11 17 6. , 14 21
1
5. −6 x 5 y 5 z 4
4
Section 4.3 1. 11 2. 18 3. 3 4. –30 5. (3, –4)
11.
4. −8a 3b6
b10 a10 21. 4.67 × 10 −6 20.
22. 1.7 × 10 −10
3a − 8b 11x − 2 y − 2 −x + y 2⋅2⋅2⋅5 4 –13
9. −
1 3
Section 5.1 4 4
1. a b
23. 4.3 × 106 24. 25. 26. 27. 28.
9.8 × 109 0.000000123 0.0000000000062 8,200,000,000,000,000 634,000
29. 2.848 × 10 −10 30. 2.08 × 107 31. 1.5 × 10 −8 32. 1.4 × 108 33. 1.14567901 × 107 34. 3 × 1010 operations 35. 1.83664508 × 103 times 36. 3.3898305 × 105 times Answers 211
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 5.2 1. –42 2. 10 y 3.
Section 5.4
3. −4 z 2 + 3z
4
−4
4
2 y 5y 6. 3 y − 4 + x 7. x + 8 8. 4 x + 7 5. −4 y + 3 −
y 4
−4
0
4
x
−4
5.
2 x−3 8 2 10. x − x+4 9. x 2 +
y 4
−4
4. t 2 − 3t + 4
x
−4
4.
0
4
x
11. − x 2 + 2 x − 1 +
−4
2
12. x − 2 x + 4 13. 2 x − 8 14. 3x + 4
6. 6 x 2 − 6 x + 5 7. –5 8. − x 2 + 1
1 x−3
15. 2 x 2 − 3x + 9
9. −3x 2 + 11x − 6
16. x 2 − 3x + 2 4 17. x − x + 2 − x+1 41 18. x 2 − 5x + 16 − x+2 19. 8 20. 1 21. –39 22. –58 23. 190 24. 302
Section 5.3
2
1. 2 x 2 − 6 x 2. −16 y 3 + 24 y 4 3. 12b4 − 48b3 − 24b 4. −2 x 2 + 5x 5. 5 y 2 − 11 y 2
6. 8 x − 8 x 2
7. 15x − 61x + 56 2
8. 4 x + 4 xy − 15 y
2
2
9. 14 x − 69 xy + 27 y 3
15. 16. 17. 18. 19. 20. 21. 22. 23. 24.
2
2
10. x − 5x + 13x − 14 11. x 3 + 8 x 2 + 7 x − 24 4
3
2
12. x + 5x − 3x − 11x + 20 13. a 2 − 16 14. 9 x 2 − 4 2
15. b − 121 16. 9 a 2 + 30 ab + 25b2 17. 25x 2 − 40 xy + 16 y 2 18. 9 a 2 − 24 ab + 16b2 19. (3x 2 − 9 x − 12) ft 2 x 20. x 2 + − 3 ft 2 2
Section 5.5 1. 3a( 2 a − 5) 2. 4b(8b + 3) 3. Nonfactorable 4. 4( 4 x 2 − 3x + 6) 5. x 2 ( 2 x 3 + 3x 2 − 4) 6. 7. 8. 9. 10. 11. 12. 13. 14.
5b2 (1 − 2b + 5b2 ) (a (x (x (x (x (a (x (x
+ − + − + + − +
2)( x − 2) 2)( a + b) 3)( x + 2) 5)( x + 4) 4)( y − 2) 7 )( b − 3) 3)( x − 5) 2)( x + 10)
( a + 1)( a + 11) ( a + 9)( a − 8) ( b + 7 )( b − 5) Nonfactorable ( 2 x + 1)( x + 3) ( 2 x + 5)( x − 8) ( 2 y + 3)(3 y − 2) ( 4 y − 3)( y − 3) ( 2b − 5)(3b + 7 ) ( 2 a + 1)( a + 6)
Section 5.6 1. ( x − 4)( x + 4) 2. ( 2 x − 1)( 2 x + 1) 3. (9 x − 2)(9 x + 2) 4. ( y − 3)2 5. ( b − 9)2 6. ( 4 x − 5)2 7. ( x − 3)( x 2 + 3x + 9) 8. ( y + 5)( y 2 − 5 y + 25) 9. ( 2 x − 1)( 4 x 2 + 2 x + 1) 10. ( 4 a + 3)(16 a 2 − 12 a + 9) 11. (3x − 2 y )(9 x 2 + 6 xy + 4 y 2 ) 12. ( 4 x + 3 y )(16 x 2 − 12 xy + 9 y 2 ) 13. ( xy − 5)( xy − 3) 14. ( xy − 11)( xy + 3) 15. ( x 2 − 6)( x 2 − 3) 16. ( b2 − 18)( b2 + 5) 17. ( a 2 + 9)( a 2 + 5) 18. ( y 2 − 8)( y 2 + 2) 19. 3( 2 x − 3)2 20. 7( x − 2)( x + 2) 21. y 3 ( y + 11)( y − 5) 22. ( x 2 + 4)( x − 2)( x + 2) 23. ( x − 1)(3x 2 + 4) 24. 3x ( x − 3)( x 2 + 3x + 9) Section 5.7 1. –4, –6 3 2. − , 7 2 3. –5, 1 4. –4, 8 5. –4, 5 6. –4, 8
212 Answers Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
7. –4, 1 8. –5, 3 2 9. − , 3 3 10. 8 in. by 21 in. 11. 5 s
Chapter 6 Prep Test 1. 50
x−4 Section 6.2 1. 1 x −1 2. − x−5 x−3 3. − 2x + 3
1 6 3 3. − 2
4.
1 24 5 5. 24 2 6. − 9
6.
2. −
1 3 8. –2 7.
Section 6.1 1. 2 2. 0
5. 6. 7. 8. 9. 10. 11.
x x x 14. x
8. −
2 a 2 − 13a ( a − 2)( a + 1)
Section 6.6
− − + +
2
2x + x + 3 ( x + 1)2 ( x − 1)
2x 2x 3x 11. 4x 10.
+3 −3 −1 +1
2x + 5 ( x + 3)( x − 2)( x + 1)
x x −1 a −1 2. − a+1 1.
3. 4. 5. 6.
2 12. − a 13.
x2 + x ( x − 3)( x − 5)
2x − 3 2x + 3
Section 6.3
x 2
2 5 x x x x x x
+2 −1 −5 +9 −2 +2
Section 6.4 1. –5 2. 0
3 5 4 3
3.
5 No solution 2 1.2 h 12 h 18 min 15 mph; 12 mph Commercial jet: 540 mph; corporate jet: 420 mph 12. Passenger train: 59 mph; freight train: 45 mph
7. −
12.
1 35 16 41 {x | x ≠ 3} {x | x ≠ −3} {x | x ≠ −5, x ≠ −1} {x | x ≠ −3, x ≠ 2} {x | x ≠ −5, x ≠ 1} All real numbers
4. 5. 6. 7. 8. 9. 10. 11.
Section 6.5 1. 4000 ducks 2. 34 additional ounces 3. 13.5 lb/in2 4. 10.2 in. 5. 32 revolutions/min 6. 16.25 amps
9.
10 9. 7 10. 110 mph, 130 mph
4.
3 2
( x − 3)2 5. ( x + 3)( x − 4)
4.
3.
x+3
15. −
3 4
P − 2L 2 5 C = ( F − 32) 9 S C = 1− r A− P t = Pr E − Ir R = I R = Pn + C
1. W = 2. 3. 4. 5. 6.
7. y = −
2 x+2 3
1 x−2 2 3 9. y = x − 6 2 3 10. y = − x + 3 4 8. y =
Chapter 7 Prep Test 1. 16 2. 32 3. 9 4.
1 12 Answers 213
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
5. −5x − 1 6.
xy 4
7. 9 x 2 − 12 x + 4 8. −12 x 2 + 14 x + 10 9. 36 x 2 − 1 10. –1, 15 Section 7.1 1. 2 2. 27 1 3. 9 4.
4. x 2 yz 2 yz 5. 3xyz 2 5 yz 6. −5 y 3 x 2 y
1
x2/5 7. y
7. −6 xy 3 3 x 2
9.
12. 9 2
20. −8 5
4
0
b
4
8t 3
0
4.
24. 4 x
26.
5 5
27.
3 3a a
28.
2 xy 2y
18. x1/ 3 20. 2 yx
3/ 2
21. xy 5 22. 11 y 6 23. a 2 b4 24. 5b5 25. –2x 26. 3a5 27. x 4 y 2 28. −2 x 3 y 4
5. 6. 7. 8.
31.
9−6 x + x 9− x
32. −5 + 2 6
x
25 72 9 16 2, 11 11.3 ft 909.8 mb; increases
9. 68 million miles Section 7.5 1. 4i 2. 8i
29. 2 5 − 4 30. 8 + 4 3 − 2 2 −
4
23. x − 6 x + 9
17. 141/ 2 19. b3 / 5
y
22. 84 + 16 5
25. b2 3a
x
4
Section 7.4 1. 9 2. 5 3. 6
16. −3 5 a 2
4
10.
21. a − 5 a + 6
5
x
y
13. −7 a 3 3a
x 12. 2 y
4
3
11. x y z
15.
y
0
17. 2 ab 4 3a 2 b 18. 6 3 9
x
4
11. 10 x 2 x
19. 9 a a − a 3
6
14.
8.
2x
16. 2 ab2 3 4b2
10. x 4 y
4
8. 2 x 2 y 2 4 4 y 2
15. 8xy x
9. b9 / 4
3
0
14. 8b 3 2b2
1 b
13.
4
10. 3 2b + 5 3b
343 5. 125
8.
Section 7.3 1. {x | x ≥ −1} 2. {x | x ∈ real numbers} 3. {x | x ≥ 2} 4. [3, ∞) 5. (-∞, 6] 6. (-∞, ∞) y 7.
3. 4 3 2
9.
1 4
6.
Section 7.2
5
6
3. 7 i 2 4. 6i 2 5. 2 3 − 3i 2 6. 2 15 − 4i 3 7. 8 − i
214 Answers Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27.
10 − 7 i −8 + 4i −5 − 3i 5−i 15 + 4i 63 -24 −4 + 12i −15 − 12i 17 − i 8 + 27 i 1 35 + 12i 21 − 20i -3i
3 2. − , 6
4. -2,
3 2
2 9 5. − , 3 2 6.
1 2 , 2 3
7. x 2 − 7 x + 10 = 0 8. x 2 + 6 x + 8 = 0 9. x 2 − 5x − 6 = 0 10. 3x 2 + 11x + 10 = 0 11. 10 x 2 − 7 x − 6 = 0 12. 50 x 2 − 25x − 3 = 0 13. –7, 7 14. –2i, 2i
4 − i 5 3 1 + i 4 2 5 16 − + i 3 3 4 2 + i 5 5 -i
9 9 15. − , 2 2 16. −3i 2 , 3i 2 17. 0, –6 18. 2 − 2 6 , 2 + 2 6
11 29 + i 13 13 3 11 29. − i 10 10
19. −1 − 2i 3, −1 + 2i 3
Chapter 8
2. −2 − 11, −2 + 11 3. 2 − 3i, 2 + 3i 4. 1 − 4i, 1 + 4i
28.
20. 3 − 3 2 , 3 + 3 2 Section 8.2 1. –1, 5
Prep Test 1. 3 2 2. 3i
1 5. -1, − 2
2x − 1 x −1 4. 8 5. Yes 3.
6.
2
6. ( 2 x − 1) 7. (3x + 2)(3x − 2) 8. –5 –4 –3 –2 –1 0 1 9. -3, 5 10. 4 Section 8.1 1. -2, 3
3
4
5
3 2
2 17. − , 6 3 18. 0, 3 Section 8.4 1. Smaller pipe: 12 min; larger pipe: 6 min 2. Smaller unit: 12 h; larger unit: 4h 3. 12 mph 4. 20 mph Section 8.5 1. {x | x ≤ 1 or x ≥ 2} – 5 – 4 – 3 – 2 – 1 0 1 2 3 2. {x | x < −3 or x > −2} – 5 – 4 – 3 – 2 – 1 0 1 3. {x | −3 < x < 4}
8. 1 − 2 2 , 1 + 2 2
4. {x | −5 < x < 4}
1 3 1 3 − , + 2 2 2 2
10. 1 − 2i, 1 + 2i 11. 1 −
3 2 3 2 i, 1 + i 2 2
12. −1 −
16. 1,
7. -2, 5
9. 2
1 9 , 2 2
Section 8.3 1. –3, –2, 2, 3 2. –2, –1, 1, 2 3. –4i, 4i, –2, 2 4. 1, 4 5. 9, 16 6. 25 7. 3 8. 5 9. –2 10. 1 11. 4 12. –2 13. –1, 10 14. –3, 1 15. –1, 0
–5 –4 –3 –2 –1
–5 –4 –3 –2 –1
0
0
1
1
4
5
2
3
4
5
2
3
4
5
2
3
4
5
5. {x | x < −2 or x > 4} – 5 – 4 – 3 – 2 – 1 0 1 6. {x | −1 < x ≤ 3}
–5 –4 –3 –2 –1
0
1
2
3
4
5
2
3
4
5
6 6 , −1 + 3 3
Answers 215
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 8.6 1. Vertex: (1, 2); Axis of symmetry: x = 1
3. 30
y
2 6. Domain: {–2, 3, 4, 6}; Range: {4, 5, 6}; yes 7. 8 y 8.
