Chapter 7 Resource Masters - KTL MATH CLASSES...©Glencoe/McGraw-Hill iv Glencoe Algebra 2 Teacher’s Guide to Using the Chapter 7 Resource Masters The Fast FileChapter Resource system

Chapter 7Resource Masters

Consumable WorkbooksMany of the worksheets contained in the Chapter Resource Masters bookletsare available as consumable workbooks.

Study Guide and Intervention Workbook 0-07-828029-XSkills Practice Workbook 0-07-828023-0Practice Workbook 0-07-828024-9

ANSWERS FOR WORKBOOKS The answers for Chapter 7 of these workbookscan be found in the back of this Chapter Resource Masters booklet.

Copyright © by The McGraw-Hill Companies, Inc. All rights reserved.Printed in the United States of America. Permission is granted to reproduce the material contained herein on the condition that such material be reproduced only for classroom use; be provided to students, teacher, and families without charge; and be used solely in conjunction with Glencoe’s Algebra 2. Any other reproduction, for use or sale, is prohibited without prior written permission of the publisher.

Send all inquiries to:The McGraw-Hill Companies8787 Orion PlaceColumbus, OH 43240-4027

ISBN: 0-07-828010-9 Algebra 2Chapter 7 Resource Masters

1 2 3 4 5 6 7 8 9 10 066 11 10 09 08 07 06 05 04 03 02

Glencoe/McGraw-Hill

© Glencoe/McGraw-Hill iii Glencoe Algebra 2

Contents

Vocabulary Builder . . . . . . . . . . . . . . . . vii

Lesson 7-1Study Guide and Intervention . . . . . . . . 375–376Skills Practice . . . . . . . . . . . . . . . . . . . . . . . 377Practice . . . . . . . . . . . . . . . . . . . . . . . . . . . 378Reading to Learn Mathematics . . . . . . . . . . 379Enrichment . . . . . . . . . . . . . . . . . . . . . . . . . 380




Lesson 7-5Study Guide and Intervention . . . . . . . 399–400Skills Practice . . . . . . . . . . . . . . . . . . . . . . . 401Practice . . . . . . . . . . . . . . . . . . . . . . . . . . . 402Reading to Learn Mathematics . . . . . . . . . . 403Enrichment . . . . . . . . . . . . . . . . . . . . . . . . . 404





Chapter 7 AssessmentChapter 7 Test, Form 1 . . . . . . . . . . . . 429–430Chapter 7 Test, Form 2A . . . . . . . . . . . 431–432Chapter 7 Test, Form 2B . . . . . . . . . . . 433–434Chapter 7 Test, Form 2C . . . . . . . . . . . 435–436Chapter 7 Test, Form 2D . . . . . . . . . . . 437–438Chapter 7 Test, Form 3 . . . . . . . . . . . . 439–440Chapter 7 Open-Ended Assessment . . . . . . 441Chapter 7 Vocabulary Test/Review . . . . . . . 442Chapter 7 Quizzes 1 & 2 . . . . . . . . . . . . . . . 443Chapter 7 Quizzes 3 & 4 . . . . . . . . . . . . . . . 444Chapter 7 Mid-Chapter Test . . . . . . . . . . . . 445Chapter 7 Cumulative Review . . . . . . . . . . . 446Chapter 7 Standardized Test Practice . . 447–448Unit 2 Test/Review (Ch. 5–7) . . . . . . . . 449–450First Semester Test (Ch. 1–7) . . . . . . . 451–452

Standardized Test Practice Student Recording Sheet . . . . . . . . . . . . . . A1

ANSWERS . . . . . . . . . . . . . . . . . . . . . . A2–A40

© Glencoe/McGraw-Hill iv Glencoe Algebra 2

Teacher’s Guide to Using theChapter 7 Resource Masters

The Fast File Chapter Resource system allows you to conveniently file the resourcesyou use most often. The Chapter 7 Resource Masters includes the core materials neededfor Chapter 7. These materials include worksheets, extensions, and assessment options.The answers for these pages appear at the back of this booklet.

All of the materials found in this booklet are included for viewing and printing in theAlgebra 2 TeacherWorks CD-ROM.

Vocabulary Builder Pages vii–viiiinclude a student study tool that presentsup to twenty of the key vocabulary termsfrom the chapter. Students are to recorddefinitions and/or examples for each term.You may suggest that students highlight orstar the terms with which they are notfamiliar.

WHEN TO USE Give these pages tostudents before beginning Lesson 7-1.Encourage them to add these pages to theirAlgebra 2 Study Notebook. Remind them to add definitions and examples as theycomplete each lesson.

Study Guide and InterventionEach lesson in Algebra 2 addresses twoobjectives. There is one Study Guide andIntervention master for each objective.

WHEN TO USE Use these masters asreteaching activities for students who needadditional reinforcement. These pages canalso be used in conjunction with the StudentEdition as an instructional tool for studentswho have been absent.

Skills Practice There is one master foreach lesson. These provide computationalpractice at a basic level.

WHEN TO USE These masters can be used with students who have weakermathematics backgrounds or needadditional reinforcement.

Practice There is one master for eachlesson. These problems more closely followthe structure of the Practice and Applysection of the Student Edition exercises.These exercises are of average difficulty.

WHEN TO USE These provide additionalpractice options or may be used ashomework for second day teaching of thelesson.

Reading to Learn MathematicsOne master is included for each lesson. Thefirst section of each master asks questionsabout the opening paragraph of the lessonin the Student Edition. Additionalquestions ask students to interpret thecontext of and relationships among termsin the lesson. Finally, students are asked tosummarize what they have learned usingvarious representation techniques.

WHEN TO USE This master can be usedas a study tool when presenting the lessonor as an informal reading assessment afterpresenting the lesson. It is also a helpfultool for ELL (English Language Learner)students.

Enrichment There is one extensionmaster for each lesson. These activities mayextend the concepts in the lesson, offer anhistorical or multicultural look at theconcepts, or widen students’ perspectives onthe mathematics they are learning. Theseare not written exclusively for honorsstudents, but are accessible for use with alllevels of students.

WHEN TO USE These may be used asextra credit, short-term projects, or asactivities for days when class periods areshortened.

© Glencoe/McGraw-Hill v Glencoe Algebra 2

Assessment OptionsThe assessment masters in the Chapter 7Resource Masters offer a wide range ofassessment tools for intermediate and finalassessment. The following lists describe eachassessment master and its intended use.

Chapter Assessment CHAPTER TESTS• Form 1 contains multiple-choice questions

and is intended for use with basic levelstudents.

• Forms 2A and 2B contain multiple-choicequestions aimed at the average levelstudent. These tests are similar in formatto offer comparable testing situations.

• Forms 2C and 2D are composed of free-response questions aimed at the averagelevel student. These tests are similar informat to offer comparable testingsituations. Grids with axes are providedfor questions assessing graphing skills.

• Form 3 is an advanced level test withfree-response questions. Grids withoutaxes are provided for questions assessinggraphing skills.

All of the above tests include a free-response Bonus question.

• The Open-Ended Assessment includesperformance assessment tasks that aresuitable for all students. A scoring rubricis included for evaluation guidelines.Sample answers are provided forassessment.

• A Vocabulary Test, suitable for allstudents, includes a list of the vocabularywords in the chapter and ten questionsassessing students’ knowledge of thoseterms. This can also be used in conjunc-tion with one of the chapter tests or as areview worksheet.

Intermediate Assessment• Four free-response quizzes are included

to offer assessment at appropriateintervals in the chapter.

• A Mid-Chapter Test provides an optionto assess the first half of the chapter. It iscomposed of both multiple-choice andfree-response questions.

Continuing Assessment• The Cumulative Review provides

students an opportunity to reinforce andretain skills as they proceed throughtheir study of Algebra 2. It can also beused as a test. This master includes free-response questions.

• The Standardized Test Practice offerscontinuing review of algebra concepts invarious formats, which may appear onthe standardized tests that they mayencounter. This practice includes multiple-choice, grid-in, and quantitative-comparison questions. Bubble-in andgrid-in answer sections are provided onthe master.

Answers• Page A1 is an answer sheet for the

Standardized Test Practice questionsthat appear in the Student Edition onpages 406–407. This improves students’familiarity with the answer formats theymay encounter in test taking.

• The answers for the lesson-by-lessonmasters are provided as reduced pageswith answers appearing in red.

• Full-size answer keys are provided forthe assessment masters in this booklet.

Reading to Learn MathematicsVocabulary Builder

NAME ______________________________________________ DATE ____________ PERIOD _____

77

© Glencoe/McGraw-Hill vii Glencoe Algebra 2

Voca

bula

ry B

uild

erThis is an alphabetical list of the key vocabulary terms you will learn in Chapter 7.As you study the chapter, complete each term’s definition or description. Rememberto add the page number where you found the term. Add these pages to your AlgebraStudy Notebook to review vocabulary at the end of the chapter.

Vocabulary Term Found on Page Definition/Description/Example

composition of functions

depressed polynomial

end behavior

Factor Theorem

Fundamental Theorem of Algebra

inverse function

inverse relation

leading coefficients

location principle

one-to-one

(continued on the next page)

© Glencoe/McGraw-Hill viii Glencoe Algebra 2

Vocabulary Term Found on Page Definition/Description/Example

polynomial function

polynomial in one variable

power function

quadratic form

Rational Zero Theorem

relative maximum

relative minimum

remainder theorem

square root function

synthetic substitution

sihn·THEH·tihk

Reading to Learn MathematicsVocabulary Builder (continued)

NAME ______________________________________________ DATE ____________ PERIOD _____

77

Study Guide and InterventionPolynomial Functions

NAME ______________________________________________ DATE ____________ PERIOD _____

7-17-1

© Glencoe/McGraw-Hill 375 Glencoe Algebra 2

Less

on

7-1

Polynomial Functions

A polynomial of degree n in one variable x is an expression of the form

Polynomial in a0xn � a1xn � 1 � … � an � 2x2 � an � 1x � an,One Variable where the coefficients a0, a1, a2, …, an represent real numbers, a0 is not zero,

and n represents a nonnegative integer.

The degree of a polynomial in one variable is the greatest exponent of its variable. Theleading coefficient is the coefficient of the term with the highest degree.

A polynomial function of degree n can be described by an equation of the form

Polynomial P(x ) � a0xn � a1xn � 1 � … � an � 2x2 � an � 1x � an,Function where the coefficients a0, a1, a2, …, an represent real numbers, a0 is not zero,

and n represents a nonnegative integer.

What are the degree and leading coefficient of 3x2 � 2x4 � 7 � x3?Rewrite the expression so the powers of x are in decreasing order.�2x4 � x3 � 3x2 � 7This is a polynomial in one variable. The degree is 4, and the leading coefficient is �2.

Find f(�5) if f(x) � x3 � 2x2 � 10x � 20.f(x) � x3 � 2x2 � 10x � 20 Original function

f(�5) � (�5)3 � 2(�5)2 � 10(�5) � 20 Replace x with �5.

� �125 � 50 � 50 � 20 Evaluate.

� �5 Simplify.

Find g(a2 � 1) if g(x) � x2 � 3x � 4.g(x) � x2 � 3x � 4 Original function

g(a2 � 1) � (a2 � 1)2 � 3(a2 � 1) � 4 Replace x with a2 � 1.

� a4 � 2a2 � 1 � 3a2 � 3 � 4 Evaluate.

� a4 � a2 � 6 Simplify.

State the degree and leading coefficient of each polynomial in one variable. If it isnot a polynomial in one variable, explain why. 8; 81. 3x4 � 6x3 � x2 � 12 4; 3 2. 100 � 5x3 � 10x7 7; 10 3. 4x6 � 6x4 � 8x8 � 10x2 � 20

4. 4x2 � 3xy � 16y2 5. 8x3 � 9x5 � 4x2 � 36 6. � � �not a polynomial in 5; �9one variable; contains 6; �two variables

Find f(2) and f(�5) for each function.

7. f(x) � x2 � 9 8. f(x) � 4x3 � 3x2 � 2x � 1 9. f(x) � 9x3 � 4x2 � 5x � 7�5; 16 23; �586 73; �1243

1�

1�72

x3�36

x6�25

x2�18

Example 1Example 1

Example 2Example 2

Example 3Example 3

ExercisesExercises


Graphs of Polynomial Functions

If the degree is even and the leading coefficient is positive, thenf(x) → �� as x → ��

f(x) → �� as x → ��

If the degree is even and the leading coefficient is negative, then

End Behaviorf(x) → �� as x → ��

of Polynomialf(x) → �� as x → ��

FunctionsIf the degree is odd and the leading coefficient is positive, then

f(x) → �� as x → ��

f(x) → �� as x → ��

If the degree is odd and the leading coefficient is negative, thenf(x) → �� as x → ��

f(x) → �� as x → ��

Real Zeros ofThe maximum number of zeros of a polynomial function is equal to the degree of the polynomial.

a PolynomialA zero of a function is a point at which the graph intersects the x-axis.

FunctionOn a graph, count the number of real zeros of the function by counting the number of times thegraph crosses or touches the x-axis.

Determine whether the graph represents an odd-degree polynomialor an even-degree polynomial. Then state the number of real zeros.

As x → ��, f(x) → �� and as x → ��, f(x) → ��,so it is an odd-degree polynomial function.The graph intersects the x-axis at 1 point,so the function has 1 real zero.

Determine whether each graph represents an odd-degree polynomial or an even-degree polynomial. Then state the number of real zeros.

1. 2. 3.

even; 6 even; 1 double zero odd; 3

x

f(x)

Ox

f(x)

Ox

f(x)

O

x

f(x)

O

Study Guide and Intervention (continued)


NAME ______________________________________________ DATE ____________ PERIOD _____

7-17-1

ExampleExample

ExercisesExercises

Skills PracticePolynomial Functions

NAME ______________________________________________ DATE ____________ PERIOD _____

7-17-1


Less

on

7-1

State the degree and leading coefficient of each polynomial in one variable. If it isnot a polynomial in one variable, explain why.

1. a � 8 1; 1 2. (2x � 1)(4x2 � 3) 3; 8

3. �5x5 � 3x3 � 8 5; �5 4. 18 � 3y � 5y2 � y5 � 7y6 6; 7

5. u3 � 4u2v2 � v4 6. 2r � r2 �

No, this polynomial contains two No, this is not a polynomialbecause

variables, u and v. �r12� cannot be written in the form rn,

where n is a nonnegative integer.

Find p(�1) and p(2) for each function.

7. p(x) � 4 � 3x 7; �2 8. p(x) � 3x � x2 �2; 10

9. p(x) � 2x2 � 4x � 1 7; 1 10. p(x) � �2x3 � 5x � 3 0; �3

11. p(x) � x4 � 8x2 � 10 �1; 38 12. p(x) � �13�x2 � �

23�x � 2 3; 2

If p(x) � 4x2 � 3 and r(x) � 1 � 3x, find each value.

13. p(a) 4a2 � 3 14. r(2a) 1 � 6a

15. 3r(a) 3 � 9a 16. �4p(a) �16a2 � 12

17. p(a2) 4a4 � 3 18. r(x � 2) 7 � 3x

For each graph,a. describe the end behavior,b. determine whether it represents an odd-degree or an even-degree polynomial

function, andc. state the number of real zeroes.

19. 20. 21.

f(x) → �� as x → ��, f(x) → �� as x → ��, f(x) → �� as x → ��,f(x) → �� as x → ��; f(x) → �� as x → ��; f(x) → �� as x → ��;

x

f(x)

Ox

f(x)

Ox

f(x)

O

1�r2


State the degree and leading coefficient of each polynomial in one variable. If it isnot a polynomial in one variable, explain why.

1. (3x2 � 1)(2x2 � 9) 4; 6 2. �15�a3 � �

35�a2 � �

45�a 3; �

15

�

3. � 3m � 12 Not a polynomial; 4. 27 � 3xy3 � 12x2y2 � 10y

�m2

2� cannot be written in the form No, this polynomial contains two

mn for a nonnegative integer n. variables, x and y.

Find p(�2) and p(3) for each function.

5. p(x) � x3 � x5 6. p(x) � �7x2 � 5x � 9 7. p(x) � �x5 � 4x3

24; �216 �29; �39 0; �135

8. p(x) � 3x3 � x2 � 2x � 5 9. p(x) � x4 � �12�x3 � �

12�x 10. p(x) � �

13�x3 � �

23�x2 � 3x

�37; 73 13; 93 �6; 24

If p(x) � 3x2 � 4 and r(x) � 2x2 � 5x � 1, find each value.

11. p(8a) 12. r(a2) 13. �5r(2a) 192a2 � 4 2a4 � 5a2 � 1 �40a2 � 50a � 5

14. r(x � 2) 15. p(x2 � 1) 16. 5[p(x � 2)]2x2 � 3x � 1 3x4 � 6x2 � 1 15x2 � 60x � 40

For each graph,a. describe the end behavior,b. determine whether it represents an odd-degree or an even-degree polynomial

function, andc. state the number of real zeroes.

17. 18. 19.

f(x) → �� as x → ��, f(x) → �� as x → ��, f(x) → �� as x → ��,f(x) → �� as x → ��; f(x) → �� as x → ��; f(x) → �� as x → ��;even; 2 even; 1 odd; 5

20. WIND CHILL The function C(s) � 0.013s2 � s � 7 estimates the wind chill temperatureC(s) at 0�F for wind speeds s from 5 to 30 miles per hour. Estimate the wind chilltemperature at 0�F if the wind speed is 20 miles per hour. about �22�F

x

f(x)

Ox

f(x)

Ox

f(x)

O

2�m2

Practice (Average)


NAME ______________________________________________ DATE ____________ PERIOD _____

7-17-1

Reading to Learn MathematicsPolynomial Functions

NAME ______________________________________________ DATE ____________ PERIOD _____

7-17-1


Less

on

7-1

Pre-Activity Where are polynomial functions found in nature?

Read the introduction to Lesson 7-1 at the top of page 346 in your textbook.

• In the honeycomb cross section shown in your textbook, there is 1 hexagonin the center, 6 hexagons in the second ring, and 12 hexagons in the thirdring. How many hexagons will there be in the fourth, fifth, and sixth rings?18; 24; 30

• There is 1 hexagon in a honeycomb with 1 ring. There are 7 hexagons ina honeycomb with 2 rings. How many hexagons are there in honeycombswith 3 rings, 4 rings, 5 rings, and 6 rings?19; 37; 61; 91

Reading the Lesson

1. Give the degree and leading coefficient of each polynomial in one variable.

degree leading coefficient

a. 10x3 � 3x2 � x � 7

b. 7y2 � 2y5 � y � 4y3

c. 100

2. Match each description of a polynomial function from the list on the left with thecorresponding end behavior from the list on the right.

a. even degree, negative leading coefficient iii i. f(x) → �� as x → ��;f(x) → �� as x → ��

b. odd degree, positive leading coefficient iv ii. f(x) → �� as x → ��;f(x) → �� as x → ��

c. odd degree, negative leading coefficient ii iii. f(x) → �� as x → ��;f(x) → �� as x → ��

d. even degree, positive leading coefficient i iv. f(x) → �� as x → ��;f(x) → �� as x → ��

Helping You Remember

3. What is an easy way to remember the difference between the end behavior of the graphsof even-degree and odd-degree polynomial functions?

Sample answer: Both ends of the graph of an even-degree functioneventually keep going in the same direction. For odd-degree functions,the two ends eventually head in opposite directions, one upward, theother downward.

1000

�25103


Approximation by Means of PolynomialsMany scientific experiments produce pairs of numbers [x, f(x)] that can be related by a formula. If the pairs form a function, you can fit a polynomial to the pairs in exactly one way. Consider the pairs given by the following table.

We will assume the polynomial is of degree three. Substitute the given values into this expression.

f(x) � A � B(x � x0) � C(x � x0)(x � x1) � D(x � x0)(x � x1)(x � x2)

You will get the system of equations shown below. You can solve this system and use the values for A, B, C, and D to find the desired polynomial.

6 � A11 � A � B(2 � 1) � A � B39 � A � B(4 � 1) � C(4 � 1)(4 � 2) � A � 3B � 6C

�54 � A � B(7 � 1) � C(7 � 1)(7 � 2) � D(7 � 1)(7 � 2)(7 � 4) � A � 6B � 30C � 90D

Solve.

1. Solve the system of equations for the values A, B, C, and D.

2. Find the polynomial that represents the four ordered pairs. Write your answer in the form y � a � bx � cx2 � dx3.

3. Find the polynomial that gives the following values.

4. A scientist measured the volume f(x) of carbon dioxide gas that can be absorbed by one cubic centimeter of charcoal at pressure x. Find the values for A, B, C, and D.

x 120 340 534 698

f (x) 3.1 5.5 7.1 8.3

x 8 12 15 20

f (x) �207 169 976 3801

x 1 2 4 7

f (x) 6 11 39 �54

Enrichment

NAME ______________________________________________ DATE ____________ PERIOD _____

7-17-1

Study Guide and InterventionGraphing Polynomial Functions

NAME ______________________________________________ DATE ____________ PERIOD _____

7-27-2


Less

on

7-2

Graph Polynomial Functions

Location PrincipleSuppose y � f(x) represents a polynomial function and a and b are two numbers such thatf(a) � 0 and f(b) � 0. Then the function has at least one real zero between a and b.

Determine the values of x between which each real zero of thefunction f(x) � 2x4 � x3 � 5 is located. Then draw the graph.Make a table of values. Look at the values of f(x) to locate the zeros. Then use the points tosketch a graph of the function.

The changes in sign indicate that there are zerosbetween x � �2 and x � �1 and between x � 1 andx � 2.

Graph each function by making a table of values. Determine the values of x atwhich or between which each real zero is located.

1. f(x) � x3 � 2x2 � 1 2. f(x) � x4 � 2x3 � 5 3. f(x) � �x4 � 2x2 � 1

between 0 and �1; between �2 and �3; at �1 at 1; between 1 and 2 between 1 and 2

4. f(x) � x3 � 3x2 � 4 5. f(x) � 3x3 � 2x � 1 6. f(x) � x4 � 3x3 � 1

at �1, 2 between 0 and 1 between 0 and 1;between 2 and 3

x

f(x)

Ox

f(x)

Ox

f(x)

O

x

f(x)

Ox

f(x)

O

x

f(x)

O 4 8–4–8

8

4

–4

–8

x

f(x)

O

x f(x)

�2 35

�1 �2

0 �5

1 �4

2 19

ExampleExample

ExercisesExercises


Maximum and Minimum Points A quadratic function has either a maximum or aminimum point on its graph. For higher degree polynomial functions, you can find turningpoints, which represent relative maximum or relative minimum points.

Graph f(x) � x3 � 6x2 � 3. Estimate the x-coordinates at which therelative maxima and minima occur.Make a table of values and graph the function.

A relative maximum occursat x � �4 and a relativeminimum occurs at x � 0.

Graph each function by making a table of values. Estimate the x-coordinates atwhich the relative maxima and minima occur.

1. f(x) � x3 � 3x2 2. f(x) � 2x3 � x2 � 3x 3. f(x) � 2x3 � 3x � 2

max. at 0, min. at 2 max. about �1, max. about �1, min. about 0.5 min. about 1

4. f(x) � x4 � 7x � 3 5. f(x) � x5 � 2x2 � 2 6. f(x) � x3 � 2x2 � 3

min. about 1 max. at 0, max. about �1, min. about 1 min. at 0

x

f(x)

Ox

f(x)

Ox

f(x)

O 4 8–4–8

8

4

–4

–8

x

f(x)

Ox

f(x)

Ox

f(x)

O

x

f(x)

O2–2–4

24

16

8

← indicates a relative maximum

← zero between x � �1, x � 0

← indicates a relative minimum

x f(x)

�5 22

�4 29

�3 24

�2 13

�1 2

0 �3

1 4

2 29


Graphing Polynomial Functions

NAME ______________________________________________ DATE ____________ PERIOD _____

7-27-2

ExampleExample

ExercisesExercises

Skills PracticeGraphing Polynomial Functions

NAME ______________________________________________ DATE ____________ PERIOD _____

7-27-2


Less

on

7-2

Complete each of the following.a. Graph each function by making a table of values.b. Determine consecutive values of x between which each real zero is located.c. Estimate the x-coordinates at which the relative maxima and minima occur.

1. f(x) � x3 � 3x2 � 1 2. f(x) � x3 � 3x � 1

zeros between �1 and 0, 0 and 1, zeros between �2 and �1, 0 and 1, and 2 and 3; rel. max. at x � 0, and 1 and 2; rel. max. at x � �1, rel. min. at x � 2 rel. min. at x � 1

3. f(x) � 2x3 � 9x2 �12x � 2 4. f(x) � 2x3 � 3x2 � 2

zero between �1 and 0; zero between �1 and 0; rel. max. at x � �2, rel. min. at x � 1, rel. max. at x � 0rel. min. at x � �1

5. f(x) � x4 � 2x2 � 2 6. f(x) � 0.5x4 � 4x2 � 4

zeros between �2 and �1, and zeros between �1 and �2, �2 and 1 and 2; rel. max. at x � 0, �3, 1 and 2, and 2 and 3; rel. max.at

x

f(x)

O

x f(x)

�3 8.5�2 �4�1 0.5

0 41 0.52 �43 8.5

x

f(x)

O

x f(x)

�3 61�2 6�1 �3

0 �21 �32 63 61

x

f(x)

O

x f(x)

�1 �30 21 12 63 29

x

f(x)

O

x f(x)

�3 �7�2 �2�1 �3

0 21 25

x

f(x)

O

x f(x)

�3 �17�2 �1�1 3

0 11 �12 33 19

x

f(x)

O

x f(x)

�2 �19�1 �3

0 11 �12 �33 14 17


Complete each of the following.a. Graph each function by making a table of values.b. Determine consecutive values of x between which each real zero is located.c. Estimate the x-coordinates at which the relative and relative minima occur.

1. f(x) � �x3 � 3x2 � 3 2. f(x) � x3 � 1.5x2 � 6x � 1

x

f(x)

O

8

4

–4

–8

2 4–2–4

x f(x)

�2 �1�1 4.5

0 11 �5.52 �93 �3.54 17

x

f(x)

O

x f(x)

�2 17�1 1

0 �31 �12 13 �34 �19

Practice (Average)

Graphing Polynomial Functions

NAME ______________________________________________ DATE ____________ PERIOD _____

7-27-2

zeros between �1 zeros between �2 and 0, 1 and 2, and �1, 0 and 1,

and 2 and 3; rel. max. at x � 2, and 3 and 4; rel. max. at x � �1, rel. min. at x � 0 rel. min. at x � 2

3. f(x) � 0.75x4 � x3 � 3x2 � 4 4. f(x) � x4 � 4x3 � 6x2 � 4x � 3

zeros between �3 and �2, and zeros between �3 and �2, �2 and �1; rel. max. at x � 0, and 0 and 1; rel. min. at x � �1rel. min. at x � �2 and x � 1

PRICES For Exercises 5 and 6, use the following information.The Consumer Price Index (CPI) gives the relative price for a fixed set of goods and services. The CPI from September, 2000 to July, 2001 is shown in the graph.Source: U. S. Bureau of Labor Statistics

5. Describe the turning points of the graph.rel max. in Nov. and June; rel. min in Dec.

6. If the graph were modeled by a polynomial equation,what is the least degree the equation could have? 4

7. LABOR A town’s jobless rate can be modeled by (1, 3.3), (2, 4.9), (3, 5.3), (4, 6.4), (5, 4.5),(6, 5.6), (7, 2.5), (8, 2.7). How many turning points would the graph of a polynomialfunction through these points have? Describe them. 4: 2 rel. max. and 2 rel. min.

Months Since September, 2000

Co

nsu

mer

Pri

ce In

dex

20 4 61 3 5 7 8 9 1011

179178177176175174173

x f(x)

�3 12�2 �3�1 �4

0 �31 122 77

x f(x)

�3 10.75�2 �4�1 0.75

0 41 2.752 12

Reading to Learn MathematicsGraphing Polynomial Functions

NAME ______________________________________________ DATE ____________ PERIOD _____

7-27-2


Less

on

7-2

Pre-Activity How can graphs of polynomial functions show trends in data?


Three points on the graph shown in your textbook are (0, 14), (70, 3.78), and(100, 9). Give the real-world meaning of the coordinates of these points.Sample answer: In 1900, 14% of the U. S. population wasforeign born. In 1970, 3.78% of the population was foreignborn. In 2000, 9% of the population was foreign born.

Reading the Lesson

1. Suppose that f(x) is a third-degree polynomial function and that c and d are realnumbers, with d � c. Indicate whether each statement is true or false. (Remember thattrue means always true.)

a. If f(c) � 0 and f(d) � 0, there is exactly one real zero between c and d. false

b. If f(c) � f(d) 0, there are no real zeros between c and d. false

c. If f(c) � 0 and f(d) � 0, there is at least one real zero between c and d. true

2. Match each graph with its description.

a. third-degree polynomial with one relative maximum and one relative minimum;leading coefficient negative iii

b. fourth-degree polynomial with two relative minima and one relative maximum i

c. third-degree polynomial with one relative maximum and one relative minimum;leading coefficient positive iv

d. fourth-degree polynomial with two relative maxima and one relative minimum ii

i. ii. iii. iv.


3. The origins of words can help you to remember their meaning and to distinguishbetween similar words. Look up maximum and minimum in a dictionary and describetheir origins (original language and meaning). Sample answer: Maximum comesfrom the Latin word maximus, meaning greatest. Minimum comes fromthe Latin word minimus, meaning least.

x

f(x)

Ox

f(x)

Ox

f(x)

Ox

f(x)

O


Golden RectanglesUse a straightedge, a compass, and the instructions below to construct a golden rectangle.

1. Construct square ABCD with sides of 2 centimeters.

2. Construct the midpoint of A�B�. Call the midpoint M.

3. Using M as the center, set your compass opening at MC. Construct an arc with center M that intersects A�B�. Call the point of intersection P.

4. Construct a line through P that is perpendicular to A�B�.

5. Extend D�C� so that it intersects the perpendicular. Call the intersection point Q.APQD is a golden rectangle. Check this

conclusion by finding the value of �QAP

P�.

A figure consisting of similar golden rectangles is shown below. Use a compass and the instructions below to draw quarter-circle arcs that form a spiral like that found in the shell of a chambered nautilus.

6. Using A as a center, draw an arc that passes through B and C.

7. Using D as a center, draw an arc that passes through C and E.

8. Using F as a center, draw an arc that passes through E and G.

9. Continue drawing arcs,using H, K, and M as the centers.

C

BA G

HJ

D E

K

M

L F

Enrichment

NAME ______________________________________________ DATE ____________ PERIOD _____

7-27-2

Study Guide and InterventionSolving Equations Using Quadratic Techniques

NAME ______________________________________________ DATE ____________ PERIOD _____

7-37-3


Less

on

7-3

Quadratic Form Certain polynomial expressions in x can be written in the quadraticform au2 � bu � c for any numbers a, b, and c, a 0, where u is an expression in x.

Write each polynomial in quadratic form, if possible.

a. 3a6 � 9a3 � 12Let u � a3.3a6 � 9a3 � 12 � 3(a3)2 � 9(a3) � 12

b. 101b � 49�b� � 42Let u � �b�.101b � 49�b� � 42 � 101(�b�)2

� 49(�b�) � 42

c. 24a5 � 12a3 � 18This expression cannot be written in quadratic form, since a5 (a3)2.

Write each polynomial in quadratic form, if possible.

1. x4 � 6x2 � 8 2. 4p4 � 6p2 � 8

(x2)2 � 6(x2) � 8 4(p2)2 � 6(p2) � 8

3. x8 � 2x4 � 1 4. x�18

�� 2x�

116�

� 1

(x4)2 � 2(x4) � 1 �x�116��2

� 2�x�116�� 1

5. 6x4 � 3x3 � 18 6. 12x4 � 10x2 � 4

not possible 12(x2)2 � 10(x2) � 4

7. 24x8 � x4 � 4 8. 18x6 � 2x3 � 12

24(x4)2 � x4 � 4 18(x3)2 � 2(x3) � 12

9. 100x4 � 9x2 � 15 10. 25x8 � 36x6 � 49

100(x2)2 � 9(x2) � 15 not possible

11. 48x6 � 32x3 � 20 12. 63x8 � 5x4 � 29

48(x3)2 � 32(x3) � 20 63(x4)2 � 5(x4) � 29

13. 32x10 � 14x5 � 143 14. 50x3 � 15x�x� � 18

32(x5)2 � 14(x5) � 143 50�x�32

��2� 15�x�

32

�� 18

15. 60x6 � 7x3 � 3 16. 10x10 � 7x5 � 7

60(x3)2 � 7(x3) � 3 10(x5)2 � 7(x5) � 7

ExampleExample

ExercisesExercises


Solve Equations Using Quadratic Form If a polynomial expression can be writtenin quadratic form, then you can use what you know about solving quadratic equations tosolve the related polynomial equation.

Solve x4 � 40x2 � 144 � 0.x4 � 40x2 � 144 � 0 Original equation

(x2)2 � 40(x2) � 144 � 0 Write the expression on the left in quadratic form.

(x2 � 4)(x2 � 36) � 0 Factor.x2 � 4 � 0 or x2 � 36 � 0 Zero Product Property

(x � 2)(x � 2) � 0 or (x � 6)(x � 6) � 0 Factor.

x � 2 � 0 or x � 2 � 0 or x � 6 � 0 or x � 6 � 0 Zero Product Property

x � 2 or x � �2 or x � 6 or x � �6 Simplify.

The solutions are 2 and 6.

Solve 2x � �x� � 15 � 0.2x � �x� � 15 � 0 Original equation

2(�x�)2 � �x� � 15 � 0 Write the expression on the left in quadratic form.

(2�x� �5)(�x� � 3) � 0 Factor.

2�x� � 5 � 0 or �x� � 3 � 0 Zero Product Property

�x� � or �x� � �3 Simplify.

Since the principal square root of a number cannot be negative, �x� � �3 has no solution.

The solution is or 6 .

Solve each equation.

1. x4 � 49 2. x4 � 6x2 � �8 3. x4 � 3x2 � 54

��7�, �i �7� �2, ��2� �3, �i �6�

4. 3t6 � 48t2 � 0 5. m6 � 16m3 � 64 � 0 6. y4 � 5y2 � 4 � 0

0, �2, �2i 2, �1 � i �3� �1, �2

7. x4 � 29x2 � 100 � 0 8. 4x4 � 73x2 � 144 � 0 9. � � 12 � 0

�5, �2 �4, � ,

10. x � 5�x� � 6 � 0 11. x � 10�x� � 21 � 0 12. x�23

�� 5x�

13

�� 6 � 0

4, 9 9, 49 27, 8

1�

1�

3�

7�x

1�x2

1�4

25�4

5�2


Solving Equations Using Quadratic Techniques

NAME ______________________________________________ DATE ____________ PERIOD _____

7-37-3

Example 1Example 1

Example 2Example 2

ExercisesExercises

Skills PracticeSolving Equations Using Quadratic Techniques

NAME ______________________________________________ DATE ____________ PERIOD _____

7-37-3


Less

on

7-3

Write each expression in quadratic form, if possible.

1. 5x4 � 2x2 � 8 5(x2)2 � 2(x2) � 8 2. 3y8 � 4y2 � 3 not possible

3. 100a6 � a3 100(a3)2 � a3 4. x8 � 4x4 � 9 (x4)2 � 4(x4) � 9

5. 12x4 � 7x2 12(x2)2 � 7(x2) 6. 6b5 � 3b3 � 1 not possible

7. 15v6 � 8v3 � 9 15(v3)2 � 8(v3) � 9 8. a9 � 5a5 � 7a a[(a4)2 � 5(a4) � 7]


9. a3 � 9a2 � 14a � 0 0, 7, 2 10. x3 � 3x2 0, 3

11. t4 � 3t3 � 40t2 � 0 0, �5, 8 12. b3 � 8b2 � 16b � 0 0, 4

13. m4 � 4 ��2�, �2�, �i�2�, i�2� 14. w3 � 6w � 0 0, �6�, ��6�

15. m4 � 18m2 � �81 �3, 3 16. x5 � 81x � 0 0, �3, 3, �3i, 3i

17. h4 � 10h2 � �9 �1, 1, �3, 3 18. a4 � 9a2 � 20 � 0 �2, 2, �5�, ��5�

19. y4 � 7y2 � 12 � 0 20. v4 � 12v2 � 35 � 02, �2, �3�, ��3� �5�, ��5�, �7�, ��7�

21. x5 � 7x3 � 6x � 0 22. c�23

�� 7c�

13

�� 12 � 0

0, �1, 1, �6�, ��6� �64, �27

23. z � 5�z� � �6 4, 9 24. x � 30�x� � 200 � 0 100, 400


Write each expression in quadratic form, if possible.

1. 10b4 � 3b2 � 11 2. �5x8 � x2 � 6 3. 28d6 � 25d3

10(b2)2 � 3(b2) � 11 not possible 28(d3)2 � 25(d3)

4. 4s8 � 4s4 � 7 5. 500x4 � x2 6. 8b5 � 8b3 � 1

4(s4)2 � 4(s4) � 7 500(x2)2 � x2 not possible

7. 32w5 � 56w3 � 8w 8. e�23

�� 7e�

13

�� 10 9. x

�15

�� 29x

�110�

� 2

8w[4(w2)2 � 7(w2) � 1] (e�13

�)2� 7(e�

13

�) � 10 (x�110�)2

� 29(x�110�) � 2


10. y4 � 7y3 � 18y2 � 0 �2, 0, 9 11. s5 � 4s4 � 32s3 � 0 �8, 0, 4

12. m4 � 625 � 0 �5, 5, �5i, 5i 13. n4 � 49n2 � 0 0, �7, 7

14. x4 � 50x2 � 49 � 0 �1, 1, �7, 7 15. t4 � 21t2 � 80 � 0 �4, 4, �5�, ��5�

16. 4r6 � 9r4 � 0 0, �32

�, ��32

� 17. x4 � 24 � �2x2 �2, 2, �i�6�, i�6�

18. d4 � 16d2 � 48 �2, 2, �2�3�, 2�3� 19. t3 � 343 � 0 7, ,

20. x�12

�� 5x

�14

�� 6 � 0 16, 81 21. x

�43

�� 29x

�23

�� 100 � 0 8, 125

22. y3 � 28y�32

�� 27 � 0 1, 9 23. n � 10�n� � 25 � 0 25

24. w � 12�w� � 27 � 0 9, 81 25. x � 2�x� � 80 � 0 100

26. PHYSICS A proton in a magnetic field follows a path on a coordinate grid modeled bythe function f(x) � x4 � 2x2 � 15. What are the x-coordinates of the points on the gridwhere the proton crosses the x-axis? ��5�, �5�

27. SURVEYING Vista county is setting aside a large parcel of land to preserve it as openspace. The county has hired Meghan’s surveying firm to survey the parcel, which is inthe shape of a right triangle. The longer leg of the triangle measures 5 miles less thanthe square of the shorter leg, and the hypotenuse of the triangle measures 13 miles lessthan twice the square of the shorter leg. The length of each boundary is a whole number.Find the length of each boundary. 3 mi, 4 mi, 5 mi

�7 � 7i�3��

�7 � 7i�3��

Practice (Average)

Solving Equations Using Quadratic Techniques

NAME ______________________________________________ DATE ____________ PERIOD _____

7-37-3

Reading to Learn MathematicsSolving Equations Using Quadratic Techniques

NAME ______________________________________________ DATE ____________ PERIOD _____

7-37-3


Less

on

7-3

Pre-Activity How can solving polynomial equations help you to find dimensions?


Explain how the formula given for the volume of the box can be obtainedfrom the dimensions shown in the figure.

Sample answer: The volume of a rectangular box is given by the formula V � �wh. Substitute 50 � 2x for �, 32 � 2x for w, and x for h to get V(x) � (50 � 2x)(32 � 2x)(x) � 4x3 � 164x2 � 1600x.