4
−4
0
x
4
1 9 2. Vertex: , − ; 4 2
4
−4
1 Axis of symmetry: x = 2 y
9.
x
4
−4
−4
3 1 3. Vertex: , − ; 4 2 Axis of symmetry: x = y
0
x
4
4
f
4
g
4. (–2, 0), (2, 0) 5. (–1, 0), (2, 0) 1 6. − 2 ,
−4
0 , (3, 0)
)(
(
7, 0
8. No x-intercepts
)(
(
9. 2 − 5 , 0 , 2 + Section 8.7 1. Minimum: 2 2. Maximum: –1 73 3. Minimum: − 8 9 4. Maximum: 4 5. 114 ft 6. $250 7. 5 days
Chapter 9 Prep Test 1. 2 2. –7
5, 0
)
0
4
y g
f
4
−4
0
4
x
−4
4.
)
x
−4
3.
7. −2 − 7 , 0 , −2 +
x
y
−4
4. f −1 ( x ) =
f
−4
2.
1 x+2 4 5. f −1 ( x ) = 2 x + 2
g
4
g
4
−4
0
Prep Test f x
5. Solid graph is answer. y 4
0
1 6. f −1 ( x ) = − x − 3 3 3 7. f −1 ( x ) = x − 6 2 8. Yes 9. Yes 10. No 11. No
Chapter 10
y
5. 6. 7. 8.
Section 9.3 1. Yes 2. No 3. No
−4
8
x
4
Section 9.1 y 1.
3 2
2 3 7 1 7 5
10. x 2 + 1
0
4.
9. x 2 − 4 x + 7
y 4
−4
−4
x
4
−4
4
0
Section 9.2 1. 5 2. 1 3. –6
4
x
1.
6. Solid graph is answer.
2. 3. 4. 5. 6. 7. 8.
1 9 16 –3 0; 108 –6 –2, 8 6326.60 y 4
y −4
4
0
4
x
0
4
x
−4
Section 10.1 1. A. 9
B. 1
C.
1 9
216 Answers Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
2. A. 16 3. A. 1
B. 4 B.
C.
1
1 8
4. A. 7.3891 B. 0.3679 C. 1.2840 5. A. 54.5982 B. 1 C. 0.1353 6. A. 0.0183 B. 0.2636 C. 54.5982 y 7. 4
−4
0
x
4
−4
8.
y
−4
0
9.
x
4
4
0
4
−4
10.
0
x
4
−4
11.
15. 16. 17. 18. 19. 20.
1 7 1 316.23 0.02 7.39 0.18
−4
0
12.
y 4
−4
0
x
4
−4
x 30. log b 2 y z x y
Chapter 11 Prep Test 1. 12
1 3. log 4 = −2 16 4. ln x = y
−4
2.
x
y
−4
1 5 y 8. 10 = x
4
4
5. 32 = 9 6. 10 −2 = 0.01
0 −4
3 4 3. 18 4. 96 5. –410 2.
4
2. log10 1000 = 3
400 99
Section 10.5 1. $1172 2. a. 21.2 mg; b. 3.5 h 3. 6.4 4. 188 decibels
33. 0.8617 34. 2.1133 35. –0.3174
1. log5 25 = 2
Section 10.4 1. 1 2. –8 3. 1.1133 4. 1.7095 5. 0.2075 6. 6 7. –2.0692 8. –0.6309 9. 8
14.
Section 10.3 y 1.
Section 10.2
x
11 2 11. –2 12. No solution 13. 5
s2 r 2 32. log 4 4 t
−4
4
10.
4
x2 z2 29. log b y
3
0 −4
27. ln x 3
x
4
x
y
−4
22. 5 log 3 x 23. ln r − ln s
31. log 8
4
4
y 4
0 −4
4.
28. log5 ( x y )
4
12. 2 13. 9 14. 64
3
y
−4
−4
26. 2 log8 x − log8 y − 2 log8 z x
y 4
24. 2 log 9 x + log 9 y + log 9 z 25. ln x + 2 ln y − 4 ln z
y
−4
3.
21. log8 x + log8 z
−4
9. 4
4
x
6.
4 9
7. x 2 + 2 xy + y 2 8. x 3 + 3x 2 y + 3xy 2 + y 3
7. 5−1 =
Answers 217
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 11.1 1. 2, 3, 4, 5 2. 0, –2, –4, –6 3. 2, 4, 8, 16 3 8 15 , , 2 3 4 5. 1, –2, 3, –4 4. 0,
1 1 1 1 , − , , − 3 4 5 2 7. 40 8. 110 6.
9. 10. 11. 12. 13. 14. 15.
12 13 45 42 20 28 91 60
Section 11.2 1. 141 2. 50 3. 71
4. 6, 4
Chapter 12 Prep Test 1. 7.21
6. –2,
4 3
7. 2186 8. –2188 2343 64 10. 62 9.
16. 17.
16 9 7 9 8 9 5 11 1 6
128 3. 243
4
2
4
2
4
x
–4
y 4
−4
−2
0
x
–2 –4
9.
y 4 2 −4
−2
0
x
–2
19. 54.9 in. 20. 2.6 ft 21. $432,194.24
2187 2. 4096
2
2
5. an = −4 n + 10
Section 11.3 1. 131,072
0
8.
18.
an = 13n + 4 26 42 21 2500 650 –760 345 420 –210 9 weeks 108 cans 2180 seats
−2
–2
381 128 13. 9
15.
y 4
−4
12.
14.
x 2 − 8 x + 16; ( x − 4)2 0; –4, 4 (1, –1) (1, –5) x = 2; (2, –2)
2
1330 11. 729
4. an = 3n − 2 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18.
2. 3. 4. 5. 6. 7.
–4
Section 12.1 1. Vertex: (1, –5); Axis of symmetry: x = 1 y
Section 11.4 1. 6 2. 40,320 3. 362,880 4. 21 5. 84 6. 1
6
–6
4
2. Vertex: (1, –2); Axis of symmetry: x = 1 y 6
3
2
2
8. x − 5x y + 10 x y − 10 x y + 5xy 4 − y 5
–6
3
9. 32r 5 − 240r 4 + 720r 3 − 1080r 2 + 810r − 243 10. −560 x
0
6
x
3. Vertex: (–4, –3); Axis of symmetry: y = –3 y
4
4
3
11. 160 x 12. –1024x
x
–6
7. x 4 + 4 x 3 y + 6 x 2 y 2 + 4 xy 3 + y 4 5
6
–4
4
x
–4
218 Answers Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
25 1 4. Vertex: − , ; Axis of 4 symmetry: y = y
2
0
6
x
−6
8.
y
5. Vertex: (1, –1); Axis of symmetry: x = 1
0
y
x
9.
y 6
–4
0
−6
8. ( x − 3)2 + ( y + 2)2 = 9
x
4
y
–4
6. Vertex: (–1, 2); Axis of symmetry: y = 2
−6
0
6
x
0
−6
6
2.
y
−6
12.
0
3.
2
2. ( x + 1) + ( y + 2) = 9
6
x
y 6
y 4
−6
0
0
4.
−4
6
x
y
2
7 1 4. x − + y − 2 2 2
2
=
−6
17 2
0
6
x
y
5. ( x − 1) + ( y + 2) = 25 y
−6
0
6
x
−6 −6
x
0
Section 12.4 1. (–2, 5), (5, 19) 2. (–1, –2), (2, 1)
5.
6.
( 6, 3) , ( 6, − 3) , ( − 6 , 3 ) , ( − 6 ,− 3 )
6. No solution
6
2
6
−6
5.
x
4. (2, –1), (8, 11)
6
3. ( x + 3)2 + ( y − 6)2 = 8
6
2 22 3. (2, 2), − , − 9 9
−6
x
4
x
y
−6
−6
−4
2
6
y
x
0
0
6
4
y
x
−6
x
6
−6
−6
1. ( x − 2)2 + ( y + 1)2 = 4
11.
6
Section 12.2
6 −6
Section 12.3 y 1.
–6
−4
y 6
6
x
x
−6
−6
6
10.
6
y
−6
6
x
0
y 6
x
6
1 7. x − + ( y + 2)2 = 1 2
–6
6 −6
2
x
6
y 6
–6
7. −6
6
6. ( x + 3)2 + ( y + 4)2 = 16
Section 12.5 y 1.
y
−8 −6
0
6
x
0
8
x
−8
Answers 219
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
2.
6.
y 8
10.
y 8
y 4 2
−8
3.
0
8
x
−8
8
0
8
4.
0
8
x
y
0
x
8
0
4
x
y 4
− 4 −2 0 −2
x
2
−8
0 −2
8
−8
8
12.
y 8
−8
2
−8
5.
0
8. No real solutions y 9.
8
−8
y
−8
y
x
−8
11.
−8
7.
y
−8
−4
2
x
0
8
x
−8
x
−8
220 Answers Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 1
Important Concepts
Review of Real Numbers
The integers is the set {…, –3, –2, –1, 0, 1, 2, 3, …}.
{x | −3 ≤ x < 2}
The negative integers are the integers …, –4, –3, –2, –1. The positive integers are the integers 1, 2, 3, 4, … . The positive integers are also called the natural numbers. p A rational number is a number of the form , where p and q are q integers and q is not equal to zero.
Example
5 −3 6 are rational numbers. , , and 6 4 1
An irrational number is a number whose decimal representation never terminates or repeats. Example
3, π , and 0.21211211121111… are irrational numbers.
The rational numbers and the irrational numbers taken together are called the real numbers. −6 Example –3, 5, π , , and 0.232332333… are real numbers. 7 [1.1.1]
Example Evaluate: − −12
Example Use the roster method to write the set of positive odd
integers less than 12.
Example Use set-builder notation to write the set of real numbers
less than 7.
Solution { x | x < 7, x ∈ real numbers}
5
1
2
3
4
5
To indicate an interval that extends forever in one or both directions using interval notation, we use the infinity symbol ∞ or the negative infinity symbol -∞. {x | x > 1} (1, ∞) {x | x ≥ 1} [1, ∞) {x | x < 1} (-∞, 1) {x | x ≤ 1} (-∞, 1]
−5 −4 −3 −2 −1
0
1
2
3
4
5
−5 −4 −3 −2 −1
0
1
2
3
4
5
−5 −4 −3 −2 −1
0
1
2
3
4
5
−5 −4 −3 −2 −1
0
1
2
3
4
5
Example Write {x | −8 ≤ x < −1} using interval notation. Solution [–8, –1)
Example Graph {x | x > −3}. 0
Addition of Real Numbers [1.2.1] • To add numbers with the same sign, add the absolute values of the numbers. Then attach the sign of the addends.
• To add numbers with different signs, find the absolute value of each number. Subtract the smaller of these absolute values from the larger. Then attach the sign of the number with the larger absolute value.
1
2
4
5
(–3, 2) Open interval 0
1
[–3, 2]
−5 −4 −3 −2 −1
Write the subtraction as the addition of the opposite number. Then add. 5 − 12 = 5 + ( −12) = −7 Multiplication of Real Numbers [1.2.1] • The product of two numbers with the same sign is positive.
3
Some sets of real numbers can also be expressed using interval notation. The examples that follow show set-builder notation, interval notation, and the graph of the set. [1.1.2]
−5 −4 −3 −2 −1
Subtraction of Real Numbers [1.2.1]
−4 − ( −5) = −4 + 5 = 1
Some sets of real numbers can be graphed on a number line. [1.1.2]
4
6 + ( −11) = −5
Set-builder notation uses a rule to write a set. [1.1.2]
{x | −3 ≤ x ≤ 2}
0
3
Half-open interval
−4 + 8 = 4
Solution {1, 3, 5, 7, 9, 11}
−5 −4 −3 −2 −1
2
4+2 = 6
The roster method of writing a set encloses the list of the elements of the set in braces. [1.1.2]
{x | −3 < x < 2}
1
−5 + ( −6) = −11
− −12 = −12
–5 –4 –3 –2 –1
0
(–3, 2]
Solution { x | x > –12}
The absolute value of a positive number is the number itself. The absolute value of a negative number is the opposite of the negative number. The absolute value of zero is zero. [1.1.1]
Solution
−5 −4 −3 −2 −1
{x | −3 < x ≤ 2}
Half-open interval
Example Write (-12, ∞) using set-builder notation.
Absolute Value
Solution
[–3, 2)
0
2
3
4
5
Closed interval 1
2
3
4
5
( 4)(5) = 20
( −3)( −7 ) = 21
• The product of two numbers with different signs is negative. ( −4)( 2) = −8
(5)( −4) = −20
Division of Real Numbers [1.2.1] • The quotient of two numbers with the same sign is positive. 8 ÷ 2 = 4
( −20) ÷ ( −5) = 4
• The quotient of two numbers with different signs is negative. ( −10) ÷ 5 = −2
15 ÷ ( −3) = −5
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
The Order of Operations Ag
The Inverse Property of Multiplication
Step 1 Perform operations insid ing symbols include parentheses ( ), brackets [ ], the absolute value symbol , and the fraction bar.
Example 4 ⋅
Step 2 Simplify exponential expressions. Step 3 Do multiplication and division as they occur from left to right. Step 4 Do addition and subtraction as they occur from left to right. [1.2.4] 12 − 2 Example Simplify: 8 − 4+1 12 − 2 Solution 8 − ÷ 22 4+1 10 = 8− ÷ 22 5 10 = 8− ÷4 5 = 8−2÷4 1 = 8− 2 15 = 2
÷2
2
Properties of Real Numbers [1.3.1] The Commutative Property of Addition If a and b are real numbers, then a + b = b + a. Example 3 + 2 = 2 + 3
The Commutative Property of Multiplication If a and b are real numbers, then ab = ba. Example 3(–2) = –2(3)
The Associative Property of Addition If a, b, and c are real numbers, then ( a + b) + c = a + ( b + c ). Example (3 + 4) + 5 = 3 + ( 4 + 5)
The Associative Property of Multiplication If a, b, and c are real numbers, then ( ab)c = a( bc ). Example (3 ⋅ 4) ⋅ 5 = 3 ⋅ ( 4 ⋅ 5)
The Addition Property of Zero
ber and a ≠ 0, then a ⋅ 4
= 1 and
1 ⋅4 = 1 4
1 1 = ⋅ a = 1. a a
The Distributive Property If a, b, and c are real numbers, then a( b + c ) = ab + ac. Example 3( 4 + 5) = 3 ⋅ 4 + 3 ⋅ 5
3 ⋅ 9 = 12 + 15 27 = 27
Replacing the variable in a variable expression by a numerical value and then simplifying the resulting expression is called evaluating the variable expression. [1.3.2] Example Evaluate ( b − c ) 2 ÷ ab when a = −3, b = 2, and
c = −4.
Solution
( b − c )2 ÷ ab
[2 − ( −4)]2 ÷ ( −3)( 2)
= [6]2 ÷ ( −3)( 2)
= 36 ÷ ( −3)( 2)
= −12( 2)
= −24
To simplify a variable expression, combine like terms using the Distributive Property. [1.3.3] Example Simplify: ( 2 x + xy − y ) − (5x − 7 xy + y ) Solution
( 2 x + xy − y ) − (5x − 7 xy + y )
= 2 x + xy − y − 5x + 7 xy − y
= –3 x + 8 xy – 2 y
Example Simplify: 2 x − 3[ y − 3( x − 2 y + 4)] Solution
2 x − 3[ y − 3( x − 2 y + 4)]
= 2 x − 3[ y − 3x + 6 y − 12]
= 2 x − 3[7 y − 3x − 12]
= 2 x − 21 y + 9 x + 36
= 11 x – 21 y + 36
Translate a verbal expression into a variable expression. [1.4.1] Example Translate and simplify “a number decreased by the
difference between eight and twice the number.”
If a is a real number, then a + 0 = 0 + a = a.
Solution the unknown number: n
Example 3 + 0 = 3 and 0 + 3 = 3
the difference between 8 and twice the unknown number: 8 − 2 n
If a is a real number, then a ⋅ 0 = 0 ⋅ a = 0.
n − (8 − 2 n)
Example 8(0) = 0 and 0(8) = 0
= n − 8 + 2n
= 3n – 8
The Multiplication Property of Zero
The Multiplication Property of One If a is a real number, then 1 ⋅ a = a ⋅ 1 = a.
Example The depth of the deep end of a swimming pool is 2 ft
more than twice the depth of the shallow end. Express the depth of the deep end in terms of the depth of the shallow end.
Example 5 ⋅ 1 = 5 and 1 ⋅ 5 = 5
The Inverse Property of Addition If a is a real number, then a + ( − a ) = ( − a ) + a = 0.
Solution the depth of the shallow end: D
Example 4 + (–4) = 0 and (–4) + 4 = 0
the depth of the deep end: 2D + 2
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
C H A P T E R 10
Important Concepts
Exponential and Logarithmic Functions
Example Write log10 0.0001 = −4 in exponential form.
Definition of an Exponential Function
Solution log10 0.0001 = −4 is equivalent to 10 −4 = 0.0001.
The exponential function with base b is defined by
Example Evaluate: log 4 64
f ( x ) = bx
Solution log 4 64 = x
where b > 0, b ≠ 1, and x is any real number. [10.1.1] 2
x
Example Evaluate f ( x ) = at x = 3 and x = −2. 3
2 Solution f ( x ) = 3
x
43 = 4 x
3= x
−2
log 4 64 = 3
Example Solve log 2 x = −4 for x.
3
64 = 4 x
2 8 f (3) = = 27 3 2 f ( −2) = 3
Solution
log 2 x = −4
9 = 4
Natural Exponential Function
The function defined by f ( x ) = e x is called the n atural exponential function. [10.1.1]
Example Evaluate f ( x ) = e x at x = 1.2. Round to the nearest
The solution is
Properties of Logarithms [10.2.2] For any positive real numbers x, y, and b, b ≠ 1,
f (1.2) = e1.2
Product Property: log b ( xy ) = log b x + log b y x Quotient Property: log b = log b x − log b y y
≈ 3.3201
Graph an Exponential Function [10.1.2]
Power Property:
Example Graph: f ( x ) = 2 + 1 x
Solution
x
y
0 1
21 + 1 = 3
–1
Solution
6
–6
0
Example Graph: f ( x ) = e −2 x − 4
y
-1
e −2( −1) − 4 ≈ 3.39
0 1
e
−2 ( 0 )
e
−2 (1)
− 4 = −3 − 4 ≈ −3.86
y
−4
4 −4
If x > 0 and b is a positive constant not equal to 1, then y = log b x is equivalent to b y = x. [10.2.1] 1 in logarithmic form. 81 1 1 = is equivalent to log 3 = −4. 81 81
Solution
3−4
xy 2 in expanded form. 1
1 log xy 2 3 b 1 = (log b x + log b y 2 ) 3 1 = (log b x + 2 log b y ) 3 1 2 = log b x + log b y 3 3 =
Example Express 2 log b x − 3 log b y − log b z as a single
Definition of Logarithm
Example Write 3−4 =
3
log b 3 xy 2 = log b ( xy 2 ) 3
x
6
–6
x
log b x r = r log b x
Example Write log b
y
5 2 −2 + 1 = 4 3 2 −1 + 1 = 2 20 + 1 = 2
–2
Solution
1 . 16
ten-thousandth.
Solution
2 −4 = x 1 = x 24 1 = x 16
x
logarithm with a coefficient of 1.
Solution 2 log b x − 3 log b y − log b z = log b x 2 − log b y 3 − log b z
= log b
x2 − log b z y3
= log b
x2 y3 z
Change-of-Base Formula: log a N =
log b N log b a
[10.2.2]
Example Evaluate log 4 2.4. Round to the nearest ten-thousandth. Solution
log 4 2.4 =
ln 2.4 ≈ 0.6315 ln 4
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Graph a Logarithmic Function [10.3.1]
32 = x 2 − 8 x x 2 − 8x
Example Graph: f ( x ) = log 2 ( x
0 = x 2 − 8 x − 9
f ( x ) = log 2 ( x − 1)
Solution
y = log 2 ( x − 1)
0 = ( x + 1)( x − 9)
y = log 2 ( x − 1) is equivalent to 2 = x − 1.
2 +1 = x
Choose values for y and find the corrresponding values of x.
y
y
x
0
5 4 3 2 −1 + 1 = 2 20 + 1 = 2
1
21 + 1 = 3
–1
Example Solve: 9
= 27
−4
0
4
x −1
9 x + 1 = 27 x − 1
(3 )
2x + 2
3
3 x −1
= (3 )
3x − 3
= 3
2 = x−3
5= x
The solution is 5.
When both sides of an exponential equation cannot easily be expressed in terms of the same base, logarithms are used to solve the exponential equation. [10.4.1] Example Solve 43x = 25. Round to the nearest ten-thousandth. Solution
43x = 25
6000 = 3000(1 + 0.01) n
6000 = 3000(1.01) n
2 = (1.01) n
log 2 = log(1.01) n
log 2 = n log 1.01 log 2 = n log 1.01
3x log 4 = log 25
3x =
log 25 log 4
log 25 x = 3 log 4
x ≈ 0.7740
The solution is 0.7740.
70 ≈ n
70 months ÷ 12 ≈ 5.8 years
In approximately 6 years, the investment will be worth $6000.
Example Find the pH of a baking soda solution, for which the
hydrogen ion concentration is 3.98 × 10 −9.
Strategy Use the formula for pH and replace H+ by 3.98 × 10 −9. Solution
pH = − log( H + )
= − log(3.98 × 10 −9 ) ≈ 8.4
The pH is approximately 8.4. Example A sample of promethium-147 contains 25 mg. One year
later, the sample contains 18.95 mg. What is the half-life of promethium-147, in years?
log 43x = log 25
12% 0.12 = = 0.01. 12 12
P = A(1 + i ) n
Solution
2 x + 2 = 3x − 3
Strategy To find the time, solve the compound formula for n. Use
P = 6000, A = 3000, and i =
the exponential expression. 2 x +1
earns 12% annual interest compounded monthly. In approximately how many years will the investment be worth twice the original amount? Round to the nearest whole number.
x
−4
Solution Express each side of the equation with 3 as the base of
Example An investment of $3000 is placed into an account that
4
An exponential equation is one in which a variable occurs in the exponent. If each side of an exponential equation can be easily expressed in terms of the same base, use the Equality of Exponents Property: If bu = bv , then u = v. [10.4.1] x +1
Application problems [10.5.1]
y
2 −2 + 1 =
–2
By the Principle of Zero Products, x = –1 and x = 9. Both values check. The solutions are –1 and 9.
y
Strategy Solve A = A0 (0.5)t / k for k, where A = 18.95 mg,
A0 = 25 mg, and t = 1.
Solution
18.95 = 25(0.5)1 / k
18.95 = (0.5)1 / k 25
18.95 1 log = log(0.5) k 25
A logarithmic equation can be solved by using the properties of logarithms. [10.4.2]
Example Solve: log 3 ( x 2 − 8 x ) = 2
k =
log(0.5) ≈ 2.5 log(0.758)
The half-life is 2.5 years.
Solution Rewrite the equation in exponential form. Then solve
for x. Check the solutions.
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
C H A P T E R 11
Important Concepts
Sequences and Series
A sequence is an ordered list of numbers. Each of the numbers of a sequence is called a term of the sequence. [11.1.1]
Formula for the Sum of n Terms of an Arithmetic Series
Example Write the first four terms of the sequence whose nth term
is given by the formula an = n( n + 1).
Solution
an = n( n + 1)
a1 = 1(1 + 1) = 2
The first term is 2.
a2 = 2( 2 + 1) = 6
The second term is 6.
a3 = 3(3 + 1) = 12
The third term is 12.
a4 = 4( 4 + 1) = 20
The fourth term is 20.
The indicated sum of the terms of a sequence is called a series. It is convenient to represent a series in a compact form called summation notation, or sigma notation. The Greek letter sigma, Σ, is used to indicate a sum. [11.1.2] Example Evaluate: Solution
6
6
∑ (i
2
− 2)
i=3
∑ (i
2
− 2)
i=3
= (32 − 2) + ( 42 − 2) + (52 − 2) + (62 − 2)
= 7 + 14 + 23 + 34 = 78
An arithmetic sequence, or arithmetic progression, is one in which the difference between any two consecutive terms is constant. The difference between consecutive terms is called the common difference of the sequence.
Let a1 be the first term of a finite arithmetic sequence, let n be the number of terms, and let an be the last term of the sequence. Then the sum of the series S n is given by n S n = ( a1 + an ). [11.2.2] 2 Example Evaluate the arithmetic series
18
∑ (3n − 2). n =1
Solution an = 3n − 2
a1 = 3(1) − 2 = 1 a18 = 3(18) − 2 = 52 n S n = ( a1 + an ) 2 18 S18 = (1 + 52) = 9(53) = 477 2
Example A contest offers 20 prizes. The first prize is $10,000, and
each successive prize is $300 less than the preceding prize. What is the value of the 20th-place prize?
Strategy To find the value of the 20th-place prize:
c Write the equation for the nth-place prize.
c Find the 20th term of the sequence.
Solution 10,000, 9700, …
d = a2 − a1 = 9700 − 10, 000 = −300
Formula for the nth Term of an Arithmetic Sequence
an = a1 + ( n − 1) d
The nth term of an arithmetic sequence with a common difference of d is given by an = a1 + ( n − 1) d . [11.2.1]
= 10, 000 + ( n − 1)( −300)
= 10, 000 − 300 n + 300
= −300 n + 10, 300
Example Find the 15th term of the arithmetic sequence 9, 3, -3, Solution
-9, … .
d = a2 − a1 = 3 − 9 = −6
an = a1 + ( n − 1) d
a15 = 9 + (15 − 1)( −6) = 9 + (14)( −6)
= 9 − 84 = −75
Example Find the number of terms in the finite arithmetic
sequence 7, 9, 11, … , 59.
Solution
d = a2 − a1 = 9 − 7 = 2
an = a1 + ( n − 1) d
59 = 7 + ( n − 1)2
59 = 7 + 2 n − 2
59 = 5 + 2 n
54 = 2 n
27 = n
There are 27 terms in the sequence.
a20 = −300( 20) + 10, 300 = −6000 + 10, 300 = 4300
The value of the 20th-place prize is $4300.
A geometric sequence, or geometric progression, is one in which each successive term of the sequence is the same nonzero constant multiple of the preceding term. The common multiple is called the common ratio of the sequence.