Reading the Lesson

1. Which of the following expressions can be written in quadratic form? b, c, d, f, g, h, i

a. x3 � 6x2 � 9 b. x4 � 7x2 � 6 c. m6 � 4m3 � 4

d. y � 2y�12

�� 15 e. x5 � x3 � 1 f. r4 � 6 � r8

g. p�14

�� 8p

�12

�� 12 h. r

�13

�� 2r

�16

�� 3 i. 5�z� � 2z � 3

2. Match each expression from the list on the left with its factorization from the list on the right.

a. x4 � 3x2 � 40 vi i. (x3 � 3)(x3 � 3)

b. x4 � 10x2 � 25 v ii. (�x� � 3)(�x� � 3)

c. x6 � 9 i iii. (�x� � 3)2

d. x � 9 ii iv. (x2 � 1)(x4 � x2 � 1)

e. x6 � 1 iv v. (x2 � 5)2

f. x � 6�x� � 9 iii vi. (x2 � 5)(x2 � 8)


3. What is an easy way to tell whether a trinomial in one variable containing one constantterm can be written in quadratic form?

Sample answer: Look at the two terms that are not constants andcompare the exponents on the variable. If one of the exponents is twicethe other, the trinomial can be written in quadratic form.


Odd and Even Polynomial Functions

Enrichment

NAME ______________________________________________ DATE ____________ PERIOD _____

7-37-3

Functions whose graphs are symmetric withrespect to the origin are called odd functions.If f(�x) � �f(x) for all x in the domain of f(x),then f (x) is odd.

Functions whose graphs are symmetric withrespect to the y-axis are called even functions.If f (�x) � f(x) for all x in the domain of f(x),then f (x) is even.

x

f(x)

O 1 2–2 –1

6

4

2f(x) � 1–4x4 � 4

x

f(x)

O 1 2–2 –1

4

2

–2

–4

f(x) � 1–2x3

ExampleExample Determine whether f(x) � x3 � 3x is odd, even, or neither.

f(x) � x3 � 3xf(�x) � (�x)3 � 3(�x) Replace x with �x.

� �x3 � 3x Simplify.

� �(x3 � 3x) Factor out �1.

� �f (x) Substutute.

Therefore, f (x) is odd.

The graph at the right verifies that f (x) is odd.The graph of the function is symmetric with respect to the origin.

Determine whether each function is odd, even, or neither by graphing or by applying the rules for odd and even functions.

1. f (x) � 4x2 2. f (x) � �7x4

3. f (x) � x7 4. f (x) � x3 � x2

5. f (x) � 3x3 � 1 6. f (x) � x8 � x5 � 6

7. f (x) � �8x5 � 2x3 � 6x 8. f (x) � x4 � 3x3 � 2x2 � 6x � 1

9. f (x) � x4 � 3x2 � 11 10. f (x) � x7 � 6x5 � 2x3 � x

11. Complete the following definitions: A polynomial function is odd if and only

if all the terms are of degrees. A polynomial function is even

if and only if all the terms are of degrees.

x

f(x)

O 1 2–2 –1

4

2

–2

–4

f(x) � x3 � 3x

Study Guide and InterventionThe Remainder and Factor Theorems

NAME ______________________________________________ DATE ____________ PERIOD _____

7-47-4


Less

on

7-4

Synthetic Substitution

Remainder The remainder, when you divide the polynomial f(x ) by (x � a), is the constant f(a).Theorem f(x) � q(x ) � (x � a) � f(a), where q(x) is a polynomial with degree one less than the degree of f(x).

If f(x) � 3x4 � 2x3 � 5x2 � x � 2, find f(�2).Example 1Example 1

Example 2Example 2

Method 1 Synthetic SubstitutionBy the Remainder Theorem, f(�2) shouldbe the remainder when you divide thepolynomial by x � 2.

�2 3 2 �5 1 �2�6 8 �6 10

3 �4 3 �5 8The remainder is 8, so f(�2) � 8.

Method 2 Direct SubstitutionReplace x with �2.

f(x) � 3x4 � 2x3 � 5x2 � x � 2f(�2) � 3(�2)4 � 2(�2)3 � 5(�2)2 � (�2) � 2

� 48 � 16 � 20 � 2 � 2 or 8So f(�2) � 8.

If f(x) � 5x3 � 2x � 1, find f(3).Again, by the Remainder Theorem, f(3) should be the remainder when you divide thepolynomial by x � 3.

3 5 0 2 �115 45 141

5 15 47 140The remainder is 140, so f(3) � 140.

Use synthetic substitution to find f(�5) and f � � for each function.

1. f(x) � �3x2 � 5x � 1 �101; 2. f(x) � 4x2 � 6x � 7 63; �3

3. f(x) � �x3 � 3x2 � 5 195; � 4. f(x) � x4 � 11x2 � 1 899;

Use synthetic substitution to find f(4) and f(�3) for each function.

5. f(x) � 2x3 � x2 � 5x � 3 6. f(x) � 3x3 � 4x � 2127; �27 178; �67

7. f(x) � 5x3 � 4x2 � 2 8. f(x) � 2x4 � 4x3 � 3x2 � x � 6258; �169 302; 288

9. f(x) � 5x4 � 3x3 � 4x2 � 2x � 4 10. f(x) � 3x4 � 2x3 � x2 � 2x � 51404; 298 627; 277

11. f(x) � 2x4 � 4x3 � x2 � 6x � 3 12. f(x) � 4x4 � 4x3 � 3x2 � 2x � 3219; 282 805; 462

29�

35�

3�

1�2

ExercisesExercises


Factors of Polynomials The Factor Theorem can help you find all the factors of apolynomial.

Factor Theorem The binomial x � a is a factor of the polynomial f(x) if and only if f(a) � 0.

Show that x � 5 is a factor of x3 � 2x2 � 13x � 10. Then find theremaining factors of the polynomial.By the Factor Theorem, the binomial x � 5 is a factor of the polynomial if �5 is a zero of thepolynomial function. To check this, use synthetic substitution.

�5 1 2 �13 10�5 15 �10

1 �3 2 0

Since the remainder is 0, x � 5 is a factor of the polynomial. The polynomial x3 � 2x2 � 13x � 10 can be factored as (x � 5)(x2 � 3x � 2). The depressed polynomial x2 � 3x � 2 can be factored as (x � 2)(x � 1).

So x3 � 2x2 � 13x � 10 � (x � 5)(x � 2)(x � 1).

Given a polynomial and one of its factors, find the remaining factors of thepolynomial. Some factors may not be binomials.

1. x3 � x2 � 10x � 8; x � 2 2. x3 � 4x2 � 11x � 30; x � 3(x � 4)(x � 1) (x � 5)(x � 2)

3. x3 � 15x2 � 71x � 105; x � 7 4. x3 � 7x2 � 26x � 72; x � 4(x � 3)(x � 5) (x � 2)(x � 9)

5. 2x3 � x2 � 7x � 6; x � 1 6. 3x3 � x2 � 62x � 40; x � 4(2x � 3)(x � 2) (3x � 2)(x � 5)

7. 12x3 � 71x2 � 57x � 10; x � 5 8. 14x3 � x2 � 24x � 9; x � 1(4x � 1)(3x � 2) (7x � 3)(2x � 3)

9. x3 � x � 10; x � 2 10. 2x3 � 11x2 � 19x � 28; x � 4(x2 � 2x � 5) (2x2 � 3x � 7)

11. 3x3 � 13x2 � 34x � 24; x � 6 12. x4 � x3 � 11x2 � 9x � 18; x � 1(3x2 � 5x � 4) (x � 2)(x � 3)(x � 3)


The Remainder and Factor Theorems

NAME ______________________________________________ DATE ____________ PERIOD _____

7-47-4

ExampleExample

ExercisesExercises

Skills PracticeThe Remainder and Factor Theorems

NAME ______________________________________________ DATE ____________ PERIOD _____

7-47-4


Less

on

7-4

Use synthetic substitution to find f(2) and f(�1) for each function.

1. f(x) � x2 � 6x � 5 21, 0 2. f(x) � x2 � x � 1 3, 3

3. f(x) � x2 � 2x � 2 �2, 1 4. f(x) � x3 � 2x2 � 5 21, 6

5. f(x) � x3 � x2 � 2x � 3 3, 3 6. f(x) � x3 � 6x2 � x � 4 30, 0

7. f(x) � x3 � 3x2 � x � 2 �4, �7 8. f(x) � x3 � 5x2 � x � 6 �8, 1

9. f(x) � x4 � 2x2 � 9 15, �6 10. f(x) � x4 � 3x3 � 2x2 � 2x � 6 2, 14

11. f(x) � x5 � 7x3 � 4x � 10 12. f(x) � x6 � 2x5 � x4 � x3 � 9x2 � 20�22, 20 �32, �26


13. x3 � 2x2 � x � 2; x � 1 14. x3 � x2 � 5x � 3; x � 1

x � 1, x � 2 x � 1, x � 3

15. x3 � 3x2 � 4x � 12; x � 3 16. x3 � 6x2 � 11x � 6; x � 3

x � 2, x � 2 x � 1, x � 2

17. x3 � 2x2 � 33x � 90; x � 5 18. x3 � 6x2 � 32; x � 4

x � 3, x � 6 x � 4, x � 2

19. x3 � x2 � 10x � 8; x � 2 20. x3 � 19x � 30; x � 2

x � 1, x � 4 x � 5, x � 3

21. 2x3 � x2 � 2x � 1; x � 1 22. 2x3 � x2 � 5x � 2; x � 2

2x � 1, x � 1 x � 1, 2x � 1

23. 3x3 � 4x2 � 5x � 2; 3x � 1 24. 3x3 � x2 � x � 2; 3x � 2

x � 1, x � 2 x2 � x � 1


Use synthetic substitution to find f(�3) and f(4) for each function.

1. f(x) � x2 � 2x � 3 6, 27 2. f(x) � x2 � 5x � 10 34, 6

3. f(x) � x2 � 5x � 4 20, �8 4. f(x) � x3 � x2 � 2x � 3 �27, 43

5. f(x) � x3 � 2x2 � 5 �4, 101 6. f(x) � x3 � 6x2 � 2x �87, �24

7. f(x) � x3 � 2x2 � 2x � 8 �31, 32 8. f(x) � x3 � x2 � 4x � 4 �52, 60

9. f(x) � x3 � 3x2 � 2x � 50 �56, 70 10. f(x) � x4 � x3 � 3x2 � x � 12 42, 280

11. f(x) � x4 � 2x2 � x � 7 73, 227 12. f(x) � 2x4 � 3x3 � 4x2 � 2x � 1 286, 377

13. f(x) � 2x4 � x3 � 2x2 � 26 181, 454 14. f(x) � 3x4 � 4x3 � 3x2 � 5x � 3 390, 537

15. f(x) � x5 � 7x3 � 4x � 10 16. f(x) � x6 � 2x5 � x4 � x3 � 9x2 � 20�430, 1446 74, 5828


17. x3 � 3x2 � 6x � 8; x � 2 18. x3 � 7x2 � 7x � 15; x � 1

x � 1, x � 4 x � 3, x � 5

19. x3 � 9x2 � 27x � 27; x � 3 20. x3 � x2 � 8x � 12; x � 3

x � 3, x � 3 x � 2, x � 2

21. x3 � 5x2 � 2x � 24; x � 2 22. x3 � x2 � 14x � 24; x � 4

x � 3, x � 4 x � 3, x � 2

23. 3x3 � 4x2 � 17x � 6; x � 2 24. 4x3 � 12x2 � x � 3; x � 3

x � 3, 3x � 1 2x � 1, 2x � 1

25. 18x3 � 9x2 � 2x � 1; 2x � 1 26. 6x3 � 5x2 � 3x � 2; 3x � 2

3x � 1, 3x � 1 2x � 1, x � 1

27. x5 � x4 � 5x3 � 5x2 � 4x � 4; x � 1 28. x5 � 2x4 � 4x3 � 8x2 � 5x � 10; x � 2

x � 1, x � 1, x � 2, x � 2 x � 1, x � 1, x2 � 5

29. POPULATION The projected population in thousands for a city over the next severalyears can be estimated by the function P(x) � x3 � 2x2 � 8x � 520, where x is thenumber of years since 2000. Use synthetic substitution to estimate the population for 2005. 655,000

30. VOLUME The volume of water in a rectangular swimming pool can be modeled by thepolynomial 2x3 � 9x2 � 7x � 6. If the depth of the pool is given by the polynomial 2x � 1, what polynomials express the length and width of the pool? x � 3 and x � 2

Practice (Average)

The Remainder and Factor Theorems

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Reading to Learn MathematicsThe Remainder and Factor Theorems

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Pre-Activity How can you use the Remainder Theorem to evaluate polynomials?


Show how you would use the model in the introduction to estimate thenumber of international travelers (in millions) to the United States in theyear 2000. (Show how you would substitute numbers, but do not actuallycalculate the result.)Sample answer: 0.02(14)3 � 0.6(14)2 � 6(14) � 25.9

Reading the Lesson

1. Consider the following synthetic division.1 3 2 �6 4

3 5 �13 5 �1 3

a. Using the division symbol �, write the division problem that is represented by thissynthetic division. (Do not include the answer.) (3x3 � 2x2 � 6x � 4) � (x � 1)

b. Identify each of the following for this division.

dividend divisor

quotient remainder

c. If f(x) � 3x3 � 2x2 � 6x � 4, what is f(1)? 3

2. Consider the following synthetic division.�3 1 0 0 27

�3 9 �271 �3 9 0

a. This division shows that is a factor of .

b. The division shows that is a zero of the polynomial function

f(x) � .

c. The division shows that the point is on the graph of the polynomial

function f(x) � .


3. Think of a mnemonic for remembering the sentence, “Dividend equals quotient timesdivisor plus remainder.”Sample answer: Definitely every quiet teacher deserves proper rewards.

x3 � 27(�3, 0)

x3 � 27�3

x3 � 27x � 3

33x3 � 5x � 1

x � 13x3 � 2x2 � 6x � 4


Using Maximum ValuesMany times maximum solutions are needed for different situations. For instance, what is the area of the largest rectangular field that can be enclosed with 2000 feet of fencing?

Let x and y denote the length and width of the field, respectively.

Perimeter: 2x � 2y � 2000 → y � 1000 � xArea: A � xy � x(1000 � x) � �x2 � 1000x

This problem is equivalent to finding the highest point on the graph of A(x) � �x2 � 1000x shown on the right.

Complete the square for �x2 � 1000x.

A � �(x2 � 1000x � 5002) � 5002

� �(x � 500)2 � 5002

Because the term �(x � 500)2 is either negative or 0, the greatest value of Ais 5002. The maximum area enclosed is 5002 or 250,000 square feet.

Solve each problem.

1. Find the area of the largest rectangular garden that can be enclosed by 300 feet of fence.

2. A farmer will make a rectangular pen with 100 feet of fence using part of his barn for one side of the pen. What is the largest area he can enclose?

3. An area along a straight stone wall is to be fenced. There are 600 meters of fencing available. What is the greatest rectangular area that can be enclosed?

A

xO 1000

x

y

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Study Guide and InterventionRoots and Zeros

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Types of Roots The following statements are equivalent for any polynomial function f(x).• c is a zero of the polynomial function f(x).• (x � c) is a factor of the polynomial f(x).• c is a root or solution of the polynomial equation f(x) � 0.If c is real, then (c, 0) is an intercept of the graph of f(x).

Fundamental Every polynomial equation with degree greater than zero has at least one root in the setTheorem of Algebra of complex numbers.

Corollary to the A polynomial equation of the form P (x) � 0 of degree n with complex coefficients hasFundamental exactly n roots in the set of complex numbers.Theorem of Algebras

If P (x) is a polynomial with real coefficients whose terms are arranged in descendingpowers of the variable,

Descartes’ Rule• the number of positive real zeros of y � P (x) is the same as the number of changes in

of Signssign of the coefficients of the terms, or is less than this by an even number, and

• the number of negative real zeros of y � P (x) is the same as the number of changes in sign of the coefficients of the terms of P (�x), or is less than this number by an evennumber.

Solve the equation 6x3 � 3x � 0 and state thenumber and type of roots.

6x3 � 3x � 03x(2x2 � 1) � 0Use the Zero Product Property.3x � 0 or 2x2 � 1 � 0x � 0 or 2x2 � �1

x �

The equation has one real root, 0,

and two imaginary roots, .i�2��2

i�2��2

State the number of positivereal zeros, negative real zeros, and imaginaryzeros for p(x) � 4x4 � 3x3 � x2 � 2x � 5.Since p(x) has degree 4, it has 4 zeros.Use Descartes’ Rule of Signs to determine thenumber and type of real zeros. Since there are threesign changes, there are 3 or 1 positive real zeros.Find p(�x) and count the number of changes insign for its coefficients.p(�x) � 4(�x)4 � 3(�x)3 � (�x)2 � 2(�x) � 5

� 4x4 � 3x3 � x2 � 2x � 5Since there is one sign change, there is exactly 1negative real zero.

Example 1Example 1 Example 2Example 2

ExercisesExercises

Solve each equation and state the number and type of roots.

1. x2 � 4x � 21� 0 2. 2x3 � 50x � 0 3. 12x3 � 100x � 0

3, �7; 2 real 0, �5; 3 real 0, � ; 1 real, 2imaginary

State the number of positive real zeros, negative real zeros, and imaginary zerosfor each function.

4. f(x) � 3x3 � x2 � 8x � 12 1; 2 or 0; 0 or 2

5. f(x) � 2x4 � x3 � 3x � 7 2 or 0; 0; 2 or 4

5i �3��


Find Zeros

Complex Conjugate Suppose a and b are real numbers with b 0. If a � bi is a zero of a polynomial Theorem function with real coefficients, then a � bi is also a zero of the function.

Find all of the zeros of f(x) � x4 � 15x2 � 38x � 60.Since f(x) has degree 4, the function has 4 zeros.f(x) � x4 � 15x2 � 38x � 60 f(�x) � x4 � 15x2 � 38x � 60Since there are 3 sign changes for the coefficients of f(x), the function has 3 or 1 positive realzeros. Since there is 1 sign change for the coefficients of f(�x), the function has 1 negativereal zero. Use synthetic substitution to test some possible zeros.

2 1 0 �15 38 �602 4 �22 32

1 2 �11 16 �28

3 1 0 �15 38 �603 9 �18 60

1 3 �6 20 0So 3 is a zero of the polynomial function. Now try synthetic substitution again to find a zeroof the depressed polynomial.

�2 1 3 �6 20�2 �2 16

1 1 �8 36

�4 1 3 �6 20�4 4 8

1 �1 �2 28

�5 1 3 �6 20�5 10 �20

1 �2 4 0

So � 5 is another zero. Use the Quadratic Formula on the depressed polynomial x2 � 2x � 4 to find the other 2 zeros, 1 i�3�.The function has two real zeros at 3 and �5 and two imaginary zeros at 1 i�3�.

Find all of the zeros of each function.

1. f(x) � x3 � x2 � 9x � 9 �1, �3i 2. f(x) � x3 � 3x2 � 4x � 12 3, �2i

3. p(a) � a3 � 10a2 � 34a � 40 4, 3 � i 4. p(x) � x3 � 5x2 � 11x � 15 3, 1 � 2i

5. f(x) � x3 � 6x � 20 6. f(x) � x4 � 3x3 � 21x2 � 75x � 100�2, 1 � 3i �1, 4, �5i


Roots and Zeros

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ExampleExample

ExercisesExercises

Skills PracticeRoots and Zeros

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Solve each equation. State the number and type of roots.

1. 5x � 12 � 0 2. x2 � 4x � 40 � 0

��152�; 1 real 2 � 6i; 2 imaginary

3. x5 � 4x3 � 0 4. x4 � 625 � 0

0, 0, 0, 2i, �2i; 3 real, 2 imaginary 5i, 5i, �5i, �5i; 4 imaginary

5. 4x2 � 4x � 1 � 0 6. x5 � 81x � 0

; 2 real 0, �3, 3, �3i, 3i; 3 real, 2 imaginary

State the possible number of positive real zeros, negative real zeros, andimaginary zeros of each function.

7. g(x) � 3x3 � 4x2 � 17x � 6 8. h(x) � 4x3 � 12x2 � x � 3

2 or 0; 1; 2 or 0 2 or 0; 1; 2 or 0

9. f(x) � x3 � 8x2 � 2x � 4 10. p(x) � x3 � x2 � 4x � 6

3 or 1; 0; 2 or 0 3 or 1; 0; 2 or 0

11. q(x) � x4 � 7x2 � 3x � 9 12. f(x) � x4 � x3 � 5x2 � 6x � 1

1; 1; 2 2 or 0; 2 or 0; 4 or 2 or 0

Find all the zeros of each function.

13. h(x) � x3 � 5x2 � 5x � 3 14. g(x) � x3 � 6x2 � 13x � 10

3, 1 � �2�, 1 � �2� 2, 2 � i, 2 � i

15. h(x) � x3 � 4x2 � x � 6 16. q(x) � x3 � 3x2 � 6x � 8

1, �2, �3 2, �1, �4

17. g(x) � x4 � 3x3 � 5x2 � 3x � 4 18. f(x) � x4 � 21x2 � 80

�1, �1, 1, 4 �4, 4, ��5�, �5�

Write a polynomial function of least degree with integral coefficients that has thegiven zeros.

19. �3, �5, 1 20. 3if(x) � x3 � 7x2 � 7x � 15 f(x) � x2 � 9

21. �5 � i 22. �1, �3�, ��3�f(x) � x2 � 10x � 26 f(x) � x3 � x2 � 3x � 3

23. i, 5i 24. �1, 1, i�6�f(x) � x4 � 26x2 � 25 f(x) � x4 � 5x2 � 6

1 � �2��


Solve each equation. State the number and type of roots.

1. �9x � 15 � 0 2. x4 � 5x2 � 4 � 0

��53

�; 1 real �1, 1, �2, 2; 4 real

3. x5 � 81x 4. x3 � x2 � 3x � 3 � 0

0, �3, 3, �3i, 3i; 3 real, 2 imaginary �1, ��3�, �3�; 3 real

5. x3 � 6x � 20 � 0 6. x4 � x3 � x2 � x � 2 � 0

�2, 1 � 3i; 1 real, 2 imaginary 2, �1, �i, i; 2 real, 2 imaginary

State the possible number of positive real zeros, negative real zeros, andimaginary zeros of each function.

7. f(x) � 4x3 � 2x2 � x � 3 8. p(x) � 2x4 � 2x3 � 2x2 � x � 1

2 or 0; 1; 2 or 0 3 or 1; 1; 2 or 0

9. q(x) � 3x4 � x3 � 3x2 � 7x � 5 10. h(x) � 7x4 � 3x3 � 2x2 � x � 1

2 or 0; 2 or 0; 4, 2, or 0 2 or 0; 2 or 0; 4, 2, or 0

Find all the zeros of each function.

11. h(x) � 2x3 � 3x2 � 65x � 84 12. p(x) � x3 � 3x2 � 9x � 7

�7, �32

�, 4 1, 1 � i�6�, 1 � i�6�

13. h(x) � x3 � 7x2 � 17x � 15 14. q(x) � x4 � 50x2 � 49

3, 2 � i, 2 � i �i, i, �7i, 7i

15. g(x) � x4 � 4x3 � 3x2 � 14x � 8 16. f(x) � x4 � 6x3 � 6x2 � 24x � 40

�1, �1, 2, �4 �2, 2, 3 � i, 3 � i

Write a polynomial function of least degree with integral coefficients that has thegiven zeros.

17. �5, 3i 18. �2, 3 � if(x) � x3 � 5x2 � 9x � 45 f(x) � x3 � 4x2 � 2x � 20

19. �1, 4, 3i 20. 2, 5, 1 � if(x) � x4 � 3x3 � 5x2 � 27x � 36 f(x) � x4 � 9x3 � 26x2 � 34x � 20

21. CRAFTS Stephan has a set of plans to build a wooden box. He wants to reduce thevolume of the box to 105 cubic inches. He would like to reduce the length of eachdimension in the plan by the same amount. The plans call for the box to be 10 inches by8 inches by 6 inches. Write and solve a polynomial equation to find out how muchStephen should take from each dimension. (10 � x)(8 � x)(6 � x) � 105; 3 in.

Practice (Average)

Roots and Zeros

NAME ______________________________________________ DATE ____________ PERIOD _____

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Reading to Learn MathematicsRoots and Zeros

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Pre-Activity How can the roots of an equation be used in pharmacology?


Using the model given in the introduction, write a polynomial equationwith 0 on one side that can be solved to find the time or times at whichthere is 100 milligrams of medication in a patient’s bloodstream.0.5t4 � 3.5t3 � 100t2 � 350t � 100 � 0

Reading the Lesson

1. Indicate whether each statement is true or false.

a. Every polynomial equation of degree greater than one has at least one root in the setof real numbers. false

b. If c is a root of the polynomial equation f(x) � 0, then (x � c) is a factor of thepolynomial f(x). true

c. If (x � c) is a factor of the polynomial f(x), then c is a zero of the polynomial function f. false

d. A polynomial function f of degree n has exactly (n � 1) complex zeros. false

2. Let f(x) � x6 � 2x5 � 3x4 � 4x3 � 5x2 � 6x � 7.

a. What are the possible numbers of positive real zeros of f ? 5, 3, or 1b. Write f(�x) in simplified form (with no parentheses).

x6 � 2x5 � 3x4 � 4x3 � 5x2 � 6x � 7What are the possible numbers of negative real zeros of f ? 1

c. Complete the following chart to show the possible combinations of positive real zeros,negative real zeros, and imaginary zeros of the polynomial function f.

Number of Number of Number of Total Number Positive Real Zeros Negative Real Zeros Imaginary Zeros of Zeros

5 1 0 6

3 1 2 6

1 1 4 6


3. It is easier to remember mathematical concepts and results if you relate them to eachother. How can the Complex Conjugates Theorem help you remember the part ofDescartes’ Rule of Signs that says, “or is less than this number by an even number.”Sample answer: For a polynomial function in which the polynomial hasreal coefficients, imaginary zeros come in conjugate pairs. Therefore,there must be an even number of imaginary zeros. For each pair ofimaginary zeros, the number of positive or negative zeros decreases by


The Bisection Method for Approximating Real ZerosThe bisection method can be used to approximate zeros of polynomial functions like f (x) � x3 � x2 � 3x � 3.

Since f (1) � �4 and f (2) � 3, there is at least one real zero between 1 and 2.

The midpoint of this interval is �1 �2

2� � 1.5. Since f(1.5) � �1.875, the zero is

between 1.5 and 2. The midpoint of this interval is �1.52� 2� � 1.75. Since

f(1.75) is about 0.172, the zero is between 1.5 and 1.75. The midpoint of this

interval is �1.5 �2

1.75� � 1.625 and f(1.625) is about �0.94. The zero is between

1.625 and 1.75. The midpoint of this interval is �1.6252� 1.75� � 1.6875. Since

f (1.6875) is about �0.41, the zero is between 1.6875 and 1.75. Therefore, the zero is 1.7 to the nearest tenth.

The diagram below summarizes the results obtained by the bisection method.

Using the bisection method, approximate to the nearest tenth the zero between the two integral values of x for each function.

1. f (x) � x3 � 4x2 � 11x � 2, f (0) � 2, f (1) � �12

2. f (x) � 2x4 � x2 � 15, f (1) � �12, f (2) � 21

3. f(x) � x5 � 2x3 � 12, f (1) � �13, f (2) � 4

4. f (x) � 4x3 � 2x � 7, f (�2) � �21, f (�1) � 5

5. f (x) � 3x3 � 14x2 � 27x � 126, f (4) � �14, f (5) � 16

1 1.5 21.625 1.75

1.6875

+ +––––sign of f (x ):

value x :

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Identify Rational Zeros

Rational Zero Let f(x) � a0xn � a1xn � 1 � … � an � 2x2 � an � 1x � an represent a polynomial function Theorem with integral coefficients. If �

pq� is a rational number in simplest form and is a zero of y � f(x),

then p is a factor of an and q is a factor of a0.

Corollary (Integral If the coefficients of a polynomial are integers such that a0 � 1 and an 0, any rational Zero Theorem) zeros of the function must be factors of an.

List all of the possible rational zeros of each function.

a. f(x) � 3x4 � 2x2 � 6x � 10

If �pq� is a rational root, then p is a factor of �10 and q is a factor of 3. The possible values

for p are 1, 2, 5, and 10. The possible values for q are 1 and 3. So all of the possible rational zeros are �

pq� � 1, 2, 5, 10, �

13�, �

23�, �

53�, and �

130�.

b. q(x) � x3 � 10x2 � 14x � 36

Since the coefficient of x3 is 1, the possible rational zeros must be the factors of theconstant term �36. So the possible rational zeros are 1, 2, 3, 4, 6, 9, 12, 18,and 36.


1. f(x) � x3 � 3x2 � x � 8 2. g(x) � x5 � 7x4 � 3x2 � x � 20

�1, �2, �4, �8 �1, �2, �4, �5, �10, �20

3. h(x) � x4 � 7x3 � 4x2 � x � 49 4. p(x) � 2x4 � 5x3 � 8x2 � 3x � 5

�1, �7, �49 �1, �5, � , �

5. q(x) � 3x4 � 5x3 � 10x � 12 6. r(x) � 4x5 � 2x � 18�1, �2, �3, �4, �6, �12, �1, �2, �3, �6, �9, �18,

� , � , � � , � , � , � , � , �

7. f(x) � x7 � 6x5 � 3x4 � x3 � 4x2 � 120 8. g(x) � 5x6 � 3x4 � 5x3 � 2x2 � 15

�1, �2, �3, �4, �5, �6, �8, �10, �12, �15, �20, �24, �30, �40, �60, �120

�1, �3, �5, �15, � , �

9. h(x) � 6x5 � 3x4 � 12x3 � 18x2 � 9x � 21 10. p(x) � 2x7 � 3x6 � 11x5 � 20x2 � 11

�1, �3, �7, �21, � , � , � , � , �1, �11, � , �

� , � , � , � 7�

1�

7�

1�

11�

1�

21�

7�

3�

1�

3�

1�

9�

3�

1�

9�

3�

1�

4�

2�

1�

5�

1�

ExampleExample

ExercisesExercises


Find Rational Zeros

Find all of the rational zeros of f(x) � 5x3 � 12x2 � 29x � 12.From the corollary to the Fundamental Theorem of Algebra, we know that there are exactly 3 complex roots. According to Descartes’ Rule of Signs there are 2 or 0 positive real roots and 1 negative real root. The possible rational zeros are 1, 2, 3, 4, 6, 12, , , , , , . Make a table and test some possible rational zeros.

Since f(1) � 0, you know that x � 1 is a zero.The depressed polynomial is 5x2 � 17x � 12, which can be factored as (5x � 3)(x � 4).By the Zero Product Property, this expression equals 0 when x � or x � �4.The rational zeros of this function are 1, , and �4.

Find all of the zeros of f(x) � 8x4 � 2x3 � 5x2 � 2x � 3.There are 4 complex roots, with 1 positive real root and 3 or 1 negative real roots. The possible rational zeros are 1, 3, , , , , , and .3

�83�4

3�2

1�8

1�4

1�2

3�5

3�5

�pq� 5 12 �29 12

1 5 17 �12 0

12�5

6�5

4�5

3�5

2�5

1�5



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Example 1Example 1

Example 2Example 2

ExercisesExercises

Make a table and test some possible values.

Since f� � � 0, we know that x �

is a root.

1�2

1�2

�pq� 8 2 5 2 �3

1 8 10 15 17 14

2 8 18 41 84 165

�12

� 8 6 8 6 0

The depressed polynomial is 8x3 � 6x2 � 8x � 6.Try synthetic substitution again. Any remainingrational roots must be negative.

x � ��34� is another rational root.

The depressed polynomial is 8x2 � 8 � 0,which has roots i.

�pq� 8 6 8 6

��14

� 8 4 7 4�14

�

��34

� 8 0 8 0

The zeros of this function are �12�, ��

34�, and i.

Find all of the rational zeros of each function.

1. f(x) � x3 � 4x2 � 25x � 28 �1, 4, �7 2. f(x) � x3 � 6x2 � 4x � 24 �6


3. f(x) � x4 � 2x3 � 11x2 � 8x � 60 4. f(x) � 4x4 � 5x3 � 30x2 � 45x � 54

3, �5, �2i , �2, �3i3�

Skills PracticeRational Zero Theorem

NAME ______________________________________________ DATE ____________ PERIOD _____

7-67-6


Less

on

7-6


1. n(x) � x2 � 5x � 3 2. h(x) � x2 � 2x � 5

�1, �3 �1, �5

3. w(x) � x2 � 5x � 12 4. f(x) � 2x2 � 5x � 3

�1, �2, �3, �4, �6, �12 ��12

�, ��32

�, �1, �3

5. q(x) � 6x3 � x2 � x � 2 6. g(x) � 9x4 � 3x3 � 3x2 � x � 27

��16

�, ��13

�, ��12

�, ��23

�, �1, �2 ��19

�, ��13

�, �1, �3, �9, �27


7. f(x) � x3 � 2x2 � 5x � 4 � 0 8. g(x) � x3 � 3x2 � 4x � 12

1 �2, 2, 3

9. p(x) � x3 � x2 � x � 1 10. z(x) � x3 � 4x2 � 6x � 4

1 2

11. h(x) � x3 � x2 � 4x � 4 12. g(x) � 3x3 � 9x2 � 10x � 8

1 4

13. g(x) � 2x3 � 7x2 � 7x � 12 14. h(x) � 2x3 � 5x2 � 4x � 3

�4, �1, �32

� �1, �12

�, 3

15. p(x) � 3x3 � 5x2 � 14x � 4 � 0 16. q(x) � 3x3 � 2x2 � 27x � 18

��13

� ��23

�

17. q(x) � 3x3 � 7x2 � 4 18. f(x) � x4 � 2x3 � 13x2 � 14x � 24

��23

�, 1, 2 �3, �1, 2, 4

19. p(x) � x4 � 5x3 � 9x2 � 25x � 70 20. n(x) � 16x4 � 32x3 � 13x2 � 29x � 6

�2, 7 �1, �14

�, �34

�, 2


21. f(x) � x3 � 5x2 � 11x � 15 22. q(x) � x3 � 10x2 � 18x � 4

�3, �1 � 2i, �1 � 2i 2, 4 � �14�, 4 � �14�

23. m(x) � 6x4 � 17x3 � 8x2 � 8x � 3 24. g(x) � x4 � 4x3 � 5x2 � 4x � 4

�13

�, �32

�, , �2, �2, �i, i1 � �5��

1 � �5��



1. h(x) � x3 � 5x2 � 2x � 12 2. s(x) � x4 � 8x3 � 7x � 14

�1, �2, �3, �4, �6, �12 �1, �2, �7, �14

3. f(x) � 3x5 � 5x2 � x � 6 4. p(x) � 3x2 � x � 7

��13

�, ��23

�, �1, �2, �3, �6 ��13

�, ��73

�, �1, �7

5. g(x) � 5x3 � x2 � x � 8 6. q(x) � 6x5 � x3 � 3

��15

�, ��25

�, ��45

�, ��85

�, �1, �2, �4, �8 ��16

�, ��13

�, ��12

�, ��32

�, �1, �3


7. q(x) � x3 � 3x2 � 6x � 8 � 0 �4, �1, 2 8. v(x) � x3 � 9x2 � 27x � 27 3

9. c(x) � x3 � x2 � 8x � 12 �3, 2 10. f(x) � x4 � 49x2 0, �7, 7

11. h(x) � x3 � 7x2 � 17x � 15 3 12. b(x) � x3 � 6x � 20 �2

13. f(x) � x3 � 6x2 � 4x � 24 6 14. g(x) � 2x3 � 3x2 � 4x � 4 �2

15. h(x) � 2x3 � 7x2 � 21x � 54 � 0�3, 2, �

92

�16. z(x) � x4 � 3x3 � 5x2 � 27x � 36 �1, 4

17. d(x) � x4 � x3 � 16 no rational zeros 18. n(x) � x4 � 2x3 � 3 �1

19. p(x) � 2x4 � 7x3 � 4x2 � 7x � 6 20. q(x) � 6x4 � 29x3 � 40x2 � 7x � 12

�1, 1, �32

�, 2 ��32

�, ��43

�


21. f(x) � 2x4 � 7x3 � 2x2 � 19x � 12 22. q(x) � x4 � 4x3 � x2 � 16x � 20

�1, �3, , �2, 2, 2 � i, 2 � i

23. h(x) � x6 � 8x3 24. g(x) � x6 � 1�1, 1, ,

0, 2, �1 � i�3�, �1 � i�3� , ,

25. TRAVEL The height of a box that Joan is shipping is 3 inches less than the width of thebox. The length is 2 inches more than twice the width. The volume of the box is 1540 in3.What are the dimensions of the box? 22 in. by 10 in. by 7 in.

26. GEOMETRY The height of a square pyramid is 3 meters shorter than the side of its base.If the volume of the pyramid is 432 m3, how tall is it? Use the formula V � �

13�Bh. 9 m

1 � i�3��

1 � i�3��

�1 � i�3��

�1 � i�3��

1 � �33��

1 � �33��

Practice (Average)


NAME ______________________________________________ DATE ____________ PERIOD _____

7-67-6

Reading to Learn MathematicsRational Zero Theorem

NAME ______________________________________________ DATE ____________ PERIOD _____

7-67-6


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Pre-Activity How can the Rational Zero Theorem solve problems involving largenumbers?


Rewrite the polynomial equation w(w � 8)(w � 5) � 2772 in the form f(x) � 0, where f(x) is a polynomial written in descending powers of x.w3 � 3w2 � 40w � 2772 � 0

Reading the Lesson

1. For each of the following polynomial functions, list all the possible values of p, all the possible values of q, and all the possible rational zeros �

pq�.

a. f(x) � x3 � 2x2 � 11x � 12

possible values of p: �1, �2, �3, �4, �6, �12

possible values of q: �1

possible values of �pq�: �1, �2, �3, �4, �6, �12

b. f(x) � 2x4 � 9x3 � 23x2 � 81x � 45

possible values of p: �1, �3, �5, �9, �15, �45

possible values of q: �1, �2

possible values of �pq�: �1, �3, �5, �9, �15, �45, ��

12

�, ��32

�, ��52

�, ��92

�, ��125�,

��425�

2. Explain in your own words how Descartes’ Rule of Signs, the Rational Zero Theorem, andsynthetic division can be used together to find all of the rational zeros of a polynomialfunction with integer coefficients.

Sample answer: Use Descartes’ Rule to find the possible numbers ofpositive and negative real zeros. Use the Rational Zero Theorem to listall possible rational zeros. Use synthetic division to test which of thenumbers on the list of possible rational zeros are actually zeros of thepolynomial function. (Descartes’ Rule may help you to limit thepossibilities.)


3. Some students have trouble remembering which numbers go in the numerators and whichgo in the denominators when forming a list of possible rational zeros of a polynomialfunction. How can you use the linear polynomial equation ax � b � 0, where a and b arenonzero integers, to remember this?Sample answer: The solution of the equation is ��

ba

�. The numerator b is a factor of the constant term in ax � b. The denominator a is a factor


Infinite Continued FractionsSome infinite expressions are actually equal to realnumbers! The infinite continued fraction at the right isone example.

If you use x to stand for the infinite fraction, then theentire denominator of the first fraction on the right isalso equal to x. This observation leads to the followingequation:

x � 1 � �1x�

Write a decimal for each continued fraction.

1. 1 � �11� 2. 1 � 3. 1 �

4. 1 � 5. 1 �

6. The more terms you add to the fractions above, the closer their value approaches the value of the infinite continued fraction. What value do the fractions seem to be approaching?

7. Rewrite x � 1 � �1x� as a quadratic equation and solve for x.

8. Find the value of the following infinite continued fraction.

3 � 1

3 � 1

3 � 1

3 � 13 � …

1

1 � 1

1 � 1

1 � 1

1 � 11

1

1 � 1

1 � 1

1 � 11

1

1 � 1

1 � 11

1

1 � 11

Enrichment

NAME ______________________________________________ DATE ____________ PERIOD _____

7-67-6

x � 1 �1

1 � 1

1 � 1

1 � 11 � …

Study Guide and InterventionOperations on Functions

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Arithmetic Operations

Sum (f � g)(x) � f(x) � g(x)Difference (f � g)(x) � f(x) � g(x)

Operations with Functions Product (f � g)(x) � f(x) � g(x)

Quotient � �(x) � , g(x) 0

Find (f � g)(x), (f � g)(x), (f g)(x), and � �(x) for f(x) � x2 � 3x � 4and g(x) � 3x � 2.(f � g)(x) � f(x) � g(x) Addition of functions

� (x2 � 3x � 4) � (3x � 2) f(x) � x2 � 3x � 4, g(x) � 3x � 2

� x2 � 6x � 6 Simplify.