Formula for the nth Term of a Geometric Sequence The nth term of a geometric sequence with first term a1 and common ratio r is given by an = a1r n − 1 . [11.3.1] Example Find the 5th term of the geometric sequence 5, 2,
a2
r =
an = a1r n − 1
2 a5 = 5 5
a1
=
2 5
Solution
5−1
2 = 5 5
4
16 16 = 5 = 1225 625
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
4 ,…. 5
Formula for the Sum of n Terms of a Finite Geometric Serie
So ut o
Let a1 be the first term of a finite geometric sequence, let n be the number of terms, and let r be the common ratio, r ≠ 1. Then the sum of the series S n is given by a (1 − r n ) Sn = 1 . [11.3.2] 1− r
n
1 Example Evaluate the geometric series ∑ . n =1 2 5
1 Solution an = 2
n
1
1 1 a1 = = 2 2
1 a2 = 2
2
1 a2 1 = 4 = r = a1 1 4 2 a (1 − r n ) Sn = 1 1− r
⋅2 1
=
1 2
5 1 1 1 − 2 2 S5 = 1 1− 2
1 1 1 31 31 1− 2 32 2 32 64 = = = 1 1 1 2 2 2
=
31 2 ⋅ = 31 64 1 32
Formula for the Sum of an Infinite Geometric Series The sum of an infinite geometric series in which r < 1 and a1 . [11.3.3] a1 is the first term is given by S = 1− r Example Find the sum of the terms of the infinite geometric
8 4 sequence 3, -2, , − , . . . . 9 3 a2 −2 2 = = − Solution r = a1 3 3
S =
=
a1
1− r 3 2 1+ 3
=
n! (which is read “n factorial”) is the product of the first n consecutive natural numbers. 0! is defined to be 1. n ! = n( n − 1)( n − 2) ⋅ ⋅ 3 ⋅ 2 ⋅ 1 [11.4.1] 12 ! Example Evaluate: 7 !5 ! 12 ! 12 ⋅ 11 ⋅ 10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 Solution = 7 !5 ! (7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1)(5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1)
n n! r = ( n − r )! r ! [11.4.1] 7
Example Evaluate: 4 Solution
3 9 = 5 5 3
7 7! 7 ⋅ 6 ⋅5⋅ 4 ⋅ 3⋅ 2 ⋅1 4 = (7 − 4)! 4 ! = (3 ⋅ 2 ⋅ 1)( 4 ⋅ 3 ⋅ 2 ⋅ 1) = 35
Binomial Expansion Formula n n n ( a + b) n = a n + a n − 1b + a n − 2 b2 + 0 1 2 n n + a n − r br + + b n r n
[11.4.1]
Example Write (3m − n)4 in expanded form. Solution
(3m − n)4
4 4 4 = (3m)4 + (3m)3 ( − n) + (3m)2 ( − n)2 + 0 1 2
4 4 3 4 3 (3m)( − n) + 4 ( − n)
= 1(81m4 ) + 4( 27 m3 )( − n) + 6(9 m2 )( n2 ) +
4(3m)( − n3 ) + 1( n4 )
= 81m 4 − 108 m 3 n + 54 m 2 n 2 − 12 mn 3 + n 4
Formula for the rth Term in a Binomial Expansion
3 2 1 − − 3
=
= 792
1 4
=
. 6 = 0.36 + 0.0036 + 0.000036 + 36 36 a1 36 4 100 100 = = = = S = 1− r 1 99 99 11 1− 100 100 4 An equivalent fraction is . 11
n n − r +1 r −1 a b . [11.4.1] The rth term of ( a + b) n is r − 1 Example Find the third term in the expansion of (t − 2s)7 . Solution
n = 7, a = t , b = −2s, r = 3
The sum of an infinite geometric series can be used to find a fraction that is equivalent to a nonterminating, repeating decimal. [11.3.3]
7 7 − 3+1 7 ( −2s)3 − 1 = (t )5 ( −2s))2 3 − 1 (t ) 2
Example Find the equivalent fraction for 0.36.
= 21t 5 ( 4s 2 ) = 84 t 5 s 2
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
C H A P T E R 12
Important Concepts
Conic Sections
Every parabola has an axis of symmetry and a vertex that is on the axis of symmetry. The graph of y = ax 2 + bx + c is a parabola that opens up when a > 0 and opens down when a < 0. The graph of x = ay 2 + by + c is a parabola that opens right with a > 0 and opens left when a < 0. [12.1.1]
Center: (1, 0)
r =
( x1 − xm )2 + ( y1 − ym )2
r =
( −2 − 1)2 + (1 − 0)2 =
Example Find the vertex and the axis of symmetry of the parabola
given by y = − x 2 + x + 3. Then sketch the graph of the parabola.
b 1 1 Solution y-coordinate: − = − = 2a 2( −1) 2
y = − x2 + x + 3
1 1 13 y = − + + 3 = 2 4 2
4
−4
2
The axis of symmetry is the line x =
0
4
1 . 2
−4
x = −1
The vertex is (-1, -1).
The axis of symmetry is the line y = -1.
x
4 −4
( x – 2)2 + ( y + 4)2 = 5
Center: (2, –4)
Radius:
y 6
5 −6
0
6
x
−6
Let r be the radius of a circle and let (h, k) be the coordinates of the center of the circle. Then the equation of the circle is given by
Standard Form of the Equation of an Ellipse with Center at the Origin The equation of an ellipse with center at the origin is x2 y2 + = 1. The x-intercepts are (a, 0) and (–a, 0). The a2 b2 y-intercepts are (0, b) and (0, –b). [12.3.1] Example Sketch a graph of the ellipse given by the equation.
( x − h)2 + ( y − k )2 = r 2 [12.2.1] Example Find the equation of the circle with radius 4 and center
C(2, -3).Then sketch its graph.
y
Solution ( x − h)2 + ( y − k )2 = r 2
6
( x − 2)2 + [ y − ( −3)]2 = 42
−6
( x − 2)2 + ( y + 3)2 = 16
6
x
−6
Example Find the equation of the circle for which a diameter has
endpoints P1 ( −2, 1) and P2 ( 4, −1).
xm
( x 2 − 4 x + 4) + ( y 2 + 8 y + 16) = −15 + 4 + 16
Standard Form of the Equation of a Circle
( x 2 − 4 x ) + ( y 2 + 8 y ) = −15
A circle can be defined as all the points (x, y) in the plane that are a fixed distance from a given point (h, k) called the center. The fixed distance is the radius of the circle.
2 −2 + 4 = = 1 2
x 2 + y 2 − 4 x + 8 y + 15 = 0
4
x = −2( −1) − 4( −1) − 3
standard form.
y
xm =
( x − 1)2 + y 2 = 10
Solution Use completing the square to write the equation in
b −4 = − = −1 2a 2( −2)
2
Solution
2
x 2 + y 2 − 4 x + 8 y + 15 = 0 in standard form. Then sketch its graph.
x = −2 y 2 − 4 y − 3
x1 + x2
( 10 )
The equation of a circle can also be expressed in standard form as x 2 + y 2 + ax + by + c = 0. [12.2.2]
given by x = −2 y 2 − 4 y − 3. Then sketch the graph of the parabola.
( x − h) + ( y − k ) 2 = r 2
Example Find the vertex and the axis of symmetry of the parabola
10 2
Example Write the equation of the circle
1 13 The vertex is , . 2 4
Radius:
10
x
−4
Solution y-coordinate: −
=
( x − 1)2 + ( y − 0)2 =
y
9+1
ym = ym
y1 + y2
2 1 + ( −1) = = 0 2
A.
x2 y2 x2 y2 + = 1 B. + =1 4 25 18 9
Solution A. x-intercepts:
y
(2, 0) and (-2, 0)
y-intercepts:
(0, 5) and (0, -5)
B. x-intercepts:
(3
)
−4
(
2 , 0 and −3 2 , 0
y-intercepts:
(0, 3) and (0, -3)
Use 3 2 ≈ 4.25.
)
0
4
x
y 4
0 −4
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
x
Standard Form of the Equa with Center at the Origin
2
The equation of a hyperbola for which the axis of symmetry x2 y2 that contains the vertices is the x-axis is 2 − 2 = 1. The a b vertices are (a, 0) and (-a, 0).
Asymptotes: 5 5 y = x and y = − x 3 3
Vertices: (0, 3) and (0, -3)
Asymptotes:
92 + y 2 = 8
92 + y 2 = 8
A.
−6
6
x
−6
−6
B. x 2 − y 2 = 10 x2 + y2 = 8
Solution A. (1) y = 2 x 2 + x − 3
(2) y = 2 x 2 − 2 x + 9
Use the substitution method.
2x2 − 2x + 9 = 2x2 + x − 3 −3x + 9 = −3
−3x = −12
x = 4
y = 2x2 + x − 3
y = 2( 4)2 + 4 − 3
y = 32 + 4 − 3
y = 33
The solution is (4, 33). 2
2
B. (1) x − y = 10
(2) x 2 + y 2 = 8
Use the addition method.
2
2 x = 18 2
x = 9
y is not a real number. Therefore, the system of equations has no real number solution.
x2 y2 + ≤ 1 B. 9 16
x2 y2 − ≤1 9 4
x2 y2 + = 1 as a solid curve. 9 16 y Shade the region of the 6 plane that includes (0, 0). −6
0
6
x
−6
6
x
−6
x2 y2 − = 1 as a solid curve. B. Graph the hyperbola 9 4
Shade the region that includes (0, 0).
y 6
−6
0
6
x
−6
Solution set of a nonlinear system of inequalities [12.5.2] Example Graph the solution set.
A.
x2 y2 x2 y2 + ≤ 1 B. + ≥1 4 9 16 25 x > y2 − 2
x2 + y2 < 9
x2 y2 + = 1 as a solid curve. 4 9 Shade inside the ellipse.
Solution A. Graph the ellipse
Graph the parabola x = y 2 − 2 as a dashed curve. Shade inside the parabola. y
Substitute the value of x into equation (1).
y = ± −1
6
y = 2x2 − 2x + 9
y = ± −1
y
Example Solve: A. y = 2 x 2 + x − 3
y 2 = −1
Solution A. Graph the ellipse
A nonlinear system of equations is one in which one or more equations of the system are not linear equations. [12.4.1]
y 2 = −1
The graph of a quadratic inequality in two variables is a region of the plane that is bounded by one of the conic sections (parabola, circle, ellipse, or hyperbola). [12.5.1]
y
y = x and y = − x
( −3)2 + y 2 = 8
6
B. Axis of symmetry: y-axis
32 + y 2 = 8
Example Graph the solution set.
Solution A. Axis of symmetry: x-axis
Vertices: (3, 0) and (-3, 0)
x2 y2 y2 x2 − = 1 B. − =1 9 25 9 9
x2 + y2 = 8
Example Sketch a graph of the hyperbola given by the equation.
A.
+ y2 = 8
The equation of a hyperbola for which the axis of symmetry y2 x2 that contains the vertices is the y-axis is 2 − 2 = 1. The b a vertices are (0, b) and (0, -b). [12.3.2]
Substitute the values of x into equation (2).
The solution set is the region 4 of the plane represented by the intersection of the −4 4 solution sets of the two −4 inequalities. 2 2 x y B. Graph the ellipse + = 1 as a solid curve. 16 25 Shade outside the ellipse.
x
Graph the circle x 2 + y 2 = 9 as a dashed curve. Shade inside the circle. y
The solution sets of the two inequalities do not intersect. This system of inequalities has no real number solution.
4 2 0
2
x
−4
x = ± 9 = ±3 Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 2
Important Concepts
First-Degree Equations and Inequalities
is three less than four times the first number. Find the numbers.
The Addition Property of Equations If a, b, and c are algebraic expressions, then the equation a = b has the same solutions as the equation a + c = b + c. [2.1.1]
The Multiplication Property of Equations If a, b, and c are algebraic expressions, and c ≠ 0, then the equation a = b has the same solutions as the equation ac = bc. [2.1.1]
Strategy c The first number: n
The second number: 2n The third number: 4n – 3
c The sum of the numbers is 81.
Solution
n + 2 n + ( 4 n − 3) = 81
7 n − 3 = 81
7 n = 84
When an equation contains parentheses, one of the steps in solving the equation requires the use of the Distributive Property. [2.1.2]
n = 12
2 n = 2(12) = 24
Example Solve: 6(5 − x ) − 12 = 2 x − 3( 4 + x )
4 n − 3 = 4(12) − 3 = 48 − 3 = 45
The numbers are 12, 24, and 45.
Solution
6(5 − x ) − 12 = 2 x − 3( 4 + x )
30 − 6 x − 12 = 2 x − 12 − 3x
18 − 6 x = − x − 12
18 − 5x = −12
−5x = −30
x = 6
Value Mixture Problems [2.3.1] Example A butcher combined hamburger that costs $4.00 per
pound with hamburger that costs $2.80 per pound. How many pounds of each were used to make a 75-pound mixture that costs $3.20 per pound?
Strategy c Pounds of $4.00 hamburger: x
The solution is 6.
Pounds of $2.80 hamburger: 75 - x
Coin and Stamp Problems [2.2.1] Example A collection of stamps contains 3¢, 10¢, and 15¢ stamps.
The number of 10¢ stamps is two more than twice the number of 3¢ stamps. There are three times as many 15¢ stamps as there are 3¢ stamps. The total value of the stamps is $1.56. Find the number of 15¢ stamps.
Strategy c The number of 3¢ stamps: x
The number of 10¢ stamps: 2x + 2 The number of 15¢ stamps: 3x Number 3¢ stamp
Solution
Value
Total value
x
3
3x
10¢ stamps
2x + 2
10
10(2x + 2)
15¢ stamps
3x
15
45x
c The sum of the total values of the individual types of stamps equals the total value of all the stamps (156 cents). 3x + 10( 2 x + 2) + 45x = 156 68 x + 20 = 156
68x = 136
x = 2
3x = 3( 2) = 6
There are six 15¢ stamps in the collection.
number is twice the first number, and the third number
Value
$4.00 hamburger
x
4.00
4.00x
$2.80 hamburger
75 - x
2.80
2.80(75 - x)
75
3.20
75(3.20)
c The sum of the values before mixing equals the value after mixing.
Solution
4.00 x + 2.80(75 − x ) = 75(3.20)
1.2 x + 210 = 240
1.2 x = 30
x = 25
75 − x = 75 − 25 = 50
The mixture must contain 25 lb of the $4.00 hamburger and 50 lb of the $2.80 hamburger.
Uniform Motion Problems [2.3.2] Example Two small planes start from the same point and fly in
opposite directions. The first plane is flying 30 mph faster than the second plane. In 4 h the planes are 1160 mi apart. Find the rate of each plane.
Strategy c Rate of the second plane: r
Rate of the first plane: r + 30 First plane Second plane
Integer Problems [2.2.2] Example The sum of three numbers is eighty-one. The second
Cost
Mixture
3x + 20 x + 20 + 45x = 156
Amount
Rate
Time
Distance
r + 30
4
4(r + 30)
r
4
4r
c The total distance traveled by the two planes is 1160 mi.
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Solution
Addition Property of Inequalities [2.5.1]
4( r + 30) + 4r = 1160
4r + 120 + 4r = 1160
8r + 120 = 1160
8r = 1040
r = 130
r can be added to each side of an inequality without tion set of the inequality. Multiplication Property of Inequalities [2.5.1] Rule 1: If each side of an inequality is multiplied by a positive number, the inequality symbol remains the same.
r + 30 = 130 + 30 = 160
The first plane is traveling 160 mph.
Rule 2: If each side of an inequality is multiplied by a negative number, the inequality symbol must be reversed.
The second plane is traveling 130 mph.
Example Solve: 2 x − 1 < 6 x + 7
Investment Problems [2.4.1]
Example An investment of $3500 is made at an annual simple
Solution 2 x − 1 < 6 x + 7
interest rate of 5.2%. How much additional money must be invested at an annual simple interest rate of 7.5% so that the total interest earned is $575?
Solution
Principal
Rate
Interest
Amount at 5.2%
3500
0.052
0.052(3500)
Amount at 7.5%
x
0.075
0.075x
c The sum of the amounts of interest earned by two investments equals the total annual interest earned ($575). 182 + 0.075x = 575
0.075x = 393
{ x | x > −2}
Example Solve: −2 ≤ 5x + 3 ≤ 13
Write the solution set in interval notation.
Percent Mixture Problems [2.4.2]
Example A butcher has some hamburger that is 22% fat and some
that is 12% fat. How many pounds of each should be mixed to make 80 lb of hamburger that is 18% fat?
−2 ≤ 5x + 3 ≤ 13
−2 − 3 ≤ 5x + 3 − 3 ≤ 13 − 3
The amount invested at 7.5% is $5240.
−4 x < 8 −4 x 8 > −4 −4 x > −2
Compound Inequalities [2.5.2]
x = 5240
Solution
0.052(3500) + 0.075x = 575
−4 x − 1 < 7
Strategy c Amount invested at 7.5%: x
Write the solution set in set-builder notation.
−5 ≤ 5x ≤ 10 −5 5x 10 ≤ ≤ 5 5 5 −1 ≤ x ≤ 2
[–1, 2]
Absolute Value Equations [2.6.1] Example Solve: 2 x − 3 = 5
2x − 3 = 5
Strategy c Pounds of 22% fat hamburger: x
Solution
Pounds of 12% fat hamburger: 80 – x
2 x − 3 = 5
2 x − 3 = −5
Amount
Percent
Quantity
2x = 8
2 x = −2
22%
x
0.22
0.22x
x = 4
x = −1
12%
80 – x
0.12
0.12(80 – x)
18%
80
0.18
0.18(80)
c The sum of the quantities before mixing is equal to the quantity after mixing.
The solutions are 4 and –1.
Absolute Value Inequalities [2.6.2] Example Solve: 3x + 2 < 8
Solution 0.22 x + 0.12(80 − x ) = 0.18(80)
Solution
0.22 x + 9.6 − 0.12 x = 14.4
0.10 x + 9.6 = 14.4
0.10 x = 4.8
x = 48
80 − x = 80 − 48 = 32
The butcher needs 48 lb of the hamburger that is 22% fat and 32 lb of the hamburger that is 12% fat.
3x + 2 < 8 −8 < 3x + 2 < 8
−8 − 2 < 3x + 2 − 2 < 8 − 2 −10 < −10 < 3 10 − < 3 10 x | − 3
3x < 6 3x 6 < 3 3 x < 2 < x < 2
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 3
Important Concepts
Linear Functions and Inequalities in Two Variables
The xy-coordinate system is used to graph equations in two variables. A solution of an equation in two variables is an ordered pair (x, y) whose coordinates make the equation a true statement. The ordered-pair solutions can be graphed on an xy-coordinate system. [3.1.1] Example Graph the ordered-pair solutions of y = x + 1 when
x = −3, -2, -1, 0, and 1. y = x +1
Solution
x
y
-3
2
-2
1
-1
0
0
1
1
2
y 4
−4
0
4
x
−4
Function notation is frequently used for those equations that define functions. The process of finding the value of f ( x ) for a given value of x is called evaluating the function. [3.2.1] Example Let f ( z ) = z + z . Find f ( −4). Solution f ( z ) = z + z
f ( −4) = −4 + −4 = −4 + 4
f ( −4) = 0
A function that can be written in the form f ( x ) = mx + b (or y = mx + b) is called a linear function. The graph of a linear function is a straight line. [3.3.1] 3 Example Graph: y = x − 4 5 Solution y x y
Distance Formula If P1 ( x1 , y1 ) and P2 ( x2 , y2 ) are two points in a plane, then the distance d between the two points is given by d =
( x2 − x1 )2 + ( y2 − y1 )2 [3.1.2]
Example Find the exact distance between the points whose
coordinates are (5, -2) and (-4, 3).
Solution
( x1 , y1 ) = (5, −2) ( x2 , y2 ) = ( −4, 3) d =
2
( x2 − x1 ) + ( y2 − y1 )
2
=
( −4 − 5)2 + (3 − ( −2))2
=
( −9)2 + (5)2
=
81 + 25 =
A relation is a set of ordered pairs. [3.2.1] The domain of a relation is the set of first coordinates of the ordered pairs of a relation. The range of a relation is the set of second coordinates of the ordered pairs of the relation. [3.2.1]
Definition of Function A function is a relation in which no two ordered pairs have the same first coordinate and different second coordinates. [3.2.1] Example What are the domain and range of the following
relation? Is the relation a function?
{(–2, 6), (–1, 3), (0, 0), (1, –3), (2, –6), (3, –9)}
Solution Domain = {–2, –1, 0, 1, 2, 3}
Range = {–9, –6, –3, 0, 3, 6}
Because there are no two ordered pairs with the same first coordinate and different second coordinates, the relation is a function.
–5
–7
0
–4
5
4
−4
0
4
x
−4
–1
Graph an equation of the form Ax + By = C . [3.3.2] Example Graph: −3x + 2 y = 4 Solution
−3x + 2 y = 4 2 y = 3x + 4
y =
x
106
Definition of Relation
Step • Replace z by –4. Then simplify.
Step • Solve for y.
3 x+2 2
y
–2
–1
0
2
2
5
y 4
−4
0
4
x
−4
The point at which a graph crosses the x-axis is called the x-intercept, and the point at which a graph crosses the y-axis is called the y-intercept. [3.3.2] Example Find the x- and y-intercepts of 3x − y = 2. Solution x-intercept:
y-intercept:
3x − y = 2
3x − y = 2
3x − 0 = 2
3(0) − y = 2
3x = 2 2 x = 3
−y = 2
2 , 0 3
y = −2 (0, –2)
Slope Formula The slope of the line containing the two points P1 ( x1 , y1 ) and P2 ( x2 , y2 ) is given by y − y1 m = 2 , x1 ≠ x2 [3.4.1] x2 − x1
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example Find the slope of the line between the points (4, –3) and
rallel Lines
(2, 7).
Solution Let P1 = ( 4, −3) and
y2 − y1
al lines in the plane with slopes of m1 and m2 are parallel if and only if m1 = m2 . Any two vertical lines are parallel. [3.6.1]
2
7 − ( −3) 10 = = = −5 m = −2 x2 − x1 2−4
The slope is –5.
Example Find the equation of the line that contains the point
(4, 1) and is parallel to the graph of y =
Slope-Intercept Form of a Straight Line The equation y = mx + b is called the slope-intercept form of a straight line. The slope of the line is m, the coefficient is x. The y-intercept is (0, b). [3.4.2]
y − y1 = m( x − x1 )
Solution
y − 1 =
3 ( x − 4) 4
Example Graph 2 x + 3 y = 6 by using the slope and y-intercept. Solution Solve for y.
2x + 3y = 6
3 y = −2 x + 6 2 y = − x+2 3 2 The slope is − . 3
Place a point at (0, 2). Move to the right 3 (change in x) and down 2 (change in y) and place another point. Draw a line between the points.
The y-intercept is (0, 2).
y
• Use the Distributive Property.
• Add 1 to each side of the equation.
−4
0
Example Find the equation of the line that contains the point
Slopes of Perpendicular Lines If m1 and m2 are the slopes of two lines, neither of which is vertical, then the lines are perpendicular if and only if m1 ⋅ m2 = −1. A vertical line is perpendicular to a horizontal line. [3.6.1] Example Find the equation of the line that contains the point
(-2, 3) and is perpendicular to the graph of x − 4 y = 3.
Solution Solve for y.
x − 4y = 3
(4, -3) and has slope -3.
m = −3
( x1 , y1 ) = ( 4, −3)
y − y1 = m( x − x1 ) y − ( −3) = −3( x − 4)
y + 3 = −3x + 12 y = −3x + 9
The equation of the line is y = −3 x + 9.
−4 y = − x + 3
1 3 y = x− 4 4 1 m1 = 4 The slope of the perpendicular line is –4.
Find the equation of a line between two points. [3.5.2] Example Find the equation of the line that contains the points
(2, 0) and (5, 3).
Solution
( x1 , y1 ) = ( 2, 0) and ( x2 , y2 ) = (5, 3) y − y1 3−0 3 = = =1 m = 2 x2 − x1 5−2 3 y − y1 = m( x − x1 )
y − 0 = 1( x − 2)
y = x−2
The equation of the line is y = x − 2.
3 x − 2. 4
x
Let m be the slope of a line, and let ( x1 , y1 ) be the coordinates of a point on the line. The equation of the line can be found using the point-slope formula: y − y1 = m( x − x1 ) [3.5.1]
The equation of the line is y =
4
Point-Slope Formula
Steps • Use the point-slope formula. 3 • m = , the slope of 4 3 the line y = x − 1. 4 ( x 1, y 1) = (4, 1), the given point.
y − 1 =
−4
Solution
3 x−3 4 3 y = x−2 4
3 x − 1. 4
Use the slope and the point (–2, 3) to find the equation of the line. y − y1 = m( x − x1 ) y − 3 = −4[ x − ( −2)] y − 3 = −4( x + 2) y − 3 = −4 x − 8 y = −4 x − 5
The equation of the line is y = −4 x − 5.
An inequality of the form y > mx + b or Ax + By > C is a linear inequality in two variables. (The inequality symbol could be replaced by ≥, <, or ≤.) [3.7.1] Example Graph the solution set of x + 3 y > 6. Solution Solve the inequality for y.
x + 3y > 6
3y > − x + 6 1 y > − x+2 3
y 4
−4
0
4
x
−4
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 4
Important Concepts
Systems of Equations and Inequalities
A system of equations is two or more equations considered together. A solution of a system of equations in two variables is an ordered pair that is a solution of each equation of the system. [4.1.1] Solve by graphing. [4.1.1] Example Solve by graphing: x + y = 1
2x + y = 0
Solution Graph each line. y
The solution is (–1, 2).
4
−4
0
x
4
−4
Solution 2 x + y = 5
4x + 2 y = 6
Eliminate y.
−2( 2 x + y ) = −2(5) 4x + 2 y = 6
−4 x − 2 y = −10 4x + 2 y = 6
Add the equations.
0 x + 0 y = −4
0 = −4
0 = −4 is not a true equation. The system is inconsistent and therefore has no solution.
An algebraic method called the substitution method can be used to find an exact solution of a system of equations. [4.1.2]
Example Solve by substitution: 3x − y = 3
A system of three equations in three unknowns can be solved by the addition method. [4.2.2]
6 x + 3 y = −4
Example Solve: (1)
Solution (1) 3x − y = 3
x− y+ z = 6
(2) 6 x + 3 y = −4
(2) 2 x + 3 y − z = 1
Solve equation (1) for y.
(3) x + 2 y + 2 z = 5
3x − y = 3
Solution Eliminate z. Add equations (1) and (2).
− y = −3x + 3
x− y+ z = 6
y = 3x − 3
2x + 3y − z = 1
(4)
6 x + 3 y = −4
Eliminate z. Multiply equation (2) by 2 and add to equation (3).
6 x + 3(3x − 3) = −4
6 x + 9 x − 9 = −4
15x − 9 = −4
x + 2 y + 2z = 5 5x + 8 y = 7
Solve the system of two equations.
(4) 3x + 2 y = 7
(5) 5x + 8 y = 7
3x − y = 3
Multiply equation (4) by –4 and add to equation (5).
1 3 − y = 3 3
−12 x − 8 y = −28
x =
5 1 = 15 3
Substitute into equation (1).
1− y = 3
−y = 2
4x + 6 y − 2z = 2
(5)
15x = 5
3x + 2 y = 7
Substitute into equation (2).
y = −2
−7 x = −21
x = 3
Replace x by 3 in equation (4).