(f � g)(x) � f(x) � g(x) Subtraction of functions

� (x2 � 3x � 4) � (3x � 2) f(x) � x2 � 3x � 4, g(x) � 3x � 2

� x2 � 2 Simplify.

(f � g)(x) � f(x) � g(x) Multiplication of functions

� (x2 � 3x � 4)(3x � 2) f(x) � x2 � 3x � 4, g(x) � 3x � 2

� x2(3x � 2) � 3x(3x � 2) � 4(3x � 2) Distributive Property

� 3x3 � 2x2 � 9x2 � 6x � 12x � 8 Distributive Property

� 3x3 � 7x2 � 18x � 8 Simplify.

� �(x) � Division of functions

� , x �23� f(x) � x2 � 3x � 4 and g(x) � 3x � 2

Find (f � g)(x), (f � g)(x), (f g)(x), and � �(x) for each f(x) and g(x).

1. f(x) � 8x � 3; g(x) � 4x � 5 2. f(x) � x2 � x � 6; g(x) � x � 2

12x � 2; 4x � 8; 32x2 � 28x � 15; x2 � 2x � 8; x2 � 4;

, x � x3 � x2 � 8x � 12; x � 3, x 2

3. f(x) � 3x2 � x � 5; g(x) � 2x � 3 4. f(x) � 2x � 1; g(x) � 3x2 � 11x � 4

3x2 � x � 2; 3x2 � 3x � 8; 3x2 � 13x � 5; �3x2 � 9x � 3; 6x3 � 11x2 � 13x � 15; 6x3 � 19x2 � 19x � 4;

, x , x , �4

5. f(x) � x2 � 1; g(x) �

x2 � 1 � ; x2 � 1 � ; x � 1; x3 � x2 � x � 1, x �11�

1�

1�x � 1

1�

2x � 1��

3�

3x2 � x � 5��

5�

8x � 3�

f�g

x2 � 3x � 4��3x � 2

f(x)�g(x)

f�g

f�g

f(x)�g(x)

f�g

ExampleExample

ExercisesExercises


Composition of Functions

Composition Suppose f and g are functions such that the range of g is a subset of the domain of f.of Functions Then the composite function f � g can be described by the equation [f � g](x) � f [g (x)].

For f � {(1, 2), (3, 3), (2, 4), (4, 1)} and g � {(1, 3), (3, 4), (2, 2), (4, 1)},find f � g and g � f if they exist.f[ g(1)] � f(3) � 3 f[ g(2)] � f(2) � 4 f[ g(3)] � f(4) � 1 f[ g(4)] � f(1) � 2f � g � {(1, 3), (2, 4), (3, 1), (4, 2)}g[f(1)] � g(2) � 2 g[f(2)] � g(4) � 1 g[f(3)] � g(3) � 4 g[f(4)] � g(1) � 3g � f � {(1, 2), (2, 1), (3, 4), (4, 3)}

Find [g � h](x) and [h � g](x) for g(x) � 3x � 4 and h(x) � x2 � 1.[g � h](x) � g[h(x)] [h � g](x) � h[ g(x)]

� g(x2 � 1) � h(3x � 4)� 3(x2 � 1) � 4 � (3x � 4)2 � 1� 3x2 � 7 � 9x2 � 24x � 16 � 1

� 9x2 � 24x � 15

For each set of ordered pairs, find f � g and g � f if they exist.

1. f � {(�1, 2), (5, 6), (0, 9)}, 2. f � {(5, �2), (9, 8), (�4, 3), (0, 4)},g � {(6, 0), (2, �1), (9, 5)} g � {(3, 7), (�2, 6), (4, �2), (8, 10)}f � g � {(2, 2), (6, 9), (9, 6)}; f � g does not exist; g � f � {(�1, �1), (0, 5), (5, 0)} g � f � {(�4, 7), (0, �2), (5, 6), (9, 10)}

Find [f � g](x) and [g � f](x).

3. f(x) � 2x � 7; g(x) � �5x � 1 4. f(x) � x2 � 1; g(x) � �4x2

[f � g](x) � �10x � 5, [f � g](x) � 16x4 � 1, [g � f ](x) � �10x � 36 [g � f ](x) � �4x4 � 8x2 � 4

5. f(x) � x2 � 2x; g(x) � x � 9 6. f(x) � 5x � 4; g(x) � 3 � x[f � g](x) � x2 � 16x � 63, [f � g](x) � 19 � 5x, [g � f ](x) � x2 � 2x � 9 [g � f ](x) � �1 � 5x


Operations on Functions

NAME ______________________________________________ DATE ____________ PERIOD _____

7-77-7

Example 1Example 1

Example 2Example 2

ExercisesExercises

Skills PracticeOperations on Functions

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Find (f � g)(x), (f � g)(x), (f g)(x), and � �(x) for each f(x) and g(x).

1. f(x) � x � 5 2x � 1; 9; 2. f(x) � 3x � 1 5x � 2; x � 4; 6x 2 � 7x �3;

g(x) � x � 4�xx

��

54

�, x 4g(x) � 2x � 3 �

32xx

��

13

�, x �32

�

3. f(x) � x2 x2 � x � 4; x2 � x � 4; 4. f(x) � 3x2 �3x3

x� 5�, x 0; �3x3

x� 5�, x

0;

g(x) � 4 � x 4x2 � x3; , x 4 g(x) � �5x� 15x, x 0; �35

x3�, x 0


5. f � {(0, 0), (4, �2)} 6. f � {(0, �3), (1, 2), (2, 2)}g � {(0, 4), (�2, 0), (5, 0)} g � {(�3, 1), (2, 0)}{(0, �2), (�2, 0), (5, 0)}; {(�3, 2), (2, �3)}; {(0, 4), (4, 0)} {(0, 1), (1, 0), (2, 0)}

7. f � {(�4, 3), (�1, 1), (2, 2)} 8. f � {(6, 6), (�3, �3), (1, 3)}g � {(1, �4), (2, �1), (3, �1)} g � {(�3, 6), (3, 6), (6, �3)}{(1, 3), (2, 1), (3, 1)}; {(�3, 6), (3, 6), (6, �3)};{(�4, �1), (�1, �4), (2, �1)} {(6, �3), (�3, 6), (1, 6)}

Find [g � h](x) and [h � g](x).

9. g(x) � 2x 2x � 4; 2x � 2 10. g(x) � �3x �12x � 3; �12x � 1h(x) � x � 2 h(x) � 4x � 1

11. g(x) � x � 6 x; x 12. g(x) � x � 3 x2 � 3; x2 � 6x � 9h(x) � x � 6 h(x) � x2

13. g(x) � 5x 5x2 � 5x � 5; 14. g(x) � x � 2 2x2 � 1; 2x2 � 8x � 5h(x) � x2 � x � 1 25x2 � 5x � 1 h(x) � 2x2 � 3

If f(x) � 3x, g(x) � x � 4, and h(x) � x2 � 1, find each value.

15. f[ g(1)] 15 16. g[h(0)] 3 17. g[f(�1)] 1

18. h[f(5)] 224 19. g[h(�3)] 12 20. h[f(10)] 899

x2�

f�g

x2 � x � 20;


Find (f � g)(x), (f � g)(x), (f g)(x), and ��gf��(x) for each f(x) and g(x).

1. f(x) � 2x � 1 2. f(x) � 8x2 3. f(x) � x2 � 7x � 12

g(x) � x � 3 g(x) � g(x) � x2 � 9

3x � 2; x � 4; �8x4

x�2

1�, x 0; 2x2 � 7x � 3; 7x � 21;

2x2 � 5x � 3; �8x4

x2� 1�, x 0; x4 � 7x3 � 3x2 � 63x � 108;

�2xx

��

31

�, x 3 8, x 0; 8x4, x 0 �xx

��

43

�, x �3


4. f � {(�9, �1), (�1, 0), (3, 4)} 5. f � {(�4, 3), (0, �2), (1, �2)}g � {(0, �9), (�1, 3), (4, �1)} g � {(�2, 0), (3, 1)}{(0, �1), (�1, 4), (4, 0)}; {(�2, �2), (3, �2)}; {(�9, 3), (�1, �9), (3, �1)} {(�4, 1), (0, 0), (1, 0)}

6. f � {(�4, �5), (0, 3), (1, 6)} 7. f � {(0, �3), (1, �3), (6, 8)}g � {(6, 1), (�5, 0), (3, �4)} g � {(8, 2), (�3, 0), (�3, 1)}{(6, 6), (�5, 3), (3, �5)}; does not exist; {(�4, 0), (0, �4), (1, 1)} {(0, 0), (1, 0), (6, 2)}

Find [g � h](x) and [h � g](x).

8. g(x) � 3x 9. g(x) � �8x 10. g(x) � x � 6h(x) � x � 4 h(x) � 2x � 3 h(x) � 3x2 3x2 � 6;3x � 12; 3x � 4 �16x � 24; �16x � 3 3x2 � 36x � 108

11. g(x) � x � 3 12. g(x) � �2x 13. g(x) � x � 2h(x) � 2x2 h(x) � x2 � 3x � 2 h(x) � 3x2 � 12x2 � 3; �2x2 � 6x � 4; 3x2 � 1; 2x2 � 12x � 18 4x2 � 6x � 2 3x2 � 12x � 13

If f(x) � x2, g(x) � 5x, and h(x) � x � 4, find each value.

14. f[ g(1)] 25 15. g[h(�2)] 10 16. h[f(4)] 20

17. f[h(�9)] 25 18. h[ g(�3)] �11 19. g[f(8)] 320

20. h[f(20)] 404 21. [f � (h � g)](�1) 1 22. [f � (g � h)](4) 1600

23. BUSINESS The function f(x) � 1000 � 0.01x2 models the manufacturing cost per itemwhen x items are produced, and g(x) � 150 � 0.001x2 models the service cost per item.Write a function C(x) for the total manufacturing and service cost per item.C(x) � 1150 � 0.011x2

24. MEASUREMENT The formula f � �1n2� converts inches n to feet f, and m � �52

f80� converts

feet to miles m. Write a composition of functions that converts inches to miles.

[m � f ]n � �63,

n360�

1�x2

Practice (Average)

Operations on Functions

NAME ______________________________________________ DATE ____________ PERIOD _____

7-77-7

Reading to Learn MathematicsOperations on Functions

NAME ______________________________________________ DATE ____________ PERIOD _____

7-77-7


Less

on

7-7

Pre-Activity Why is it important to combine functions in business?


Describe two ways to calculate Ms. Coffmon’s profit from the sale of 50 birdhouses. (Do not actually calculate her profit.) Sample answer: 1. Find the revenue by substituting 50 for x in the expression125x. Next, find the cost by substituting 50 for x in theexpression 65x � 5400. Finally, subtract the cost from therevenue to find the profit. 2. Form the profit function p(x) � r(x) � c(x) � 125x � (65x � 5400) � 60x � 5400.Substitute 50 for x in the expression 60x � 5400.

Reading the Lesson

1. Determine whether each statement is true or false. (Remember that true means always true.)

a. If f and g are polynomial functions, then f � g is a polynomial function. true

b. If f and g are polynomial functions, then is a polynomial function. false

c. If f and g are polynomial functions, the domain of the function f � g is the set of allreal numbers. true

d. If f(x) � 3x � 2 and g(x) � x � 4, the domain of the function is the set of all realnumbers. false

e. If f and g are polynomial functions, then (f � g)(x) � (g � f)(x). false

f. If f and g are polynomial functions, then (f � g)(x) � (g � f)(x) true

2. Let f(x) � 2x � 5 and g(x) � x2 � 1.

a. Explain in words how you would find ( f � g)(�3). (Do not actually do any calculations.)Sample answer: Square �3 and add 1. Take the number you get,multiply it by 2, and subtract 5.

b. Explain in words how you would find (g � f)(�3). (Do not actually do anycalculations.) Sample answer: Multiply �3 by 2 and subtract 5. Take thenumber you get, square it, and add 1.


3. Some students have trouble remembering the correct order in which to apply the twooriginal functions when evaluating a composite function. Write three sentences, each ofwhich explains how to do this in a slightly different way. (Hint: Use the word closest inthe first sentence, the words inside and outside in the second, and the words left andright in the third.) Sample answer: 1. The function that is written closest tothe variable is applied first. 2. Work from the inside to the outside. 3. Work from right to left.

f�g

f�g


Relative Maximum ValuesThe graph of f (x) � x3 � 6x � 9 shows a relative maximum value somewhere between f (�2) and f (�1). You can obtain a closer approximation by comparing values such as those shown in the table.

To the nearest tenth a relative maximum value for f (x) is �3.3.

Using a calculator to find points, graph each function. To the nearest tenth, find a relative maximum value of the function.

1. f (x) � x(x2 � 3) 2. f (x) � x3 � 3x � 3

3. f (x) � x3 � 9x � 2 4. f (x) � x3 � 2x2 � 12x � 24

5

x

f(x)

O 1

2

x

f(x)

O 2

x

f(x)

O

x

f(x)

O

x f (x)

�2 �5

�1.5 �3.375

�1.4 �3.344

�1.3 �3.397

�1 �4

x

f(x)

O 2–2–4

–8

–12

–16

–20

–4 4

f(x) � x3 � 6x � 9

Enrichment

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Study Guide and InterventionInverse Functions and Relations

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Find Inverses

Inverse RelationsTwo relations are inverse relations if and only if whenever one relation contains the element (a, b), the other relation contains the element (b, a).

Property of Inverse Suppose f and f�1 are inverse functions.Functions Then f(a) � b if and only if f�1(b) � a.

Find the inverse of the function f(x) � x � . Then graph thefunction and its inverse.Step 1 Replace f(x) with y in the original equation.

f(x) � �25�x � → y � �

25�x �

Step 2 Interchange x and y.

x � �25�y �

Step 3 Solve for y.

x � �25�y � Inverse

5x � 2y � 1 Multiply each side by 5.

5x � 1 � 2y Add 1 to each side.

(5x � 1) � y Divide each side by 2.

The inverse of f(x) � �25�x � is f�1(x) � (5x � 1).

Find the inverse of each function. Then graph the function and its inverse.

1. f(x) � x � 1 2. f(x) � 2x � 3 3. f(x) � x � 2

f�1(x) � x � f�1(x) � x � f�1(x) � 4x � 8

x

f(x)

Ox

f(x)

O

x

f(x)

O

3�

1�

3�

3�

1�4

2�3

1�2

1�5

1�2

1�5

1�5

1�5

1�5

x

f(x)

O

f(x) � 2–5x � 1–5

f –1(x) � 5–2x � 1–2

1�5

2�5

ExampleExample

ExercisesExercises


Inverses of Relations and Functions

Inverse Functions Two functions f and g are inverse functions if and only if [f � g](x) � x and [g � f ](x) � x.

Determine whether f(x) � 2x � 7 and g(x) � (x � 7) are inversefunctions.

[ f � g](x) � f[ g(x)] [ g � f ](x) � g[ f(x)]

� f��12�(x � 7)� � g(2x � 7)

� 2��12�(x � 7)� � 7 � �

12�(2x � 7 � 7)

� x � 7 � 7 � x� x

The functions are inverses since both [ f � g](x) � x and [ g � f ](x) � x.

Determine whether f(x) � 4x � and g(x) � x � 3 are inversefunctions.

[ f � g](x) � f[ g(x)]

� f��14�x � 3�

� 4��14�x � 3� � �

13�

� x � 12 � �13�

� x � 11�23�

Since [ f � g](x) x, the functions are not inverses.

Determine whether each pair of functions are inverse functions.

1. f(x) � 3x � 1 2. f(x) � �14�x � 5 3. f(x) � �

12�x � 10

g(x) � �13�x � �

13� yes g(x) � 4x � 20 yes g(x) � 2x � �1

10� no

4. f(x) � 2x � 5 5. f(x) � 8x � 12 6. f(x) � �2x � 3

g(x) � 5x � 2 no g(x) � �18�x � 12 no g(x) � ��

12�x � �

32� yes

7. f(x) � 4x � �12� 8. f(x) � 2x � �

35� 9. f(x) � 4x � �

12�

g(x) � �14�x � �

18� yes g(x) � �1

10�(5x � 3) yes g(x) � �

12�x � �

32� no

10. f(x) � 10 � �2x

� 11. f(x) � 4x � �45� 12. f(x) � 9 � �

32�x

g(x) � 20 � 2x yes g(x) � �4x

� � �15� yes g(x) � �

23�x � 6 yes

1�4

1�3

1�2


Inverse Functions and Relations

NAME ______________________________________________ DATE ____________ PERIOD _____

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Example 1Example 1

Example 2Example 2

ExercisesExercises

Skills PracticeInverse Functions and Relations

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on

7-8

Find the inverse of each relation.

1. {(3, 1), (4, �3), (8, �3)} 2. {(�7, 1), (0, 5), (5, �1)}{(1, 3), (�3, 4), (�3, 8)} {(1, �7), (5, 0), (�1, 5)}

3. {(�10, �2), (�7, 6), (�4, �2), (�4, 0)} 4. {(0, �9), (5, �3), (6, 6), (8, �3)}{(�2, �10), (6, �7), (�2, �4), (0, �4)} {(�9, 0), (�3, 5), (6, 6), (�3, 8)}

5. {(�4, 12), (0, 7), (9, �1), (10, �5)} 6. {(�4, 1), (�4, 3), (0, �8), (8, �9)}{(12, �4), (7, 0), (�1, 9), (�5, 10)} {(1, �4), (3, �4), (�8, 0), (�9, 8)}


7. y � 4 8. f(x) � 3x 9. f(x) � x � 2

x � 4 f �1(x) � �13

�x f �1(x) � x � 2

10. g(x) � 2x � 1 11. h(x) � �14�x 12. y � �

23�x � 2

g�1(x) � �x �

21

� h�1(x) � 4x y � �32

�x � 3


13. f(x) � x � 1 no 14. f(x) � 2x � 3 yes 15. f(x) � 5x � 5 yesg(x) � 1 � x g(x) � �

12�(x � 3) g(x) � �

15�x � 1

16. f(x) � 2x yes 17. h(x) � 6x � 2 no 18. f(x) � 8x � 10 yesg(x) � �

12�x g(x) � �

16�x � 3 g(x) � �

18�x � �

54�

x

y

Ox

h(x)

Ox

g(x)

O

x

f(x)

Ox

f(x)

Ox

y

O


Find the inverse of each relation.

1. {(0, 3), (4, 2), (5, �6)} 2. {(�5, 1), (�5, �1), (�5, 8)}{(3, 0), (2, 4), (�6, 5)} {(1, �5), (�1, �5), (8, �5)}

3. {(�3, �7), (0, �1), (5, 9), (7, 13)} 4. {(8, �2), (10, 5), (12, 6), (14, 7)}{(�7, �3), (�1, 0), (9, 5), (13, 7)} {(�2, 8), (5, 10), (6, 12), (7, 14)}

5. {(�5, �4), (1, 2), (3, 4), (7, 8)} 6. {(�3, 9), (�2, 4), (0, 0), (1, 1)}{(�4, �5), (2, 1), (4, 3), (8, 7)} {(9, �3), (4, �2), (0, 0), (1, 1)}


7. f(x) � �34�x 8. g(x) � 3 � x 9. y � 3x � 2

f�1(x) � �43

�x g�1(x) � x � 3 y � �x �

32

�


10. f(x) � x � 6 yes 11. f(x) � �4x � 1 yes 12. g(x) � 13x � 13 nog(x) � x � 6 g(x) � �

14�(1 � x) h(x) � �1

13�x � 1

13. f(x) � 2x no 14. f(x) � �67�x yes 15. g(x) � 2x � 8 yes

g(x) � �2x g(x) � �76�x h(x) � �

12�x � 4

16. MEASUREMENT The points (63, 121), (71, 180), (67, 140), (65, 108), and (72, 165) givethe weight in pounds as a function of height in inches for 5 students in a class. Give thepoints for these students that represent height as a function of weight.(121, 63), (180, 71), (140, 67), (108, 65), (165, 72)

REMODELING For Exercises 17 and 18, use the following information.The Clearys are replacing the flooring in their 15 foot by 18 foot kitchen. The new flooringcosts $17.99 per square yard. The formula f(x) � 9x converts square yards to square feet.

17. Find the inverse f�1(x). What is the significance of f�1(x) for the Clearys? f�1(x) � �x9

�; It will allow them to convert the square footage of their kitchen floor tosquare yards, so they can then calculate the cost of the new flooring.

18. What will the new flooring cost the Cleary’s? $539.70

x

g(x)

Ox

f(x)

O

Practice (Average)

Inverse Functions and Relations

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Reading to Learn MathematicsInverse Functions and Relations

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Pre-Activity How are inverse functions related to measurement conversions?


A function multiplies a number by 3 and then adds 5 to the result. What doesthe inverse function do, and in what order? Sample answer: It firstsubtracts 5 from the number and then divides the result by 3.

Reading the Lesson

1. Complete each statement.

a. If two relations are inverses, the domain of one relation is the ofthe other.

b. Suppose that g(x) is a relation and that the point (4, �2) is on its graph. Then a point

on the graph of g�1(x) is .

c. The test can be used on the graph of a function to determine

whether the function has an inverse function.

d. If you are given the graph of a function, you can find the graph of its inverse by

reflecting the original graph over the line with equation .

e. If f and g are inverse functions, then (f � g)(x) � and

(g � f)(x) � .

f. A function has an inverse that is also a function only if the given function is

.

g. Suppose that h(x) is a function whose inverse is also a function. If h(5) � 12, thenh�1(12) � .

2. Assume that f(x) is a one-to-one function defined by an algebraic equation. Write the foursteps you would follow in order to find the equation for f�1(x).

1. Replace f(x) with y in the original equation.

2. Interchange x and y.

3. Solve for y.

4. Replace y with f �1(x).


3. A good way to remember something new is to relate it to something you already know.How are the vertical and horizontal line tests related? Sample answer: The verticalline test determines whether a relation is a function because the orderedpairs in a function can have no repeated x-values. The horizontal linetest determines whether a function is one-to-one because a one-to-onefunction cannot have any repeated y-values.

5

one-to-one

xx

y � x

horizontal line

(�2, 4)

range


Miniature GolfIn miniature golf, the object of the game is to roll the golf ball into the hole in as few shots as possible. As in the diagram at the right,the hole is often placed so that a direct shot is impossible. Reflectionscan be used to help determine the direction that the ball should berolled in order to score a hole-in-one.

Using wall E�F�, find the path to use to score a hole-in-one.

Find the reflection image of the “hole” with respect to E�F� and label it H . The intersection of B�H� � with wall E�F� is the point at which the shot should be directed.

For the hole at the right, find a path to score a hole-in-one.

Find the reflection image of H with respect to E�F� and label it H .In this case, B�H� � intersects J�K� before intersecting E�F�. Thus, thispath cannot be used. To find a usable path, find the reflection image of H with respect to G�F� and label it H�. Now, the intersection of B�H�� with wall G�F� is the point at which the shotshould be directed.

Copy each figure. Then, use reflections to determine a possible path for a hole-in-one.

1. 2. 3.

B

G F

J K H'

H"

E

H

Ball

Hole

E

H'

F

Ball

Hole

Enrichment

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Example 2Example 2

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Square Root Functions A function that contains the square root of a variableexpression is a square root function.

Graph y � �3x ��2�. State its domain and range.

Since the radicand cannot be negative, 3x � 2 � 0 or x � �23�.

The x-intercept is �23�. The range is y � 0.

Make a table of values and graph the function.

Graph each function. State the domain and range of the function.

1. y � �2x� 2. y � �3�x� 3. y � ��2x�

D: x � 0; R: y � 0 D: x � 0; R: y � 0 D: x � 0; R: y � 0

4. y � 2�x � 3� 5. y � ��2x � 3� 6. y � �2x � 5�

D: x � 3; R: y � 0 D: x � �32

�; R: y � 0 D: x � ��52

�; R: y � 0

x

y

O

x

y

O

x

y

O

x

y

O

xy

O

x

y

O

x y

�23

� 0

1 1

2 2

3 �7�

x

y

O

y � ��3x � 2

ExampleExample

ExercisesExercises


Square Root Inequalities A square root inequality is an inequality that containsthe square root of a variable expression. Use what you know about graphing square rootfunctions and quadratic inequalities to graph square root inequalities.

Graph y � �2x ��1� � 2.Graph the related equation y � �2x � 1� � 2. Since the boundary should be included, the graph should be solid.

The domain includes values for x � �12�, so the graph is to the right

of x � �12�. The range includes only numbers greater than 2, so the

graph is above y � 2.

Graph each inequality.

1. y � 2�x� 2. y � �x � 3� 3. y � 3�2x � 1�

4. y � �3x � 4� 5. y � �x � 1� � 4 6. y � 2�2x � 3�

7. y � �3x � 1� � 2 8. y � �4x � 2� � 1 9. y � 2�2x � 1� � 4

x

y

O

x

y

O

x

y

O

x

y

O

x

y

O

x

y

O

x

y

O

x

y

Ox

y

O

x

y

O

y � ��2x � 1 � 2


Square Root Functions and Inequalities

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ExampleExample

ExercisesExercises

Skills PracticeSquare Root Functions and Inequalities

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Graph each function. State the domain and range of each function.

1. y � �2x� 2. y � ��3x� 3. y � 2�x�

D: x � 0, R: y � 0 D: x � 0, R: y � 0 D: x � 0, R: y � 0

4. y � �x � 3� 5. y � ��2x � 5� 6. y � �x � 4� � 2

D: x � �3, R: y � 0 D: x � 2.5, R: y � 0 D: x � �4, R: y � �2


7. y � �4x� 8. y � �x � 1� 9. y � �4x � 3�

x

y

Ox

y

Ox

y

O

x

y

Ox

y

O

x

y

O

x

y

O

x

y

Ox

y

O


Graph each function. State the domain and range of each function.

1. y � �5x� 2. y � ��x � 1� 3. y � 2�x � 2�

D: x � 0, R: y � 0 D: x � 1, R: y � 0 D: x � �2, R: y � 0

4. y � �3x � 4� 5. y � �x � 7� � 4 6. y � 1 � �2x � 3�

D: x � �43

�, R: y � 0 D: x � �7, R: y � �4 D: x � ��32

�, R: y � 1


7. y � ��6x� 8. y � �x � 5� � 3 9. y � �2�3x � 2�

10. ROLLER COASTERS The velocity of a roller coaster as it moves down a hill is v � �v0

2 ��64h�, where v0 is the initial velocity and h is the vertical drop in feet. If v � 70 feet per second and v0 � 8 feet per second, find h. about 75.6 ft

11. WEIGHT Use the formula d � �� 3960, which relates distance from Earth d

in miles to weight. If an astronaut’s weight on Earth WE is 148 pounds and in space Ws is115 pounds, how far from Earth is the astronaut? about 532 mi

39602 WE��Ws

x

y

O

x

y

O

x

y

O

x

y

O

x

y

O

Practice (Average)

Square Root Functions and Inequalities

NAME ______________________________________________ DATE ____________ PERIOD _____

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Reading to Learn MathematicsSquare Root Functions

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Pre-Activity How are square root functions used in bridge design?


If the weight to be supported by a steel cable is doubled, should thediameter of the support cable also be doubled? If not, by what numbershould the diameter be multiplied?

no; �2�

Reading the Lesson

1. Match each square root function from the list on the left with its domain and range fromthe list on the right.

a. y � �x� iv i. domain: x � 0; range: y � 3

b. y � �x � 3� viii ii. domain: x � 0; range: y � 0

c. y � �x� � 3 i iii. domain: x � 0; range: y � �3

d. y � �x � 3� v iv. domain: x � 0; range: y � 0

e. y � ��x� ii v. domain: x � 3; range: y � 0

f. y � ��x � 3� vii vi. domain: x � 3; range: y � 3

g. y � �3 � x� � 3 vi vii. domain: x � 3; range: y � 0

h. y � ��x� � 3 iii viii. domain: x � �3; range: y � 0

2. The graph of the inequality y � �3x � 6� is a shaded region. Which of the followingpoints lie inside this region?

(3, 0) (2, 4) (5, 2) (4, �2) (6, 6)

(3, 0), (5, 2), (4, �2)


3. A good way to remember something is to explain it to someone else. Suppose you arestudying this lesson with a classmate who thinks that you cannot have square rootfunctions because every positive real number has two square roots. How would youexplain the idea of square root functions to your classmate?

Sample answer: To form a square root function, choose either the positive or negative square root. For example, y � �x� and y � ��x� aretwo separate functions.


Reading AlgebraIf two mathematical problems have basic structural similarities,they are said to be analogous. Using analogies is one way ofdiscovering and proving new theorems.

The following numbered sentences discuss a three-dimensionalanalogy to the Pythagorean theorem.

01 Consider a tetrahedron with three perpendicular faces thatmeet at vertex O.

02 Suppose you want to know how the areas A, B, and C of the three faces that meet at vertex O are related to the area Dof the face opposite vertex O.

03 It is natural to expect a formula analogous to the Pythagorean theorem z2 � x2 � y2, which is true for a similar situation in two dimensions.

04 To explore the three-dimensional case, you might guess a formula and then try to prove it.

05 Two reasonable guesses are D3 � A3 � B3 � C3 and D2 � A2 � B2 � C2.

Refer to the numbered sentences to answer the questions.

1. Use sentence 01 and the top diagram. The prefix tetra- means four. Write aninformal definition of tetrahedron.

2. Use sentence 02 and the top diagram. What are the lengths of the sides ofeach face of the tetrahedron?

3. Rewrite sentence 01 to state a two-dimensional analogue.

4. Refer to the top diagram and write expressions for the areas A, B, and C

5. To explore the three-dimensional case, you might begin by expressing a, b,and c in terms of p, q, and r. Use the Pythagorean theorem to do this.

6. Which guess in sentence 05 seems more likely? Justify your answer.

y

O

z

x

b

c

O

p

a

qr

Enrichment

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Chapter 7 Test, Form 1

NAME DATE PERIOD

SCORE


Ass

essm

ent

Write the letter for the correct answer in the blank at the right of each question.

1. Find p(�3) if p(x) � 4 � x.A. 12 B. 4 C. 1 D. 7 1.

2. State the number of real zeros for the function whose graph is shown at the right.A. 0 B. 1C. 2 D. 3 2.

For Questions 3 and 4, use the graph shown at the right.

3. Determine the values of x between which a real zero is located.A. between �1 and 0B. between 6 and 7C. between �2 and �1D. between 2 and 3 3.

4. Estimate the x-coordinate at which a relative minimum occurs.A. 3 B. 2 C. 0 D. �1 4.

5. Write the expression x4 � 5x2 � 8 in quadratic form, if possible.A. (x2)2 � 5(x2) � 8 B. (x4)2 � 5(x4) � 8C. (x2)2 � 5(x2) � 8 D. not possible 5.

6. Solve x4 � 13x2 � 36 � 0.A. �3, �2, 2, 3 B. �9, �4, 4, 9 C. 2, 3, 2i, 3i D. �2, �3, 2i, 3i 6.

7. Use synthetic substitution to find f(3) for f(x) � x2 � 9x � 5.A. �23 B. �16 C. �13 D. 41 7.

8. One factor of x3 � 4x2 � 11x � 30 is x � 2. Find the remaining factors.A. x � 5, x � 3 B. x � 3, x � 5 C. x � 6, x � 5 D. x � 5, x � 6 8.

9. Which describes the number and type of roots of the equation 4x � 7 � 0?A. 1 imaginary root B. 1 real root and 1 imaginary rootC. 2 real roots D. 1 real root 9.

10. Which is not a root of the equation x3 � x2 � 10x � 8 � 0?A. 1 B. 4 C. �2 D. �1 10.

11. List all of the possible rational zeros of f(x) � x3 � 7x2 � 8x � 6.

A. �1, ��12�, � �

13�, �

16� B. 0, �1, �2, �3, �6

C. �1, �2, �3, �4, �6 D. �1, �2, �3, �6 11.

77

xO

f(x )

xO

f(x )


Chapter 7 Test, Form 1 (continued)

12. Find all of the rational zeros of p(x) � x3 � 12x � 16.A. �2, 4 B. 2, �4 C. 4 D. �2 12.

For Questions 13 and 14, use f(x) � x � 5 and g(x) � 2x.

13. Find (f � g)(x).A. 3x � 5 B. x � 5 C. 2x � 10 D. 2x2 � 5 13.

14. Find (f � g)(x).A. 2x2 � 5 B. 3x2 � 10x C. 2x2 � 10x D. 2x � 10 14.

15. If f(x) � 3x � 7 and g(x) � 2x � 5, find g[f(�3)].A. �26 B. �9 C. �1 D. 10 15.

16. If f(x) � x2 and g(x) � 3x � 1 find [ g � f](x).A. x2 � 3x � 1 B. 9x2 � 1C. 9x2 � 6x � 1 D. 3x2 � 1 16.

17. Find the inverse of g(x) � �3x.A. g�1(x) � x � 1 B. g�1(x) � �3x � 3

C. g�1(x) � x � 1 D. g�1(x) � ��13�x 17.

18. Determine which pair of functions are inverse functions.A. f(x) � x � 4 B. f(x) � x � 4

g(x) � x � 4 g(x) � 4x � 1C. f(x) � x � 4 D. f(x) � 4x � 1 18.

g(x) � �x �

44

� g(x) � 4x � 1

19. State the domain and range of the function graphed.A. D: x � 2, R: y � 0B. D: x 2, R: y � 0C. D: x 2, R: y 0D. D: x 2, R: y 0 19.

20. Which inequality is graphed?

A. y � �4x � 8�B. y � �4x � 8�C. y �4x � 8�D. y �4x � 8� 20.

Bonus If g(x) � 2x � 1, find g[g(x)]. B:

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Chapter 7 Test, Form 2A

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1. Find p(�4) if p(x) � 3x3 � 2x2 � 6x � 4.A. �188 B. �252 C. �140 D. �204 1.

2. If r(x) � x3 � 2x � 1, find r(2a3).A. 8a6 � 4a3 � 1 B. 4a6 � 4a3 � 1C. 6a6 � 4a3 � 1 D. 8a9 � 4a3 � 1 2.

3. State the number of real zeros for the function whose graph is shown at the right.A. 0 B. 2C. 3 D. 1 3.

For Questions 4 and 5, use the graph shown.

4. Determine the values of x between which a real zero is located.A. between 1 and 2B. between �4 and �3C. between �2 and �1D. between 2 and 3 4.

5. Estimate the x-coordinate at which a relative maximum occurs.A. 1 B. �1 C. 2 D. �2 5.

6. Write the expression 10x8 � 6x4 � 20 in quadratic form, if possible.A. 10(x4)2 � 6(x2)2 � 20 B. 10(x4)2 � 6(x4) � 20C. 10(x2)4 � 6(x2)2 � 20 D. not possible 6.

7. Solve x4 � 6x2 � 27 � 0.A. �3�, 3, 3i, i�3� B. �3, ��3�, �3�, 3C. �3, 3, i�3�, �i�3� D. ��3�, 3, 3i, �3i 7.

8. Use synthetic substitution to find f(�2) for f(x) � 2x4 � 3x3 � x2 � x � 5.A. 15 B. 67 C. 63 D. 19 8.


10. Which describes the number and type of roots of the equation x4 � 64 � 0?A. 2 real roots, 2 imaginary roots B. 4 real rootsC. 3 real roots, 1 imaginary root D. 4 imaginary roots 10.

11. State the possible number of imaginary zeros of f(x) � 7x3 � x2 � 10x � 4.A. exactly 1 B. exactly 3 C. 3 or 1 D. 2 or 0 11.

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Chapter 7 Test, Form 2A (continued)

12. Write a polynomial function of least degree with integral coefficients whose zeros include 4 and 2i.A. f(x) � x2 � 4 B. f(x) � x3 � 4x2 � 4x � 16C. f(x) � x3 � 4x2 � 4x � 16 D. f(x) � x3 � 4x2 � 4x � 16 12.

13. List all of the possible rational zeros of f(x) � 3x3 � 2x2 � 7x � 6.

A. �1, �2, �3, �6 B. 0, �1, �2, �3, �6, ��13�, ��

23�

C. �1, �2, �3, �6, ��13�, ��

23� D. �1, �3, ��

16�, ��

13�, ��

12�, ��

32� 13.

14. Find all of the rational zeros of f(x) � 4x3 � 3x2 � 22x � 15.

A. ��52�, �1, �3 B. �1, 3 C. �1, 3 D. �5, �1, 3 14.

15. Find ( f � g)(x) for f(x) � 3x2 and g(x) � 5 � x.A. 3x2 � x � 5 B. 75 � 30x � 3x2

C. 3x2 � 15x2 D. 15x2 � 3x3 15.

16. If f(x) � x2 � 1, and g(x) � x � 2, find [f � g](x).A. x2 � 4x � 5 B. x2 � 1C. x2 � 3 D. x3 � 2x2 � x � 2 16.

17. State the domain and range of the function graphed at the right.A. D: x � �3, R: y � 0B. D: x � �3, R: y 0C. D: x �3, R: y 0D. D: x �3, R: y � 0 17.

18. Find the inverse of f(x) � 2x � 7.

A. f�1(x) � 7x � 2 B. f�1(x) � �12�x � 7

C. f�1(x) � �x �

27

� D. f�1(x) � x � �72� 18.

19. Determine which pair of functions are inverse functions.A. f(x) � 3x � 1 B. f(x) � 2x � 5 C. f(x) � 2x � 2 D. f(x) � 3x � 8 19.

g(x) � �3x1� 1� g(x) � �

x �2

5� g(x) � 2x � 2 g(x) � �

13�x � 8

20. Which inequality is graphed at the right?A. y � �x � 4� B. y �x � 4�C. y �x � 4� D. y � �x � 4� 20.

Bonus If f(x) � 3x � 4, solve f [f(x)] � f(x) for x. B:

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2

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2�2

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8

�8 4

�4

�4

Chapter 7 Test, Form 2B

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1. Find p(�3) if p(x) � 4x3 � 5x2 � 7x � 10.A. �94 B. 32 C. �184 D. �142 1.

2. If r(x) � 4x2 � 3x � 7, find r(3a2).A. 36a4 � 9a2 � 7 B. 144a4 � 9a2 � 7C. 36a4 � 9a2 � 7 D. 12a4 � 9a2 � 7 2.

3. State the number of real zeros for the function whose graph is shown.A. 1 B. 4C. 3 D. 2 3.


4. Determine the values of x between which a real zero is located.A. between �2 and �1B. between �1 and 0C. between 0 and 1D. between �3 and �2 4.

5. Estimate the x-coordinate at which a relative minimum occurs.A. �1 B. 0 C. 1 D. 2 5.

6. Write the expression 9n6 � 7n3 � 6 in quadratic form, if possible.A. 9(n3)3 � 7(n3) � 6 B. 9(n2)3 � 7(n2) � 6C. 9(n3)2 � 7(n3) � 6 D. not possible 6.

7. Solve b4 � 2b2 � 24 � 0.A. �2, ��6�, �6�, 2 B. ��6�, 2, 2i, i�6�C. �2, 2, �i�6�, i�6� D. �2i, 2i, ��6�, �6� 7.

8. Use synthetic substitution to find f(�3) for f(x) � x4 � 4x3 � 2x2 � 4x � 6.A. 9 B. 225 C. 201 D. �15 8.


10. Which describes the number and type of roots of the equation x3 � 121x � 0?A. 1 real root, 2 imaginary roots B. 2 real roots, 1 imaginary rootC. 3 real roots D. 3 imaginary roots 10.

11. State the possible number of imaginary zeros of g(x) � x4 � 3x3 � 7x2 � 6x � 13.A. 3 or 1 B. 2 or 0 C. exactly 1 D. exactly 3 11.

77

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Chapter 7 Test, Form 2B (continued)

12. Write a polynomial function of least degree with integral coefficients whose zeros include �3 and i.A. g(x) � x3 � 3x2 � x � 3 B. g(x) � x3 � 3x2 � x � 3C. g(x) � x2 � 1 D. g(x) � x3 � 3x2 � x � 3 12.