3x + 2 y = 7
The addition method is an alternative method for solving a system of equations. [4.2.1] Example Solve by the addition method: 2 x + y = 5
1 The solution is , − 2 . 3
5x + 8 y = 7
4x + 2 y = 6
3(3) + 2 y = 7
9 + 2 y = 7
2 y = −2
y = −1
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Replace x by 3 and y by –1 in equation (1).
5
x − y + z = 6 3 − ( −1) + z = 6
4+ z = 6
z = 2
The solution is (3, –1, 2).
Cramer’s Rule is a method of solving a system of equations using determinants. [4.3.2] Example Solve by using Cramer’s Rule: 6 x − 6 y = 5
2 x − 10 y = −1
6 −6 = −48 2 −10
Solution
D =
5 −6 Dx = = −56 −1 −10
6 5 Dy = = −16 2 −1
−56 7 x = = = D −48 6
7 1 The solution is , . 6 3
−16 1 y = = = D −48 3 Dy
3. Replace a row by the sum of that row and a multiple of any other row. [4.3.3] These operations are used to write a matrix in row echelon form. The Gaussian elimination method is a procedure to solve a system of equations using matrices and row echelon form. [4.3.3] Example Solve by using the Gaussian elimination method:
Solution
4 −5 17 3 2 7
Change a11 to 1. 1 R 4 1
1 − 5 17 4 4 2 7 3
5 17 5 17 Change a to 0. 1 − 4 21 4 1 − 4 4 −3R1 + R2 23 23 0 − 2 7 3 4 4
• We multiplied R1 by the opposite of a21.
5 17 Change a22 to 1. 1 − 4 4 4 0 23 − 23 R 2 4 4 23
Example A rowing team rowing with the current traveled 18 mi
in 2 h. Against the current, the team rowed 10 mi in 2 h. Find the rate of the current. c Rate of the rowing team in calm water: t
Rate of the current: c Rate
Time
Distance
With current
t+c
2
2(t + c )
Against current
t-c
2
2(t − c )
c The distance traveled with the current is 18 mi.
The distance traveled against the current is 10 mi.
2(t − c ) = 10 → t − c = 5
2t = 14
t = 7
t + c = 9
7+c = 9 c = 2
The rate of the rowing team in calm water is 7 mph.
The rate of the current is 2 mph.
Two or more inequalities considered together are called a system of inequalities. The solution set of a system of inequalities is the intersection of the solution sets of the individual inequalities. [4.5.1] Example Graph the solution set: y ≥ 2 x − 3
Steps • We multiplied R1 by the reciprocal of a11.
The solution is (3, –1).
Rate-of-wind and water-current problems [4.4.1]
2. Multiply all the elements in a row by the same nonzero number.
3x + 2 y = 7
5 17 ( −1) = 4 4 5 17 x+ = 4 4 x = 3
Solution 2(t + c ) = 18 → t + c = 9
1. Interchange two rows.
Elementary Row Operations
4 x − 5 y = 17
Strategy
Dx
x−
17 4 y = −1 y =
y > −3x
Solution Shade above the solid line y = 2 x − 3.
Shade above the dashed line y = −3x.
The solution set of the system is the intersection of the solution sets of the individual inequalities. y
4
−4
0
4
x
5 17 1 − 4 4 1 −1 0
• This is row echelon form.
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 5
Important Concepts
Polynomials and Exponents
Example Subtract: (3x 2 − 2 x + 4) − (7 x 2 + 3x − 12)
Rules of Exponents
Solution Rewrite the subtraction as the addition of the opposite.
If m, n, and p are integers, then xm
⋅ xn
= xm+ n
( x m ) n = x mn p
2 −1 x 2 y −2 a −1b4 A. ( 2 x y )(5x y ) B. −2 −2 C. a b 4 x −2 y −5
4
−2
−3
= 10 x −1 y −2 =
a −1b4 B. −2 −2 = a −1−( −2 ) b4 −( −2 ) = ab 6 a b
2 −1 x 2 y −2 C. −2 −5 4x y
−2
x4 y3 = 8
−7 x 2 − 3x + 12
−2
To multiply a polynomial by a monomial, use the Distributive Property and the Rule for Multiplying Exponential Expressions. [5.3.1] Example Multiply: −4 y ( y 2 − 3 y + 2) Solution
Solution A. ( 2 x −5 y )(5x 4 y −3 ) = 10 x −5+ 4 y1+ ( −3)
−4 x 2 − 5 x + 16
x 0 = 1, x ≠ 0 [5.1.1 and 5.1.2]
Example Simplify.
−5
(3x 2 − 2 x + 4) + ( −7 x 2 − 3x + 12) 3x 2 − 2 x + 4
( x m y n ) p = x mp y np
xm x mp = , y ≠ 0 n y np y
xm = xm− n , x ≠ 0 xn 1 x−n = n , x ≠ 0 x
10 xy 2
= −4 y ( y 2 ) − ( −4 y )(3 y ) + ( −4 y )( 2)
= –4 y 3 + 12 y 2 – 8 y
A convenient method of multiplying two polynomials is to use a vertical format similar to that used for multiplication of whole numbers. [5.3.2] Example Multiply: ( −2b2 + 5b − 4)( −3b + 2)
82 x −8 y −6 = 8 6 −2 8 x y 64 = 8 6 x y =
−2b2 +
Solution
Example Write 942,000,000 in scientific notation.
3
6b − 15b2 + 12b 6b 3 − 19b 2 + 22b − 8 The product of two binomials can be found by using a method called FOIL. [5.3.2]
942, 000, 000 = 9.42 × 108
Example Write 2.7 × 10 −5 in decimal notation. Solution
5b − 4 −3b + 2
−4b2 + 10b − 8
A number written in the form a × 10 n , where 1 ≤ a < 10 and n is an integer, is in scientific notation. [5.1.3] Solution
−4 y ( y 2 − 3 y + 2)
Example Multiply: (5a – 3b)(2a + 7b)
2.7 × 10 –5 = 0.000027
Solution
(5a − 3b)( 2 a + 7 b)
A polynomial function is an expression whose terms are monomials. To evaluate a polynomial function, replace the variable by its value and simplify. [5.2.1]
= 10 a 2 + 35ab − 6 ab − 21b2
= 10a 2 + 29ab − 21b 2
Example Given P( x ) = 2 x 3 − 4 x − 5, evaluate P( −2).
Product of the Sum and Difference of Two Terms and the Square of a Binomial [5.3.3]
Solution P( x ) = 2 x 3 − 4 x − 5
Example Multiply: A. (3x − 7 )(3x + 7 ) B. (3x − 4 y ) 2
P( −2) = 2( −2)3 − 4( −2) − 5
= 2( −8) − 4( −2) − 5
= −16 + 8 − 5 = −13
Solution A. (3x − 7 )(3x + 7 ) = 9 x 2 − 49
To add or subtract polynomials, add or subtract the coefficients of the like terms. [5.2.2] 2
2
2
Example Add: (5x + 3x − 1) + ( 2 x + 4 x − 6) + (8 − 7 x + x )
5x 2 + 3x − 1
Solution
2x2 + 4x − 6 x2 − 7x + 8 8x2
+ 1
B. (3x − 4 y )2 = 9 x 2 − 24 xy + 16 y 2
To divide a polynomial by a monomial, divide each term of the numerator by the denominator. [5.4.1] Example Divide:
12 x 2 y 2 − 16 xy 2 − 8 x 4x2 y
12 x 2 y 2 − 16 xy 2 − 8 x 12 x 2 y 2 16 xy 2 8x = − − 2 4x2 y 4x y 4x2 y 4x2 y 4y 2 = 3y − − x xy Solution
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Divide polynomials [5.4.2] 3x 3 + 8 x 2 − x 3x − 1
Example Divide: Solution
Trinomials of the form ax 2 + bx + c can be factored by using trial uping. Factoring by grouping is shown here. See thods.
)
Example Factor: 6 x 2 + 7 x − 20
x 2 + 3x − 1
Solution Find a ⋅ c = 6( −20) = −120. Find two factors of –120
3x − 1 3x 3 + 8 x 2 − 6 x + 2 3
3x −
x
whose sum is 7. The factors are –8 and 15.
2
9 x 2 − 3x − 3x + 2 − 3x + 1 1 1 3x 3 + 8 x 2 − 6 x + 2 2 = x + 3x − 1 + 3x − 1 3x − 1
= 6 x 2 + 15x − 8 x − 20
= (6 x 2 + 15x ) − (8 x + 20)
= 3x ( 2 x + 5) − 4( 2 x + 5)
= (2 x + 5)(3 x − 4)
Factors of the Difference of Two Perfect Squares
Synthetic division is a shorter method of dividing a polynomial by a binomial of the form x - a. [5.4.3]
Solution
−2 6
8 −5 −12 8 6 −4 3
4 x 2 − 9 = ( 2 x )2 − 32 = (2 x − 3)(2 x + 3)
Factors of a Perfect-Square Trinomial
3 (6 x 2 + 8 x − 5) ÷ ( x + 2) = 6 x − 4 + x+2
a 2 − b2 = ( a + b)( a − b) [5.6.1]
Example Factor: 4 x 2 − 9
Example Divide: (6 x 2 + 8 x − 5) ÷ ( x + 2) Solution
6 x 2 + 7 x − 20
9x2 − 6x
a 2 + 2 ab + b2 = ( a + b)2 a 2 − 2 ab + b2 = ( a − b)2 [5.6.1]
Remainder Theorem
Example Factor: 9 x 2 + 12 x + 4
If the polynomial P( x ) is divided by x − a, the remainder is P( a ). [5.4.4]
Solution
Factors of the Sum or Difference of Two Perfect Cubes
Example Use the Remainder Theorem to evaluate P(3) when 3
P( x ) = 2 x − 4 x − 5. 3 2 0 −4 −5 6 18 42 2 6 14 37
Solution
a 3 + b3 = ( a + b)( a 2 − ab + b2 )
Example Factor: 8 x 3 + y 3 z 3
By the Remainder Theorem, P(3) = 37.
Example Factor the greatest common factor (GCF) from
3x 3 y − 6 x 2 y 2 − 3xy 3.
Solution The GCF of 3x 3 y, 6 x 2 y 2 , and 3xy 3 is 3xy.
3
2
2
a 3 − b3 = ( a − b)( a 2 + ab + b2 ) [5.6.2]
To factor a polynomial means to write the polynomial as a product of other polynomials. [5.5.1 and 5.5.2]
9 x 2 + 12 x + 4 = (3 x + 2)2
3
2
Solution 8 x 3 + y 3 z 3 = ( 2 x )3 + ( yz )3
= (2 x + yz )(4 x 2 − 2 xyz + y 2 z 2 )
A trinomial is quadratic in form if it can be written as au 2 + bu + c. [5.6.3] Example Factor: 3x 4 + 4 x 2 − 4
2
3x y − 6 x y − 3xy = 3 xy ( x − 2 xy − y )
Solution
3x 4 + 4 x 2 − 4 = ( x 2 + 2)(3 x 2 − 2)
To factor a quadratic trinomial means to express the trinomial as the product of two binomials. [5.5.3]
Principle of Zero Products If ab = 0, then a = 0 or b = 0. This principle is used to solve some equations. [5.7.1]
Example Factor: A. x 2 + 13x + 42 B. x 2 − x − 20
Example Solve: ( x − 2)( x + 5) = 8
Solution A. Find two factors of 42 whose sum is 13.
Solution
The factors are 6 and 7.
2
x + 13x + 42 = ( x + 6)( x + 7)
B. Find two factors of –20 whose sum is –1. The factors are –5 and 4.
x 2 − x − 20 = ( x – 5)( x + 4)
( x − 2)( x + 5) = 8
x 2 + 3x − 10 = 8 x 2 + 3x − 18 = 0
( x + 6)( x − 3) = 0
x +6 = 0
x = −6
x−3= 0 x = 3
The solutions are –6 and 3.
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 6
Important Concepts
Rational Expressions
The domain of a rational function must exclude those numbers for which the value of the polynomial in the denominator is zero. [6.1.1] 5− x Example Find the domain of g ( x ) = 2 . x −4 Solution The domain excludes values for which the denominator is zero. x2 − 4 = 0
( x − 2)( x + 2) = 0
x − 2 = 0
x = 2
The domain is { x | x ≠ −2, 2}.
=
( 2 x − 1)(3x − 2)( x + 1)(5x − 4) ( x + 1)(3x − 2)( 2 x − 1)( 2 x − 3) 1
=
1
=
( x + 1) (3x − 2) ( 2 x − 1) ( 2 x − 3) 1
x = −2
5x − 4 2x − 3
Example Subtract:
A rational expression is in simplest form when the numerator and denominator have no common factors other than 1. [6.1.2] 2
x − 16 x + 11x + 28 x 2 − 16 ( x + 4)( x − 4) Solution 2 = ( x + 4)( x + 7 ) x + 11x + 28
2x x −1 − x+2 x−3
Solution The LCM of the denominators is ( x + 2)( x − 3).