13. List all of the possible rational zeros of p(x) � 2x3 � 6x2 � 7x � 6.

A. �1, �2, �3, �6, ��12�, ��

32� B. �1, �2, �3, �6

C. �1, �2, ��16�, ��

13�, ��

12�, ��

23� D. �1, �2, �3, �6, ��

13�, ��

12�, ��

23� 13.

14. Find all of the rational zeros of g(x) � 2x3 � 11x2 � 8x � 21.

A. �1, 3, �72� B. �1, �3, ��

72� C. �1, 3 D. �1, 3, 7 14.

15. Find (f � g)(x) for f(x) � x2 � 8x and g(x) � 3x � 5.A. �x2 � 5x � 5 B. x2 � 5x � 5 C. x2 � 5x � 5 D. x2 � 11x � 5 15.

16. If f(x) � x2 � 3, and g(x) � 2x � 1, find [g � f ](x).A. 2x3 � x2 � 6x � 3 B. 4x2 � 4x � 2C. x2 � 2x � 4 D. 2x2 � 7 16.

17. State the domain and range of the function graphed at the right.A. D: x � �4, R: y � 0B. D: x �4, R: y 0C. D: x �4, R: y � 0D. D: x � �4, R: y 0 17.

18. Find the inverse of f(x) � 3 � 5x.

A. f �1(x) � 5 � 3x B. f �1(x) � �x �

53

�

C. f �1(x) � �3 �

55x

� D. f �1(x) � �3 � �15�x 18.

19. Determine which pair of functions are not inverse functions.A. g(x) � 2x � 9 B. g(x) � x � 1 C. g(x) � 3x � 6 D. g(x) � 3x � 4 19.

h(x) � �12�x � 9 h(x) � x � 1 h(x) � �

13�x � 2 h(x) � �

x �3

4�

20. Which inequality is graphed at the right?A. y �x � 3� B. y �x � 3�C. y � �x � 3� D. y � �x � 3� 20.

Bonus If g(x) � 4x � 9, solve g [g(x)] � g(x) B:for x.

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Chapter 7 Test, Form 2C


1. Find p(�5) if p(x) � x3 � 2x2 � x � 4. 1.

2. Find p(x � 1) if p(x) � x2 � 3x � 1. 2.

3. Determine whether the graph 3.represents an odd-degree or an even-degree polynomial function. Then state the number of real zeros.

4. Graph f(x) � x3 � 3x � 1 by making a table of values. 4.Then determine consecutive values of x between which each real zero is located.

5. For the graph in Question 4, estimate the x-coordinates 5.at which the relative maxima and relative minima occur.

6. Write the expression 9n6 � 36n3 in quadratic form, if 6.possible.

7. Solve x4 � 12x2 � 45 � 0. 7.

8. Use synthetic substitution to find f(�4) for 8.f(x) � x3 � 3x2 � 5x � 7.

9. One factor of x3 � 2x2 � 23x � 60 is x � 4. Find the 9.remaining factors.

10. State the possible number of positive real zeros, 10.negative real zeros, and imaginary zeros for f (x) � 3x4 � 2x3 � 5x2 � 6x � 2.

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Chapter 7 Test, Form 2C (continued)

11. Find all the zeros of the function h(x) � x3 � 5x2 � 4x � 20. 11.

12. List all of the possible rational zeros of 12.f(x) � 2x3 � x2 � 4x � 8.

13. Find all of the rational zeros of g(x) � 2x3 � x2 � 7x � 6. 13.

14. Find ( f � g)(x) for f(x) � x2 � 4 and g(x) � 7 � x. 14.

15. If f(x) � x � 5 and g(x) � x2 � 3, find f [g(�2)]. 15.

16. If f(x) � 2x � 5 and g(x) � x2 � 3, find [f � g](x). 16.

17. Find the inverse of f(x) � 5x � 10. 17.

18. Determine whether f(x) � 5x � 3 and g(x) � �x �

53

� are 18.

inverse functions.

19. Graph y � �2x � 8�. Then state the domain and range 19.of the function.

20. Graph y �x � 2�. 20.

Bonus If g(x) � 5x � 8, solve g[ g(x)] � g(x) for x. B:

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Chapter 7 Test, Form 2D


1. Find p(�4) if p(x) � x3 � 3x2 � 7x � 6. 1.

2. Find p(x � 1) if p(x) � x2 � 4x � 2. 2.

3. Determine whether the graph represents 3.an odd-degree or an even-degree polynomial function. Then state the number of real zeros.

4. Graph f(x) � �x3 � 3x � 1 by making a table of values. 4.Then determine consecutive values of x between which each real zero is located.

5. For the graph in Question 4, estimate the x-coordinates at 5.which the relative maxima and relative minima occur.

6. Write the expression 5x10 � 4x5 � 3 in quadratic form, if 6.possible.

7. Solve x4 � 4x2 � 12 � 0. 7.

8. Use synthetic substitution to find f(�4) for 8.f(x) � x4 � 7x2 � 12.

9. One factor of g(x) � x3 � x2 � 9x � 9 is x � 3. Find the 9.remaining factors.

10. State the number of positive real zeros, negative real zeros, 10.and imaginary zeros for f(x) � 2x4 � 5x3 � 3x2 � x � 6.

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Chapter 7 Test, Form 2D (continued)

11. Find all the zeros of the function p(x) � x3 � 2x2 � 9x � 18. 11.

12. List all of the possible rational zeros of 12.g(x) � 2x3 � 2x2 � 7x � 14.

13. Find all of the rational zeros of h(x) � 3x3 � 4x2 � 13x � 6. 13.

14. Find (f � g)(x) for f(x) � x2 � 4 and g(x) � 6 � x. 14.

15. If f(x) � 2x � 7 and g(x) � x2 � 5, find g[f(5)]. 15.

16. If f(x) � 3 � x and g(x) � x2 � 4, find [ g � f ](x). 16.

17. Find the inverse of g(x) � �2x � 4. 17.

18. Determine whether f(x) � 4x � 8 and g(x) � �14�x � 2 are 18.

inverse functions.

19. Graph y � �3x � 6�. Then state the domain and range of 19.the function.

20. Graph y �2x � 2�. 20.

Bonus If g(x) � 3x � 8, solve g[ g(x)] � g(x) for x. B:

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Chapter 7 Test, Form 3


1. Find p(�2) if p(x) � �18�x3 � �

34�x2 � �

12�x � �

43�. 1.

2. If p(x) � 2x2 � 3x � 1 and r(x) � x2 � 5x, find 2.r(x2) � p(x � 1).

3. Describe the end behavior and 3.determine whether the graph represents an odd-degree or an even-degree polynomial function.Then state the number of real zeros.

4. Graph f(x) � �x4 � 3x2 � x � 2 by making a table of 4.values. Then determine the values of x between which the real zeros are located.

5. For the graph in Question 4, estimate the x-coordinates at 5.which the relative maxima and relative minima occur.

6. Write the expression 9b5 � 3b3 � 8b in quadratic form, if 6.possible.

7. Solve x�12�

� 5x�14�

� 6 � 0. 7.

8. Use synthetic substitution to find f(�4) for 8.f(x) � 2x6 � 4x4 � 2x3 � 5x � 6.

9. Find the value of k so that the remainder is 3 for 9.(x2 � x � k) � (x � 1).

10. State the possible number of positive real zeros, negative 10.real zeros, and imaginary zeros for f(x) � 2x10 � 3x8 � 4x6 � x4 � 3x2 � 2.

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Chapter 7 Test, Form 3 (continued)

11. Find all of the zeros of the function 11.q(x) � x4 � 8x3 � 22x2 � 8x � 39.

12. List all of the possible rational zeros of 12.h(x) � 9x6 � 12x3 � 15.

13. Find all of the rational zeros of 13.h(x) � 24x4 � 38x3 � 23x2 � 5x � 2.

14. Find (f � g)(x) for f(x) � x2 � 4 and g(x) � �x �x

2�. 14.

15. If g(x) � 3x and h(x) � x3 � x2 � x � 1, find [h � g](x). 15.

16. If f(x) � 5x, g(x) � 2x � 1, and h(x) � x2 � 1, find 16.[h � ( g � f )](�3).

17. Find the inverse of h(x) � �2x

5� 6�. 17.

18. Determine whether f(x) � �12�x � �

73� and g(x) � 2x � �

134� 18.

are inverse functions.

19. Graph y � �x � 4� � 2. Then state the domain and 19.range of the function.

20. Graph y �x � 3� � 3. 20.

Bonus If f(�3) � �120, for f(x) � x4 � x3 � 19x2 � kx � 30, B:find f(1).

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Chapter 7 Open-Ended Assessment


Demonstrate your knowledge by giving a clear, concise solutionto each problem. Be sure to include all relevant drawings andjustify your answers. You may show your solutions in more thanone way or investigate beyond the requirements of the problem.

1. a. Sketch a graph of a polynomial function f(x) of degree 5 thathas the maximum number of real zeros possible for a functionof its degree. Label the zeros z1, z2, … .

b. Label relative maximum points of the graph, if any, A1, A2, …and label relative minimum points of the graph, if any,B1, B2, … .

c. State the domain and range of the function.d. Use the notation “As x → ___ , f(x) → ___” to describe the

end behavior of your graph.

2. The domain of a polynomial function g(x) is all real numbers andthe range of this function is g(x) 2. What do you know aboutthe degree, the leading coefficient, and the zeros of this function?Explain your reasoning.

3. a. Write a fourth-degree polynomial P(x) where no coefficient iszero, that is an 0 for any n.

b. Find P(�2) in two different ways.c. Determine whether x � 1 is a factor of P(x).d. Explain what information Descartes’ Rule of Signs provides

about P(x).e. Explain how to find, then list, all of the possible rational zeros

of P(x).f. Explain how to find, then state, the rational zeros of P(x).

4. a. Write a first-degree function g(x) and a second-degree function h(x).

Find g(2x � 3), h(3a), ( g � h)(x), ( g � h)(x), ( g � h)(x), ��hg��(x),

(h � g)(x), g[h(x)], [h � ( g � g)](2), and g�1(x).b. Explain, then show, how to prove that g(x) and g�1(x) are, in

fact, inverse functions. Then explain the relationship betweenthe graphs of these two functions.

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Chapter 7 Vocabulary Test/Review

Underline the correct word or phase that best completes each sentence.

1. (End behavior, Composition of functions, Synthetic substitution) is a methodfor evaluating a polynomial function f(x) at a particular value of x.

2. If a function has an inverse that is also a function, then it must be a (one-to-one function, square root function, power function).

3. Writing the polynomial 2x4 � 9x2 � 15 as 2(x2)2 � 9(x2) � 15 uses the ideaof (leading coefficients, quadratic form, end behavior).

4. The (rational zero theorem, remainder theorem, fundamental theorem of algebra)says that every polynomial equation with degree one or greater has at least oneroot in the set of complex numbers.

5. y � �3x � 5� is a(n) (square root, polynomial, inverse) function.

6. If f(x) is a polynomial function such that f(�2) � 8 and f(�3) � �5, then the (depressed polynomial, Location Principle, relative minimum) tells you thatf(x) has at least one real zero between �2 and �3.

7. If a point is on a graph of a polynomial function and no other nearby points of the graph have a lesser y-coordinate, the point is a relative(maximum, minimum) of the function.

8. If x3 � 3x2 � 4x � 12 is divided by x � 2, the quotient will be x2 � 5x � 6and the remainder will be 0. In this case, x2 � 5x � 6 is called the (quadratic form, depressed polynomial, power function).

9. The process of forming a new function from two given functions by performing the two functions in succession is called (synthetic substitution, end behavior, composition of functions).

10. The expression 4x3 � 3x2 � 5x � 6 is a(n) (polynomial function, inverse relation, polynomial in one variable).

In your own words—Define each term.

11. end behavior

12. Factor Theorem

Complex Conjugates Theorem

composition of functionsdegree of a polynomialdepressed polynomialDescartes’ Rule of Signsend behavior

Factor TheoremFundamental Theorem of Algebra

identity functionIntegral Zero Theoreminverse functioninverse relation

leading coefficientLocation Principleone-to-onepolynomial functionpolynomial in one variable

quadratic form

Rational Zero Theoremrelative maximumrelative minimumRemainder Theoremsquare root functionsquare root inequalitysynthetic substitution

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Chapter 7 Quiz (Lessons 7–1 through 7–3)

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Chapter 7 Quiz (Lessons 7–4 and 7–5)

1. Use synthetic substitution to find f(3) and f(�4) for 1.f(x) � x4 � 8x � 11.

2. One factor of x3 � x2 � 14x � 24 is x � 4. Find the 2.remaining factors.

3. State the possible number of positive real zeros, 3.negative real zeros, and imaginary zeros for g(x) � 3x5 � 2x3 � 4x2 � 8x � 1.

4. Find all of the zeros of f(x) � x3 � 5x2 � 8x � 6. 4.

5. Standardized Test Practice Write a polynomial function of least degree with integral coefficients whose zeros include 4 and 1 � i.A. f(x) � x3 � 2x2 � 6x � 8B. f(x) � x3 � 6x2 � 10x � 8C. f(x) � x3 � 6x2 � 10x � 8D. f(x) � x3 � 6x2 � 10x � 8 5.

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1. If p(x) � 3x2 � 2x � 1, find p(�4). 1.

2. Determine whether the graph at the right represents an odd-degree polynomial or an even-degree polynomial function.Then state the number of real zeros.

3. Graph f(x) � x3 � 5x2 � 4x � 3 by making a table of values.Then determine consecutive values of x between which each real zero is located. Estimate the x-coordinates atwhich the relative maxima and relative minima occur.


4. x4 � 14x2 � 45 � 0 5. a4 � 49

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3.

4.

5.

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77


1. List all of the possible rational zeros of 1.h(x) � 2x4 � 5x3 � 3x2 � 4x � 6. Then find all of the rational zeros of the function.

2. Find (f � g)(x), (f � g)(x), (f � g)(x), and ��gf��(x) for 2.

f(x) � x2 � 3x � 2 and g(x) � 2x � 4.

3. For f(x) � {(2, 3), (4, 4), (5, 8)} and g(x) � {(2, 4), (3, 5), (4, 2), 3.(8, 4)}, find f � g and g � f if they exist.

4. Find [g � h](x) and [h � g](x) for g(x) � x2 � 2x � 1 and 4.h(x) � x � 4.

5. If f(x) � 3x � 2 and g(x) � x2 � 1, find f[g(�3)] and g[f(�3)]. 5.


1. Find the inverse of the relation {(�2, 5), (0, 4), (1, �8), (4, 7)}. 1.

2. Find the inverse of the function f(x) � 4x � 2.Then graph the function and its inverse.

3. Determine whether g(x) � 3x � 6 and f(x) � �13�x � 2 are

inverse functions.

4. Graph y � �3x � 9�. Then state the domain and range of the function.

NAME DATE PERIOD

SCORE


77

NAME DATE PERIOD

SCORE

77

1.

2.

3.

4.y

xO

xO

f(x )

Chapter 7 Mid-Chapter Test (Lessons 7–1 through 7–5)



1. Find p(�4) if p(x) � 3x2 � 4x � 7.A. 7 B. 71 C. 57 D. 39 1.

2. State the degree of 2x2 � 5x3 � 7x4 � 9.A. 4 B. 7 C. �9 D. 3 2.


3. State the number of real zeros of the function.A. 2 B. 4C. 1 D. 3 3.

4. As x → � �, f(x) →�?��

describes the end behavior of the graph.A. �∞ B. 0 C. �∞ D. x 4.

5. Use synthetic substitution to find f(�2) for f(x) � x3 � 6x2 � 5x � 1.A. �41 B. 7 C. �21 D. 27 5.

6. One factor of x3 � 6x2 � x � 6 is x � 6. Find the remaining factors.A. x � 1, x � 1 B. x, x � 1 C. x � 1, x � 1 D. x � 1, x � 1 6.

7. Graph f(x) � x3 � 4x2 � 5 by using a table of values. 7.Then determine consecutive values of x between which each real zero is located.

8. Solve t5 � 81t � 0. 8.

9. State the possible number of positive real zeros, 9.negative real zeros, and imaginary zeros of f(x) � x4 � 3x3 � 2x2 � x � 1.

10. Write a polynomial function of least degree with integral 10.coefficients whose zeros include 3 and 3i.

11. Find all the zeros of the function f(x) � x3 � x2 � 16x � 16. 11.

xO

f(x )

Part I

NAME DATE PERIOD

SCORE 77

Ass

essm

ent

xO

f(x )

Part II


Chapter 7 Cumulative Review (Chapters 1–7)

1. Define a variable and write an inequality. Then solve. 1.Marlea received an inheritance of $10,000. She plans to invest some in a stock that pays 7% interest annually. She will deposit the remainder in a savings account that pays 5% interest annually. What is the least amount that Marlea can invest in stock if she wants to earn at least $550 on her investments for the year? (Lesson 1-5)

2. Describe the system of equations as consistent and 2.independent, consistent and dependent, or inconsistent.6x � 2y � 49x � 3y � 6 (Lesson 3-1)

3. Triangle ABC with vertices at A(�1, �3), B(2, 3), and 3.C(�4, 1) is translated 5 units right and 3 units down.Find the coordinates of A�, B�, and C�. (Lesson 4-4)

4. Use Cramer’s Rule to solve the system of equations. 4.2x � y � �1�3x � y � 4 (Lesson 4-6)

5. Factor 4n2 � 20n � 25 completely. If the polynomial is not 5.factorable, write prime. (Lesson 5-4)

6. Solve �2x � 1�0� � 1 � 5. (Lesson 5-8) 6.

7. Graph the quadratic function f(x) � x2 � 2x � 8, labeling 7.the y-intercept, vertex, and axis of symmetry. (Lesson 6-1)

8. Write a quadratic equation with 3 and �2 as its roots.Write the equation in the form ax2 � bx � c � 0, where a, b, and c are integers. (Lesson 6-3) 8.

9. Find p(�3) if p(x) � x4 � 8x3 � 5x � 4. (Lesson 7-1) 9.

10. State the number of possible positive real zeros, 10.negative real zeros, and imaginary zeros for h(x) � x4 � 2x3 � 5x � 4. (Lesson 7-5)

11. Find all of the rational zeros of f(x) � 2x3 � x2 � 13x � 6. 11.(Lesson 7-6)

12. Find the inverse of f(x) � 5x � 4. (Lesson 7-8) 12.

NAME DATE PERIOD

77

xOf(x )

Standardized Test Practice (Chapters 1–7)


1. If r2 � 1 � �2r, then �r � �12��2

� .

A. ��14� B. �

14�

C. 1 D. cannot be determined 1.

2. How many fourths is 26�23�%?

E. �14� F. 4 G. �

1165�

H. �145�

2.

3. Find p in terms of m if �mp� � q, q � p, p 0, and m 0.

A. ��m� B. ��mq� C. m D. ��p� 3.

4. Find the average of �a3�, �

a6�, and �

a9�.

E. �1514a

� F. �1118a

� G. �5a4�

H. �161a� 4.

5. What is the ones digit in 350?A. 1 B. 3 C. 7 D. 9 5.

6. What is the value of a2 � b2 if a � b � 6 and a � b � �3?E. �18 F. 3 G. 9 H. 18 6.

7. If �n� is an irrational number, which of the following must be irrational?

A. �n2� B. 2�n� C. ��n2�� D. �2n� 7.

8. Evaluate 4m3 � 3m2 � 2m � 2 if m � �1.E. 1 F. �11 G. �1 H. 7 8.

9. If the slope of the line through A(�7, 4) and B(5, y) is ��14�, what is

the value of y?

A. �1 B. 7 C. �92� D. 1 9.

10. Find the area of square ABCD.

E. �53� units2 F. 25 units2

G. 625 units2 H. �295� units2 10. HGFE

DCBA

HGFE

DCBA

HGFE

DCBA

HGFE

DCBA

HGFE

DCBA

?

NAME DATE PERIOD

77

Ass

essm

ent

Part 1: Multiple Choice

Instructions: Fill in the appropriate oval for the best answer.

9x2

30x � 25

B C

A D


Standardized Test Practice (continued)

11. Of the 24 socks in a drawer, there are three 11. 12.times as many black socks as brown socks.Some of the black socks are plain and some are patterned. There are five times as many plain socks as there are patterned socks.What is the probability that, without looking,you select a plain black sock from the drawer?

12. Let d� and d* be defined for any positive integer d as follows: d� is the number obtained by dividing d by its first digit and d* is the sum of the digits of d. What is

the value of �335544�*�?

13. Point X lies between points P and Q on a 13. 14.number line. If XQ � 15 and PQ � 24,then PX � .

14. If �18� � �0

n.4�

, what is the value of n?

Column A Column B

15. 15.

16. 16.

17. 17. DCBA(x � 1)2(x � 2)2

DCBA8 � 6 � (9 � 3)8 � 6 � 9 � 3

DCBAthe number of

faces of a cubethe number of

vertices of a pentagon

Part 3: Quantitative Comparison

Instructions: Compare the quantities in columns A and B. Shade in if the quantity in column A is greater; if the quantity in column B is greater; if the quantities are equal; or if the relationship cannot be determined from the information given.

?0 0 0

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.

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NAME DATE PERIOD

77

NAME DATE PERIOD

A

D

C

B

Part 2: Grid In

Instructions: Enter your answer by writing each digit of the answer in a column boxand then shading in the appropriate oval that corresponds to that entry.

Unit 2 Test (Chapters 5–7)


For Questions 1–7, simplify. Assume that no denominator equals 0.

1. (7x2 � 3x � 9) � (�x2 � 8x � 3)

2. 5x3(7x)2 3. (2x � 3)2

4. 5. �16x2y�4�

6. �12� � �18� � 3�50� � �75� 6.

7. �12

��

3ii� 7.

8. Use synthetic division to find (2x3 � 5x2 � 7x � 1) � (x � 1). 8.

9. Write the expression m�79�

in radical form. 9.

10. Solve �3x � 6� � 4 � 7. 10.

11. Graph f(x) � �x2 � 4x � 3, labeling the y-intercept, vertex, 11.and axis of symmetry.

12. The shape of a supporting arch can be modeled by h(x) � �0.03x2 � 3x, where h(x) represents the height of the 12.arch and x represents the horizontal distance from one end of the base of the arch in meters. Find the maximum height of the arch.

13. Solve 2x2 � 3x � 2 by graphing. If exact roots cannot be 13.found, state the consecutive integers between which the roots are located.

14. Solve x2 � 2x � 24 by factoring. 14.

15. Write a quadratic equation with ��34

� and 4 as its roots. 15.

Write the equation in the form ax2 � bx � c � 0, where a, b, and c are integers.

16. Find the exact solutions to 6x2 � x � 4 � 0 by using the 16.Quadratic Formula.

17. Find the value of the discriminant for 9x2 � 1 � 6x. Then 17.describe the number and type of roots for the equation.

xO

f(x )

8y3 � 27��2xy � 10y � 3x � 15

NAME DATE PERIOD

SCORE

Ass

essm

ent

1.

2.

3.

4.

5.


Unit 2 Test (continued)(Chapters 5–7)

18. Identify the vertex, axis of symmetry, and direction of opening 18.for y � 2(x � 3)2 � 5.

19. Write y � �4x2 � 8x � 1 in vertex form. 19.

20. Graph y � x2 � 2x � 1. 20.

21. Find p(�3) if p(x) � x5 � 3x2. 21.

22. Graph f(x) � (�x)4 � 4x2 � 2x by making a table of values. 22.Then estimate the x-coordinates at which the relative maxima and relative minima occur.

23. Solve �x4 � 200 � 102x2. 23.

24. Use synthetic substitution to find f(�3) for 24.f(x) � 2x3 � 6x2 � 5x � 7.

25. One factor of f(x) � x3 � x2 � 22x � 40 is x � 4. Find the 25.other factors.

26. State the number of positive real zeros, negative real zeros, 26.and imaginary zeros for g(x) � 9x3 � 7x2 � 10x � 4.

27. List all of the possible rational zeros of 27.f(x) � 3x5 � 7x3 � 2x � 15.

28. If f(x) � 3x and g(x) � 4x � 3, find f [g(5)] and g[ f(5)]. 28.

29. Find the inverse of f(x) � 7x � 2. 29.

30. Graph y �3x � 1�2�. 30.y

xO

xO

f(x )

y

xO

NAME DATE PERIOD

First Semester Test (Chapters 1–7)

NAME DATE PERIOD

SCORE


For Questions 1–18, write the letter for the correct answer in the blank at the right of each question.

1. Name the sets of numbers to which �14� belongs.A. rational numbers B. irrational numbersC. rational numbers, real numbers D. irrational numbers, real numbers 1.

2. Find the range of the relation {(�3, 0), (�2, 0), (1, 4)}. Then determine whether the relation is a function.A. {0, 4}; function B. {0, 4}; not a functionC. {�3, �2, 1}; function D. {�3, �2, 1}; not a function 2.

3. The graph of which equation is a line with undefined slope that passes through (5, 1)?A. y � 1 B. y � 5 C. x � 1 D. x � 5 3.

4. Which point does not satisfy the inequality y � 2x � 3 �?A. (0, 2) B. (�1, �3) C. (1, 3) D. (2, 0) 4.

5. To solve the system of equations 3x � y � 5 and 2x � 3y � 18, which expression could be substituted for y into the second equation?

A. 5 � 3x B. 3x � 5 C. 6 � �23�x D. 18 � 2x 5.

6. Which system of inequalities is graphed?A. y 2x � 4 B. y 2x � 4

y � ��23�x � 2 y ��

23�x � 2

C. y � 2x � 4 D. y � 2x � 4 6.

y � ��23�x � 2 y ��

23�x � 2

7. Find the maximum value of the function f(x, y) � 2x � 3y for the feasible region shown.A. 14 B. 6C. 11 D. �6 7.

8. Which statement is not true for the matrices

A � � , B � � , C � � , and the constant c � �2?

A. A(B � C) � (B � C)A B. A(B � C) � A(C � B)C. c(A � B) � (A � B)c D. A(B � C) � AB � AC 8.

7 59 1

4 7�6 3

2 �30 1

y

xO

y

xO

(1, 4)

(3, 0)(�3, 0)A

sses

smen

t


First Semester Test (continued)(Chapters 1–7)

9. Find the value of .A. 58 B. 47 C. 23 D. �1 9.

10. Cramer’s Rule is used to solve the system of 3x � y � 2z � 1equations at the right. Which determinant 4x � 2y � z � �2represents the numerator for y? 2x � 3y � 3z � 4

A. B. C. D. 10.

11. Evaluate �260

��

1100�

�

2

5�. Express the result in scientific notation.

A. 0.3 � 10 B. 3 � 102 C. 0.3 � 10�3 D. 3 � 10�4 11.

12. Simplify (5 � 2�3�)(2 � 4�3�).A. 10 � 8�3� B. �62 � 16�3� C. �14 D. �14 � 16�3� 12.

13. Solve �3

y � 3� � 6 � 4.A. 1003 B. 103 C. �5 D. 11 13.

14. The quadratic equation 9x2 � 6x � 1 � 9 is to be solved by completing the square. Which equation would not be a step in that solution?

A. �x � �13��2

� 1 B. x � ��13� � 1

C. 9x2 � 6x � 8 � 0 D. x2 � �23�x � �

19� � 1 14.

15. Solve the inequality �x2 � 25 0.A. {x � x �5 or x � 5} B. {x � x � �5 or x � 5}C. {x � �5 x 5} D. � 15.

16. Write the expression 2n�23�

� 3n�13�

� 5 in quadratic form, if possible.

A. �2n�13��

2

� 3�n�13�� 5 B. 2�n�

13��

2

� 3�n�13�� 5

C. 2(n2)�13�

� 3(n)�13�

� 5 D. not possible 16.

17. Find all of the rational zeros of f(x) � x5 � 2x3 � 24x.A. 0, �2, � �6� B. 0, �2 C. �6, 0, 4D. 0, �1, �2, �3, �4, �6, �8, �12, �24 17.

18. Determine which pair of functions are not inverses.

A. g(x) � 3x � 5 B. g(x) � �43�x C. g(x) � 2x � 8 D. g(x) � �3

x� � 1 18.

h(x) � �x �

35

� h(x) � �34�x h(x) � �2

x� � 8 h(x) � 3x � 3

3 2 14 1 �22 3 4

3 �1 14 2 �22 3 4

3 1 24 �2 12 4 3

3 �1 24 2 12 3 3

7 112 5

NAME DATE PERIOD



19. The formula for the area A of a circle with diameter d is 19.

A � ��d2��2

. Find the area of a circle with a diameter of

40 centimeters. Use 3.14 for �.

20. Simplify 7(4 � x) � 5(x � 1). 20.

21. Define a variable and write an inequality. Then solve. A 21.shelf in a lumber yard will safely hold up to 1000 pounds.A crate on the shelf is marked 270 pounds. What is the greatest number of sheets of plywood, each weighing 7 pounds, that may safely be stacked on the shelf?

22. Solve 10 � 9 � 2(1 � m) 19. Describe the solution set 22.using set builder or interval notation. Then graph the solution set on a number line.

23. If f(x) � 4(x � 3) � x2, find f(2a). 23.

24. What is the slope of a line that is perpendicular to the 24.graph of the line passing through (2, 1) and (3, 5)?

25. Graph the piecewise function f(x) � � 3x if x � 1 25.

Identify the domain and range. �2x � 2 if x � 1.

26. Solve the system of equations 7x � 3y � �1 and 2x � y � 9 by using elimination.

At a school-sponsored car wash, the fees charged were:$5 per car, $8 per pickup truck, $10 per full-size van. 26.Twice as many cars were washed as pickup trucks. The amount collected for washing cars and pickup trucks was $360. A total of $410 was collected at the car wash.

27. Let c represent the number of cars washed, t represent the 27.number of pickup trucks washed, and v represent the number of vans washed. Write a system of three equations that represents the number of vehicles washed.

28. Find the number of cars washed. 28.

For Questions 29 and 30, perform the indicated matrix operations. If the matrix does not exist, write impossible. 29.

29. �4� � 3� 30. � � � 30.2 0

�3 �41 5

12 4 �9�5 0 3

7 �41 4

6 �42 3

xO

f(x )

3 40 1 2�1

NAME DATE PERIOD

Ass

essm

ent



31. Evaluate using diagonals. 31.

32. Find the inverse of A � � , if it exists. 32.

For Questions 33–35, simplify. Assume that no variable 33.equals 0.

33. (3x2)2(4y5)(2x0y�2) 34. (2x � 3y)(x � 5y) 34.

35. 2i(3 � 4i) � (�1 � i) 35.

36. Use a calculator to approximate �3

287� to three decimal 36.places.

37. Determine whether the function f(x) � 3x2 � 6x � 11 has a 37.minimum or a maximum value and find that value.

38. Solve the equation x2 � 2x � 63 � 0 by factoring. 38.

39. Solve the equation x2 � 4x � 13 � 0 by completing the 39.square.

40. Find the exact solutions to 7x2 � 6x � 1 by using the 40.Quadratic Formula.

41. Write an equation for the parabola with vertex at (�2, 3) 41.and y-intercept �1.

42. One factor of x3 � 5x2 � 8x � 12 is x � 2. Find the 42.remaining factors.

43. Find all the zeros of the function f(x) � x3 � 6x2 � 16x � 96. 43.

44. Find (f � g)(x) and (f � g)(x) for f(x) � 4x � 9 and g(x) � 3x2. 44.

45. Find the inverse of the function p(x) � 4x � 8. 45.

46. Graph y � ��2x � 5�. Then state the domain and range of 46.the function. y

xO

8 2�2 1

3 �4 51 �1 �20 2 3

NAME DATE PERIOD

Standardized Test PracticeStudent Record Sheet (Use with pages 406–407 of the Student Edition.)

© Glencoe/McGraw-Hill A1 Glencoe Algebra 2

NAME DATE PERIOD

77

An

swer

s

Select the best answer from the choices given and fill in the corresponding oval.

1 4 7 10

2 5 8 11

3 6 9 12

Solve the problem and write your answer in the blank.

Also enter your answer by writing each number or symbol in a box. Then fill inthe corresponding oval for that number or symbol.

13 15 17 19

14 16 18

Select the best answer from the choices given and fill in the corresponding oval.

20 22 24

21 23 DCBADCBA

DCBADCBADCBA

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Part 2 Short Response/Grid InPart 2 Short Response/Grid In

Part 1 Multiple ChoicePart 1 Multiple Choice

Part 3 Quantitative ComparisonPart 3 Quantitative Comparison


Answers (Lesson 7-1)

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Fin

d f

(�5)

if

f(x)

�x3

�2x

2�

10x

�20

.f(

x) �

x3�

2x2

�10

x�

20O

rigin

al f

unct

ion

f(�

5) �

(�5)

3�

2(�

5)2

�10

(�5)

�20

Rep

lace

xw

ith �

5.

��

125

�50

�50

�20

Eva

luat

e.

��

5S

impl

ify.

Fin

d g

(a2

�1)

if

g(x)

�x2

�3x

�4.

g(x)

�x2

�3x

�4

Orig

inal

fun

ctio

n

g(a2

�1)

�(a

2�

1)2

�3(

a2�

1) �

4R

epla

ce x

with

a2

�1.

�a4

�2a

2�

1 �

3a2

�3

�4

Eva

luat

e.

�a4

�a2

�6

Sim

plify

.

Sta

te t

he

deg

ree

and

lea

din

g co

effi

cien

t of

eac

h p

olyn

omia

l in

on

e va

riab

le.I

f it

is

not

a p

olyn

omia

l in

on

e va

riab

le,e

xpla

in w

hy.

8;8

1.3x

4�

6x3

�x2

�12

4;3

2.10

0 �

5x3

�10

x77;

103.

4x6

�6x

4�

8x8

�10

x2�

20

4.4x

2�

3xy

�16

y25.

8x3

�9x

5�

4x2

�36

6.�

��

no

t a

po

lyn

om

ial i

n

5;�

9o

ne

vari

able

;co

nta

ins

6;�

two

var

iab

les

Fin

d f

(2)

and

f(�

5) f

or e

ach

fu

nct

ion

.

7.f(

x) �

x2�

98.

f(x)

�4x

3�

3x2

�2x

�1

9.f(

x) �

9x3

�4x

2�

5x�

7�

5;16

23;

�58

673

;�

1243

1 � 25

1 � 72x3� 36

x6� 25

x2� 18

Exam

ple1

Exam

ple1

Exam

ple2

Exam

ple2

Exam

ple3

Exam

ple3

Exer

cises

Exer

cises

©G

lenc

oe/M

cGra

w-H

ill37

6G

lenc

oe A

lgeb

ra 2

Gra

ph

s o

f Po

lyn

om

ial F

un

ctio

ns

If th

e de

gree

is e

ven

and

the

lead

ing

coef

ficie

nt is

pos

itive

, th

enf(

x) →

��

as x

→�

�

f(x)

→�

�as

x→

��

If th

e de

gree

is e

ven

and

the

lead

ing

coef

ficie

nt is

neg

ativ

e, t

hen

En

d B

ehav

ior

f(x)

→�

�as

x→

��

of

Po

lyn

om

ial

f(x)

→�

�as

x→

��

Fu

nct

ion

sIf

the

degr

ee is

odd

and

the

lead

ing

coef

ficie

nt is

pos

itive

, th

enf(

x) →

��

as x

→�

�

f(x)

→�

�as

x→

��

If th

e de

gree

is o

dd a

nd t

he le

adin

g co

effic

ient

is n

egat

ive,

the

nf(

x) →

��

as x

→�

�

f(x)

→�

�as

x→

��

Rea

l Zer

os

of

The

max

imum

num

ber

of z

eros

of

a po

lyno

mia

l fun

ctio

n is

equ

al t

o th

e de

gree

of

the

poly

nom

ial.

a P

oly

no

mia

lA

zero

of

a fu

nctio

n is

a p

oint

at

whi

ch t

he g

raph

inte

rsec

ts t

he x

-axi

s.

Fu

nct

ion

On

a gr

aph,

cou

nt t

he n

umbe

r of

rea

l zer

os o

f th

e fu

nctio

n by

cou

ntin

g th

e nu

mbe

r of

tim

es t

hegr

aph

cros

ses

or t

ouch

es t

he x

-axi

s.

Det

erm

ine

wh

eth

er t

he

grap

h r

epre

sen

ts a

n o

dd

-deg

ree

pol

ynom

ial

or a

n e

ven

-deg

ree

pol

ynom

ial.

Th

en s

tate

th

e n

um

ber

of

real

zer

os.

As

x→

��

,f(x

) →

��

and

as x

→�

�,f

(x)

→�

�,

so i

t is

an

odd

-deg

ree

poly

nom

ial

fun

ctio

n.

Th

e gr

aph

in

ters

ects

th

e x-

axis

at

1 po

int,

so t

he

fun

ctio

n h

as 1

rea

l ze

ro.

Det

erm

ine

wh

eth

er e

ach

gra

ph

rep

rese

nts

an

od

d-d

egre

e p

olyn

omia

l or

an

eve

n-

deg

ree

pol

ynom

ial.

Th

en s

tate

th

e n

um

ber

of

real

zer

os.

1.2.

3.

even

;6

even

;1

do

ub

le z

ero

od

d;

3

x

f (x)

Ox

f (x)

Ox

f (x)

O

x

f (x)

O

Stu

dy G

uid

e a

nd I

nte

rven

tion

(c

onti

nued

)

Po

lyn

om

ial F

un

ctio

ns

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

7-1

7-1

Exam

ple

Exam

ple

Exer

cises

Exer

cises


An

swer

s


Skil

ls P

ract

ice

Po

lyn

om

ial F

un

ctio

ns

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

7-1

7-1

©G

lenc

oe/M

cGra

w-H

ill37

7G

lenc

oe A

lgeb

ra 2

Lesson 7-1

Sta

te t

he

deg

ree

and

lea

din

g co

effi

cien

t of

eac

h p

olyn

omia

l in

on

e va

riab

le.I

f it

is

not

a p

olyn

omia

l in

on

e va

riab

le,e

xpla

in w

hy.

1.a

�8

1;1

2.(2

x�

1)(4

x2�

3)3;

8

3.�

5x5

�3x

3�

85;

�5

4.18

�3y

�5y

2�

y5�

7y6

6;7

5.u

3�

4u2 v

2�

v46.

2r�

r2�

No

,th

is p

oly

no

mia

lco

nta

ins

two

N

o,t

his

is n

ot

a p

oly

no

mia

l bec

ause

vari

able

s,u

and

v.

� r1 2�ca

nn

ot

be

wri

tten

in t

he

form

rn,

wh

ere

nis

a n

on

neg

ativ

e in

teg

er.

Fin

d p

(�1)

an

d p

(2)

for

each

fu

nct

ion

.

7.p(

x) �

4 �

3x7;

�2

8.p(

x) �

3x�

x2�

2;10

9.p(

x) �

2x2

�4x

�1

7;1

10.p

(x)

��

2x3

�5x

�3

0;�

3

11.p

(x)

�x4

�8x

2�

10�

1;38

12.p

(x)

��1 3� x

2�

�2 3� x�

23;

2

If p

(x)

�4x

2�

3 an

d r

(x)

�1

�3x

,fin

d e

ach

val

ue.

13.p

(a)

4a2

�3

14.r

(2a)

1 �

6a

15.3

r(a)

3 �

9a16

.�4p

(a)

�16

a2�

12

17.p

(a2 )

4a4

�3

18.r

(x�

2)7

�3x

For

eac

h g

rap

h,

a.d

escr

ibe

the

end

beh

avio

r,b

.det

erm

ine

wh

eth

er i

t re

pre

sen

ts a

n o

dd

-deg

ree

or a

n e

ven

-deg

ree

pol

ynom

ial

fun

ctio

n,a

nd

c.st

ate

the

nu

mb

er o

f re

al z

eroe

s.

19.

20.

21.

f(x

) →

��

as x

→�

�,

f(x

) →

��

as x

→�

�,

f(x

) →

��

as x

→�

�,

f(x

) →

��

as x

→�

�;

f(x

) →

��

as x

→�

�;

f(x

) →

��

as x

→�

�;

od

d;

1ev

en;

4o

dd

;3

x

f (x)

Ox

f (x)

Ox

f (x)

O

1 � r2

©G

lenc

oe/M

cGra

w-H

ill37

8G

lenc

oe A

lgeb

ra 2

Sta

te t

he

deg

ree

and

lea

din

g co

effi

cien

t of

eac

h p

olyn

omia

l in

on

e va

riab

le.I

f it

is

not

a p

olyn

omia

l in

on

e va

riab

le,e

xpla

in w

hy.