2
2x x −1 2x − = x+2 x−3 x+2
x−4 = x+7
Solution
2 x 2 − 13x + 20 x 2 − 16
2x2 + 9x + 4 6x2 − 7 x − 5
⋅
6x − 7 x − 5
=
x2 − 7x + 2 ( x + 2)( x − 3)
=
=
( 2 x − 5)( x − 4)( 2 x + 1)( x + 4) ( x − 4)( x + 4)(3x − 5)( 2 x + 1)
=
1
( x − 4) ( x + 4) (3x − 5) ( 2 x + 1) 1
Solution
16 + x 5 6+ − x
16 16 16 3+ + 2 2 x x2 = x ⋅x 4 5 4 x2 6+ − 2 2 x x x 16 3 ⋅ x2 + ⋅ x 2 + 162 ⋅ x 2 x x = 5 2 4 2 6 ⋅ x + ⋅ x − 2 ⋅ x2 x x 3x 2 + 16 x + 16 = 6 x 2 + 5x − 4
3+
=
(3x + 4)( x + 4) ( 2 x − 1)(3x + 4) 1
To divide two rational expressions, multiply by the reciprocal of the divisor. [6.2.1] 6 x 2 − 7 x + 2 4 x 2 − 8x + 3 Example Divide: ÷ 3x 2 + x − 2 5x 2 + x − 4 2
1
2x − 5 3x − 5
16 x2 4 x2
Solution The LCM of x and x2 is x2.
1
( 2 x − 5) ( x − 4) ( 2 x + 1) ( x + 4) 1
16 + x Example Simplify: 5 6+ − x 3+
( 2 x − 5)( x − 4) ( 2 x + 1)( x + 4) = ⋅ ( x − 4)( x + 4) (3x − 5)( 2 x + 1)
x+2 x+2
( 2 x 2 − 6 x ) − ( x 2 + x − 2) ( x + 2)( x − 3)
2
⋅ 2x 2 + 9x + 4
1
⋅
A complex fraction is a fraction whose numerator or denominator contains one or more fractions. [6.3.1]
The product of two rational expressions can often be simplified by factoring the numerator and the denominator. [6.2.1] 2 x 2 − 13x + 20 x 2 − 16
x −3 x −1 − x−3 x−3
=
( x + 4) ( x + 7 ) 1
Example Multiply:
⋅
1
( x + 4) ( x − 4)
=
1
To add or subtract rational expressions, write each rational expression in terms of the LCM of the denominators of the rational expressions. [6.2.2]
x+2 = 0
1
( 2 x − 1) (3x − 2) ( x + 1) (5x − 4) 1
Example Simplify:
2
6 x − 7 x + 2 4 x − 8x + 3 ÷ 3x 2 + x − 2 5x 2 + x − 4
=
6x2 − 7 x + 2 3x 2 + x − 2
=
( 2 x − 1)(3x − 2) ( x + 1)(5x − 4) ⋅ ( x + 1)(3x − 2) ( 2 x − 1)( 2 x − 3)
⋅
=
(3x + 4) ( x + 4) ( 2 x − 1) (3x + 4)
=
x+4 2x − 1
1
To solve an equation containing fractions, clear denominators by multiplying each side of the equation by the LCM of the denominators. Then solve for the variable. [6.4.1]
5x 2 + x − 4 4 x 2 − 8x + 3
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
4x + 1 3 = 2+ 2x − 1 x−3 4x + 1 3 Solution = 2+ 2x − 1 x−3 4x + 1 3 ( 2 x − 1)( x − 3) = ( 2 x − 1)( x − 3) 2 + 2x − 1 x − 3 Example Solve:
Pillaged by LibraryPirateWith wind
( x − 3)( 4 x + 1) = ( 2 x − 1)( x − 3)2 + ( 2 x − 1)3 4 x 2 − 11x − 3 = 4 x 2 − 14 x + 6 + 6 x − 3 −11x − 3 = −8 x + 3 −3x = 6
x = −2
The solution is –2.
Distance
Rate
Time
700
150 + r
700 150 + r
500
150 - r
500 150 − r
Against wind
c The time flying with the wind equals the time flying against the wind. 700 500 Solution = 150 + r 150 − r 700 (150 + r )(150 − r ) = 150 + r 500 (150 + r )(150 − r ) 150 − r
Rate of work is that part of a task that is completed in one unit of time. The basic equation that is used to solve work problems is
(150 − r )700 = (150 + r )500
105, 000 − 700r = 75, 000 + 500r
Rate of work × Time worked = Part of task completed [6.4.2]
30, 000 = 1200r
Example Two water pipes can fill a tank with water in 6 h. The
25 = r
larger pipe working alone can fill the tank in 9 h. How long would it take the smaller pipe, working alone, to fill the tank?
Strategy c Time required for the small pipe to fill the tank: x
Rate Large pipe Small pipe
1 9 1 x
Time 6 6
6 9 6 x
6 6 + =1 9 x 2 6 + =1 3 x 2 6 3x + = 3x ⋅ 1 3 x Solution
2 x + 18 = 3x 18 = x
The small pipe working alone will fill the tank in 18 h.
Uniform motion means that the speed of an object does not change. The basic equation used to solve uniform motion problems is distance = rate × time. An alternative form of this equation is distance = time. This form of the equation is used to solve some rate uniform motion problems. [6.4.3] Example A plane can fly at a rate of 150 mph in calm air.
Traveling with the wind, the plane flew 700 mi in the same amount of time it took to fly 500 mi against the wind. Find the rate of the wind.
Strategy c Rate of the wind: r
The rate of the wind is 25 mph.
Variation Problems [6.5.2] Example The strength (s) of a rectangular beam varies jointly
as its width (W) and as the square of its depth (d) and inversely as its length (L). If the strength of a beam 2 in. wide, 12 in. deep, and 12 ft long is 1200 lb, find the strength of a beam 4 in. wide, 8 in. deep, and 16 ft long.
Part
c The sum of the part of the task completed by the large pipe and the part of the task completed by the small pipe is 1.
Strategy To find the strength of the beam:
c Write the basic combined variation equation, replace the variables by the given values, and solve for k.
c Write the basic combined variation equation, replacing k by its value and substituting 4 for W, 8 for d, and 16 for L. Solve for s.
kWd 2 L k( 2)(12)2 1200 = 12 k( 288) 1200 = 12 14, 400 = 288k Solution
s =
50 = k
50Wd 2 L 50( 4)82 s = 16 12, 800 s = 16 s = 800 s =
The strength of the beam is 800 lb.
A literal equation is an equation that contains more than one variable. [6.6.1] r Example Solve t = for r. r +1 r t = Solution r +1 r ( r + 1)t = ( r + 1) r + 1 tr + t = r
t = r − tr
t = r (1 − t ) t = r 1−t
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 7
Important Concepts
Rational Exponents and Radicals
If m and n are positive integers and a1/ n is a real number, then a m / n = ( a1 / n ) m . [7.1.1] 2
Example Simplify. A. 64 3 B. 16 2 3
−
3 4
Example Multiply:
2 6 3
Solution
Solution A. 64 = ( 2 ) = 24 = 16
B. 16
3 4
−
= ( 24 )
−
3 4
= 2 −3 =
1
x2 y
Example Simplify. A.
−
−
4
3 4
1 1 = 8 23
B. ( x y z ) 3 4
1
1 2
−
2 3
−
4 3
x 3 y3 Solution A.
1 2
x y
−
1 3
3 4
1 2 − 2 3
x y
y
B. x y z
x
=
4 3
−
(
3 4
)
−
1 4 + 2 3
=
1 3 + 3 4
4 3
x y
2 3
8
8 9
z9 2
If a is a real number, then a
= a
1 m⋅ n
= (a ) =
n
a . [7.1.2] m
expression.
5
3
A. ( 2 x 3 ) 4 B. −5a 6 3 3 4
Solution A. ( 2 x ) =
4
3 3
(2 x ) =
4
5 6
8x
1 5 6
Solution
3
3
−8 x12 y 3
−8 x12 y 3 = –2 x 4 y
Product Property of Radicals If a and b are positive real numbers, then
n
a
⋅nb
=
n
ab . [7.2.1] 5
Example Simplify: 5
7
128 x 7 5
=
32 x ( 4 x )
=
5
32 x 5 5 4 x 2
= 2x 5 4x2
Solution
128 x
Solution
5
5
8
3xy 81x y − 192 x y 3
3
3 3
2
3
8
192 x y
4
( (3 5 − 6) ( 2
)( 5 + 3)
)
= 6 ⋅ 25 + 9 5 − 12 5 − 18
= 30 − 3 5 − 18 = 12 − 3 5
n
a
a = n . The procedure used to remove a radical from a b b denominator is called rationalizing the denominator. [7.2.4] y Example Simplify: 3y n
y 3y
y
=
3y
3y
⋅
3y
=
3x y −
3 3
= 3xy ⋅ 3x 3x y − 4 x y 3x y
= 9 x 2 y 3 3x 2 y − 4 x 2 y 3 3x 2 y
= 5x2 y 3 3x2 y
3 2
64 x y 3
2
2
3x y
y 3y 9y
=
2
y 3y 3y
3y 3
=
The expressions a + b and a – b, which are the sum and difference of two terms, are called conjugates of each other. To simplify a fraction that has a square-root radical expression with two terms in the denominator, multiply the numerator and denominator by the conjugate of the denominator. Then simplify. [7.2.4] Example Simplify:
4+
2
3−
3
4+
2
3−
3
= =
=
4+
2
3−
3
⋅3+
3
3+
3
12 + 4 3 + 3 2 +
( 3)
6
2
12 + 4 3 + 3 2 + 6
6
Graph a radical function [7.3.2]
Solution 6
= 3xy 27 x
2
= b 15 − 5 2b
Example Graph: F ( x ) =
4
3
)
25 2b
9−
Example Subtract: 3xy 81x y − 3
b2 15 −
The Distributive Property is used to simplify the sum or difference of radical expressions that have the same radicand and the same index. [7.2.2] 3
=
Solution
2
5
A variable expression is a perfect power if the exponents on the factors are evenly divisible by the index of the radical. [7.1.3] Example Simplify:
)
15b2 − 50b
Solution
9
B. −5a = −5( a ) = −5 6 a 5
3b − 10
=
then
1 m n
Example Rewrite the exponential expression as a radical
3b − 10
Quotient Property of Radical If a and b are positive real numbers,
xy 3 m n
(
(
Solution
13 12
= x y z =
1 n
5b
5b
Example Multiply: 3 5 − 6 2 5 + 3
11 6
−
−1
The Product Property of Radicals is used to multiply radical expressions with the same index. [7.2.3]
x
y
2 3 6
0 1 2
11
3
x−2
y 8 4 –4 0 –4
4
8
12
16
x
–8
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1
Property of Raising Each Si of an Equation to a Power
=
If two numbers are equal, then the same powers of the numbers are equal. If a = b, then a n = bn . [7.4.1]
When you raise both sides of an equation to an even power, the resulting equation may have a solution that is not a solution of the original equation. Therefore, it is necessary to check the solution of a radical equation. [7.4.1] Example Solve:
4 x + 5 − 12 = −5
(
)
2
= 72
4 x + 5 = 49
4 x = 44
x = 11
Example Solve: x + 3 x + 2 = 8 Solution
(
)
2
9( x + 2) = 64 − 16 x + x 2
2
9 x + 18 = 64 − 16 x + x 2
0 = x − 25x + 46
0 = ( x − 2)( x − 23)
x − 2 = 0
x − 23 = 0
x = 2
negative a is the imaginary number i a. − a = i a [7.5.1] Example Simplify:
−45
−45 = i 45 = i 9 ⋅ 5 = 3i 5
To add two complex numbers, add the real parts and add the imaginary parts. To subtract two complex numbers, subtract the real parts and subtract the imaginary parts. [7.5.2]
( −4 + 2i ) − (6 − 8i ) = −10 + 10i
Multiply complex numbers [7.5.3]
= (8 − x )2
If a is a positive real number, then the principal square root of
Solution
3 x+2 = 8− x
3 x + 2
The diagonal is 6.7 cm.
Example Subtract: ( −4 + 2i ) − (6 − 8i )
x+3 x+2 = 8
c ≈ 6.7
Addition and Subtraction of Complex Numbers
11 checks as the solution. The solution is 11.
45
A complex number is a number of the form a + bi, where a and b are real numbers and i = −1. The number a is the real part of a + bi, and b is the imaginary part.
4x + 5 = 7 4x + 5
Solution
4 x + 5 − 12 = −5
Solution
1
= ( 45) 2
x = 23
23 does not check as a solution. 2 checks as a solution. The solution is 2.
Example Multiply: (4 – 3i)(2 – i) Solution
( 4 − 3i )( 2 − i ) = 8 − 4i − 6i + 3i 2
= 8 − 10i + 3i 2 = 8 − 10i + 3( −1)
= 5 − 10i
Divide complex numbers [7.5.4] 2 − 3i Example Simplify: 4i Solution
2 − 3i 2 − 3i = 4i 4i
⋅i i
=
2i − 3i 2 2i − 3( −1) = 2 4( −1) 4i
3 + 2i 3 1 = = − − i −4 4 2
Pythagorean Theorem
The square of the hypotenuse, c, of a right triangle is equal to the sum of the squares of the two legs, a and b.
The conjugate of the complex number a + bi is a – bi. To divide complex numbers, multiply the numerator and denominator by the conjugate of the denominator. [7.5.4]
c 2 = a 2 + b2 [7.4.2] Example Find the diagonal of a rectangle that is 6 cm in length
and 3 cm in width. Round to the nearest tenth.
Strategy To find the diagonal, use the Pythagorean Theorem. One
leg is the length of the rectangle. The second leg is the width of the rectangle. The hypotenuse is the diagonal of the rectangle.