1.(3

x2�

1)(2

x2�

9)4;

62.

�1 5� a3

��3 5� a

2�

�4 5� a3;

�1 5�

3.�

3m�

12N

ot

a p

oly

no

mia

l;4.

27 �

3xy3

�12

x2y2

�10

y

� m2 2�ca

nn

ot

be

wri

tten

in t

he

form

No

,th

is p

oly

no

mia

l co

nta

ins

two

mn

for

a n

on

neg

ativ

e in

teg

er n

.va

riab

les,

xan

d y

.

Fin

d p

(�2)

an

d p

(3)

for

each

fu

nct

ion

.

5.p(

x) �

x3�

x56.

p(x)

��

7x2

�5x

�9

7.p(

x) �

�x5

�4x

3

24;

�21

6�

29;

�39

0;�

135

8.p(

x) �

3x3

�x2

�2x

�5

9.p(

x) �

x4�

�1 2� x3

��1 2� x

10.p

(x)

��1 3� x

3�

�2 3� x2

�3x

�37

;73

13;

93�

6;24

If p

(x)

�3x

2�

4 an

d r

(x)

�2x

2�

5x�

1,fi

nd

eac

h v

alu

e.

11.p

(8a)

12.r

(a2 )

13.�

5r(2

a)

192a

2�

42a

4�

5a2

�1

�40

a2�

50a

�5

14.r

(x�

2)15

.p(x

2�

1)16

.5[p

(x�

2)]

2x2

�3x

�1

3x4

�6x

2�

115

x2�

60x

�40

For

eac

h g

rap

h,

a.d

escr

ibe

the

end

beh

avio

r,b

.det

erm

ine

wh

eth

er i

t re

pre

sen

ts a

n o

dd

-deg

ree

or a

n e

ven

-deg

ree

pol

ynom

ial

fun

ctio

n,a

nd

c.st

ate

the

nu

mb

er o

f re

al z

eroe

s.

17.

18.

19.

f(x

) →

��

as x

→�

�,

f(x

) →

��

as x

→�

�,

f(x

) →

��

as x

→�

�,

f(x

) →

��

as x

→�

�;

f(x

) →

��

as x

→�

�;

f(x

) →

��

as x

→�

�;

even

;2

even

;1

od

d;

5

20.W

IND

CH

ILL

Th

e fu

nct

ion

C(s

) �

0.01

3s2

�s

�7

esti

mat

es t

he

win

d ch

ill

tem

pera

ture

C(s

) at

0�F

for

win

d sp

eeds

sfr

om 5

to

30 m

iles

per

hou

r.E

stim

ate

the

win

d ch

ill

tem

pera

ture

at

0�F

if

the

win

d sp

eed

is 2

0 m

iles

per

hou

r.ab

ou

t �

22�F

x

f (x)

Ox

f (x)

Ox

f (x)

O

2� m

2

Pra

ctic

e (

Ave

rag

e)

Po

lyn

om

ial F

un

ctio

ns

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

7-1

7-1



Readin

g t

o L

earn

Math

em

ati

csP

oly

no

mia

l Fu

nct

ion

s

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

7-1

7-1

©G

lenc

oe/M

cGra

w-H

ill37

9G

lenc

oe A

lgeb

ra 2

Lesson 7-1

Pre-

Act

ivit

yW

her

e ar

e p

olyn

omia

l fu

nct

ion

s fo

un

d i

n n

atu

re?

Rea

d th

e in

trod

uct

ion

to

Les

son

7-1

at

the

top

of p

age

346

in y

our

text

book

.

•In

the

hon

eyco

mb

cros

s se

ctio

n sh

own

in y

our

text

book

,the

re i

s 1

hexa

gon

in t

he

cen

ter,

6 h

exag

ons

in t

he

seco

nd

rin

g,an

d 12

hex

agon

s in

th

e th

ird

ring

.How

man

y he

xago

ns w

ill t

here

be

in t

he f

ourt

h,fi

fth,

and

sixt

h ri

ngs?

18;

24;

30•

Th

ere

is 1

hex

agon

in

a h

oney

com

b w

ith

1 r

ing.

Th

ere

are

7 h

exag

ons

ina

hon

eyco

mb

wit

h 2

rin

gs.H

ow m

any

hex

agon

s ar

e th

ere

in h

oney

com

bsw

ith

3 r

ings

,4 r

ings

,5 r

ings

,an

d 6

rin

gs?

19;

37;

61;

91

Rea

din

g t

he

Less

on

1.G

ive

the

degr

ee a

nd

lead

ing

coef

fici

ent

of e

ach

pol

ynom

ial

in o

ne

vari

able

.

deg

ree

lead

ing

coef

fici

ent

a.10

x3�

3x2

�x

�7

b.

7y2

�2y

5�

y�

4y3

c.10

0

2.M

atch

eac

h d

escr

ipti

on o

f a

poly

nom

ial

fun

ctio

n f

rom

th

e li

st o

n t

he

left

wit

h t

he

corr

espo

ndi

ng

end

beh

avio

r fr

om t

he

list

on

th

e ri

ght.

a.ev

en d

egre

e,n

egat

ive

lead

ing

coef

fici

ent

iiii.

f(x)

→�

�as

x→

��

;f(

x) →

��

as x

→�

�

b.

odd

degr

ee,p

osit

ive

lead

ing

coef

fici

ent

ivii

.f(x

) →

��

as x

→�

�;

f(x)

→�

�as

x→

��

c.od

d de

gree

,neg

ativ

e le

adin

g co

effi

cien

tii

iii.

f(x)

→�

�as

x→

��

;f(

x) →

��

as x

→�

�

d.

even

deg

ree,

posi

tive

lea

din

g co

effi

cien

ti

iv.

f(x)

→�

�as

x→

��

;f(

x) →

��

as x

→�

�

Hel

pin

g Y

ou

Rem

emb

er

3.W

hat

is

an e

asy

way

to

rem

embe

r th

e di

ffer

ence

bet

wee

n t

he

end

beh

avio

r of

th

e gr

aph

sof

eve

n-d

egre

e an

d od

d-de

gree

pol

ynom

ial

fun

ctio

ns?

Sam

ple

an

swer

:B

oth

en

ds

of

the

gra

ph

of

an e

ven

-deg

ree

fun

ctio

nev

entu

ally

kee

p g

oin

g in

th

e sa

me

dir

ecti

on

.Fo

r o

dd

-deg

ree

fun

ctio

ns,

the

two

en

ds

even

tual

ly h

ead

in o

pp

osi

te d

irec

tio

ns,

on

e u

pw

ard

,th

eo

ther

do

wn

war

d.

100

0

�2

510

3

©G

lenc

oe/M

cGra

w-H

ill38

0G

lenc

oe A

lgeb

ra 2

Ap

pro

xim

atio

n b

y M

ean

s o

f P

oly

no

mia

lsM

any

scie

nti

fic

expe

rim

ents

pro

duce

pai

rs o

f n

um

bers

[x,

f(x)

] th

at c

an

be r

elat

ed b

y a

form

ula

.If

the

pair

s fo

rm a

fu

nct

ion

,you

can

fit

a

poly

nom

ial

to t

he

pair

s in

exa

ctly

on

e w

ay.C

onsi

der

the

pair

s gi

ven

by

the

foll

owin

g ta

ble.

We

wil

l as

sum

e th

e po

lyno

mia

l is

of

degr

ee t

hree

.Sub

stit

ute

the

give

n va

lues

int

o th

is e

xpre

ssio

n.

f(x)

�A

�B

(x�

x 0)

�C

(x�

x 0)(

x�

x 1)

�D

(x�

x 0)(

x�

x 1)(

x�

x 2)

You

wil

l get

the

sys

tem

of

equa

tion

s sh

own

belo

w.Y

ou c

an s

olve

thi

s sy

stem

an

d us

e th

e va

lues

for

A,B

,C,a

nd D

to f

ind

the

desi

red

poly

nom

ial.

6 �

A11

�A

�B

(2 �

1) �

A�

B39

�A

�B

(4 �

1) �

C(4

�1)

(4 �

2) �

A�

3B�

6C�

54 �

A�

B(7

�1)

�C

(7 �

1)(7

�2)

�D

(7 �

1)(7

�2)

(7 �

4) �

A�

6B�

30C

�90

D

Sol

ve.

1.S

olve

the

sys

tem

of

equa

tion

s fo

r th

e va

lues

A,B

,C,a

nd D

.A

�6,

B�

5,C

�3,

D�

�2

2.F

ind

the

poly

nom

ial

that

rep

rese

nts

the

four

ord

ered

pai

rs.W

rite

you

r an

swer

in

the

form

y�

a�

bx�

cx2

�d

x3.

y�

�2x

3�

17x

2�

32x

�23

3.F

ind

the

poly

nom

ial

that

giv

es t

he f

ollo

win

g va

lues

.

A�

�20

7,B

�94

,C�

25,D

�1;

y�

x3�

10x2

�10

x�

1

4.A

sci

enti

st m

easu

red

the

volu

me

f(x)

of

carb

on d

ioxi

de g

as t

hat

can

be

abso

rbed

by

one

cubi

c ce

ntim

eter

of

char

coal

at

pres

sure

x.F

ind

the

valu

es f

or A

,B,C

,and

D.

A�

3.1,

B�

0.01

091,

C�

�0.

0000

0643

,D�

0.00

0000

0066

x12

034

053

469

8

f(x

)3.

15.

57.

18.

3

x8

1215

20

f(x

)�

207

169

976

3801

x1

24

7

f(x

)6

1139

�54

En

rich

men

t

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

7-1

7-1


An

swer

s


Stu

dy G

uid

e a

nd I

nte

rven

tion

Gra

ph

ing

Po

lyn

om

ial F

un

ctio

ns

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

7-2

7-2

©G

lenc

oe/M

cGra

w-H

ill38

1G

lenc

oe A

lgeb

ra 2

Lesson 7-2

Gra

ph

Po

lyn

om

ial F

un

ctio

ns

Lo

cati

on

Pri

nci

ple

Sup

pose

y�

f(x)

rep

rese

nts

a po

lyno

mia

l fun

ctio

n an

d a

and

bar

e tw

o nu

mbe

rs s

uch

that

f(a)

�0

and

f(b)

�0.

The

n th

e fu

nctio

n ha

s at

leas

t on

e re

al z

ero

betw

een

aan

d b.

Det

erm

ine

the

valu

es o

f x

bet

wee

n w

hic

h e

ach

rea

l ze

ro o

f th

efu

nct

ion

f(x

) �

2x4

�x3

�5

is l

ocat

ed.T

hen

dra

w t

he

grap

h.

Mak

e a

tabl

e of

val

ues

.Loo

k at

th

e va

lues

of

f(x)

to

loca

te t

he

zero

s.T

hen

use

th

e po

ints

to

sket

ch a

gra

ph o

f th

e fu

nct

ion

.T

he

chan

ges

in s

ign

in

dica

te t

hat

th

ere

are

zero

sbe

twee

n x

��

2 an

d x

��

1 an

d be

twee

n x

�1

and

x�

2.

Gra

ph

eac

h f

un

ctio

n b

y m

akin

g a

tab

le o

f va

lues

.Det

erm

ine

the

valu

es o

f x

atw

hic

h o

r b

etw

een

wh

ich

eac

h r

eal

zero

is

loca

ted

.

1.f(

x) �

x3�

2x2

�1

2.f(

x) �

x4�

2x3

�5

3.f(

x) �

�x4

�2x

2�

1

bet

wee

n 0

an

d �

1;b

etw

een

�2

and

�3;

at �

1 at

1;

bet

wee

n 1

an

d 2

bet

wee

n 1

an

d 2

4.f(

x) �

x3�

3x2

�4

5.f(

x) �

3x3

�2x

�1

6.f(

x) �

x4�

3x3

�1

at �

1,2

bet

wee

n 0

an

d 1

bet

wee

n 0

an

d 1

;b

etw

een

2 a

nd

3x

f (x)

Ox

f (x)

Ox

f (x)

O

x

f (x)

Ox

f (x)

O

x

f (x)

O4

8–4

–8

8 4 –4 –8

x

f (x)

O

xf(

x)

�2

35

�1

�2

0�

5

1�

4

219

Exam

ple

Exam

ple

Exer

cises

Exer

cises

©G

lenc

oe/M

cGra

w-H

ill38

2G

lenc

oe A

lgeb

ra 2

Max

imu

m a

nd

Min

imu

m P

oin

tsA

qu

adra

tic

fun

ctio

n h

as e

ith

er a

max

imu

m o

r a

min

imu

m p

oin

t on

its

gra

ph.F

or h

igh

er d

egre

e po

lyn

omia

l fu

nct

ion

s,yo

u c

an f

ind

turn

ing

poin

ts,w

hic

h r

epre

sen

t re

lati

ve m

axim

um

or r

elat

ive

min

imu

mpo

ints

.

Gra

ph

f(x

) �

x3�

6x2

�3.

Est

imat

e th

e x-

coor

din

ates

at

wh

ich

th

ere

lati

ve m

axim

a an

d m

inim

a oc

cur.

Mak

e a

tabl

e of

val

ues

an

d gr

aph

th

e fu

nct

ion

.A

rel

ativ

e m

axim

um

occ

urs

at x

��

4 an

d a

rela

tive

min

imu

m o

ccu

rs a

t x

�0.

Gra

ph

eac

h f

un

ctio

n b

y m

akin

g a

tab

le o

f va

lues

.Est

imat

e th

e x-

coor

din

ates

at

wh

ich

th

e re

lati

ve m

axim

a an

d m

inim

a oc

cur.

1.f(

x) �

x3�

3x2

2.f(

x) �

2x3

�x2

�3x

3.f(

x) �

2x3

�3x

�2

max

.at

0,m

in.a

t 2

max

.ab

ou

t �

1,m

ax.a

bo

ut

�1,

min

.ab

ou

t 0.

5m

in.a

bo

ut

14.

f(x)

�x4

�7x

�3

5.f(

x) �

x5�

2x2

�2

6.f(

x) �

x3�

2x2

�3

min

.ab

ou

t 1

max

.at

0,m

ax.a

bo

ut

�1,

min

.ab

ou

t 1

min

.at

0

x

f (x)

Ox

f (x)

Ox

f (x)

O4

8–4

–8

8 4 –4 –8

x

f (x)

Ox

f (x)

Ox

f (x)

O

x

f (x)

O2

–2–4

24 16 8

← in

dica

tes

a re

lativ

e m

axim

um

← z

ero

betw

een

x�

�1,

x�

0

← in

dica

tes

a re

lativ

e m

inim

um

xf(

x)

�5

22

�4

29

�3

24

�2

13

�1

2

0�

3

14

229

Stu

dy G

uid

e a

nd I

nte

rven

tion

(c

onti

nued

)

Gra

ph

ing

Po

lyn

om

ial F

un

ctio

ns

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

7-2

7-2

Exam

ple

Exam

ple

Exer

cises

Exer

cises



Skil

ls P

ract

ice

Gra

ph

ing

Po

lyn

om

ial F

un

ctio

ns

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

7-2

7-2

©G

lenc

oe/M

cGra

w-H

ill38

3G

lenc

oe A

lgeb

ra 2

Lesson 7-2

Com

ple

te e

ach

of

the

foll

owin

g.a.

Gra

ph

eac

h f

un

ctio

n b

y m

akin

g a

tab

le o

f va

lues

.b

.Det

erm

ine

con

secu

tive

val

ues

of

xb

etw

een

wh

ich

eac

h r

eal

zero

is

loca

ted

.c.

Est

imat

e th

e x-

coor

din

ates

at

wh

ich

th

e re

lati

ve m

axim

a an

d m

inim

a oc

cur.

1.f(

x) �

x3�

3x2

�1

2.f(

x) �

x3�

3x�

1

zero

s b

etw

een

�1

and

0,0

an

d 1

,ze

ros

bet

wee

n �

2 an

d �

1,0

and

1,

and

2 a

nd

3;

rel.

max

.at

x�

0,an

d 1

an

d 2

;re

l.m

ax.a

t x

��

1,re

l.m

in.a

t x

�2

rel.

min

.at

x�

1

3.f(

x) �

2x3

�9x

2�

12x

�2

4.f(

x) �

2x3

�3x

2�

2

zero

bet

wee

n �

1 an

d 0

;ze

ro b

etw

een

�1

and

0;

rel.

max

.at

x�

�2,

rel.

min

.at

x�

1,re

l.m

ax.a

t x

�0

rel.

min

.at

x�

�1

5.f(

x) �

x4�

2x2

�2

6.f(

x) �

0.5x

4�

4x2

�4

zero

s b

etw

een

�2

and

�1,

and

ze

ros

bet

wee

n �

1 an

d �

2,�

2 an

d

1 an

d 2

;re

l.m

ax.a

t x

�0,

�3,

1 an

d 2

,an

d 2

an

d 3

;rel

.max

.at

rel.

min

.at

x�

�1

and

x�

1x

�0,

rel.

min

.at

x�

�2

and

x�

2

x

f (x)

O

xf(

x)

�3

8.5

�2

�4

�1

0.5

04

10.

52

�4

38.

5

x

f (x)

O

xf(

x)

�3

61�

26

�1

�3

0�

21

�3

26

361

x

f (x)

O

xf(

x)

�1

�3

02

11

26

329

x

f (x)

O

xf(

x)

�3

�7

�2

�2

�1

�3

02

125

x

f (x)

O

xf(

x)

�3

�17

�2

�1

�1

30

11

�1

23

319

x

f (x)

O

xf(

x)

�2

�19

�1

�3

01

1�

12

�3

31

417

©G

lenc

oe/M

cGra

w-H

ill38

4G

lenc

oe A

lgeb

ra 2

Com

ple

te e

ach

of

the

foll

owin

g.a.

Gra

ph

eac

h f

un

ctio

n b

y m

akin

g a

tab

le o

f va

lues

.b

.Det

erm

ine

con

secu

tive

val

ues

of

xb

etw

een

wh

ich

eac

h r

eal

zero

is

loca

ted

.c.

Est

imat

e th

e x-

coor

din

ates

at

wh

ich

th

e re

lati

ve a

nd

rel

ativ

e m

inim

a oc

cur.

1.f(

x) �

�x3

�3x

2�

32.

f(x)

�x3

�1.

5x2

�6x

�1

x

f (x)

O8 4 –4 –8

24

–2–4

xf(

x)

�2

�1

�1

4.5

01

1�

5.5

2�

93

�3.

54

17

x

f (x)

O

xf(

x)

�2

17�

11

0�

31

�1

21

3�

34

�19Pra

ctic

e (

Ave

rag

e)

Gra

ph

ing

Po

lyn

om

ial F

un

ctio

ns

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

7-2

7-2

zero

s b

etw

een

�1

zero

s b

etw

een

�2

and

0,1

an

d 2

,an

d �

1,0

and

1,

and

2 a

nd

3;

rel.

max

.at

x�

2,an

d 3

an

d 4

;re

l.m

ax.a

t x

��

1,re

l.m

in.a

t x

�0

rel.

min

.at

x�

2

3.f(

x) �

0.75

x4�

x3�

3x2

�4

4.f(

x) �

x4�

4x3

�6x

2�

4x�

3

zero

s b

etw

een

�3

and

�2,

and

ze

ros

bet

wee

n �

3 an

d �

2,�

2 an

d �

1;re

l.m

ax.a

t x

�0,

and

0 a

nd

1;

rel.

min

.at

x�

�1

rel.

min

.at

x�

�2

and

x�

1

PR

ICE

SF

or E

xerc

ises

5 a

nd

6,u

se t

he

foll

owin

g in

form

atio

n.

Th

e C

onsu

mer

Pri

ce I

nde

x (C

PI)

giv

es t

he

rela

tive

pri

ce

for

a fi

xed

set

of g

oods

an

d se

rvic

es.T

he

CP

I fr

om

Sep

tem

ber,

2000

to

July

,200

1 is

sh

own

in

th

e gr

aph

.So

urce

: U. S

. Bur

eau

of L

abor

Sta

tistic

s

5.D

escr

ibe

the

turn

ing

poin

ts o

f th

e gr

aph

.re

l max

.in

Nov

.an

d J

un

e;re

l.m

in in

Dec

.6.

If t

he

grap

h w

ere

mod

eled

by

a po

lyn

omia

l eq

uat

ion

,w

hat

is

the

leas

t de

gree

th

e eq

uat

ion

cou

ld h

ave?

4

7.LA

BO

RA

tow

n’s

jobl

ess

rate

can

be

mod

eled

by

(1,3

.3),

(2,4

.9),

(3,5

.3),

(4,6

.4),

(5,4

.5),

(6,5

.6),

(7,2

.5),

(8,2

.7).

How

man

y tu

rnin

g po

ints

wou

ld t

he

grap

h o

f a

poly

nom

ial

fun

ctio

n t

hro

ugh

th

ese

poin

ts h

ave?

Des

crib

e th

em.

4:2

rel.

max

.an

d 2

rel

.min

.

Mo

nth

s Si

nce

Sep

tem

ber

, 200

0

Consumer Price Index

20

46

13

57

89

1011

179

178

177

176

175

174

173

f(x)

xO

xf(

x)

�3

12�

2�

3�

1�

40

�3

112

277

f(x)

xO

xf(

x)

�3

10.7

5�

2�

4�

10.

750

41

2.75

212


An

swer

s


Readin

g t

o L

earn

Math

em

ati

csG

rap

hin

g P

oly

no

mia

l Fu

nct

ion

s

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

7-2

7-2

©G

lenc

oe/M

cGra

w-H

ill38

5G

lenc

oe A

lgeb

ra 2

Lesson 7-2

Pre-

Act

ivit

yH

ow c

an g

rap

hs

of p

olyn

omia

l fu

nct

ion

s sh

ow t

ren

ds

in d

ata?

Rea

d th

e in

trod

uct

ion

to

Les

son

7-2

at

the

top

of p

age

353

in y

our

text

book

.

Th

ree

poin

ts o

n t

he

grap

h s

how

n i

n y

our

text

book

are

(0,

14),

(70,

3.78

),an

d(1

00,9

).G

ive

the

real

-wor

ld m

ean

ing

of t

he

coor

din

ates

of

thes

e po

ints

.S

amp

le a

nsw

er:

In 1

900,

14%

of

the

U.S

.po

pu

lati

on

was

fore

ign

bo

rn.I

n 1

970,

3.78

% o

f th

e p

op

ula

tio

n w

as f

ore

ign

bo

rn.I

n 2

000,

9% o

f th

e p

op

ula

tio

n w

as f

ore

ign

bo

rn.

Rea

din

g t

he

Less

on

1.S

upp

ose

that

f(x

) is

a t

hir

d-de

gree

pol

ynom

ial

fun

ctio

n a

nd

that

can

d d

are

real

nu

mbe

rs,w

ith

d�

c.In

dica

te w

het

her

eac

h s

tate

men

t is

tru

eor

fal

se.(

Rem

embe

r th

attr

ue

mea

ns

alw

ays

tru

e.)

a.If

f(c

) �

0 an

d f(

d)

�0,

ther

e is

exa

ctly

on

e re

al z

ero

betw

een

can

d d

.fa

lse

b.

If f

(c)

�f(

d)

0,

ther

e ar

e n

o re

al z

eros

bet

wee

n c

and

d.

fals

e

c.If

f(c

) �

0 an

d f(

d)

�0,

ther

e is

at

leas

t on

e re

al z

ero

betw

een

can

d d

.tr

ue

2.M

atch

eac

h g

raph

wit

h i

ts d

escr

ipti

on.

a.th

ird-

degr

ee p

olyn

omia

l w

ith

on

e re

lati

ve m

axim

um

an

d on

e re

lati

ve m

inim

um

;le

adin

g co

effi

cien

t n

egat

ive

iii

b.

fou

rth

-deg

ree

poly

nom

ial

wit

h t

wo

rela

tive

min

ima

and

one

rela

tive

max

imu

mi

c.th

ird-

degr

ee p

olyn

omia

l w

ith

on

e re

lati

ve m

axim

um

an

d on

e re

lati

ve m

inim

um

;le

adin

g co

effi

cien

t po

siti

veiv

d.

fou

rth

-deg

ree

poly

nom

ial

wit

h t

wo

rela

tive

max

ima

and

one

rela

tive

min

imu

mii

i.ii

.ii

i.iv

.

Hel

pin

g Y

ou

Rem

emb

er

3.T

he

orig

ins

of w

ords

can

hel

p yo

u t

o re

mem

ber

thei

r m

ean

ing

and

to d

isti

ngu

ish

betw

een

sim

ilar

wor

ds.L

ook

up

max

imu

man

d m

inim

um

in a

dic

tion

ary

and

desc

ribe

thei

r or

igin

s (o

rigi

nal

lan

guag

e an

d m

ean

ing)

.S

amp

le a

nsw

er:

Max

imu

mco

mes

fro

m t

he

Lat

in w

ord

max

imu

s,m

ean

ing

gre

ates

t.M

inim

um

com

es f

rom

the

Lat

in w

ord

min

imu

s,m

ean

ing

leas

t.

x

f (x)

Ox

f (x)

Ox

f (x)

Ox

f (x)

O

©G

lenc

oe/M

cGra

w-H

ill38

6G

lenc

oe A

lgeb

ra 2

Go

lden

Rec

tan

gle

sU

se a

str

aigh

ted

ge,a

com

pas

s,an

d t

he

inst

ruct

ion

s b

elow

to

con

stru

ct

a go

lden

rec

tan

gle.

1.C

onst

ruct

squ

are

AB

CD

wit

h s

ides

of

2 ce

nti

met

ers.

2.C

onst

ruct

th

e m

idpo

int

of A �

B�.C

all

the

mid

poin

t M

.

3.U

sin

g M

as t

he

cen

ter,

set

you

r co

mpa

ss

open

ing

at M

C.C

onst

ruct

an

arc

wit

h

cen

ter

Mth

at i

nte

rsec

ts A �

B�.C

all

the

poin

t of

in

ters

ecti

on P

.

4.C

onst

ruct

a l

ine

thro

ugh

Pth

at i

s pe

rpen

dicu

lar

to A �

B�.

5.E

xten

d D�

C�so

th

at i

t in

ters

ects

th

e pe

rpen

dicu

lar.

Cal

l th

e in

ters

ecti

on p

oin

t Q

.A

PQ

Dis

a g

olde

n r

ecta

ngl

e.C

hec

k th

is

con

clu

sion

by

fin

din

g th

e va

lue

of �Q A

PP �.

0.62

A f

igu

re c

onsi

stin

g of

sim

ilar

gol

den

rec

tan

gles

is

show

n b

elow

.Use

a

com

pas

s an

d t

he

inst

ruct

ion

s b

elow

to

dra

w q

uar

ter-

circ

le a

rcs

that

fo

rm a

sp

iral

lik

e th

at f

oun

d i

n t

he

shel

l of

a c

ham

ber

ed n

auti

lus.

6.U

sin

g A

as a

cen

ter,

draw

an

arc

th

at p

asse

s th

rou

gh

Ban

d C

.

7.U

sin

g D

as a

cen

ter,

draw

an

arc

th

at p

asse

s th

rou

gh

Can

d E

.

8.U

sin

g F

as a

cen

ter,

draw

an

arc

th

at p

asse

s th

rou

gh

Ean

d G

.

9.C

onti

nu

e dr

awin

g ar

cs,

usi

ng

H,K

,an

d M

as

the

cen

ters

.

C

BA

G

HJD

E

K

M

LF

D AM

QC

PB

En

rich

men

t

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

7-2

7-2



Stu

dy G

uid

e a

nd I

nte

rven

tion

So

lvin

g E

qu

atio

ns

Usi

ng

Qu

adra

tic

Tech

niq

ues

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

7-3

7-3

©G

lenc

oe/M

cGra

w-H

ill38

7G

lenc

oe A

lgeb

ra 2

Lesson 7-3

Qu

adra

tic

Form

Cer

tain

pol

ynom

ial

expr

essi

ons

in x

can

be

wri

tten

in

th

e qu

adra

tic

form

au

2�

bu�

cfo

r an

y n

um

bers

a,b

,an

d c,

a

0,w

her

e u

is a

n e

xpre

ssio

n i

n x

.

Wri

te e

ach

pol

ynom

ial

in q

uad

rati

c fo

rm,i

f p

ossi

ble

.

a.3a

6�

9a3

�12

Let

u�

a3.

3a6

�9a

3�

12 �

3(a3

)2�

9(a3

) �

12

b.

101b

�49

�b�

�42

Let

u�

�b�.

101b

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�42

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1(�

b�)2

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(�b�)

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a5

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a3

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is e

xpre

ssio

n c

ann

ot b

e w

ritt

en i

n q

uad

rati

c fo

rm,s

ince

a5

(a

3 )2 .

Wri

te e

ach

pol

ynom

ial

in q

uad

rati

c fo

rm,i

f p

ossi

ble

.

1.x4

�6x

2�

82.

4p4

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2�

8

(x2 )

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2 ) �

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p2 )

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2 ) �

8

3.x8

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x�1 8��

2x� 11 6�

�1

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4 ) �

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� �1

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t p

oss

ible

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2 )2

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(x2 )

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x4�

48.

18x6

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3�

12

24(x

4 )2

�x

4�

418

(x3 )

2�

2(x

3 ) �

12

9.10

0x4

�9x

2�

1510

.25x

8�

36x6

�49

100(

x2 )

2�

9(x

2 ) �

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ot

po

ssib

le

11.4

8x6

�32

x3�

2012

.63x

8�

5x4

�29

48(x

3 )2

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(x3 )

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13.3

2x10

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143

14.5

0x3

�15

x�x�

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350

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0x6

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316

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60(x

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5 )2

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x5 )

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Exam

ple

Exam

ple

Exer

cises

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cises

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lenc

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cGra

w-H

ill38

8G

lenc

oe A

lgeb

ra 2

Solv

e Eq

uat

ion

s U

sin

g Q

uad

rati

c Fo

rmIf

a p

olyn

omia

l ex

pres

sion

can

be

wri

tten

in q

uad

rati

c fo

rm,t

hen

you

can

use

wh

at y

ou k

now

abo

ut

solv

ing

quad

rati

c eq

uat

ion

s to

solv

e th

e re

late

d po

lyn

omia

l eq

uat

ion

.

Sol

ve x

4�

40x

2�

144

�0.

x4�

40x2

�14

4 �

0O

rigin

al e

quat

ion

(x2 )

2�

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2 ) �

144

�0

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e th

e ex

pres

sion

on

the

left

in q

uadr

atic

for

m.

(x2

�4)

(x2

�36

) �

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acto

r.x2

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orx2

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o P

rodu

ct P

rope

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(x�

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(x�

6) �

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acto

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x�

2 �

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2 �

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6 �

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pert

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impl

ify.

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e so

luti

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and

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ve 2

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inal

equ

atio

n

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��

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rite

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expr

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ft in

qua

drat

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orm

.

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prin

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uar

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ot o

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nu

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r ca

nn

ot b

e n

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�x�

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3 h

as n

o so

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e so

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Sol

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ach

eq

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.

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0

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0

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7 � x1 � x2

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5 � 2Stu

dy G

uid

e a

nd I

nte

rven

tion

(c

onti

nued

)

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lvin

g E

qu

atio

ns

Usi

ng

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adra

tic

Tech

niq

ues

NA

ME

____

____

____

____

____

____

____

____

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____

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AT

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____

____

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ER

IOD

____

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7-3

7-3

Exam

ple1

Exam

ple1

Exam

ple2

Exam

ple2

Exer

cises

Exer

cises


An

swer

s


Skil

ls P

ract

ice

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lvin

g E

qu

atio

ns

Usi

ng

Qu

adra

tic

Tech

niq

ues

NA

ME

____

____

____

____

____

____

____

____

____

____

____

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AT

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____

____

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ER

IOD

____

_

7-3

7-3

©G

lenc

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cGra

w-H

ill38

9G

lenc

oe A

lgeb

ra 2

Lesson 7-3

Wri

te e

ach

exp

ress

ion

in

qu

adra

tic

form

,if

pos

sib

le.

1.5x

4�

2x2

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5(x

2 )2

�2(

x2 )

�8

2.3y

8�

4y2

�3

no

t p

oss

ible

3.10

0a6

�a3

100(

a3)2

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4.x8

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4�

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4 )2

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x4 )

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6.6b

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3b3

�1

no

t p

oss

ible

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v6�

8v3

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15(v

3 )2

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v3 )

�9

8.a9

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7aa[

(a4 )

2�

5(a4

) �

7]

Sol

ve e

ach

eq

uat

ion

.

9.a3

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210

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Wri

te e

ach

exp

ress

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in

qu

adra

tic

form

,if

pos

sib

le.

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b4�

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t p

oss

ible

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0x4

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) �

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ssib

le

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w5

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HY

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SA

pro

ton

in

a m

agn

etic

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ld f

ollo

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a pa

th o

n a

coo

rdin

ate

grid

mod

eled

by

the

fun

ctio

n f

(x)

�x4

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15.W

hat

are

th

e x-

coor

din

ates

of

the

poin

ts o

n t

he

grid

wh

ere

the

prot

on c

ross

es t

he

x-ax

is?

��

5� ,�

5�

27.S

UR

VEY

ING

Vis

ta c

oun

ty i

s se

ttin

g as

ide

a la

rge

parc

el o

f la

nd

to p

rese

rve

it a

s op

ensp

ace.

Th

e co

un

ty h

as h

ired

Meg

han

’s s

urv

eyin

g fi

rm t

o su

rvey

th

e pa

rcel

,wh

ich

is

inth

e sh

ape

of a

rig

ht

tria

ngl

e.T

he

lon

ger

leg

of t

he

tria

ngl

e m

easu

res

5 m

iles

les

s th

anth

e sq

uar

e of

th

e sh

orte

r le

g,an

d th

e h

ypot

enu

se o

f th

e tr

ian

gle

mea

sure

s 13

mil

es l

ess

than

tw

ice

the

squ

are

of t

he

shor

ter

leg.

Th

e le

ngt

h o

f ea

ch b

oun

dary

is

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hol

e n

um

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d th

e le

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oun

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.3

mi,

4 m

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mi

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e (

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rag

e)

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lvin

g E

qu

atio

ns

Usi

ng

Qu

adra

tic

Tech

niq

ues

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

7-3

7-3



Readin

g t

o L

earn

Math

em

ati

csS

olv

ing

Eq

uat

ion

s U

sin

g Q

uad

rati

c Te

chn

iqu

es

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

7-3

7-3

©G

lenc

oe/M

cGra

w-H

ill39

1G

lenc

oe A

lgeb

ra 2

Lesson 7-3

Pre-

Act

ivit

yH

ow c

an s

olvi

ng

pol

ynom

ial

equ

atio

ns

hel

p y

ou t

o fi

nd

dim

ensi

ons?

Rea

d th

e in

trod

uct

ion

to

Les

son

7-3

at

the

top

of p

age

360

in y

our

text

book

.

Exp

lain

how

th

e fo

rmu

la g

iven

for

th

e vo

lum

e of

th

e bo

x ca

n b

e ob

tain

edfr

om t

he

dim

ensi

ons

show

n i

n t

he

figu

re.

Sam

ple

an

swer

:Th

e vo

lum

e o

f a

rect

ang

ula

r b

ox is

giv

en

by t

he

form

ula

V�

�wh

.Su

bst

itu

te 5

0 �

2xfo

r �,

32 �

2xfo

r w

,an

d x

for

hto

get

V

(x)

�(5

0 �

2x)(

32 �

2x)(

x) �

4x3

�16

4x2

�16

00x.

Rea

din

g t

he

Less

on

1.W

hic

h o

f th

e fo

llow

ing

expr

essi

ons

can

be

wri

tten

in

qu

adra

tic

form

?b

,c,d

,f,g

,h,i

a.x3

�6x

2�

9b

.x4

�7x

2�

6c.

m6

�4m

3�

4

d.

y�

2y�1 2�

�15

e.x5

�x3

�1

f.r4

�6

�r8

g.p�1 4�

�8p

�1 2��

12h

.r�1 3�

�2r

�1 6��

3i.

5�z�

�2z

�3

2.M

atch

eac

h e

xpre

ssio

n f

rom

th

e li

st o

n t

he

left

wit

h i

ts f

acto

riza

tion

fro

m t

he

list

on

th

e ri

ght.

a.x4

�3x

2�

40vi

i.(x

3�

3)(x

3�

3)

b.

x4�

10x2

�25

vii

.(�

x��

3)(�

x��

3)

c.x6

�9

iii

i.(�

x��

3)2

d.

x�

9ii

iv.

(x2

�1)

(x4

�x2

�1)

e.x6

�1

ivv.

(x2

�5)

2

f.x

�6�

x��

9iii

vi.

(x2

�5)

(x2

�8)

Hel

pin

g Y

ou

Rem

emb

er

3.W

hat

is

an e

asy

way

to

tell

wh

eth

er a

tri

nom

ial

in o

ne

vari

able

con

tain

ing

one

con

stan

tte

rm c

an b

e w

ritt

en i

n q

uad

rati

c fo

rm?

Sam

ple

an

swer

:L

oo

k at

th

e tw

o t

erm

s th

at a

re n

ot

con

stan

ts a

nd

com

par

e th

e ex

po

nen

ts o

n t

he

vari

able

.If

on

e o

f th

e ex

po

nen

ts is

tw

ice

the

oth

er,t

he

trin

om

ial c

an b

e w

ritt

en in

qu

adra

tic

form

.

©G

lenc

oe/M

cGra

w-H

ill39

2G

lenc

oe A

lgeb

ra 2

Od

d a

nd

Eve

n P

oly

no

mia

l Fu

nct

ion

s

En

rich

men

t

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

7-3

7-3

Fu

nct

ion

s w

hos

e gr

aph

s ar

e sy

mm

etri

c w

ith

resp

ect

to t

he o

rigi

n ar

e ca

lled

odd

func

tion

s.If

f(�

x) �

�f(

x) f

or a

ll x

in t

he d

omai

n of

f(x

),th

en f

(x)

is o

dd.

Fu

nct

ion

s w

hos

e gr

aph

s ar

e sy

mm

etri

c w

ith

resp

ect

to t

he y

-axi

s ar

e ca

lled

even

func

tion

s.If

f(�

x) �

f(x)

for

all

xin

th

e do

mai

n o

f f(

x),

then

f(x

) is

eve

n.

x

f (x)

O1

2–2

–1

6 4 2f(

x) �

1 – 4x4 �

4

x

f (x)

O1

2–2

–1

4 2 –2 –4

f(x)

� 1 – 2x3

Exam

ple

Exam

ple

Det

erm

ine

wh

eth

er f

(x)

�x3

�3x

is o

dd

,eve

n,o

r n

eith

er.

f(x)

�x3

�3x

f(�

x) �

(�x)

3�

3(�

x)R

epla

ce x

with

�x.

��

x3�

3xS

impl

ify.

��

(x3

�3x

)F

acto

r ou

t �

1.

��

f(x)

Sub

stut

ute.

Th

eref

ore,

f(x)

is

odd.

Th

e gr

aph

at

the

righ

t ve

rifi

es t

hat

f(x

) is

odd

.T

he

grap

h o

f th

e fu

nct

ion

is

sym

met

ric

wit

h

resp

ect

to t

he

orig

in.

Det

erm

ine

wh

eth

er e

ach

fu

nct

ion

is

odd

,eve

n,o

r n

eith

erb

y gr

aph

ing

or b

y ap

ply

ing

the

rule

s fo

r od

d a

nd

eve

n f

un

ctio

ns.

1.f(

x) �

4x2

even

2.f(

x) �

�7x

4ev

en

3.f(

x) �

x7o

dd

4.f(

x) �

x3�

x2n

eith

er

5.f(

x) �

3x3

�1

nei

ther

6.f(

x) �

x8�

x5�

6n

eith

er

7.f(

x) �

�8x

5�

2x3

�6x

od

d8.

f(x)

�x4

�3x

3�

2x2

�6x

�1

nei

ther

9.f(

x) �

x4�

3x2

�11

even

10.f

(x)

�x7

�6x

5�

2x3

�x

od

d

11.C

ompl

ete

the

foll

owin

g de

fin

itio

ns:

A p

olyn

omia

l fu

nct

ion

is

odd

if a

nd

only

if a

ll t

he

term

s ar

e of

de

gree

s.A

pol

ynom

ial

fun

ctio

n i

s ev

en

if a

nd

only

if

all

the

term

s ar

e of

de

gree

s.ev

eno

dd

x

f (x)

O1

2–2

–1

4 2 –2 –4

f(x)

� x

3 �

3x


An

swer

s


Stu

dy G

uid

e a

nd I

nte

rven

tion

Th

e R

emai

nd

er a

nd

Fac

tor T

heo

rem

s

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

7-4

7-4

©G

lenc

oe/M

cGra

w-H

ill39

3G

lenc

oe A

lgeb

ra 2

Lesson 7-4

Syn

thet

ic S

ub

stit

uti

on

Rem

ain

der

The

rem

aind

er,

whe

n yo

u di

vide

the

pol

ynom

ial f

(x)

by (

x�

a),

is t

he c

onst

ant

f(a)

.T

heo

rem

f(x)

�q

(x)

�(x

�a)

�f(

a),

whe

re q

(x)

is a

pol

ynom

ial w

ith d

egre

e on

e le

ss t

han

the

degr

ee o

f f(

x).

If f

(x)

�3x

4�

2x3

�5x

2�

x�

2,fi

nd

f(�

2).

Exam

ple1

Exam

ple1

Exam

ple2

Exam

ple2

Met

hod

1S

ynth

etic

Su

bsti

tuti

onB

y th

e R

emai

nde

r T

heo

rem

,f(�

2) s

hou

ldbe

th

e re

mai

nde

r w

hen

you

div

ide

the

poly

nom

ial

by x

�2.

�2

32

�5

1�

2�

68

�6

103

�4

3�

58

Th

e re

mai

nde

r is

8,s

o f(

�2)

�8.

Met

hod

2D

irec

t S

ubs

titu

tion

Rep

lace

xw

ith

�2.

f(x)

�3x

4�

2x3

�5x

2�

x�

2f(

�2)

�3(

�2)

4�

2(�

2)3

�5(

�2)

2�

(�2)

�2

�48

�16

�20

�2

�2

or 8

So

f(�

2) �

8.

If f

(x)

�5x

3�

2x�

1,fi

nd

f(3

).A

gain

,by

the

Rem

ain

der

Th

eore

m,f

(3)

shou

ld b

e th

e re

mai

nde

r w

hen

you

div

ide

the

poly

nom

ial

by x

�3.

35

02

�1

1545

141

515

4714

0T

he

rem

ain

der

is 1

40,s

o f(

3) �

140.

Use

syn

thet

ic s

ub

stit

uti

on t

o fi

nd

f(�

5) a

nd

f�

�for

each

fu

nct

ion

.

1.f(

x) �

�3x

2�

5x�

1�

101;

2.f(

x) �

4x2

�6x

�7

63;

�3

3.f(

x) �

�x3

�3x

2�

519

5;�

4.f(

x) �

x4�

11x2

�1

899;

Use

syn

thet

ic s

ub

stit

uti

on t

o fi

nd

f(4

) an

d f

(�3)

for

eac

h f

un

ctio

n.

5.f(

x) �

2x3

�x2

�5x

�3

6.f(

x) �

3x3

�4x

�2

127;

�27

178;

�67

7.f(

x) �

5x3

�4x

2�

28.

f(x)

�2x

4�

4x3

�3x

2�

x�

625

8;�

169

302;

288

9.f(

x) �

5x4

�3x

3�

4x2

�2x

�4

10.f

(x)

�3x

4�

2x3

�x2

�2x

�5

1404

;29

862

7;27

7

11.f

(x)

�2x

4�

4x3

�x2

�6x

�3

12.f

(x)

�4x

4�

4x3

�3x

2�

2x�

321

9;28

280

5;46

2

29 � 1635 � 83 � 4

1 � 2

Exer

cises

Exer

cises

©G

lenc

oe/M

cGra

w-H

ill39

4G

lenc

oe A

lgeb

ra 2

Fact

ors

of

Poly

no

mia

lsT

he

Fac

tor

Th

eore

mca

n h

elp

you

fin

d al

l th

e fa

ctor

s of

apo

lyn

omia

l.

Fact

or T

heo

rem

The

bin

omia

l x�

ais

a f

acto

r of

the

pol

ynom

ial f

(x)

if an

d on

ly if

f(a

) �

0.

Sh

ow t

hat

x�

5 is

a f

acto

r of

x3

�2x

2�

13x

�10

.Th

en f

ind

th

ere

mai

nin

g fa

ctor

s of

th

e p

olyn

omia

l.B

y th

e Fa

ctor

Th

eore

m,t

he

bin

omia

l x

�5

is a

fac

tor

of t

he

poly

nom

ial

if �

5 is

a z

ero

of t

he

poly

nom

ial

fun

ctio

n.T

o ch

eck

this

,use

syn

thet

ic s

ubs

titu

tion

.

�5

12

�13

10�

515

�10

1�

32

0

Sin

ce t

he

rem

ain

der

is 0

,x�

5 is

a f

acto

r of

th

e po

lyn

omia

l.T

he

poly

nom

ial

x3�

2x2

�13

x�

10 c

an b

e fa

ctor

ed a

s (x

�5)

(x2

�3x

�2)

.Th

e de

pres

sed

poly

nom

ial

x2�

3x�

2 ca

n b

e fa

ctor

ed a

s (x

�2)

(x�

1).

So

x3�

2x2

�13

x�

10 �

(x�

5)(x

�2)

(x�

1).

Giv

en a

pol

ynom

ial

and

on

e of

its

fac

tors

,fin

d t

he

rem

ain

ing

fact

ors

of t

he

pol

ynom

ial.

Som

e fa

ctor

s m

ay n

ot b

e b

inom

ials

.

1.x3

�x2

�10

x�

8;x

�2

2.x3

�4x

2�

11x

�30

;x�

3(x

�4)

(x�

1)(x

�5)

(x�

2)

3.x3

�15

x2�

71x

�10

5;x

�7

4.x3

�7x

2�

26x

�72

;x�

4(x

�3)

(x�

5)(x

�2)

(x�

9)

5.2x

3�

x2�

7x�

6;x

�1

6.3x

3�

x2�

62x

�40

;x�

4(2

x�

3)(x

�2)

(3x

�2)

(x�

5)

7.12

x3�

71x2

�57

x�

10;x

�5

8.14

x3�

x2�

24x

�9;

x�

1(4

x�

1)(3

x�

2)(7

x�

3)(2

x�

3)

9.x3

�x

�10

;x�

210

.2x3

�11

x2�

19x

�28

;x�

4(x

2�

2x�

5)(2

x2

�3x

�7)

11.3

x3�

13x2

�34

x�

24;x

�6

12.x

4�

x3�

11x2

�9x

�18

;x�

1(3

x2

�5x

�4)

(x�

2)(x

�3)

(x�

3)

Stu

dy G

uid

e a

nd I

nte

rven

tion

(c

onti

nued

)

Th

e R

emai

nd

er a

nd

Fac

tor T

heo

rem

s

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

7-4

7-4

Exam

ple

Exam

ple

Exer

cises

Exer

cises



Skil

ls P

ract

ice

Th

e R

emai

nd

er a

nd

Fac

tor T

heo

rem

s

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

7-4

7-4

©G

lenc

oe/M

cGra

w-H

ill39

5G

lenc

oe A

lgeb

ra 2

Lesson 7-4

Use

syn

thet

ic s

ub

stit

uti

on t

o fi

nd

f(2

) an

d f

(�1)

for

eac

h f

un

ctio

n.

1.f(

x) �

x2�

6x�

521

,02.

f(x)

�x2

�x

�1

3,3

3.f(

x) �

x2�

2x�

2�

2,1

4.f(

x) �

x3�

2x2

�5

21,6

5.f(

x) �

x3�

x2�

2x�

33,

36.

f(x)

�x3

�6x

2�

x�

430

,0

7.f(

x) �

x3�

3x2

�x

�2

�4,

�7

8.f(

x) �

x3�

5x2

�x

�6

�8,

1

9.f(

x) �

x4�

2x2

�9

15,�

610

.f(x

) �

x4�

3x3

�2x

2�

2x�

62,

14

11.f

(x)

�x5

�7x

3�

4x�

1012

.f(x

) �

x6�

2x5

�x4

�x3

�9x

2�

20�

22,2

0�

32,�

26

Giv

en a

pol

ynom

ial

and

on

e of

its

fac

tors

,fin

d t

he

rem

ain

ing

fact

ors

of t

he

pol

ynom

ial.

Som

e fa

ctor

s m

ay n

ot b

e b

inom

ials

.

13.x

3�

2x2

�x

�2;

x�

114

.x3

�x2

�5x

�3;

x�

1

x�

1,x

�2

x�

1,x

�3

15.x

3�

3x2

�4x

�12

;x�

316

.x3

�6x

2�

11x

�6;

x�

3

x�

2,x

�2

x�

1,x

�2

17.x

3�

2x2

�33

x�

90;x

�5

18.x

3�

6x2

�32

;x�

4

x�

3,x

�6

x�

4,x

�2

19.x

3�

x2�

10x

�8;

x�

220

.x3

�19

x�

30;x

�2

x�

1,x

�4

x�

5,x

�3

21.2

x3�

x2�

2x�

1;x

�1

22.2

x3�

x2�

5x�

2;x

�2

2x�

1,x

�1

x�

1,2x

�1

23.3

x3�

4x2

�5x

�2;

3x�

124

.3x3

�x2

�x

�2;

3x�

2

x�

1,x

�2

x2

�x

�1

©G

lenc

oe/M

cGra

w-H

ill39

6G

lenc

oe A

lgeb

ra 2

Use

syn

thet

ic s

ub

stit

uti

on t

o fi

nd

f(�

3) a

nd

f(4

) fo

r ea

ch f

un

ctio

n.

1.f(

x) �

x2�

2x�

36,

272.

f(x)

�x2

�5x

�10

34,6

3.f(

x) �

x2�

5x�

420

,�8

4.f(

x) �

x3�

x2�

2x�

3�

27,4

3

5.f(

x) �

x3�

2x2

�5

�4,

101

6.f(

x) �

x3�

6x2

�2x

�87

,�24

7.f(

x) �

x3�

2x2

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�31

,32

8.f(

x) �

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x2�

4x�

4�

52,6

0

9.f(

x) �

x3�

3x2

�2x

�50

�56

,70

10.f

(x)

�x4

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�3x

2�

x�

1242

,280

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x�

773

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128

6,37

7

13.f

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339

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7

15.f

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�7x

3�

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1016

.f(x

) �

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20�

430,

1446

74,5

828

Giv

en a

pol

ynom

ial

and

on

e of

its

fac

tors

,fin

d t

he

rem

ain

ing

fact

ors

of t

he

pol

ynom

ial.

Som

e fa

ctor

s m

ay n

ot b

e b

inom

ials

.

17.x

3�

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x�

218

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15;x

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x�

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5

29.P

OPU

LATI

ON

Th

e pr

ojec

ted

popu

lati

on i

n t

hou

san

ds f

or a

cit

y ov

er t

he

nex

t se

vera

lye

ars

can

be

esti

mat

ed b

y th

e fu

nct

ion

P(x

) �

x3�

2x2

�8x

�52

0,w

her

e x

is t

he

nu

mbe

r of

yea

rs s

ince

200

0.U

se s

ynth

etic

su

bsti

tuti

on t

o es

tim

ate

the

popu

lati

on

for

2005

.65

5,00

0

30.V

OLU

ME

Th

e vo

lum

e of

wat

er i

n a

rec

tan

gula

r sw

imm

ing

pool

can

be

mod

eled

by

the

poly

nom

ial

2x3

�9x

2�

7x�

6.If

th

e de

pth

of

the

pool

is

give

n b

y th

e po

lyn

omia

l 2x

�1,

wh

at p

olyn

omia

ls e

xpre

ss t

he

len

gth

an

d w

idth

of

the

pool

?x

�3

and

x�

2

Pra

ctic

e (

Ave

rag

e)

Th

e R

emai

nd

er a

nd

Fac

tor T

heo

rem

s

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

7-4

7-4


An

swer

s


Readin

g t

o L

earn

Math

em

ati

csT

he

Rem

ain

der

an

d F

acto

r Th

eore

ms

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

7-4

7-4

©G

lenc

oe/M

cGra

w-H

ill39

7G

lenc

oe A

lgeb

ra 2

Lesson 7-4

Pre-

Act

ivit

yH

ow c

an y

ou u

se t

he

Rem

ain

der

Th

eore

m t

o ev

alu

ate

pol

ynom

ials

?

Rea

d th

e in

trod

uct

ion

to

Les

son

7-4

at

the

top

of p

age

365

in y

our

text

book

.

Sh

ow h

ow y

ou w

ould

use

th

e m

odel

in

th

e in

trod

uct

ion

to

esti

mat

e th

en

um

ber

of i

nte

rnat

ion

al t

rave

lers

(in

mil

lion

s) t

o th

e U

nit

ed S

tate

s in

th

eye

ar 2

000.

(Sh

ow h

ow y

ou w

ould

su

bsti

tute

nu

mbe

rs,b

ut

do n

ot a

ctu

ally

calc

ula

te t

he

resu

lt.)

Sam

ple

an

swer

:0.

02(1

4)3

�0.

6(14

)2�

6(14

) �

25.9

Rea

din

g t

he

Less

on

1.C

onsi

der

the

foll

owin

g sy

nth

etic

div

isio

n.

13

2�

64

35

�1

35

�1

3

a.U

sin

g th

e di

visi

on s

ymbo

l �

,wri

te t

he

divi

sion

pro

blem

th

at i

s re

pres

ente

d by

th

issy

nth

etic

div

isio

n.(

Do

not

in

clu

de t

he

answ

er.)

(3x3

�2x

2�

6x�

4) �

(x�

1)

b.

Iden

tify

eac

h o

f th

e fo

llow

ing

for

this

div

isio

n.

divi

den

ddi

viso

r

quot

ien

tre

mai

nde

r

c.If

f(x

) �

3x3

�2x

2�

6x�

4,w

hat

is

f(1)

?3

2.C

onsi

der

the

foll

owin

g sy

nth

etic

div

isio

n.

�3

10

027

�3

9�

271

�3

90

a.T

his

div

isio

n s

how

s th

at

is a

fac

tor

of

.

b.

Th

e di

visi

on s

how

s th

at

is a

zer

o of

th

e po

lyn

omia

l fu

nct

ion

f(x)

�.

c.T

he

divi

sion

sh

ows

that

th

e po

int

is o

n t

he

grap

h o

f th

e po

lyn

omia

l

fun

ctio

n f

(x)

�.

Hel

pin

g Y

ou

Rem

emb

er

3.T

hin

k of

a m

nem

onic

for

rem

embe

rin

g th

e se

nte

nce

,“D

ivid

end

equ

als

quot

ien

t ti

mes

divi

sor

plu

s re

mai

nde

r.”S

amp

le a

nsw

er:

Def

init

ely

ever

y q

uie

t te

ach

er d

eser

ves

pro

per

rew

ard

s.

x3�

27(�

3,0)

x3�

27�

3

x3�

27x

� 3

33x

3�

5x�

1

x �

13x

3�

2x2

�6x

�4

©G

lenc

oe/M

cGra

w-H

ill39

8G

lenc

oe A

lgeb

ra 2

Usi

ng

Max

imu

m V

alu

esM

any

tim

es m

axim

um

sol

uti

ons

are

nee

ded

for

diff

eren

t si

tuat

ion

s.F

or

inst

ance

,wh

at i

s th

e ar

ea o

f th

e la

rges

t re

ctan

gula

r fi

eld

that

can

be

encl

osed

w

ith

200

0 fe

et o

f fe

nci

ng?

Let

xan

d y

den

ote

the

len

gth

an

d w

idth

of

th

e fi

eld,

resp

ecti

vely

.

Per

imet

er:2

x�

2y�

2000

→y

�10

00 �

xA

rea:

A�

xy�

x(10

00 �

x) �

�x2

�10

00x

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is p

robl

em i

s eq

uiv

alen

t to

fin

din

g

the

hig

hes

t po

int

on t

he

grap

h o

f A

(x)

��

x2�

1000

xsh

own

on

th

e ri

ght.

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plet

e th

e sq

uar

e fo

r �

x2�

1000

x.

A�

�(x

2�

1000

x�

5002

) �

5002

��

(x�

500)

2�

5002

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ause

th

e te

rm �

(x�

500)

2is

eit

her

n

egat

ive

or 0

,th

e gr

eate

st v

alu

e of

Ais

500

2 .T

he

max

imu

m a

rea

encl

osed

is

5002

or 2

50,0

00 s

quar

e fe

et.

Sol

ve e

ach

pro

ble

m.

1.F

ind

the

area

of

the

larg

est

rect

angu

lar

gard

en t

hat

can

be

encl

osed

by

300

feet

of

fen

ce.

5625

ft2

2.A

far

mer

wil

l m

ake

a re

ctan

gula

r pe

n w

ith

100

fee

t of

fen

ce u

sin

g pa

rt

of h

is b

arn

for

on

e si

de o

f th

e pe

n.W

hat

is

the

larg

est

area

he

can

en

clos

e?

1250

ft2

3.A

n a

rea

alon

g a

stra

igh

t st

one

wal

l is

to

be f

ence

d.T

her

e ar

e 60

0 m

eter

s of

fen

cin

g av

aila

ble.

Wh

at i

s th

e gr

eate

st r

ecta

ngu

lar

area

th

at c

an b

e en

clos

ed?

45,0

00 m

2

A

xO

1000

x

y

En

rich

men

t

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

7-4

7-4



Stu

dy G

uid

e a

nd I

nte

rven

tion

Ro

ots

an

d Z

ero

s

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

7-5

7-5

©G

lenc

oe/M

cGra

w-H

ill39

9G

lenc

oe A

lgeb

ra 2

Lesson 7-5

Typ

es o

f R

oo

tsT

he f

ollo

win

g st

atem

ents

are

equ

ival

ent

for

any

poly

nom

ial

func

tion

f(x

).•

cis

a z

ero

of t

he

poly

nom

ial

fun

ctio

n f

(x).

•(x

�c)

is

a fa

ctor

of

the

poly

nom

ial

f(x)

.•

cis

a r

oot

or s

olu

tion

of

the

poly

nom

ial

equ

atio

n f

(x)

�0.

If c

is r

eal,

then

(c,

0) i

s an

in

terc

ept

of t

he

grap

h o

f f(

x).

Fu

nd

amen

tal

Eve

ry p

olyn

omia

l equ

atio

n w

ith d

egre

e gr

eate

r th

an z

ero

has

at le

ast

one

root

in t

he s

etT

heo

rem

of

Alg

ebra

of c

ompl

ex n

umbe

rs.

Co

rolla

ry t

o t

he

Apo

lyno

mia

l equ

atio

n of

the

for

m P

(x)

�0

of d

egre

e n

with

com

plex

coe

ffici

ents

has

Fu

nd

amen

tal

exac

tly n

root

s in

the

set

of

com

plex

num

bers

.T

heo

rem

of

Alg

ebra

s

If P

(x)

is a

pol

ynom

ial w

ith r

eal c

oeffi

cien

ts w

hose

ter

ms

are

arra

nged

in d

esce

ndin

gpo

wer

s of

the

var

iabl

e,

Des

cart

es’R

ule

•th

e nu

mbe

r of

pos

itive

rea

l zer

os o

f y

�P

(x)

is t

he s

ame

as t

he n

umbe

r of

cha

nges

in

of

Sig

ns

sign

of

the

coef

ficie

nts

of t

he t

erm

s, o

r is

less

tha

n th

is b

y an

eve

n nu

mbe

r, an

d•

the

num

ber

of n

egat

ive

real

zer

os o

f y

�P

(x)

is t

he s

ame

as t

he n

umbe

r of

cha

nges

in

sign

of

the

coef

ficie

nts

of t

he t

erm

s of

P(�

x),

or is

less

tha

n th

is n

umbe

r by

an

even

num

ber.

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ve t

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equ

atio

n

6x3

�3x

�0

and

sta

te t

he

nu

mb

er a

nd

typ

e of

roo

ts.

6x3

�3x

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x2�

1) �

0U

se t

he

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o P

rodu

ct P

rope

rty.

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2x2

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x�

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��

1

x�

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e eq

uat

ion

has

on

e re

al r

oot,

0,

and

two

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inar

y ro

ots,

.

i�2�

�2

i�2�

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te t

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nu

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er o

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osit

ive

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ive

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os,a

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im

agin

ary

zero

s fo

r p

(x)

�4x

4�

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ce p

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ree

4,it

has

4 z

eros

.U

se D

esca

rtes

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ign

s to

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erm

ine

the

num

ber

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type

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os.S

ince

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s,th

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are

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osit

ive

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os.

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d p(

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an

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un

t th

e n

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ges

insi

gn f

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ts c

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icie

nts

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ce t

her

e is

on

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gn c

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her

e is

exa

ctly

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egat

ive

real

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o.

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ple1

Exam

ple1

Exam

ple2

Exam

ple2

Exer

cises

Exer

cises

Sol

ve e

ach

eq

uat

ion

an

d s

tate

th

e n

um

ber

an

d t

ype

of r

oots

.

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te t

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ive

real

zer

os,a

nd

im

agin

ary

zero

sfo

r ea

ch f

un

ctio

n.

4.f(

x) �

3x3

�x2

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�12

1;2

or

0;0

or

2

5.f(

x) �

2x4

�x3

�3x

�7

2 o

r 0;

0;2

or

4

6.f(

x) �

3x5

�x4

�x3

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2�

53

or

1;2

or

0;0,

2,o

r 4

5i�

3��

3

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0G

lenc

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Fin

d Z

ero

s

Co

mp

lex

Co

nju

gat

eS

uppo

se a

and

bar

e re

al n

umbe

rs w

ith b

0.

If

a�

biis

a z

ero

of a

pol

ynom

ial

Th

eore

mfu

nctio

n w

ith r

eal c

oeffi

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then

a�

biis

als

o a

zero

of

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func

tion.

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d a

ll o

f th

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ros

of f

(x)

�x4

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38x

�60

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ince

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as d

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the

fun

ctio

n h

as 4

zer

os.

f(x)

�x4

�15

x2�

38x

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f(�

x) �

x4�

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x�

60S

ince

th

ere

are

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gn c

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ges

for

the

coef

fici

ents

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f(x)

,th

e fu

nct

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r 1

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tive

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ros.

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ce t

her

e is

1 s

ign

ch

ange

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f f(

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,th

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nct

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real

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o.U

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etic

su

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st s

ome

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21

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31

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39

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601

3�

620

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o 3

is a

zer

o of

th

e po

lyn

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l fu

nct

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.Now

try

syn

thet

ic s

ubs

titu

tion

aga

in t

o fi

nd

a ze

roof

th

e de

pres

sed

poly

nom

ial.

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11

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81

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he

Qu

adra

tic

For

mu

la o

n t

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esse

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lyn

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l x2

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to f

ind

the

oth

er 2

zer

os,1

i �

3�.T

he

fun

ctio

n h

as t

wo

real

zer

os a

t 3

and

�5

and

two

imag

inar

y ze

ros

at 1

i �

3�.

Fin

d a

ll o

f th

e ze

ros

of e

ach

fu

nct

ion

.

1.f(

x) �

x3�

x2�

9x�

9�

1,�

3i2.

f(x)

�x3

�3x

2�

4x�

123,

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3.p(

a) �

a3�

10a2

�34

a�

404,

3 �

i4.

p(x)

�x3

�5x

2�

11x

�15

3,1

�2i

5.f(

x) �

x3�

6x�

206.

f(x)

�x4

�3x

3�

21x2

�75

x�

100

�2,

1 �

3i�

1,4,

�5i

Stu

dy G

uid

e a

nd I

nte

rven

tion

(c

onti

nued

)

Ro

ots

an

d Z

ero

s

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

7-5

7-5

Exam

ple

Exam

ple

Exer

cises

Exer

cises


An

swer

s


Skil

ls P

ract

ice

Ro

ots

an

d Z

ero

s

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

7-5

7-5

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lenc

oe/M

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w-H

ill40

1G

lenc

oe A

lgeb

ra 2

Lesson 7-5

Sol

ve e

ach

eq

uat

ion

.Sta

te t

he

nu

mb

er a

nd

typ

e of

roo

ts.

1.5x

�12

�0

2.x2

�4x

�40

�0

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;1

real

2 �

6i;

2 im

agin

ary

3.x5

�4x

3�

04.

x4�

625

�0

0,0,

0,2i

,�2i

;3

real

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agin

ary

5i,5

i,�

5i,�

5i;

4 im

agin

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2�

4x�

1 �

06.

x5�

81x

�0

;2

real

0,�

3,3,

�3i

,3i;

3 re

al,2

imag

inar

y

Sta

te t

he

pos

sib

le n

um

ber

of

pos

itiv

e re

al z

eros

,neg

ativ

e re

al z

eros

,an

dim

agin

ary

zero

s of

eac

h f

un

ctio

n.

7.g(

x) �

3x3

�4x

2�

17x

�6

8.h

(x)

�4x

3�

12x2

�x

�3

2 o

r 0;

1;2

or

02

or

0;1;

2 o

r 0

9.f(

x) �

x3�

8x2

�2x

�4

10.p

(x)

�x3

�x2

�4x

�6

3 o

r 1;

0;2

or

03

or

1;0;

2 o

r 0

11.q

(x)

�x4

�7x

2�

3x�

912

.f(x

) �

x4�

x3�

5x2

�6x

�1

1;1;

22

or

0;2

or

0;4

or

2 o

r 0

Fin

d a

ll t

he

zero

s of

eac

h f

un

ctio

n.

13.h

(x)

�x3

�5x

2�

5x�

314

.g(x

) �

x3�

6x2

�13

x�

10

3,1

��

2�,1

��

2�2,

2 �

i,2

�i

15.h

(x)

�x3

�4x

2�

x�

616

.q(x

) �

x3�

3x2

�6x

�8

1,�

2,�

32,

�1,

�4

17.g

(x)

�x4

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3�

5x2

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�4

18.f

(x)

�x4

�21

x2�

80

�1,

�1,

1,4

�4,

4,�

�5�,

�5�

Wri

te a

pol

ynom

ial

fun

ctio

n o

f le

ast

deg

ree

wit

h i

nte

gral

coe

ffic

ien

ts t

hat

has

th

egi

ven

zer

os.

19.�

3,�

5,1

20.3

if(

x)

�x

3�

7x2

�7x

�15

f(x

) �

x2

�9

21.�

5 �

i22

.�1,

�3�,

��

3�f(

x)

�x

2�

10x

�26

f(x

) �

x3

�x

2�

3x�

3

23.i

,5i

24.�

1,1,

i�6�

f(x

) �

x4

�26

x2

�25

f(x

) �

x4

�5x

2�

6

1 �

�2�

�2

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lenc

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cGra

w-H

ill40

2G

lenc

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lgeb

ra 2

Sol

ve e

ach

eq

uat

ion

.Sta

te t

he

nu

mb

er a

nd

typ

e of

roo

ts.

1.�

9x�

15 �

02.

x4�

5x2

�4

�0

��5 3� ;

1 re

al�

1,1,

�2,

2;4

real

3.x5

�81

x4.

x3�

x2�

3x�

3 �

0

0,�

3,3,

�3i

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3 re

al,2

imag

inar

y�

1,�

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�3�;

3 re

al

5.x3

�6x

�20

�0

6.x4

�x3

�x2

�x

�2

�0

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1 �

3i;

1 re

al,2

imag

inar

y2,

�1,

�i,

i;2

real

,2 im

agin

ary

Sta

te t

he

pos

sib

le n

um

ber

of

pos

itiv

e re

al z

eros

,neg

ativ

e re

al z

eros

,an

dim

agin

ary

zero

s of

eac

h f

un

ctio

n.

7.f(

x) �

4x3

�2x

2�

x�

38.

p(x)

�2x

4�

2x3

�2x

2�

x�

1

2 o

r 0;

1;2

or

03

or

1;1;

2 o

r 0

9.q(

x) �

3x4

�x3

�3x

2�

7x�

510

.h(x

) �

7x4

�3x

3�

2x2

�x

�1

2 o

r 0;

2 o

r 0;

4,2,

or

02

or

0;2

or

0;4,

2,o

r 0

Fin

d a

ll t

he

zero

s of

eac

h f

un

ctio

n.

11.h

(x)

�2x

3�

3x2

�65

x�

8412

.p(x

) �

x3�

3x2

�9x

�7

�7,

�3 2� ,4

1,1

�i�

6�,1

�i�

6�

13.h

(x)

�x3

�7x

2�

17x

�15

14.q

(x)

�x4

�50

x2�

49

3,2

�i,

2 �

i�

i,i,

�7i

,7i

15.g

(x)

�x4

�4x

3�

3x2

�14

x�

816

.f(x

) �

x4�

6x3

�6x

2�

24x

�40

�1,

�1,

2,�

4�

2,2,

3 �

i,3

�i

Wri

te a

pol

ynom

ial

fun

ctio

n o

f le

ast

deg

ree

wit

h i

nte

gral

coe

ffic

ien

ts t

hat

has

th

egi

ven

zer

os.

17.�

5,3i

18.�

2,3

�i

f(x

) �

x3

�5x

2�

9x�

45f(

x)

�x

3�

4x2

�2x

�20

19.�

1,4,

3i20

.2,5

,1 �

if(

x)

�x

4�

3x3

�5x

2�

27x

�36

f(x

) �

x4

�9x

3�

26x

2�

34x

�20

21.C

RA

FTS

Ste

phan

has

a s

et o

f pl

ans

to b

uil

d a

woo

den

box

.He

wan

ts t

o re

duce

th

evo

lum

e of

th

e bo

x to

105

cu

bic

inch

es.H

e w

ould

lik

e to

red

uce

th

e le

ngt

h o

f ea

chdi

men

sion

in

th

e pl

an b

y th

e sa

me

amou

nt.

Th

e pl

ans

call

for

th

e bo

x to

be

10 i

nch

es b

y8

inch

es b

y 6

inch

es.W

rite

an

d so

lve

a po

lyn

omia

l eq

uat

ion

to

fin

d ou

t h

ow m

uch

Ste

phen

sh

ould

tak

e fr

om e

ach

dim

ensi

on.

(10

�x)

(8 �

x)(6

�x)

�10

5;3

in.

Pra

ctic

e (

Ave

rag

e)

Ro

ots

an

d Z

ero

s

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

7-5

7-5



Readin

g t

o L

earn

Math

em

ati

csR

oo

ts a

nd

Zer

os

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

7-5

7-5

©G

lenc

oe/M

cGra

w-H

ill40

3G

lenc

oe A

lgeb

ra 2

Lesson 7-5

Pre-

Act

ivit

yH

ow c

an t

he

root

s of

an

eq

uat

ion

be

use

d i

n p

har

mac

olog

y?

Rea

d th

e in

trod

uct

ion

to

Les

son

7-5

at

the

top

of p

age

371

in y

our

text

book

.

Usi

ng

the

mod

el g

iven

in

th

e in

trod

uct

ion

,wri

te a

pol

ynom

ial

equ

atio

nw

ith

0 o

n o

ne

side

th

at c

an b

e so

lved

to

fin

d th

e ti

me

or t

imes

at

wh

ich

ther

e is

100

mil

ligr

ams

of m

edic

atio

n i

n a

pat

ien

t’s b

lood

stre

am.

0.5t

4�

3.5t

3�

100t

2�

350t

�10

0 �

0

Rea

din

g t

he

Less

on

1.In

dica

te w

het

her

eac

h s

tate

men

t is

tru

eor

fal

se.

a.E

very

pol

ynom

ial

equ

atio

n o

f de

gree

gre

ater

th

an o

ne

has

at

leas

t on

e ro

ot i

n t

he

set

of r

eal

nu

mbe

rs.

fals

eb

.If

cis

a r

oot

of t

he

poly

nom

ial

equ

atio

n f

(x)

�0,

then

(x

�c)

is

a fa

ctor

of

the

poly

nom

ial

f(x)

.tr

ue

c.If

(x

�c)

is

a fa

ctor

of

the

poly

nom

ial

f(x)

,th

en c

is a

zer

o of

th

e po

lyn

omia

l fu

nct

ion

f.

fals

ed

.A

pol

ynom

ial

fun

ctio

n f

of d

egre

e n

has

exa

ctly

(n

�1)

com

plex

zer

os.

fals

e

2.L

et f

(x)

�x6

�2x

5�

3x4

�4x

3�

5x2

�6x

�7.

a.W

hat

are

th

e po

ssib

le n

um

bers

of

posi

tive

rea

l ze

ros

of f

?5,

3,o

r 1

b.

Wri

te f

(�x)

in

sim

plif

ied

form

(w

ith

no

pare

nth

eses

).x

6�

2x5

�3x

4�

4x3

�5x

2�

6x�

7W

hat

are

th

e po

ssib

le n

um

bers

of

neg

ativ

e re

al z

eros

of

f?1

c.C

ompl

ete

the

foll

owin

g ch

art

to s

how

th

e po

ssib

le c

ombi

nat

ion

s of

pos

itiv

e re

al z

eros

,n

egat

ive

real

zer

os,a

nd

imag

inar

y ze

ros

of t

he

poly

nom

ial

fun

ctio

n f

.

Nu

mb

er o

fN

um

ber

of

Nu

mb

er o

f To

tal N

um

ber

P

osi

tive

Rea

l Zer

os

Neg

ativ

e R

eal Z

ero

sIm

agin

ary

Zer

os

of

Zer

os

51

06

31

26

11

46

Hel

pin

g Y

ou

Rem

emb

er

3.It

is

easi

er t

o re

mem

ber

mat

hem

atic

al c

once

pts

and

resu

lts

if y

ou r

elat

e th

em t

o ea

chot

her

.How

can

th

e C

ompl

ex C

onju

gate

s T

heo

rem

hel

p yo

u r

emem

ber

the

part

of

Des

cart

es’ R

ule

of

Sig

ns

that

say

s,“o

r is

les

s th

an t

his

nu

mbe

r by

an

eve

n n

um

ber.”

Sam

ple

an

swer

:F

or

a p

oly

no

mia

l fu

nct

ion

in w

hic

h t

he

po

lyn

om

ial h

asre

al c

oef

ficie

nts

,im

agin

ary

zero

s co

me

in c

on

jug

ate

pai

rs.T

her

efo

re,t

her

em

ust

be

an e

ven

nu

mb

er o

f im

agin

ary

zero

s.F

or

each

pai

r o

f im

agin

ary

zero

s,th

e n

um

ber

of

po

siti

ve o

r n

egat

ive

zero

s d

ecre

ases

by

2.

©G

lenc

oe/M

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w-H

ill40

4G

lenc

oe A

lgeb

ra 2

Th

e B

isec

tio

n M

eth

od

fo

r A

pp

roxi

mat

ing

Rea

l Zer

os

Th

e b

isec

tion

met

hod

can

be

use

d to

app

roxi

mat

e ze

ros

of p

olyn

omia

l fu

nct

ion

s li

ke f

(x)

�x3

�x2

�3x

�3.

Sin

ce f

(1)

� �

4 an

d f(

2) �

3,th

ere

is a

t le

ast

one

real

zer

o be

twee

n 1

an

d 2.

Th

e m

idpo

int

of t

his

in

terv

al i

s �1

� 22

��

1.5.

Sin

ce f

(1.5

) �

�1.

875,

the

zero

is

betw

een

1.5

an

d 2.

Th

e m

idpo

int

of t

his

in

terv

al i

s �1.

5 2�2

��

1.75

.Sin

ce

f(1.

75)

is a

bout

0.1

72,t

he z

ero

is b

etw

een

1.5

and

1.75

.Th

e m

idpo

int

of t

his

inte

rval

is �1.

5� 2

1.75

��

1.62

5 an

d f(

1.62

5) i

s ab

out

�0.

94.T

he

zero

is

betw

een

1.62

5 an

d 1.

75.T

he

mid

poin

t of

th

is i

nte

rval

is �1.

625

2�1.

75�

�1.

6875

.Sin

ce

f(1.

6875

) is

abo

ut

�0.

41,t

he

zero

is

betw

een

1.6

875

and

1.75

.Th

eref

ore,

the

zero

is

1.7

to t

he

nea

rest

ten

th.

Th

e di

agra

m b

elow

su

mm

ariz

es t

he

resu

lts

obta

ined

by

the

bise

ctio

n m

eth

od.

Usi

ng

the

bis

ecti

on m

eth

od,a

pp

roxi

mat

e to

th

e n

eare

st t

enth

th

e ze

ro b

etw

een

th

e tw

o in

tegr

al v

alu

es o

f x

for

each

fu

nct

ion

.

1.f(

x) �

x3�

4x2

�11

x �

2,f(

0) �

2,f(

1) �

�12

0.2

2.f(

x) �

2x4

�x2

�15

,f(1

) �

�12

,f(2

) �

211.

6

3.f(

x) �

x5�

2x3

�12

,f(1

) �

�13

,f(2

) �

41.

9

4.f(

x) �

4x3

�2x

�7,

f(�

2) �

�21

,f(�

1) �

5�

1.3

5.f(

x) �

3x3

�14

x2�

27x

�12

6,f(

4) �

�14

,f(5

) �

164.

7

11.

52

1.62

51.

75

1.68

75

++

––

––

sign

of f

(x):

valu

e x

:

En

rich

men

t

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

7-5

7-5


An

swer

s


Stu

dy G

uid

e a

nd I

nte

rven

tion

Rat

ion

al Z

ero

Th

eore

m

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

7-6

7-6

©G

lenc

oe/M

cGra

w-H

ill40

5G

lenc

oe A

lgeb

ra 2

Lesson 7-6

Iden

tify

Rat

ion

al Z

ero

s

Rat

ion

al Z

ero

Le

t f(

x) �

a 0xn

�a 1

xn�

1�

… �

a n�

2x2

�a n

�1x

�an

repr

esen

t a

poly

nom

ial f

unct

ion

Th

eore

mw

ith in

tegr

al c

oeffi

cien

ts.

If �p q�

is a

rat

iona

l num

ber

in s

impl

est

form

and

is a

zer

o of

y�

f(x)

,th

en p

is a

fac

tor

of a

nan

d q

is a

fac

tor

of a

0.

Co

rolla

ry (

Inte

gra

l If

the

coef

ficie

nts

of a

pol

ynom

ial a

re in

tege

rs s

uch

that

a0

�1

and

a n

0, a

ny r

atio

nal

Zer

o T

heo

rem

)ze

ros

of t

he f

unct

ion

mus

t be

fac

tors

of

a n.

Lis

t al

l of

th

e p

ossi

ble

rat

ion

al z

eros

of

each

fu

nct

ion

.

a.f(

x) �

3x4

�2x

2�

6x�

10

If �p q�

is a

rat

ion

al r

oot,

then

pis

a f

acto

r of

�10

an

d q

is a

fac

tor

of 3

.Th

e po

ssib

le v

alu

es

for

par

e

1,

2,

5,an

d

10.T

he

poss

ible

val

ues

for

qar

e

1 an

d

3.S

o al

l of

th

e po

ssib

le r

atio

nal

zer

os a

re �p q�

�

1,

2,

5,

10,

�1 3� ,

�2 3� ,

�5 3� ,an

d

�1 30 �.

b.

q(x

) �

x3�

10x2

�14

x�

36

Sin

ce t

he

coef

fici

ent

of x

3is

1,t

he

poss

ible

rat

ion

al z

eros

mu

st b

e th

e fa

ctor

s of

th

eco

nst

ant

term

�36

.So

the

poss

ible

rat

ion

al z

eros

are

1,

2,

3,

4,

6,

9,

12

,18

,an

d

36.

Lis

t al

l of

th

e p

ossi

ble

rat

ion

al z

eros

of

each

fu

nct

ion

.