Solution
c 2 = a 2 + b2
c 2 = (6)2 + (3)2
c 2 = 36 + 9
c 2 = 45
Example Simplify: Solution
2 + 5i 3 − 2i
2 + 5i ( 2 + 5i ) (3 + 2i ) ⋅ = 3 − 2i (3 − 2i ) (3 + 2i )
=
6 + 4i + 15i + 10i 2 6 + 19i + 10( −1) = 13 32 + 22
=
−4 + 19i 4 19 = − + i 13 13 13
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 8
Important Concepts
Quadratic Equations and Inequalities
A quadratic equation is an equation of the form ax 2 + bx + c = 0, where a ≠ 0. The Principle of Zero Products states that if the product of two factors is zero, then at least one of the factors must be zero. This principle is used to solve a quadratic equation by factoring. [8.1.1/8.1.2]
Complete the square. 1 1 5 1 x2 + x + = + 2 16 2 16 1 x + 4
2
1 x + 4
2
Example Solve: 2 x 2 − 3x − 2 = 0 Solution 2 x 2 − 3x − 2 = 0
( 2 x + 1)( x − 2) = 0
Steps • Factor.
2x + 1 = 0 x − 2 = 0 • Principle of Zero Products. 1 x = − x = 2 2 1 The solutions are − and 2. 2 Example Write a quadratic equation that has integer coefficients 2 1 and has solutions − and . 3 6 Solution ( x − r1 )( x − r2 ) = 0
2 1 x − − x − = 0 3 6
2 1 x + 3 x − 6 = 0
3 2 x + x− = 0 6 18
18 x 2 + 9 x − 2 = 0
Example Solve by taking square roots: 2( x + 1)2 + 24 = 0 2
2( x + 1) + 24 = 0 2( x + 1)2 = −24
( x + 1)2 = −12
( x + 1)2 =
−12
x + 1 = 2i 3
x + 1 = ±2i 3
x + 1 = 2i 3
x = −1 + 2i 3
1 = 4
x+
1 41 = ± 4 4
x+
1 = 4
41 4
x = −
1 + 4
The solutions are
x+ 41 4
x = −
x + 1 = −2i 3 x = −1 − 2i 3
2
Example Solve by completing the square: 2 x + x − 5 = 0
x2 + 6x − 9 = 0
a = 1, b = 6, c = −9
x =
= =
−b ±
b2 − 4 ac 2a
−6 ±
62 − 4(1)( −9) 2 ⋅1
−6 ±
36 + 36 2
−6 ± 6 2 = −3 ± 3 2 2 The solutions are −3 + 3 2 and −3 − 3 2.
An equation is quadratic in form if it can be written as au 2 + bu + c = 0. [8.3.1] 1
Example Solve: x − 5x 2 + 6 = 0 1
x − 5x 2 + 6 = 0
Solution
(x ) 1 2
2
( ) 1
− 5 x2 + 6 = 0
u 2 − 5u + 6 = 0
2x2 + x = 5
(u − 2)(u − 3) = 0
2x + x 5 = 2 2 1 5 2 x + x = 2 2
41 4
−6 ± 72 2
2x2 + x − 5 = 0 2
1 − 4
−1 + 41 −1 − 41 and . 4 4
Solution
The solutions are −1 + 2i 3 and −1 − 2i 3.
1 41 = − 4 4
=
Solve by completing the square. [8.2.1] Solution
41 4
x+
=
41 16
Example Solve by using the quadratic formula: x 2 + 6 x − 9 = 0
2
=
Solve by using the quadratic formula. [8.2.2]
An equation containing the square of a binomial can be solved by taking square roots. [8.1.3] Solution
41 16
3 2 18 x 2 + x − = 0 6 18
=
u − 2 = 0
u−3= 0
u = 2
u = 3
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1
dinate: −
Replace u with x 2 .
x
1 2
(x ) 1 2
x
(x )
2
1 2
= 22
x = 4
= 2
1 2
x2 − 2
3 = 32
x = 9
Certain equations containing a radical can be solved by first solving the equation for the radical expression and then squaring each side of the equation. [8.3.2] 2y + 1 −
2y + 1 −
Solution
(
y +1
) ( 2
=
)
y +1
y = 2 y
y2 = 2 y
y2 = 4 y y2 − 4 y = 0
y ( y − 4) = 0 y = 0
2
The vertex is (0, –2).
The equation of the axis of symmetry is x = 0.
Example Find the x-intercepts of the parabola whose equation is Solution
y = 2 x 2 − 5x + 2
0 = 2 x 2 − 5x + 2
0 = ( 2 x − 1)( x − 2)
2x − 1 = 0
x−2 = 0
2x = 1 1 x = 2
x = 2
1 The x-intercepts are , 0 and (2, 0). 2
When a > 0, the graph of y = ax 2 + bx + c opens up and the y-coordinate of the vertex is the minimum value of the function. When a < 0, the graph of y = ax 2 + bx + c opens down and the y-coordinate of the vertex is the maximum value of the function. [8.7.1]
y = 4
Example Find the maximum or minimum value of
The solutions are 0 and 4.
f ( x ) = 2 x 2 − 3x + 1.
A quadratic inequality in one variable is one that can be written in the form ax 2 + bx + c > 0, where a ≠ 0. The symbols <, ≤, and ≥ can also be used. [8.5.1]
Solution Find the coordinates of the vertex.
Example Solve and graph the solution set of 2 x 2 − x − 10 ≤ 0.
f ( x ) = 2 x 2 − 3x + 1
Solution 2 x 2 − x − 10 ≤ 0
( 2 x − 5)( x + 2) ≤ 0
2x − 5 − − −
−−−−−−−−
+++
x + 2 −−−
++++++++
+++
0
1
2
b −3 3 = − = 2a 2( 2) 4 2
3 3 3 9 9 1 f = 2 − 3 + 1 = − + 1 = − 8 4 8 4 4 4
Applications of Maximum and Minimum [8.7.2]
5 x | −2 ≤ x ≤ 2 –5 –4 –3 –2 –1
x = −
Because a is positive, the function has a minimum value. The 1 minimum value of the function is − . 8
−3 −2 −1 0 1 2 3
x
The x-intercepts of a parabola are the points where the parabola crosses the x-axis. [8.6.2]
y−4 = 0
4
−4
( )
2
2y + 1 = y + 2 y + 1
0
y = 2 x 2 − 5x + 2.
y =1
2y + 1
−4
= −2
y =1
2y + 1 =
y 4
= 02 − 2
2
The solutions are 4 and 9.
Example Solve:
b 0 = − = 0 2a 2(1)
Example The height in feet (s) of a ball thrown straight up is
3
4
given by s(t ) = −16t 2 + 64t , where t is the time in seconds. Find the time it takes the ball to reach its maximum height. What is the maximum height?
5
Vertex and Axis of Symmetry of a Parabola Let f ( x ) = ax 2 + bx + c be the equation of a parabola. The b b coordinates of the vertex are − , f − . The equa2 a 2a tion of the axis of symmetry is x = −
b . [8.6.1] 2a
Example Find the vertex and axis of symmetry of the parabola
whose equation is y = x 2 − 2. Then sketch the graph.
Strategy c To find the time it takes the ball to reach its maximum
height, find the t-coordinate of the vertex.
c To find the maximum height, evaluate the function at the t-coordinate of the vertex. b 64 = − = 2 Solution t = − 2a 2( −16)
The ball reaches its maximum height in 2 s.
s(t ) = −16t 2 + 64t
s( 2) = −16( 2)2 + 64( 2) = −64 + 128 = 64
The maximum height is 64 ft.
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 9
Important Concepts
Functions and Relations
Vertical Translation
Operations on Functions
If f is a function and c is a positive constant, then
If f and g are functions and x is an element of the domain of each function, then
y = f ( x ) + c is the graph of y = f ( x ) shifted up c units.
y = f ( x ) − c is the graph of y = f ( x ) shifted down c units. [9.1.1]
Example Given the graph of
y
f ( x ) = x − 2 x shown at the right, graph g ( x ) = x 2 − 2 x − 3 by using a vertical translation.
2 −4 −2
( f − g )( x ) = f ( x ) − g ( x )
(f
2
x
f f ( x) g ( x ) = g ( x ) , g ( x ) ≠ 0 [9.2.1]
4
−4
is the graph of f ( x ) = x 2 − 2 x shifted down 3 units.
f
4
( f + g )( −2).
g
–4
Solution
x
4 –4
Horizontal Translation
= [( −2)2 + 2( −2)] + [5( −2) − 2]
= ( 4 − 4) + ( −10 − 2)
= −12
( f + g )( −2) = −12
Example Given f ( x ) = 4 − x 2 and g ( x ) = 3x − 4, find
y = f ( x − c ) is the graph of y = f ( x ) shifted to the right c units.
y = f ( x + c ) is the graph of y = f ( x ) shifted to the left c units. [9.1.1]
Solution The graph of g ( x ) =
(f
Solution
y
Example Given the graph of the function 3
f
2
0
x
2
x
6
4
= ( −5)(5)
= −25
(f
2
4
4 −4 −2 0 −4
2
4
y
3
4
−4
g
42 − 4 42 + 2 ⋅ 4 + 1
=
16 − 4 16 + 8 + 1
=
12 25
x
f 12 g (4) = 25
Definition of the Composition of Two Functions
V
4 −4
=
−8
Solution The graph of V ( x ) = ( x − 2)3 + 1
g
8
= −25
f f ( 4) g ( 4) = g ( 4)
x
6
y
Example Given the graph of the function
⋅ g )(3)
Example Given f ( x ) = x 2 − 4 and g ( x ) = x 2 + 2 x + 1, find
Solution
It is possible for a graph to involve both a horizontal and a vertical translation. [9.1.1]
is the graph of g ( x ) = x shifted up 1 unit and shifted 2 units to the right.
= ( 4 − 9) ⋅ (9 − 4)
f g ( 4).
g
1 0
g ( x ) = x 3 shown at the right, graph V ( x ) = ( x − 2)3 + 1 by using both a horizontal and a vertical translation.
f
2
f (3) ⋅ g (3)
= ( 4 − 32 ) ⋅ [3(3) − 4]
y 3
(f
⋅ g )(3). ⋅ g )(3) =
1
x−3
is the graph of f ( x ) = shifted 3 units to the right.
( f + g )( −2) = f ( −2) + g ( −2)
If f is a function and c is a positive constant, then
right, graph g ( x ) = x − 3 by using a horizontal translation.
= f ( x) ⋅ g( x)
Example Given f ( x ) = x 2 + 2 x and g ( x ) = 5x − 2, find
y
Solution The graph of g ( x ) = x − 2 x − 3
x shown at the
⋅ g )( x )
−2
2
f ( x) =
( f + g )( x ) = f ( x ) + g ( x )
f
4
2
x
Let f and g be two functions such that g ( x ) is in the domain of f for all x in the domain of g. Then the composition of the two functions, denoted by f g , is the function whose value at x is given by ( f g )( x ) = f [ g ( x )]. [9.2.2]
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Example Given g ( x ) = 3x − 2 and h( x ) = x 2 + 1, evaluate
each composite function.
The inverse of a function is the set of ordered pairs formed by rdinates of each ordered pair of the function.
A. g[ h(0)] B. h[ g
Solution A.
h( x ) = x 2 + 1
Condition for an Inverse Function
h(0) = 0 + 1 = 1
g ( x ) = 3x − 2
A function f has an inverse function if and only if f is a 1–1 function. [9.3.2]
g(1) = 3(1) − 2 = 1
Example Find the inverse of the function defined by the equation
f ( x ) = 4 x + 2.
g[ h(0)] = 1
Solution To find the inverse, replace f(x) by y, interchange x and
B. h[ g ( x )] = h(3x − 2)
y, and then solve for y.
= (3x − 2)2 + 1
= 9 x 2 − 12 x + 4 + 1
= 9 x 2 − 12 x + 5
f ( x) = 4x + 2
y = 4x + 2
x = 4y + 2
4y = x − 2 1 y = x− 4 1 f −1 ( x ) = x − 4
Horizontal Line Test A graph of a function is the graph of a 1–1 function if any horizontal line intersects the graph at no more than one point. [9.3.1] Example Determine whether or not the graph represents the graph
of a 1–1 function. A.
B.
y
1 2 1 2
Example Are the functions defined by the equations
h( x ) = 4 x + 2 and g ( x ) =
y
other?
1 1 x − inverses of each 4 2
Solution Find h[ g ( x )] and g[ h( x )]. If the result of each x
Solution A. Because any vertical line will intersect the graph at
no more than one point, and any horizontal line will intersect the graph at no more than one point, the graph is the graph of a 1–1 function.
calculation is x, the functions are inverses of one another.
x
B. Because any vertical line will intersect the graph at no more than one point, and any horizontal line will intersect the graph at no more than one point, the graph is the graph of a 1–1 function.
1 1 h[ g ( x )] = 4 x − + 2 2 4
g[ h( x )] =
= x−2+2 = x 1 1 ( 4 x + 2) − 4 2
= x+
1 1 − = x 2 2
The functions are inverses of each other.
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Intermediate Algebra with Applications, 7e
eBook Aufmann, Barker, and Lockwood’s eBook for Intermediate Algebra with Applica tions, Seventh Edition, contains the same content found in the print version, but in a format that offers increased flexibility, affordability, and functionality to students. The eBook is accessible to students who have accounts with Enhanced WebAssign.
Within Enhanced WebAssign, students can access individual sections of the eBook using the “Read It” buttons that accompany questions. This version of the eBook is static and available free of charge. Students can also purchase a searchable eBook, with links to videos and tutorials. This eBook can be accessed through the “My eBooks” link in the Student View of Enhanced WebAssign. For more information, visit
www.webassign.net/cengage or
www.cengage.com/math/aufmann
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