1.f(

x) �

x3�

3x2

�x

�8

2.g(

x) �

x5�

7x4

�3x

2�

x�

20

�1,

�2,

�4,

�8

�1,

�2,

�4,

�5,

�10

,�20

3.h

(x)

�x4

�7x

3�

4x2

�x

�49

4.p(

x) �

2x4

�5x

3�

8x2

�3x

�5

�1,

�7,

�49

�1,

�5,

�,�

5.q(

x) �

3x4

�5x

3�

10x

�12

6.r(

x) �

4x5

�2x

�18

�1,

�2,

�3,

�4,

�6,

�12

,�

1,�

2,�

3,�

6,�

9,�

18,

�,�

,��

,�,�

,�,�

,�

7.f(

x) �

x7�

6x5

�3x

4�

x3�

4x2

�12

08.

g(x)

�5x

6�

3x4

�5x

3�

2x2

�15

�1,

�2,

�3,

�4,

�5,

�6,

�8,

�10

,�12

,�

15,�

20,�

24,�

30,�

40,�

60,�

120

�1,

�3,

�5,

�15

,�,�

9.h

(x)

�6x

5�

3x4

�12

x3�

18x2

�9x

�21

10.p

(x)

�2x

7�

3x6

�11

x5�

20x2

�11

�1,

�3,

�7,

�21

,�,�

,�,�

,�

1,�

11,�

,�

�,�

,�,�

7 � 61 � 6

7 � 31 � 3

11 � 21 � 2

21 � 27 � 2

3 � 21 � 2

3 � 51 � 5

9 � 43 � 4

1 � 49 � 2

3 � 21 � 2

4 � 32 � 3

1 � 3

5 � 21 � 2

Exam

ple

Exam

ple

Exer

cises

Exer

cises

©G

lenc

oe/M

cGra

w-H

ill40

6G

lenc

oe A

lgeb

ra 2

Fin

d R

atio

nal

Zer

os

Fin

d a

ll o

f th

e ra

tion

al z

eros

of

f(x)

�5x

3�

12x2

�29

x�

12.

Fro

m t

he

coro

llar

y to

th

e F

un

dam

enta

l Th

eore

m o

f Alg

ebra

,we

know

th

at t

her

e ar

e ex

actl

y 3

com

plex

roo

ts.A

ccor

din

g to

Des

cart

es’ R

ule

of

Sig

ns

ther

e ar

e 2

or 0

pos

itiv

e re

al r

oots

an

d 1

neg

ativ

e re

al r

oot.

Th

e po

ssib

le r

atio

nal

zer

os a

re

1,

2,

3,

4,

6,

12,

,

,,

,,

.Mak

e a

tabl

e an

d te

st s

ome

poss

ible

rat

ion

al z

eros

.

Sin

ce f

(1)

�0,

you

kn

ow t

hat

x�

1 is

a z

ero.

Th

e de

pres

sed

poly

nom

ial

is 5

x2�

17x

�12

,wh

ich

can

be

fact

ored

as

(5x

�3)

(x�

4).

By

the

Zer

o P

rodu

ct P

rope

rty,

this

exp

ress

ion

equ

als

0 w

hen

x�

or x

��

4.T

he

rati

onal

zer

os o

f th

is f

un

ctio

n a

re 1

,,a

nd

�4.

Fin

d a

ll o

f th

e ze

ros

of f

(x)

�8x

4�

2x3

�5x

2�

2x�

3.T

her

e ar

e 4

com

plex

roo

ts,w

ith

1 p

osit

ive

real

roo

t an

d 3

or 1

neg

ativ

e re

al r

oots

.Th

e po

ssib

le r

atio

nal

zer

os a

re

1,

3,

,,

,,

,an

d

.3 � 8

3 � 43 � 2

1 � 81 � 4

1 � 2

3 � 5

3 � 5

�p q�5

12�

2912

15

17�

120

12 � 56 � 5

4 � 53 � 5

2 � 51 � 5

Stu

dy G

uid

e a

nd I

nte

rven

tion

(c

onti

nued

)

Rat

ion

al Z

ero

Th

eore

m

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

7-6

7-6

Exam

ple1

Exam

ple1

Exam

ple2

Exam

ple2

Exer

cises

Exer

cises

Mak

e a

tabl

e an

d te

st s

ome

poss

ible

val

ues

.

Sin

ce f�

��0,

we

know

tha

t x

�

is a

roo

t.

1 � 21 � 2

�p q�8

25

2�

3

18

1015

1714

28

1841

8416

5

�1 2�8

68

60

The

dep

ress

ed p

olyn

omia

l is

8x3

�6x

2�

8x�

6.T

ry s

ynth

etic

su

bsti

tuti

on a

gain

.An

y re

mai

nin

gra

tion

al r

oots

mu

st b

e n

egat

ive.

x�

��3 4�

is a

noth

er r

atio

nal r

oot.

The

dep

ress

ed p

olyn

omia

l is

8x2

�8

�0,

whi

ch h

as r

oots

i.

�p q�8

68

6

��1 4�

84

74�

1 4�

��3 4�

80

80

Th

e ze

ros

of t

his

fu

nct

ion

are

�1 2� ,�

�3 4� ,an

d

i.

Fin

d a

ll o

f th

e ra

tion

al z

eros

of

each

fu

nct

ion

.

1.f(

x) �

x3�

4x2

�25

x�

28�

1,4,

�7

2.f(

x) �

x3�

6x2

�4x

�24

�6

Fin

d a

ll o

f th

e ze

ros

of e

ach

fu

nct

ion

.

3.f(

x) �

x4�

2x3

�11

x2�

8x�

604.

f(x)

�4x

4�

5x3

�30

x2�

45x

�54

3,�

5,�

2i,�

2,�

3i3 � 4



Skil

ls P

ract

ice

Rat

ion

al Z

ero

Th

eore

m

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

7-6

7-6

©G

lenc

oe/M

cGra

w-H

ill40

7G

lenc

oe A

lgeb

ra 2

Lesson 7-6

Lis

t al

l of

th

e p

ossi

ble

rat

ion

al z

eros

of

each

fu

nct

ion

.

1.n

(x)

�x2

�5x

�3

2.h

(x)

�x2

�2x

�5

�1,

�3

�1,

�5

3.w

(x)

�x2

�5x

�12

4.f(

x) �

2x2

�5x

�3

�1,

�2,

�3,

�4,

�6,

�12

��1 2� ,

��3 2� ,

�1,

�3

5.q(

x) �

6x3

�x2

�x

�2

6.g(

x) �

9x4

�3x

3�

3x2

�x

�27

��1 6� ,

��1 3� ,

��1 2� ,

��2 3� ,

�1,

�2

��1 9� ,

��1 3� ,

�1,

�3,

�9,

�27

Fin

d a

ll o

f th

e ra

tion

al z

eros

of

each

fu

nct

ion

.

7.f(

x) �

x3�

2x2

�5x

�4

�0

8.g(

x) �

x3�

3x2

�4x

�12

1�

2,2,

3

9.p(

x) �

x3�

x2�

x�

110

.z(x

) �

x3�

4x2

�6x

�4

12

11.h

(x)

�x3

�x2

�4x

�4

12.g

(x)

�3x

3�

9x2

�10

x�

8

14

13.g

(x)

�2x

3�

7x2

�7x

�12

14.h

(x)

�2x

3�

5x2

�4x

�3

�4,

�1,

�3 2��

1,�1 2� ,

3

15.p

(x)

�3x

3�

5x2

�14

x�

4 �

016

.q(x

) �

3x3

�2x

2�

27x

�18

��1 3�

��2 3�

17.q

(x)

�3x

3�

7x2

�4

18.f

(x)

�x4

�2x

3�

13x2

�14

x�

24

��2 3� ,

1,2

�3,

�1,

2,4

19.p

(x)

�x4

�5x

3�

9x2

�25

x�

7020

.n(x

) �

16x4

�32

x3�

13x2

�29

x�

6

�2,

7�

1,�1 4� ,

�3 4� ,2

Fin

d a

ll o

f th

e ze

ros

of e

ach

fu

nct

ion

.

21.f

(x)

�x3

�5x

2�

11x

�15

22.q

(x)

�x3

�10

x2�

18x

�4

�3,

�1

�2i

,�1

�2i

2,4

��

14�,4

��

14�

23.m

(x)

�6x

4�

17x3

�8x

2�

8x�

324

.g(x

) �

x4�

4x3

�5x

2�

4x�

4

�1 3� ,�3 2� ,

,�

2,�

2,�

i,i

1 �

�5�

�2

1 �

�5�

�2

©G

lenc

oe/M

cGra

w-H

ill40

8G

lenc

oe A

lgeb

ra 2

Lis

t al

l of

th

e p

ossi

ble

rat

ion

al z

eros

of

each

fu

nct

ion

.

1.h

(x)

�x3

� 5

x2�

2x�

122.

s(x)

�x4

� 8

x3�

7x�

14

�1,

�2,

�3,

�4,

�6,

�12

�1,

�2,

�7,

�14

3.f(

x) �

3x5

�5x

2�

x�

64.

p(x)

�3x

2�

x�

7

��1 3� ,

��2 3� ,

�1,

�2,

�3,

�6

��1 3� ,

��7 3� ,

�1,

�7

5.g(

x) �

5x3

�x2

�x

�8

6.q(

x) �

6x5

�x3

�3

��1 5� ,

��2 5� ,

��4 5� ,

��8 5� ,

�1,

�2,

�4,

�8

��1 6� ,

��1 3� ,

��1 2� ,

��3 2� ,

�1,

�3

Fin

d a

ll o

f th

e ra

tion

al z

eros

of

each

fu

nct

ion

.

7.q(

x) �

x3�

3x2

�6x

�8

�0

�4,

�1,

28.

v(x)

�x3

�9x

2�

27x

�27

3

9.c(

x) �

x3�

x2�

8x�

12�

3,2

10.f

(x)

�x4

�49

x20,

�7,

7

11.h

(x)

�x3

�7x

2�

17x

�15

312

.b(x

) �

x3�

6x�

20�

2

13.f

(x)

�x3

�6x

2�

4x�

246

14.g

(x)

�2x

3�

3x2

�4x

�4

�2

15.h

(x)

�2x

3�

7x2

�21

x�

54 �

0�3,

2,�9 2�

16.z

(x)

�x4

�3x

3�

5x2

�27

x�

36�

1,4

17.d

(x)

�x4

�x3

�16

no

rat

ion

al z

ero

s18

.n(x

) �

x4�

2x3

�3

�1

19.p

(x)

�2x

4�

7x3

�4x

2�

7x�

620

.q(x

) �

6x4

�29

x3�

40x2

�7x

�12

�1,

1,�3 2� ,

2�

�3 2� ,�

�4 3�

Fin

d a

ll o

f th

e ze

ros

of e

ach

fu

nct

ion

.

21.f

(x)

�2x

4�

7x3

�2x

2�

19x

�12

22.q

(x)

�x4

�4x

3�

x2�

16x

�20

�1,

�3,

,�

2,2,

2 �

i,2

�i

23.h

(x)

�x6

�8x

324

.g(x

) �

x6�

1�

1,1,

,

0,2,

�1

�i�

3�,�

1 �

i�3�

,,

25.T

RA

VEL

Th

e h

eigh

t of

a b

ox t

hat

Joa

n i

s sh

ippi

ng

is 3

in

ches

les

s th

an t

he

wid

th o

f th

ebo

x.T

he

len

gth

is

2 in

ches

mor

e th

an t

wic

e th

e w

idth

.Th

e vo

lum

e of

th

e bo

x is

154

0 in

3 .W

hat

are

th

e di

men

sion

s of

th

e bo

x?22

in.b

y 10

in.b

y 7

in.

26.G

EOM

ETRY

The

hei

ght

of a

squ

are

pyra

mid

is 3

met

ers

shor

ter

than

the

sid

e of

its

base

.If

th

e vo

lum

e of

th

e py

ram

id i

s 43

2 m

3 ,h

ow t

all

is i

t? U

se t

he

form

ula

V�

�1 3� Bh

.9

m

1 �

i�3�

�� 2

1 �

i�3�

�� 2

�1

�i�

3��

� 2

�1

�i�

3��

� 2

1 �

�33�

�� 4

1 �

�33�

�� 4

Pra

ctic

e (

Ave

rag

e)

Rat

ion

al Z

ero

Th

eore

m

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

7-6

7-6


An

swer

s


Readin

g t

o L

earn

Math

em

ati

csR

atio

nal

Zer

o T

heo

rem

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

7-6

7-6

©G

lenc

oe/M

cGra

w-H

ill40

9G

lenc

oe A

lgeb

ra 2

Lesson 7-6

Pre-

Act

ivit

yH

ow c

an t

he

Rat

ion

al Z

ero

Th

eore

m s

olve

pro

ble

ms

invo

lvin

g la

rge

nu

mb

ers?

Rea

d th

e in

trod

uct

ion

to

Les

son

7-6

at

the

top

of p

age

378

in y

our

text

book

.

Rew

rite

th

e po

lyn

omia

l eq

uat

ion

w(w

�8)

(w�

5) �

2772

in

th

e fo

rm

f(x)

�0,

wh

ere

f(x)

is

a po

lyn

omia

l w

ritt

en i

n d

esce

ndi

ng

pow

ers

of x

.w

3�

3w2

�40

w�

2772

�0

Rea

din

g t

he

Less

on

1.F

or e

ach

of

the

foll

owin

g po

lyn

omia

l fu

nct

ion

s,li

st a

ll t

he

poss

ible

val

ues

of

p,al

l th

e po

ssib

le v

alu

es o

f q,

and

all

the

poss

ible

rat

ion

al z

eros

�p q�.

a.f(

x) �

x3�

2x2

�11

x�

12

poss

ible

val

ues

of

p:�

1,�

2,�

3,�

4,�

6,�

12

poss

ible

val

ues

of

q:�

1

poss

ible

val

ues

of

�p q�:�

1,�

2,�

3,�

4,�

6,�

12

b.

f(x)

�2x

4�

9x3

�23

x2�

81x

�45

poss

ible

val

ues

of

p:�

1,�

3,�

5,�

9,�

15,�

45

poss

ible

val

ues

of

q:�

1,�

2

poss

ible

val

ues

of

�p q�:�

1,�

3,�

5,�

9,�

15,�

45,�

�1 2� ,�

�3 2� ,�

�5 2� ,�

�9 2� ,�

�1 25 �,�

�4 25 �

2.E

xpla

in i

n yo

ur o

wn

wor

ds h

ow D

esca

rtes

’ Rul

e of

Sig

ns,t

he R

atio

nal

Zero

The

orem

,and

syn

thet

ic d

ivis

ion

can

be

use

d to

geth

er t

o fi

nd

all

of t

he

rati

onal

zer

os o

f a

poly

nom

ial

fun

ctio

n w

ith

in

tege

r co

effi

cien

ts.

Sam

ple

an

swer

:U

se D

esca

rtes

’Ru

le t

o f

ind

th

e p

oss

ible

nu

mb

ers

of

po

siti

ve a

nd

neg

ativ

e re

al z

ero

s.U

se t

he

Rat

ion

al Z

ero

Th

eore

m t

o li

st a

llp

oss

ible

rat

ion

al z

ero

s.U

se s

ynth

etic

div

isio

n t

o t

est

wh

ich

of

the

nu

mb

ers

on

th

e lis

t o

f p

oss

ible

rat

ion

al z

ero

s ar

e ac

tual

ly z

ero

s o

f th

ep

oly

no

mia

l fu

nct

ion

.(D

esca

rtes

’Ru

le m

ay h

elp

yo

u t

o li

mit

th

ep

oss

ibili

ties

.)

Hel

pin

g Y

ou

Rem

emb

er

3.S

ome

stud

ents

hav

e tr

oubl

e re

mem

beri

ng w

hich

num

bers

go

in t

he n

umer

ator

s an

d w

hich

go i

n t

he

den

omin

ator

s w

hen

for

min

g a

list

of

poss

ible

rat

ion

al z

eros

of

a po

lyn

omia

lfu

nct

ion

.How

can

you

use

th

e li

nea

r po

lyn

omia

l eq

uat

ion

ax

�b

�0,

wh

ere

aan

d b

are

non

zero

in

tege

rs,t

o re

mem

ber

this

?S

amp

le a

nsw

er:T

he

solu

tio

n o

f th

e eq

uat

ion

is �

�b a� .T

he

nu

mer

ato

r b

is a

fac

tor

of

the

con

stan

t te

rm in

ax

�b

.Th

e d

eno

min

ato

r a

is a

fac

tor

of

the

lead

ing

co

effi

cien

t in

ax

�b

.

©G

lenc

oe/M

cGra

w-H

ill41

0G

lenc

oe A

lgeb

ra 2

Infi

nit

e C

on

tin

ued

Fra

ctio

ns

Som

e in

fin

ite

expr

essi

ons

are

actu

ally

equ

al t

o re

aln

um

bers

! Th

e in

fin

ite

con

tin

ued

fra

ctio

n a

t th

e ri

ght

ison

e ex

ampl

e.

If y

ou u

se x

to s

tan

d fo

r th

e in

fin

ite

frac

tion

,th

en t

he

enti

re d

enom

inat

or o

f th

e fi

rst

frac

tion

on

th

e ri

ght

isal

so e

qual

to

x.T

his

obs

erva

tion

lea

ds t

o th

e fo

llow

ing

equ

atio

n:

x�

1 �

�1 x�

Wri

te a

dec

imal

for

eac

h c

onti

nu

ed f

ract

ion

.

1.1

��1 1�

22.

1 �

1.5

3.1

�1.

66

4.1

�1.

65.

1 �

1.62

5

6.T

he

mor

e te

rms

you

add

to

the

frac

tion

s ab

ove,

the

clos

er t

hei

r va

lue

appr

oach

es t

he

valu

e of

th

e in

fin

ite

con

tin

ued

fra

ctio

n.W

hat

val

ue

do t

he

frac

tion

s se

em t

o be

app

roac

hin

g?ab

ou

t 1.

6

7.R

ewri

te x

�1

��1 x�

as a

qu

adra

tic

equ

atio

n a

nd

solv

e fo

r x.

x2

�x

�1

�0;

x�

;x

�1.

618

or

�0.

618

(Th

e p

osi

tive

ro

ot

is t

he

valu

e o

f th

e in

fin

ite

frac

tio

n,

bec

ause

th

e o

rig

inal

fra

ctio

n is

cle

arly

no

t n

egat

ive.

)

8.F

ind

the

valu

e of

th

e fo

llow

ing

infi

nit

e co

nti

nu

ed f

ract

ion

.

3 �

x�

3 �

�1 x� ;x

�o

r ab

ou

t 3.

303

��

13��

� 2

1

3 �

1

3 �

1

3 �

13

�…

1 �

�5�

�2

1

1 �

1

1 �

1

1 �

1

1 �

1 1

1

1 �

1

1 �

1

1 �

1 1

61

1 �

1

1 �

1 1

1

1 �

1 1

En

rich

men

t

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

7-6

7-6

x�

1 �

1

1 �

1

1 �

1

1 �

11

�…



Stu

dy G

uid

e a

nd I

nte

rven

tion

Op

erat

ion

s o

n F

un

ctio

ns

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

7-7

7-7

©G

lenc

oe/M

cGra

w-H

ill41

1G

lenc

oe A

lgeb

ra 2

Lesson 7-7

Ari

thm

etic

Op

erat

ion

s

Sum

(f�

g)(x

) �

f(x)

�g

(x)

Diff

eren

ce(f

�g)

(x)

�f(

x) �

g(x

)O

per

atio

ns

wit

h F

un

ctio

ns

Pro

duct

(f�

g)(x

) �

f(x)

�g

(x)

Quo

tient

��(x

) �

, g

(x)

0

Fin

d (

f�

g)(x

),(f

�g)

(x),

(f

g)(x

),an

d �

�(x)

for

f(x)

�x2

�3x

�4

and

g(x

) �

3x�

2.(f

�g)

(x)

�f(

x) �

g(x)

Add

ition

of

func

tions

�(x

2�

3x�

4) �

(3x

�2)

f(x)

�x2

�3x

�4,

g(x

) �

3x�

2

�x2

�6x

�6

Sim

plify

.

(f�

g)(x

) �

f(x)

�g(

x)S

ubtr

actio

n of

fun

ctio

ns

�(x

2�

3x�

4) �

(3x

�2)

f(x)

�x2

�3x

�4,

g(x

) �

3x�

2

�x2

�2

Sim

plify

.

(f�

g)(x

)�

f(x)

�g(

x)M

ultip

licat

ion

of f

unct

ions

�(x

2�

3x�

4)(3

x�

2)f(

x) �

x2�

3x�

4, g

(x)

�3x

�2

�x2

(3x

�2)

�3x

(3x

�2)

�4(

3x�

2)D

istr

ibut

ive

Pro

pert

y

�3x

3�

2x2

�9x

2�

6x�

12x

�8

Dis

trib

utiv

e P

rope

rty

�3x

3�

7x2

�18

x�

8S

impl

ify.

��(x

)�

Div

isio

n of

fun

ctio

ns

�,x

�2 3�

f(x)

�x2

�3x

�4

and

g(x

) �

3x�

2

Fin

d (

f�

g)(x

),(f

�g)

(x),

(f

g)(x

),an

d �

�(x)

for

each

f(x

) an

d g

(x).

1.f(

x) �

8x�

3;g(

x) �

4x�

52.

f(x)

�x2

�x

�6;

g(x)

�x

�2

12x

�2;

4x�

8;32

x2

�28

x�

15;

x2

�2x

�8;

x2

�4;

,x

�x

3�

x2

�8x

�12

;x

�3,

x

2

3.f(

x) �

3x2

�x

�5;

g(x)

�2x

�3

4.f(

x) �

2x�

1;g(

x) �

3x2

�11

x�

4

3x2

�x

�2;

3x2

�3x

�8;

3x2

�13

x�

5;�

3x2

�9x

�3;

6x3

�11

x2

�13

x�

15;

6x3

�19

x2

�19

x�

4;

,x

,x

,�4

5.f(

x) �

x2�

1;g(

x) �

x2

�1

�;

x2

�1

�;

x�

1;x

3�

x2

�x

�1,

x

�1

1� x

�1

1� x

�1

1� x

�1

1 � 32x

�1

��

(3x

�1)

(x�

4)3 � 2

3x2

�x

�5

��

2x�

3

5 � 48x

�3

� 4x�

5

f � g

x2�

3x�

4�

�3x

�2

f(x)

� g(x)

f � g

f � g

f(x)

� g(x

)f � g

Exam

ple

Exam

ple

Exer

cises

Exer

cises

©G

lenc

oe/M

cGra

w-H

ill41

2G

lenc

oe A

lgeb

ra 2

Co

mp

osi

tio

n o

f Fu

nct

ion

s

Co

mp

osi

tio

n

Sup

pose

fan

d g

are

func

tions

suc

h th

at t

he r

ange

of

gis

a s

ubse

t of

the

dom

ain

of f

.o

f F

un

ctio

ns

The

n th

e co

mpo

site

fun

ctio

n f

�g

can

be d

escr

ibed

by

the

equa

tion

[f�

g](

x) �

f[g

(x)]

.

For

f�

{(1,

2),(

3,3)

,(2,

4),(

4,1)

} an

d g

�{(

1,3)

,(3,

4),(

2,2)

,(4,

1)},

fin

d f

�g

and

g�

fif

th

ey e

xist

.f[

g(1)

] �

f(3)

�3

f[g(

2)]

�f(

2) �

4f[

g(3)

] �

f(4)

�1

f[g(

4)]

�f(

1) �

2f

�g

�{(

1,3)

,(2,

4),(

3,1)

,(4,

2)}

g[f(

1)]

�g(

2) �

2g[

f(2)

] �

g(4)

�1

g[f(

3)]

�g(

3) �

4g[

f(4)

] �

g(1)

�3

g�

f�

{(1,

2),(

2,1)

,(3,

4),(

4,3)

}

Fin

d [

g�

h](

x) a

nd

[h

�g]

(x)

for

g(x)

�3x

�4

and

h(x

) �

x2�

1.[g

�h

](x)

�g[

h(x

)][h

�g]

(x)

�h

[g(x

)]�

g(x2

�1)

�h

(3x

�4)

�3(

x2�

1) �

4�

(3x

�4)

2�

1�

3x2

�7

�9x

2�

24x

�16

�1

�9x

2�

24x

�15

For

eac

h s

et o

f or

der

ed p

airs

,fin

d f

�g

and

g�

fif

th

ey e

xist

.

1.f

�{(

�1,

2),(

5,6)

,(0,

9)},

2.f

�{(

5,�

2),(

9,8)

,(�

4,3)

,(0,

4)},

g�

{(6,

0),(

2,�

1),(

9,5)

}g

�{(

3,7)

,(�

2,6)

,(4,

�2)

,(8,

10)}

f�

g�

{(2,

2),(

6,9)

,(9,

6)};

f�

gd

oes

no

t ex

ist;

g�

f�

{(�

1,�

1),(

0,5)

,(5,

0)}

g�

f�

{(�

4,7)

,(0,

�2)

,(5,

6),(

9,10

)}

Fin

d [

f�

g](x

) an

d [

g�

f](x

).

3.f(

x) �

2x�

7;g(

x) �

�5x

�1

4.f(

x) �

x2�

1;g(

x) �

�4x

2

[f�

g](

x)

��

10x

�5,

[f�

g](

x)

�16

x4

�1,

[g�

f](x

) �

�10

x�

36[g

�f]

(x)

��

4x4

�8x

2�

4

5.f(

x) �

x2�

2x;g

(x)

�x

�9

6.f(

x) �

5x�

4;g(

x) �

3 �

x[f

�g

](x

) �

x2

�16

x�

63,

[f�

g](

x)

�19

�5x

,[g

�f]

(x)

�x

2�

2x�

9[g

�f]

(x)

��

1 �

5x

Stu

dy G

uid

e a

nd I

nte

rven

tion

(c

onti

nued

)

Op

erat

ion

s o

n F

un

ctio

ns

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

7-7

7-7

Exam

ple1

Exam

ple1

Exam

ple2

Exam

ple2

Exer

cises

Exer

cises


An

swer

s


Skil

ls P

ract

ice

Op

erat

ion

s o

n F

un

ctio

ns

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

7-7

7-7

©G

lenc

oe/M

cGra

w-H

ill41

3G

lenc

oe A

lgeb

ra 2

Lesson 7-7

Fin

d (

f�

g)(x

),(f

�g)

(x),

(f

g)(x

),an

d �

�(x)

for

each

f(x

) an

d g

(x).

1.f(

x) �

x�

52x

�1;

9;2.

f(x)

�3x

�1

5x�

2;x

�4;

6x2

�7x

�3;

g(x)

�x

�4

�x x� �

5 4�

,x

4g(

x) �

2x�

3�3 2x x

� �1 3

�,x

�3 2�

3.f(

x) �

x2x

2�

x�

4;x2

�x

�4;

4.f(

x) �

3x2�3x

3 x�5

�,x

0;

�3x3 x�

5�

,x

0;

g(x)

�4

�x

4x2

�x3

;,x

4

g(x)

��5 x�

15x,

x

0;�3 5x3 �

,x

0

For

eac

h s

et o

f or

der

ed p

airs

,fin

d f

�g

and

g�

fif

th

ey e

xist

.

5.f

�{(

0,0)

,(4,

�2)

}6.

f�

{(0,

�3)

,(1,

2),(

2,2)

}g

�{(

0,4)

,(�

2,0)

,(5,

0)}

g�

{(�

3,1)

,(2,

0)}

{(0,

�2)

,(�

2,0)

,(5,

0)};

{(�

3,2)

,(2,

�3)

};{(

0,4)

,(4,

0)}

{(0,

1),(

1,0)

,(2,

0)}

7.f

�{(

�4,

3),(

�1,

1),(

2,2)

}8.

f�

{(6,

6),(

�3,

�3)

,(1,

3)}

g�

{(1,

�4)

,(2,

�1)

,(3,

�1)

}g

�{(

�3,

6),(

3,6)

,(6,

�3)

}{(

1,3)

,(2,

1),(

3,1)

};{(

�3,

6),(

3,6)

,(6,

�3)

};{(

�4,

�1)

,(�

1,�

4),(

2,�

1)}

{(6,

�3)

,(�

3,6)

,(1,

6)}

Fin

d [

g�

h](

x) a

nd

[h

�g]

(x).

9.g(

x) �

2x2x

�4;

2x�

210

.g(x

) �

�3x

�12

x�

3;�

12x

�1

h(x

) �

x�

2h

(x)

�4x

�1

11.g

(x)

�x

�6

x;x

12.g

(x)

�x

�3

x2

�3;

x2

�6x

�9

h(x

) �

x�

6h

(x)

�x2

13.g

(x)

�5x

5x2

�5x

�5;

14.g

(x)

�x

�2

2x2

�1;

2x2

�8x

�5

h(x

) �

x2�

x�

125

x2

�5x

�1

h(x

) �

2x2

�3

If f

(x)

�3x

,g(x

) �

x�

4,an

d h

(x)

�x2

�1,

fin

d e

ach

val

ue.

15.f

[g(1

)]15

16.g

[h(0

)]3

17.g

[f(�

1)]

1

18.h

[f(5

)]22

419

.g[h

(�3)

]12

20.h

[f(1

0)]

899

21.f

[h(8

)]18

922

.[f

�(h

�g)

](1)

7223

.[f

�(g

�h

)](�

2)21

x2

� 4 �

x

f � g

x2

�x

�20

;

©G

lenc

oe/M

cGra

w-H

ill41

4G

lenc

oe A

lgeb

ra 2

Fin

d (

f�

g)(x

),(f

�g)

(x),

(f

g)(x

),an

d �� gf � �(

x) f

or e

ach

f(x

) an

d g

(x).

1.f(

x) �

2x�

12.

f(x)

�8x

23.

f(x)

�x2

�7x

�12

g(x)

�x

�3

g(x)

�g(

x) �

x2�

9

3x�

2;x

�4;

�8x4 x

� 21

�,x

0;

2x2

�7x

�3;

7x�

21;

2x2

�5x

�3;

�8x4 x2�

1�

,x

0;x

4�

7x3

�3x

2�

63x

�10

8;

�2 xx ��31

�,x

3

8,x

0;

8x4 ,

x

0�x x

� �4 3

�,x

�

3

For

eac

h s

et o

f or

der

ed p

airs

,fin

d f

�g

and

g�

fif

th

ey e

xist

.

4.f

�{(

�9,

�1)

,(�

1,0)

,(3,

4)}

5.f

�{(

�4,

3),(

0,�

2),(

1,�

2)}

g�

{(0,

�9)

,(�

1,3)

,(4,

�1)

}g

�{(

�2,

0),(

3,1)

}{(

0,�

1),(

�1,

4),(

4,0)

};{(

�2,

�2)

,(3,

�2)

};{(

�9,

3),(

�1,

�9)

,(3,

�1)

}{(

�4,

1),(

0,0)

,(1,

0)}

6.f

�{(

�4,

�5)

,(0,

3),(

1,6)

}7.

f�

{(0,

�3)

,(1,

�3)

,(6,

8)}

g�

{(6,

1),(

�5,

0),(

3,�

4)}

g�

{(8,

2),(

�3,

0),(

�3,

1)}

{(6,

6),(

�5,

3),(

3,�

5)};

do

es n

ot

exis

t;{(

�4,

0),(

0,�

4),(

1,1)

}{(

0,0)

,(1,

0),(

6,2)

}

Fin

d [

g�

h](

x) a

nd

[h

�g]

(x).

8.g(

x) �

3x9.

g(x)

��

8x10

.g(x

) �

x�

6h

(x)

�x

�4

h(x

) �

2x�

3h

(x)

�3x

23x

2�

6;3x

�12

;3x

�4

�16

x�

24;

�16

x�

33x

2�

36x

�10

8

11.g

(x)

�x

�3

12.g

(x)

��

2x13

.g(x

) �

x�

2h

(x)

�2x

2h

(x)

�x2

�3x

�2

h(x

) �

3x2

�1

2x2

�3;

�2x

2�

6x�

4;3x

2�

1;2x

2�

12x

�18

4x2

�6x

�2

3x2

�12

x�

13

If f

(x)

�x2

,g(x

) �

5x,a

nd

h(x

) �

x�

4,fi

nd

eac

h v

alu

e.

14.f

[g(1

)]25

15.g

[h(�

2)]

1016

.h[f

(4)]

20

17.f

[h(�

9)]

2518

.h[g

(�3)

]�

1119

.g[f

(8)]

320

20.h

[f(2

0)]

404

21.[

f�

(h�

g)](

�1)

122

.[f

�(g

�h

)](4

)16

00

23.B

USI

NES

ST

he

fun

ctio

n f

(x)

�10

00 �

0.01

x2m

odel

s th

e m

anu

fact

uri

ng

cost

per

ite

mw

hen

xit

ems

are

prod

uce

d,an

d g(

x) �

150

�0.

001x

2m

odel

s th

e se

rvic

e co

st p

er i

tem

.W

rite

a f

un

ctio

n C

(x)

for

the

tota

l m

anu

fact

uri

ng

and

serv

ice

cost

per

ite

m.

C(x

) �

1150

�0.

011x

2

24.M

EASU

REM

ENT

Th

e fo

rmu

la f

�� 1n 2�

con

vert

s in

ches

nto

fee

t f,

and

m�

� 52f 80�

con

vert

s fe

et t

o m

iles

m.W

rite

a c

ompo

siti

on o

f fu

nct

ion

s th

at c

onve

rts

inch

es t

o m

iles

.

[m�

f]n

�� 63

,n 360

�

1 � x2

Pra

ctic

e (

Ave

rag

e)

Op

erat

ion

s o

n F

un

ctio

ns

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

7-7

7-7



Readin

g t

o L

earn

Math

em

ati

csO

per

atio

ns

on

Fu

nct

ion

s

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

7-7

7-7

©G

lenc

oe/M

cGra

w-H

ill41

5G

lenc

oe A

lgeb

ra 2

Lesson 7-7

Pre-

Act

ivit

yW

hy

is i

t im

port

ant

to c

ombi

ne

fun

ctio

ns

in b

usi

nes

s?

Rea

d th

e in

trod

uct

ion

to

Les

son

7-7

at

the

top

of p

age

383

in y

our

text

book

.

Des

crib

e tw

o w

ays

to c

alcu

late

Ms.

Cof

fmon

’s p

rofi

t fr

om t

he

sale

of

50 b

irdh

ouse

s.(D

o n

ot a

ctu

ally

cal

cula

te h

er p

rofi

t.)

Sam

ple

an

swer

:1.

Fin

d t

he

reve

nu

e by

su

bst

itu

tin

g 5

0 fo

r x

in t

he

exp

ress

ion

125x

.Nex

t,fi

nd

th

e co

st b

y su

bst

itu

tin

g 5

0 fo

r x

in t

he

exp

ress

ion

65x

�54

00.F

inal

ly,s

ub

trac

t th

e co

st f

rom

th

ere

ven

ue

to f

ind

th

e p

rofi

t.2.

Fo

rm t

he

pro

fit

fun

ctio

n

p(x

) �

r(x

) �

c(x

) �

125x

�(6

5x�

5400

) �

60x

�54

00.

Su

bst

itu

te 5

0 fo

r x

in t

he

exp

ress

ion

60x

�54

00.

Rea

din

g t

he

Less

on

1.D

eter

min

e w

het

her

eac

h s

tate

men

t is

tru

eor

fal

se.(

Rem

embe

r th

at t

rue

mea

ns

alw

ays

tru

e.)

a.If

fan

d g

are

poly

nom

ial

fun

ctio

ns,

then

f�

gis

a p

olyn

omia

l fu

nct

ion

.tr

ue

b.

If f

and

gar

e po

lyn

omia

l fu

nct

ion

s,th

en

is a

pol

ynom

ial

fun

ctio

n.

fals

e

c.If

fan

d g

are

poly

nom

ial

fun

ctio

ns,

the

dom

ain

of

the

fun

ctio

n f

�g

is t

he

set

of a

llre

al n

um

bers

.tr

ue

d.

If f

(x)

�3x

�2

and

g(x)

�x

�4,

the

dom

ain

of

the

fun

ctio

n

is t

he

set

of a

ll r

eal

nu

mbe

rs.

fals

e

e.If

fan

d g

are

poly

nom

ial

fun

ctio

ns,

then

(f

�g)

(x)

�(g

�f)

(x).

fals

e

f.If

fan

d g

are

poly

nom

ial

fun

ctio

ns,

then

(f

�g)

(x)

�(g

� f)

(x)

tru

e

2.L

et f

(x)

�2x

�5

and

g(x)

�x2

�1.

a.E

xpla

in i

n w

ords

how

you

wou

ld f

ind

(f�

g)(�

3).(

Do

not

act

ual

ly d

o an

y ca

lcu

lati

ons.

)S

amp

le a

nsw

er:

Sq

uar

e �

3 an

d a

dd

1.T

ake

the

nu

mb

er y

ou

get

,m

ult

iply

it b

y 2,

and

su

btr

act

5.

b.

Exp

lain

in

wor

ds h

ow y

ou w

ould

fin

d (g

�f)

(�3)

.(D

o n

ot a

ctu

ally

do

any

calc

ula

tion

s.)

Sam

ple

an

swer

:M

ult

iply

�3

by 2

an

d s

ub

trac

t 5.

Take

th

en

um

ber

yo

u g

et,s

qu

are

it,a

nd

ad

d 1

.

Hel

pin

g Y

ou

Rem

emb

er

3.S

ome

stu

den

ts h

ave

trou

ble

rem

embe

rin

g th

e co

rrec

t or

der

in w

hic

h t

o ap

ply

the

two

orig

inal

fu

nct

ion

s w

hen

eva

luat

ing

a co

mpo

site

fu

nct

ion

.Wri

te t

hre

e se

nte

nce

s,ea

ch o

fw

hic

h e

xpla

ins

how

to

do t

his

in

a s

ligh

tly

diff

eren

t w

ay.(

Hin

t:U

se t

he

wor

d cl

oses

tin

the

firs

t se

nte

nce

,th

e w

ords

in

sid

ean

d ou

tsid

ein

th

e se

con

d,an

d th

e w

ords

lef

tan

dri

ght

in t

he

thir

d.)

Sam

ple

an

swer

:1.

Th

e fu

nct

ion

th

at is

wri

tten

clo

sest

to

the

vari

able

is a

pp

lied

fir

st.2

.Wo

rk f

rom

th

e in

sid

e to

th

e o

uts

ide.

3.W

ork

fro

m r

igh

t to

left

.

f � g

f � g

©G

lenc

oe/M

cGra

w-H

ill41

6G

lenc

oe A

lgeb

ra 2

Rel

ativ

e M

axim

um

Val

ues

Th

e gr

aph

of

f(x)

�x3

�6x

�9

show

s a

rela

tive

max

imu

m v

alu

e so

mew

her

e be

twee

n f

(�2)

an

d f(

�1)

.You

can

obt

ain

a

clos

er a

ppro

xim

atio

n b

y co

mpa

rin

g va

lues

su

ch a

s th

ose

show

n i

n t

he

tabl

e.

To

the

nea

rest

ten

th a

rel

ativ

e m

axim

um

va

lue

for

f(x)

is

�3.

3.

Usi

ng

a ca

lcu

lato

r to

fin

d p

oin

ts,g

rap

h e

ach

fu

nct

ion

.To

the

nea

rest

te

nth

,fin

d a

rel

ativ

e m

axim

um

val

ue

of t

he

fun

ctio

n.

1.f(

x) �

x(x2

�3)

rel.

max

.of

2.0

2.f(

x) �

x3�

3x�

3re

l.m

ax.o

f �

1.0

3.f(

x) �

x3�

9x�

2re

l.m

ax.o

f 8.

44.

f(x)

�x3

�2x

2�

12x

�24

rel.

max

.of

3.3

5

x

f(x)

O1

2

x

f(x)

O2

x

f(x)

O

x

f(x)

O

xf(

x)

�2

�5

�1.

5�

3.37

5

�1.

4�

3.34

4

�1.

3�

3.39

7

�1

�4

x

f(x)

O2

–2–4 –8 –12

–16

–20

–44

f(x)

� x

3 �

6x

� 9

En

rich

men

t

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

7-7

7-7


An

swer

s


Stu

dy G

uid

e a

nd I

nte

rven

tion

Inve

rse

Fu

nct

ion

s an

d R

elat

ion

s

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

7-8

7-8

©G

lenc

oe/M

cGra

w-H

ill41

7G

lenc

oe A

lgeb

ra 2

Lesson 7-8

Fin

d In

vers

es

Inve

rse

Rel

atio

ns

Two

rela

tions

are

inve

rse

rela

tions

if a

nd o

nly

if w

hene

ver

one

rela

tion

cont

ains

the

el

emen

t (a

, b

), t

he o

ther

rel

atio

n co

ntai

ns t

he e

lem

ent

(b,

a).

Pro

per

ty o

f In

vers

e S

uppo

se f

and

f�1

are

inve

rse

func

tions

.F

un

ctio

ns

The

n f(

a) �

bif

and

only

if f

�1 (

b) �

a.

Fin

d t

he

inve

rse

of t

he

fun

ctio

n f

(x)

�x

�.T

hen

gra

ph

th

efu

nct

ion

an

d i

ts i

nve

rse.

Ste

p 1

Rep

lace

f(x

) w

ith

yin

th

e or

igin

al e

quat

ion

.

f(x)

��2 5� x

�→

y�

�2 5� x�

Ste

p 2

Inte

rch

ange

xan

d y.

x�

�2 5� y�

Ste

p 3

Sol

ve f

or y

.

x�

�2 5� y�

Inve

rse

5x�

2y�

1M

ultip

ly e

ach

side

by

5.

5x�

1 �

2yA

dd 1

to

each

sid

e.

(5x

�1)

�y

Div

ide

each

sid

e by

2.

Th

e in

vers

e of

f(x

) �

�2 5� x�

is f

�1 (

x) �

(5x

�1)

.

Fin

d t

he

inve

rse

of e

ach

fu

nct

ion

.Th

en g

rap

h t

he

fun

ctio

n a

nd

its

in

vers

e.

1.f(

x) �

x�

12.

f(x)

�2x

�3

3.f(

x) �

x�

2

f�1 (

x)

�x

�f�

1 (x

) �

x�

f�1 (

x)

�4x

�8

f–1(x

) � 4

x �

8

f(x)

� 1 – 4x

� 2 x

f (x)

O

f(x)

� 2

x �

3

f–1(x

) � 1 – 2x

� 3 – 2

x

f (x)

O

f(x)

� 2 – 3x

� 1

f–1(x

) � 3 – 2x

� 3 – 2

x

f (x)

O

3 � 21 � 2

3 � 23 � 2

1 � 42 � 3

1 � 21 � 5

1 � 2

1 � 51 � 5

1 � 51 � 5

x

f (x)

O

f(x)

� 2 – 5x

� 1 – 5

f–1(x

) � 5 – 2x

� 1 – 2

1 � 52 � 5

Exam

ple

Exam

ple

Exer

cises

Exer

cises

©G

lenc

oe/M

cGra

w-H

ill41

8G

lenc

oe A

lgeb

ra 2

Inve

rses

of

Rel

atio

ns

and

Fu

nct

ion

s

Inve

rse

Fu

nct

ion

sTw

o fu

nctio

ns f

and

gar

e in

vers

e fu

nctio

ns if

and

onl

y if

[f�

g](

x) �

xan

d [g

�f]

(x)

�x.

Det

erm

ine

wh

eth

er f

(x)

�2x

�7

and

g(x

) �

(x�

7) a

re i

nve

rse

fun

ctio

ns.

[f�

g](x

) �

f[g(

x)]

[g�

f](x

) �

g[f(

x)]

�f ��1 2� (

x�

7)�

�g(

2x�

7)

�2 ��1 2� (

x�

7)��

7�

�1 2� (2x

�7

�7)

�x

�7

�7

�x

�x

Th

e fu

nct

ion

s ar

e in

vers

es s

ince

bot

h [

f�

g](x

) �

xan

d [g

�f]

(x)

�x.

Det

erm

ine

wh

eth

er f

(x)

�4x

�an

d g

(x)

�x

�3

are

inve

rse

fun

ctio

ns.

[f�

g](x

) �

f[g(

x)]

�f ��1 4� x

�3 �

�4 ��1 4� x

�3 �

��1 3�

�x

�12

��1 3�

�x

�11

�2 3�

Sin

ce [

f�

g](x

)

x,th

e fu

nct

ion

s ar

e n

ot i

nve

rses

.

Det

erm

ine

wh

eth

er e

ach

pai

r of

fu

nct

ion

s ar

e in

vers

e fu

nct

ion

s.

1.f(

x) �

3x�

12.

f(x)

��1 4� x

�5

3.f(

x) �

�1 2� x�

10

g(x)

��1 3� x

��1 3�

yes

g(x)

�4x

�20

yes

g(x)

�2x

�� 11 0�

no

4.f(

x) �

2x�

55.

f(x)

�8x

�12

6.f(

x) �

�2x

�3

g(x)

�5x

�2

no

g(x)

��1 8� x

�12

no

g(x)

��

�1 2� x�

�3 2�ye

s

7.f(

x) �

4x�

�1 2�8.

f(x)

�2x

��3 5�

9.f(

x) �

4x�

�1 2�

g(x)

��1 4� x

��1 8�

yes

g(x)

�� 11 0�

(5x

�3)

yes

g(x)

��1 2� x

��3 2�

no

10.f

(x)

�10

�� 2x �

11.f

(x)

�4x

��4 5�

12.f

(x)

�9

��3 2� x

g(x)

�20

�2x

yes

g(x)

�� 4x �

��1 5�

yes

g(x)

��2 3� x

�6

yes

1 � 41 � 3

1 � 2

Stu

dy G

uid

e a

nd I

nte

rven

tion

(c

onti

nued

)

Inve

rse

Fu

nct

ion

s an

d R

elat

ion

s

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

7-8

7-8

Exam

ple1

Exam

ple1

Exam

ple2

Exam

ple2

Exer

cises

Exer

cises



Skil

ls P

ract

ice

Inve

rse

Fu

nct

ion

s an

d R

elat

ion

s

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

7-8

7-8

©G

lenc

oe/M

cGra

w-H

ill41

9G

lenc

oe A

lgeb

ra 2

Lesson 7-8

Fin

d t

he

inve

rse

of e

ach

rel

atio

n.

1.{(

3,1)

,(4,

�3)

,(8,

�3)

}2.

{(�

7,1)

,(0,

5),(

5,�

1)}

{(1,

3),(

�3,

4),(

�3,

8)}

{(1,

�7)

,(5,

0),(

�1,

5)}

3.{(

�10

,�2)

,(�

7,6)

,(�

4,�

2),(

�4,

0)}

4.{(

0,�

9),(

5,�

3),(

6,6)

,(8,

�3)

}{(

�2,

�10

),(6

,�7)

,(�

2,�

4),(

0,�

4)}

{(�

9,0)

,(�

3,5)

,(6,

6),(

�3,

8)}

5.{(

�4,

12),

(0,7

),(9

,�1)

,(10

,�5)

}6.

{(�

4,1)

,(�

4,3)

,(0,

�8)

,(8,

�9)

}{(

12,�

4),(

7,0)

,(�

1,9)

,(�

5,10

)}{(

1,�

4),(

3,�

4),(

�8,

0),(

�9,

8)}

Fin

d t

he

inve

rse

of e

ach

fu

nct

ion

.Th

en g

rap

h t

he

fun

ctio

n a

nd

its

in

vers

e.

7.y

�4

8.f(

x) �

3x9.

f(x)

�x

�2

x�

4f�

1 (x

) �

�1 3� xf�

1 (x

) �

x�

2

10.g

(x)

�2x

�1

11.h

(x)

��1 4� x

12.y

��2 3� x

�2

g�

1 (x

) �

�x� 2

1�

h�

1 (x

) �

4xy

��3 2� x

�3

Det

erm

ine

wh

eth

er e

ach

pai

r of

fu

nct

ion

s ar

e in

vers

e fu

nct

ion

s.

13.f

(x)

�x

�1

no

14.f

(x)

�2x

�3

yes

15.f

(x)

�5x

�5

yes

g(x)

�1

�x

g(x)

��1 2� (

x�

3)g(

x) �

�1 5� x�

1

16.f

(x)

�2x

yes

17.h

(x)

�6x

�2

no

18.f

(x)

�8x

�10

yes

g(x)

��1 2� x

g(x)

��1 6� x

�3

g(x)

��1 8� x

��5 4�

x

y

Ox

h (x)

Ox

g (x)

O

x

f (x)

Ox

f (x)

Ox

y

O

©G

lenc

oe/M

cGra

w-H

ill42

0G

lenc

oe A

lgeb

ra 2

Fin

d t

he

inve

rse

of e

ach

rel

atio

n.

1.{(

0,3)

,(4,

2),(

5,�

6)}

2.{(

�5,

1),(

�5,

�1)

,(�

5,8)

}{(

3,0)

,(2,

4),(

�6,

5)}

{(1,

�5)

,(�

1,�

5),(

8,�

5)}

3.{(

�3,

�7)

,(0,

�1)

,(5,

9),(

7,13

)}4.

{(8,

�2)

,(10

,5),

(12,

6),(

14,7

)}{(

�7,

�3)

,(�

1,0)

,(9,

5),(

13,7

)}{(

�2,

8),(

5,10

),(6

,12)

,(7,

14)}

5.{(

�5,

�4)

,(1,

2),(

3,4)

,(7,

8)}

6.{(

�3,

9),(

�2,

4),(

0,0)

,(1,

1)}

{(�

4,�

5),(

2,1)

,(4,

3),(

8,7)

}{(

9,�

3),(

4,�

2),(

0,0)

,(1,

1)}

Fin

d t

he

inve

rse

of e

ach

fu

nct

ion

.Th

en g

rap

h t

he

fun

ctio

n a

nd

its

in

vers

e.

7.f(

x) �

�3 4� x8.

g(x)

�3

�x

9.y

�3x

�2

f�1 (

x)

��4 3� x

g�

1 (x

) �

x�

3y

��x

� 32

�

Det

erm

ine

wh

eth

er e

ach

pai

r of

fu

nct

ion

s ar

e in

vers

e fu

nct

ion

s.

10.f

(x)

�x

�6

yes

11.f

(x)

��

4x�

1ye

s12

.g(x

) �

13x

�13

no

g(x)

�x

�6

g(x)

��1 4� (

1 �

x)h

(x)

�� 11 3�

x�

1

13.f

(x)

�2x

no

14.f

(x)

��6 7� x

yes

15.g

(x)

�2x

�8

yes

g(x)

��

2xg(

x) �

�7 6� xh

(x)

��1 2� x

�4

16. M

EASU

REM

ENT

Th

e po

ints

(63

,121

),(7

1,18

0),(

67,1

40),

(65,

108)

,an

d (7

2,16

5) g

ive

the

wei

ght

in p

oun

ds a

s a

fun

ctio

n o

f h

eigh

t in

in

ches

for

5 s

tude

nts

in

a c

lass

.Giv

e th

epo

ints

for

th

ese

stu

den

ts t

hat

rep

rese

nt

hei

ght

as a

fu

nct

ion

of

wei

ght.

(121

,63)

,(18

0,71

),(1

40,6

7),(

108,

65),

(165

,72)

REM

OD

ELIN

GF

or E

xerc

ises

17

and

18,

use

th

e fo

llow

ing

info

rmat

ion

.T

he

Cle

arys

are

rep

laci

ng

the

floo

rin

g in

th

eir

15 f

oot

by 1

8 fo

ot k

itch

en.T

he

new

flo

orin

gco

sts

$17.

99 p

er s

quar

e ya

rd.T

he

form

ula

f(x

) �

9xco

nve

rts

squ

are

yard

s to

squ

are

feet

.

17.F

ind

the

inve

rse

f�1 (

x).W

hat

is

the

sign

ific

ance

of

f�1 (

x) f

or t

he

Cle

arys

?f�

1 (x

) �

�x 9� ;It

will

allo

w t

hem

to

co

nver

t th

e sq

uar

e fo

ota

ge

of

thei

r ki

tch

en f

loo

r to

squ

are

yard

s,so

th

ey c

an t

hen

cal

cula

te t

he

cost

of

the

new

flo

ori

ng

.

18.W

hat

wil

l th

e n

ew f

loor

ing

cost

th

e C

lear

y’s?

$539

.70

x

y

Ox

g (x)

Ox

f (x)

OPra

ctic

e (

Ave

rag

e)

Inve

rse

Fu

nct

ion

s an

d R

elat

ion

s

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

7-8

7-8


An

swer

s


Readin

g t

o L

earn

Math

em

ati

csIn

vers

e F

un

ctio

ns

and

Rel

atio

ns

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

7-8

7-8

©G

lenc

oe/M

cGra

w-H

ill42

1G

lenc

oe A

lgeb

ra 2

Lesson 7-8

Pre-

Act

ivit

yH

ow a

re i

nve

rse

fun

ctio

ns

rela

ted

to

mea

sure

men

t co

nve

rsio

ns?

Rea

d th

e in

trod

uct

ion

to

Les

son

7-8

at

the

top

of p

age

390

in y

our

text

book

.

A f

unct

ion

mul

tipl

ies

a nu

mbe

r by

3 a

nd t

hen

adds

5 t

o th

e re

sult

.Wha

t do

esth

e in

vers

e fu

nct

ion

do,

and

in w

hat

ord

er?

Sam

ple

an

swer

:It

fir

stsu

btr

acts

5 f

rom

th

e n

um

ber

an

d t

hen

div

ides

th

e re

sult

by

3.

Rea

din

g t

he

Less

on

1.C

ompl

ete

each

sta

tem

ent.

a.If

tw

o re

lati

ons

are

inve

rses

,th

e do

mai

n o

f on

e re

lati

on i

s th

e of

the

oth

er.

b.

Su

ppos

e th

at g

(x)

is a

rel

atio

n a

nd

that

th

e po

int

(4,�

2) i

s on

its

gra

ph.T

hen

a p

oin

t

on t

he

grap

h o

f g�

1 (x)

is

.

c.T

he

test

can

be

use

d on

th

e gr

aph

of

a fu

nct

ion

to

dete

rmin

e

wh

eth

er t

he

fun

ctio

n h

as a

n i

nve

rse

fun

ctio

n.

d.

If y

ou a

re g

iven

th

e gr

aph

of

a fu

nct

ion

,you

can

fin

d th

e gr

aph

of

its

inve

rse

by

refl

ecti

ng

the

orig

inal

gra

ph o

ver

the

lin

e w

ith

equ

atio

n

.

e.If

fan

d g

are

inve

rse

fun

ctio

ns,

then

(f

�g)

(x)

�

and

(g�

f)(x

) �

.

f.A

fu

nct

ion

has

an

in

vers

e th

at i

s al

so a

fu

nct

ion

on

ly i

f th

e gi

ven

fu

nct

ion

is

.

g.S

upp

ose

that

h(x

) is

a f

un

ctio

n w

hos

e in

vers

e is

als

o a

fun

ctio

n.I

f h

(5)

�12

,th

enh

�1 (

12)

�.

2.A

ssu

me

that

f(x

) is

a o

ne-

to-o

ne

fun

ctio

n d

efin

ed b

y an

alg

ebra

ic e

quat

ion

.Wri

te t

he

fou

rst

eps

you

wou

ld f

ollo

w i

n o

rder

to

fin

d th

e eq

uat

ion

for

f�

1 (x)

.

1.R

epla

ce f

(x)

wit

h y

in t

he

ori

gin

al e

qu

atio

n.

2.In

terc

han

ge

xan

d y

.

3.S

olv

e fo

r y.

4.R

epla

ce y

wit

h f

�1 (

x).

Hel

pin

g Y

ou

Rem

emb

er

3.A

goo

d w

ay t

o re

mem

ber

som

eth

ing

new

is

to r

elat

e it

to

som

eth

ing

you

alr

eady

kn

ow.

How

are

the

ver

tica

l an

d ho

rizo

ntal

lin

e te

sts

rela

ted?

Sam

ple

an

swer

:Th

e ve

rtic

allin

e te

st d

eter

min

es w

het

her

a r

elat

ion

is a

fu

nct

ion

bec

ause

th

e o

rder

edp

airs

in a

fu

nct

ion

can

hav

e n

o r

epea

ted

x-v

alu

es.T

he

ho

rizo

nta

l lin

e te

std

eter

min

es w

het

her

a f

un

ctio

n is

on

e-to

-on

e b

ecau

se a

on

e-to

-on

efu

nct

ion

can

no

t h

ave

any

rep

eate

d y

-val

ues

.

5

on

e-to

-on

e

xx

y �

x

ho

rizo

nta

l lin

e

(�2,

4)

ran

ge

©G

lenc

oe/M

cGra

w-H

ill42

2G

lenc

oe A

lgeb

ra 2

Min

iatu

re G

olf

In m

inia

ture

gol

f,th

e ob

ject

of

the

gam

e is

to

roll

th

e go

lf b

all

into

th

e h

ole

in a

s fe

w s

hot

s as

pos

sibl

e.A

s in

th

e di

agra

m a

t th

e ri

ght,

the

hol

e is

oft

en p

lace

d so

th

at a

dir

ect

shot

is

impo

ssib

le.R

efle

ctio

ns

can

be

use

d to

hel

p de

term

ine

the

dire

ctio

n t

hat

th

e ba

ll s

hou

ld b

ero

lled

in

ord

er t

o sc

ore

a h

ole-

in-o

ne.

Usi

ng

wal

l E �

F�,f

ind

th

e p

ath

to

use

to

sc

ore

a h

ole-

in-o

ne.

Fin

d th

e re

flec

tion

im

age

of t

he

“hol

e”w

ith

res

pect

to

E�F�

and

labe

l it

H .

Th

e in

ters

ecti

on o

f B �

H� �w

ith

wal

l E�

F�is

th

e po

int

at w

hic

h t

he

shot

sh

ould

be

dire

cted

.

For

th

e h

ole

at t

he

righ

t,fi

nd

a p

ath

to

scor

e a

hol

e-in

-on

e.

Fin

d th

e re

flec

tion

im

age

of H

wit

h r

espe

ct t

o E�

F�an

d la

bel

it H

.In

th

is c

ase,

B �H�

�in

ters

ects

J�K�

befo

re i

nte

rsec

tin

g E�

F�.T

hu

s,th

ispa

th c

ann

ot b

e u

sed.

To

fin

d a

usa

ble

path

,fin

d th

e re

flec

tion

im

age

of H

w

ith

res

pect

to

G �F�

and

labe

l it

H�.

Now

,th

e in

ters

ecti

on o

f B �

H��

wit

h w

all

G�F�

is t

he

poin

t at

wh

ich

th

e sh

otsh

ould

be

dire

cted

.

Cop

y ea

ch f

igu

re.T

hen

,use

ref

lect

ion

s to

det

erm

ine

a p

ossi

ble

p

ath

for

a h

ole-

in-o

ne.

1.2.

3.

H

B

H

B

H

B

B GF

JK

H' H"

E

H

Bal

l

Hol

e

E

H'

F

Bal

l

Hol

e

En

rich

men

t

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

7-8

7-8

Exam

ple1

Exam

ple1

Exam

ple2

Exam

ple2



Stu

dy G

uid

e a

nd I

nte

rven

tion

Sq

uar

e R

oo

t F

un

ctio

ns

and

Ineq

ual

itie

s

NA

ME

____

____

____

____

____

____

____

____

____

____

____

__D

AT

E__

____

____

__P

ER

IOD

____

_

7-9

7-9

©G

lenc

oe/M

cGra

w-H

ill42

3G

lenc

oe A

lgeb

ra 2

Lesson 7-9

Squ

are

Ro

ot

Fun

ctio

ns

A f

un

ctio

n t

hat

con

tain

s th

e sq

uar

e ro

ot o

f a

vari

able

expr

essi

on i

s a

squ

are

root

fu

nct

ion

.

Gra

ph

y�

�3x

��

2�.S

tate

its

dom

ain

an

d r

ange

.

Sin

ce t

he

radi

can

d ca

nn

ot b

e n

egat

ive,

3x�

2 �

0 or

x�

�2 3� .

Th

e x-

inte

rcep

t is

�2 3� .T

he

ran

ge i

s y

�0.

Mak

e a

tabl

e of

val

ues

an

d gr

aph

th

e fu

nct

ion

.

Gra

ph

eac

h f

un

ctio

n.S

tate

th

e d

omai

n a

nd

ran

ge o

f th

e fu

nct

ion

.

1.y

��

2x�2.

y�

�3�

x�3.

y�

�� 2x �

D:

x�

0;R

:y

�0

D:

x�

0;R

:y

�0

D:

x�

0;R

:y

�0

4.y

�2�

x�

3�

5.y

��

�2x

�3

�6.

y�

�2x

�5

�

D:

x�

3;R

:y

�0

D:

x�

�3 2� ;R

:y

�0

D:

x�

��5 2� ;

R:

y�

0

y �

��

�2x

� 5

x

y

O

y �

��

��

�2x

� 3

x

y

O

y �

2�

��

x �

3

x

y

O

y �

��x – 2

x

y

Oy

� �

3��x

xy

O

y �

��2x

x

y

O

xy

�2 3�0

11

22

3�

7�

x

y

O

y �

��

��

3x �

2

Exam

ple

Exam

ple

Exer

cises

Exer

cises

©G

lenc

oe/M

cGra

w-H

ill42

4G

lenc

oe A

lgeb

ra 2

Squ

are

Ro

ot

Ineq

ual

itie

sA

sq

uar

e ro

ot i

neq

ual

ity

is a

n i

neq

ual

ity

that

con

tain

sth

e sq

uar

e ro

ot o

f a

vari

able

exp

ress

ion

.Use

wh

at y

ou k

now

abo

ut

grap

hin

g sq

uar

e ro

otfu

nct

ion

s an

d qu

adra

tic

ineq

ual

itie

s to

gra

ph s

quar

e ro

ot i

neq

ual

itie

s.

Gra

ph

y�

�2x

��

1��

2.G

raph

th

e re

late

d eq

uat

ion

y�

�2x

�1

��

2.S

ince

th

e bo

un

dary

sh

ould

be

incl

ude

d,th

e gr

aph

sh

ould

be

soli

d.

Th

e do

mai

n i

ncl

ude

s va

lues

for

x�

�1 2� ,so

th

e gr

aph

is

to t

he

righ

t

of x

��1 2� .

Th

e ra

nge

in

clu

des

only

nu

mbe

rs g

reat

er t

han

2,s

o th

e

grap

h i

s ab

ove

y�

2.

Gra

ph

eac

h i

neq

ual

ity.

1.y

�2�

x�2.

y�

�x

�3

�3.

y�

3�2x

�1

�

4.y

��

3x�

4�

5.y

��

x�

1�

�4

6.y

�2�

2x�

3�

7.y

��

3x�

1�

�2

8.y

��

4x�

2�

�1

9.y

�2�

2x�

1�

�4

y �

2�

��

2x �

1 �

4 x

y

O

y �

��

�4x

� 2

� 1 x

y

Oy

� �

��

�3x

� 1

� 2

x

y

O

y �

2�

��

�2x

� 3 x

y

Oy

� �

��

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7-9

7-9

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ple

Exam

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7-9

7-9

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Lesson 7-9

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ph

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tate

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f ea

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7-9

7-9



Readin

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qu

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7-9

7-9

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Lesson 7-9

Pre-

Act

ivit

yH

ow a

re s

qu

are

root

fu

nct

ion

s u

sed

in

bri

dge

des

ign

?

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d th

e in

trod

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to

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son

7-9

at

the

top

of p

age

395

in y

our

text

book

.

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he

wei

ght

to b

e su

ppor

ted

by a

ste

el c

able

is

dou

bled

,sh

ould

th

edi

amet

er o

f th

e su

ppor

t ca

ble

also

be

dou

bled

? If

not

,by

wh

at n

um

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shou

ld t

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diam

eter

be

mu

ltip

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?

no

;�

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on

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atch

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ot f

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rom

th

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st o

n t

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left

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omai

n a

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ge f

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the

list

on

th

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ght.

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)(2

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(5,2

)(4

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(6,6

)

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),(5

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(4,�

2)

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ou

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er

3.A

goo

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ay t

o re

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som

eth

ing

is t

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plai

n i

t to

som

eon

e el

se.S

upp

ose

you

are

stu

dyin

g th

is l

esso

n w

ith

a c

lass

mat

e w

ho

thin

ks t

hat

you

can

not

hav

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uar

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nct

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s be

cau

se e

very

pos

itiv

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al n

um

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has

tw

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uar

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How

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ld y

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plai

n t

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idea

of

squ

are

root

fu

nct

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s to

you

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assm

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ple

an

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fo

rm a

sq

uar

e ro

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fun

ctio

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ho

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the

po

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ple

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o m

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pro

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s h

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they

are

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anal

ogou

s.U

sin

g an

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ies

is o

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way

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and

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th

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t at

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02S

uppo

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now

how

the

are

as A

,B,a

nd C

ofth

e th

ree

face

s th

at m

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at v

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x O

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thre

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ase,

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ht

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form

ula

an

d th

en t

ry t

o pr

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it.

05T

wo

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D3

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um

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1.U

se s

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nce

01

and

the

top

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ram

.Th

e pr

efix

tet

ra-

mea

ns

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r.W

rite

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def

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a th

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se s

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and

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ram

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len

gth

s of

th

e si

des

ofea

ch f

ace

of t

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ron

?a,

b,a

nd

c;

a,q

,an

d r

;b

,p,a

nd

r;

c,p

,an

d q

3.R

ewri

te s

ente

nce

01

to s

tate

a t

wo-

dim

ensi

onal

an

alog

ue.

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nsi

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a t

rian

gle

wit

h t

wo

per

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dic

ula

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des

th

at m

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.

4.R

efer

to

the

top

diag

ram

an

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rite

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ress

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s fo

r th

e ar

eas

A,B

,an

d C

men

tion

ed i

n s

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nce

02.

Po

ssib

le a

nsw

er:

A�

�1 2� pr,

B�

�1 2� pq

,C�

�1 2� rq

5.T

o ex

plor

e th

e th

ree-

dim

ensi

onal

cas

e,yo

u m

igh

t be

gin

by

expr

essi

ng

a,b,

and

cin

ter

ms

of p

,q,a

nd

r.U

se t

he

Pyt

hag

orea

n t

heo

rem

to

do t

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.

a2

�q

2�

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2�

r2�

p2,c

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p2

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2

6.W

hic

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ues

s in

sen

ten

ce 0

5 se

ems

mor

e li

kely

? Ju

stif

y yo

ur

answ

er.

See

stu

den

ts’e

xpla

nat

ion

s.

y O

z

x

b

c

Op

a

qr

En

rich

men

t

NA

ME

____

____

____

____

____

____

____

____

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____

____

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____

____

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ER

IOD

____

_

7-9

7-9


1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

B:

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11. D

A

D

B

C

B

A

C

C

D

B

4x � 3

A

D

A

D

D

B

C

A

A

D

A

D

B

C

A

A

B

C

A

D

Chapter 7 Assessment Answer Key Form 1 Form 2APage 429 Page 430 Page 431

An

swer

s



12.

13.

14.

15.

16.

17.

18.

19.

20.

B:

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

B: 3

D

A

B

B

D

C

A

A

D

B

A

C

B

C

C

B

A

D

A

C

�2

B

B

C

C

A

D

B

C

B

Chapter 7 Assessment Answer Key Form 2A (continued) Form 2BPage 432 Page 433 Page 434


1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

B: 2

y

xO

y

xO

yes

f �1(x) � �15

�x � 2

2x2 � 1

2

�x 3 � 7x2 � 4x � 28

�2, 1, �32

�

�1, �2, �4, �8, ��12

�

5, �2i, 2i

3 or 1; 1; 2 or 0

x � 3, x � 5

�3

�15�, ��15�,i �3�, �i �3�

9(n3)2 � 36(n3)

Sample answer: rel. max. at x � �1,

rel. min. at x � 1

xO

f(x )

between �2 and �1,between 0 and 1,between 1 and 2

even; 4

x2 � x � 3

�176

Chapter 7 Assessment Answer Key Form 2CPage 435 Page 436

An

swer

s


1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

B: �4

y

xO

y

xO

D: x � �2, R: y � 0

yes

g�1(x) � ��12

�x � 2

x2 � 6x � 5

4

�x3 � 6x2 � 4x � 24

�3, �23

�, 1

�1, �2, �7, �14, ��12

�, ��72

�

2, �3i, 3i

2 or 0; 2 or 0; 4, 2, or 0

x � 3, x � 1

132

��6�, �6�, i �2�, �i �2�

5(x5)2 � 4(x5) � 3

Sample answer: rel. max. at x � 1, rel. min. at x � �1

xO

f(x )

between �2 and �1,between �1 and 0,between 1 and 2

even; 4

x2 � 2x � 1

�134

Chapter 7 Assessment Answer Key Form 2DPage 437 Page 438


1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

B: 24

y

x

O

y

xO

D: x � �4, R: y � �2

yes

h�1(x) � �5x

2� 6�

960

27x3 � 9x2 � 3x � 1

x2 � 2x

��14

�, ��12

�, �13

�, 2

�1, �3, �5, �15,

��19

�, ��13

�, ��59

�, ��53

�

�1, 3, 3 � 2i, 3 � 2i

5, 3, or 1; 5, 3, or 1; 10, 8, 6, 4, 2, or 0

�3

7014

16, 81

b[9(b2)2 � 3(b2) � 8]

Sample answer: rel.max. at x � �1 and

x � 1, rel. min. at x � 0

xO

f(x )

between 0 and 1,between 1 and 2

f (x ) → �� as x → ��,

f (x) → �� as x → ��; odd; 4

x4 � 7x2 � x

�133�

An

swer

s

Chapter 7 Assessment Answer Key Form 3Page 439 Page 440

An

swer

s


Chapter 7 Assessment Answer KeyPage 441, Open-Ended Assessment

Scoring Rubric

Score General Description Specific Criteria

• Shows thorough understanding of the concepts ofpolynomial functions; graphing polynomial functions;determining number and type of roots of a polynomialequation; finding rational zeros of a polynomial function;operations with functions; and finding inverse of a function.

• Uses appropriate strategies to solve problems.• Computations are correct.• Written explanations are exemplary.• Graphs are accurate and appropriate.• Goes beyond requirements of some or all problems.

• Shows an understanding of the concepts of polynomialfunctions; graphing polynomial functions; determiningnumber and type of roots of a polynomial equation; findingrational zeros of a polynomial function; operations withfunctions; and finding inverse of a function.

• Uses appropriate strategies to solve problems.• Computations are mostly correct.• Written explanations are effective.• Graphs are mostly accurate and appropriate.• Satisfies all requirements of problems.

• Shows an understanding of most of the concepts ofpolynomial functions; graphing polynomial functions;determining number and type of roots of a polynomialequation; finding rational zeros of a polynomial function;operations with functions; and finding inverse of a function.

• May not use appropriate strategies to solve problems.• Computations are mostly correct.• Written explanations are satisfactory.• Graphs are mostly accurate.• Satisfies the requirements of most of the problems.

• Final computation is correct.• No written explanations or work is shown to substantiate

the final computation.• Graphs may be accurate but lack detail or explanation.• Satisfies minimal requirements of some of the problems.

• Shows little or no understanding of most of the concepts ofpolynomial functions; graphing polynomial functions;determining number and type of roots of a polynomialequation; finding rational zeros of a polynomial function;operations with functions; and finding inverse of a function.

• Does not use appropriate strategies to solve problems.• Computations are incorrect.• Written explanations are unsatisfactory.• Graphs are inaccurate or inappropriate.• Does not satisfy requirements of problems.• No answer may be given.

0 UnsatisfactoryAn incorrect solutionindicating no mathematicalunderstanding of theconcept or task, or nosolution is given

1 Nearly Unsatisfactory A correct solution with nosupporting evidence orexplanation

2 Nearly SatisfactoryA partially correctinterpretation and/orsolution to the problem

3 SatisfactoryA generally correct solution,but may contain minor flawsin reasoning or computation

4 SuperiorA correct solution that is supported by well-developed, accurateexplanations

Chapter 7 Assessment Answer Key Page 441, Open-Ended Assessment

Sample Answers


1a. and b. Students must sketch a polynomial function having exactly 5 zeros, opposite end behavior,2 relative maxima, and 2 relative minima, labeled as shown.

1c. Regardless of the function sketched,D: all real numbers and R: all real numbers.

1d. Students must state either “As x → ��, f(x)→ �� and as x → ��, f(x) → ��” (as for the samplefunction shown) or “As x → ��, f(x) → �� and as x → ��,f(x) → ��.”

2. Since the range is g(x) � 2, the graph ofg(x) lies entirely on or above thehorizontal line g(x) � 2. Students shouldindicate that this would require that thedegree of g(x) must be even, that theleading coefficient must be positive, andthat the function has no zeros.

3a. Answers must be of the formP(x) � a0x4 � a1x3 � a2x2 � a3x � a4,where an � 0 for any n. Sampleanswer: P(x) � x4 � x3 � x2 � 2x � 3.

3b. Students should show by directsubstitution, and by syntheticsubstitution, how to find P(�2).For the sample function in a,P(�2) � 11.

3c. Students should indicate that x � 1 isa factor of P(x) if and only if P(�1) � 0.For the sample function in a,P(�1) � 2 � 0, so x � 1 is not a factorof P(x).

3d. Students should use Descartes’ Rule ofSigns to determine the number ofpositive and negative real zeros ofP(x). For the sample function in a,P(x) has no positive real zeros and has4, 2 or 0 negative real zeros.

3e. Students must explain that anyrational zeros of P(x) must be of the form �

pq�, where p is a factor of

a4 and q is a factor of a0.For the sample function in a, a0 � 1and a4 � 3, so the only possiblerational zeros are �1 and �3.

3f. For each of the possible rational zeroszn found in part e, students must showwhether x � zn is a factor of P(x). Forthe sample function in a, there are norational zeros.

4a. Sample answers: For g(x) � x � 1 andh(x) � x2, the answers would be:2x � 4; 9a2; x2 � x � 1; 1 � x � x2;

x3 � x2; �xx�

2

1� for x � �1; x2 � 2x � 1;

x2 � 1; 16; x � 1.4b. Students should show that, for their

functions g(x) and g�1(x),g�1 � g(x) � x. Students should thenindicate that the graphs of inversefunctions are reflections of one anotherover the line y � x.

xO

A1

z1 z2 z3

z4 z5

A2

B1

B2

f(x )

In addition to the scoring rubric found on page A34, the following sample answers may be used as guidance in evaluating open-ended assessment items.

An

swer

s


Chapter 7 Assessment Answer Key Vocabulary Test/Review Quiz (Lessons 7–1 through 7–3) Quiz (Lessons 7–6 and 7–7)

Page 442 Page 443 Page 444

1. syntheticsubstitution

2. one-to-one function

3. quadratic form

4. fundamentaltheorem of algebra

5. square root

6. location principle

7. minimum

8. depressedpolynomial

9. composition offunctions

10. polynomial in onevariable

11. Sample answer: Theend behavior of agraph is adescription of howthe graph behaveswhen the value of xbecomes very smallor very large.

12. Sample answer: TheFactor Theorem isthe theorem thatstates that thebinomial x � a is afactor of thepolynomial f(x) ifand only if f(a) � 0.

1.

2.

3.

4.

5.

Quiz (Lessons 7–4 and 7–5)

Page 443

1.

2.

3.

4.

5.

1.

2.

3.

4.

5.

Quiz (Lessons 7–8 and 7–9)

Page 444

1.

2.

3.

4.y

xO

D: x � �3, R: y � 0

yes

x

f(x)

f –1(x)

f (x)

O

f�1(x) � �x �

42

�

{(5, �2), (4, 0), (�8, 1), (7, 4)}

28; 122

x2 � 6x � 7; x2 � 2x � 5

{(2, 4), (3, 8), (4, 3), (8, 4)}; {(2, 5), (4, 2), (5, 4)}

x2 � x � 6; x2 � 5x � 2;2x3 � 2x2 � 8x � 8; �x2

2�x

3�x

4� 2

�, x �2

�1, �2, �3, �6,

��12

�, ��32

�; �1, �32

�

B

3, 1 � i, 1 � i

3 or 1; 2 or 0; 4, 2, or 0

x � 2, x � 3

46, 277

��7�, �7�,�i �7�, i �7�

�3, ��5�, �5�, 3

xO

f(x )

between �1 and 0,between 1 and 2,between 3 and 4

even; 2

57


1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.f�1(x) � �

x �5

4�

�3, �12

�, 2

278

x2 � x � 6 � 0

x � 13

(2n � 5)2

(�3, 5)

�1, �4i, 4i

f(x) � x3 � 3x2 � 9x � 27

2 or 0; 2 or 0; 4, 2, or 0

�3, 0, 3, �3i, 3i

xO

f(x )

at x � 1, between �4 and�3, between �2 and �1

D

A

C

D

A

B

Chapter 7 Assessment Answer Key Mid-Chapter Test Cumulative ReviewPage 445 Page 446

An

swer

s

xOf(x )

f (x) � x 2 � 2x � 8

(�1, �9)

(�4, 0) (2, 0)

x � �1

s � amount invested instock:

0.07s � 0.05(10,000 � s)� 550; at least $2500

consistent and dependent

A(4, �6), B(7, 0), C(1, �2)

2 or 0; 2 or 0; 4, 2, or 0


1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11. 12.

13. 14.

15.

16.

17. DCBA

DCBA

DCBA

0 0 0

.. ./ /

.

99 9 987654321

87654321

87654321

87654321

0 5.

0 0 0

.. ./ /

.

99 9 987654321

87654321

87654321

87654321

9

0 0 0

.. ./ /

.

99 9 987654321

87654321

87654321

87654321

5 9 6/

0 0 0

.. ./ /

.

99 9 987654321

87654321

87654321

87654321

5 / 8

HGFE

DCBA

HGFE

DCBA

HGFE

DCBA

HGFE

DCBA

HGFE

DCBA

Chapter 7 Assessment Answer KeyStandardized Test Practice

Page 447 Page 448


1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

21.

22.

23.

24.

25.

26.

27.

28.

29.

30.

1.

2.

3.

4.

5.

6.

7.

8. A

A

D

B

C

D

A

D

y

xO

f�1 (x) � �x �

72

�

51; 57

�1, �3, �5, �15,

��13

�, ��53

�

3 or 1; 0; 2 or 0

x � 2; x � 5

�86

10, �10, �2�, ��2�

Sample answer: rel.max. at x � �2 and x � 1, rel. min. at x � 0

xO

f(x )

�216

y

xO

y � �4(x � 1)2 � 3

(�3, �5); x � �3; up

0; 1 real root

��1 �

12i�95��

4x2 � 13x � 12 � 0

{�4, 6}

between �1 and 0; 2

75 m

xO

(2, 1)

(0, �3)

f(x )

f (x ) � –x 2 � 4x – 3x � 2

�2 � x � 1

�9m7�

2x2 � 3x � 4 � �x �

31

�

��110� � �

170�i

7�3� � 12�2�

4 � x � y2

�4y2

x�

�6y

5� 9

�

4x2 � 12x � 9

245x5

8x2 � 5x � 6

Chapter 7 Assessment Answer Key Unit 2 Test Semester TestPage 449 Page 450 Page 451

An

swer

s



9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

21.

22.

23.

24.

25.

26.

27.

28.

29.

30.

31.

32.

33.

34.

35.

36.

37.

38.

39.

40.

41.

42.

43.

44.

45.

46.y

xO

D: x � ��52

�; R: y � 0

p�1(x) � �x �

48

�

�3x2 � 4x � 9;12x3 � 27x2

�6, �4i, 4i

y � �(x � 2)2 � 3

��1, �17

��

{�9, 7}

minimum; �14

6.596

2x2 � 7xy � 15y2

72x4y3

25

40 cars

c � 2t; 5c � 8t � 360;

5c � 8t � 10v � 410

(2, �5)

xO

f(x )

D � all reals; R � {y � y � 3}

��14

�

4a2 � 8a � 12

3 40 1 2�1

�m � ��12

� � m � 4�or ��

12

�, 4�

p � the number ofsheets of plywood; 270 � 7p � 1000; no

more than 104 sheets ofplywood

1256 cm2

C

B

B

A

C

A

D

B

B

C

Chapter 7 Assessment Answer Key Semester Test (continued)Page 452 Page 453 Page 454

� ��3 4�5 0

� �3 �61�7 15

�112�� 1 �2

2 8

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