Chapter 7 Resource Masters - Math Class · PDF file©Glencoe/McGraw-Hill iv Glencoe Algebra 2 Teacher’s Guide to Using the Chapter 7 Resource Masters The Fast File Chapter Resource

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 formPolynomial 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 formPolynomial 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 Behavior f(x) → "% as x → "%

of Polynomial f(x) → "% as x → !%

Functions If the degree is odd and the leading coefficient is positive, thenf(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 of The maximum number of zeros of a polynomial function is equal to the degree of the polynomial.

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

Function On 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

Brad and Anissa Stuckey




11-19 all

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 698f (x) 3.1 5.5 7.1 8.3

x 8 12 15 20f (x) "207 169 976 3801

x 1 2 4 7f (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 Principle Suppose 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. falseb. If f(c) # f(d) ) 0, there are no real zeros between c and d. falsec. 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 ic. third-degree polynomial with one relative maximum and one relative minimum;

leading coefficient positive ivd. 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 $QAPP$.

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

KM

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








9-23 odd


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$1

10$)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, ia. 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 " 1x ! 1, x " 2 x ! 1, x " 3

15. x3 ! 3x2 " 4x " 12; x ! 3 16. x3 " 6x2 ! 11x " 6; x " 3x ! 2, x " 2 x ! 1, x ! 2

17. x3 ! 2x2 " 33x " 90; x ! 5 18. x3 " 6x2 ! 32; x " 4x " 3, x ! 6 x ! 4, x " 2

19. x3 " x2 " 10x " 8; x ! 2 20. x3 " 19x ! 30; x " 2x " 1, x ! 4 x " 5, x ! 3

21. 2x3 ! x2 " 2x " 1; x ! 1 22. 2x3 ! x2 " 5x ! 2; x ! 22x " 1, x ! 1 x ! 1, 2x ! 1

23. 3x3 ! 4x2 " 5x " 2; 3x ! 1 24. 3x3 ! x2 ! x " 2; 3x " 2x ! 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 " 1x " 1, x " 4 x " 3, x " 5

19. x3 " 9x2 ! 27x " 27; x " 3 20. x3 " x2 " 8x ! 12; x ! 3x ! 3, x ! 3 x ! 2, x ! 2

21. x3 ! 5x2 " 2x " 24; x " 2 22. x3 " x2 " 14x ! 24; x ! 4x " 3, x " 4 x ! 3, x ! 2

23. 3x3 " 4x2 " 17x ! 6; x ! 2 24. 4x3 " 12x2 " x ! 3; x " 3x ! 3, 3x ! 1 2x ! 1, 2x " 1

25. 18x3 ! 9x2 " 2x " 1; 2x ! 1 26. 6x3 ! 5x2 " 3x " 2; 3x " 23x " 1, 3x ! 1 2x " 1, x " 1

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

NAME ______________________________________________ DATE ____________ PERIOD _____

7-47-4

Reading to Learn MathematicsThe Remainder and Factor Theorems

NAME ______________________________________________ DATE ____________ PERIOD _____

7-47-4


Less

on

7-4

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 ! 1x ! 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

Enrichment

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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 Signs sign 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 25. 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

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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 ! 32 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 " 63 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 ! 11; 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 " 103, 1 " !2", 1 ! !2" 2, 2 " i, 2 ! i

15. h(x) # x3 ! 4x2 ! x " 6 16. q(x) # x3 ! 3x2 " 6x " 81, !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 " 12 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 ! 12 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

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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 63 1 2 61 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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Study Guide and InterventionRational Zero Theorem

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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






ODDSonly


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



NAME ______________________________________________ DATE ____________ PERIOD _____

7-67-6

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

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1. n(x) # x2 ! 5x ! 3 2. h(x) # x2 " 2x " 5

'1, '3 '1, '53. 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, 39. p(x) # x3 " x2 ! x " 1 10. z(x) # x3 " 4x2 ! 6x " 4

1 211. h(x) # x3 " x2 ! 4x " 4 12. g(x) # 3x3 " 9x2 " 10x " 8

1 413. 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, '143. 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

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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, '12possible values of q: '1possible 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, '45possible values of q: '1, '2possible 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

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x # 1 !1

1 ! 1

1 ! 1

1 ! 11 ! …

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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 " 43x2 " 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

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Example 1Example 1

Example 2Example 2

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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 * 4

g(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 ! 12g(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)] 2017. f[h("9)] 25 18. h[ g("3)] !11 19. g[f(8)] 32020. h[f(20)] 404 21. [f ! (h ! g)]("1) 1 22. [f ! (g ! h)](4) 160023. BUSINESS The function f(x) # 1000 " 0.01x2 models the manufacturing cost per item

when 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

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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. trueb. If f and g are polynomial functions, then is a polynomial function. falsec. If f and g are polynomial functions, the domain of the function f + g is the set of all

real numbers. trued. If f(x) # 3x ! 2 and g(x) # x " 4, the domain of the function is the set of all real

numbers. falsee. If f and g are polynomial functions, then (f ! g)(x) # (g ! f)(x). falsef. 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

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Find Inverses

Inverse Relations Two 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

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Example 1Example 1

Example 2Example 2

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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 "3

2$


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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7-8

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 Lesson1. 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).

Helping You Remember3. 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

NAME ______________________________________________ DATE ____________ PERIOD _____

7-87-8

Example 1Example 1

Example 2Example 2

Study Guide and InterventionSquare Root Functions and Inequalities

NAME ______________________________________________ DATE ____________ PERIOD _____

7-97-9


Less

on

7-9

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 0 "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

NAME ______________________________________________ DATE ____________ PERIOD _____

7-97-9

ExampleExample

ExercisesExercises

Skills PracticeSquare Root Functions and Inequalities

NAME ______________________________________________ DATE ____________ PERIOD _____

7-97-9


Less

on

7-9

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 0 "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 0 "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 _____

7-97-9

Reading to Learn MathematicsSquare Root Functions

NAME ______________________________________________ DATE ____________ PERIOD _____

7-97-9


Less

on

7-9

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 0

c. y # "x! ! 3 i iii. domain: x / 0; range: y 0 "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 0 3; range: y / 3

g. y # "3 " x! ! 3 vi vii. domain: x / 3; range: y 0 0

h. y # ""x! " 3 iii viii. domain: x / "3; range: y / 0

2. The graph of the inequality y 0 "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

NAME ______________________________________________ DATE ____________ PERIOD _____

7-97-9

© Glencoe/McGraw-Hill A2 Glencoe Algebra 2

Answers (Lesson 7-1)

Stu

dy

Gu

ide

and I

nte

rven

tion

Poly

nom

ial F

unct

ions

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-1

7-1

©G

lenc

oe/M

cGra

w-Hi

ll37

5G

lenc

oe A

lgeb

ra 2

Lesson 7-1

Poly

no

mia

l Fu

nct

ion

s

Apo

lynom

ial o

f deg

ree

nin

one

var

iabl

e x

is an

exp

ress

ion

of th

e fo

rmPo

lyno

mia

l in

a 0xn

!a 1

xn"

1!

… !

a n"

2x2

!a n

"1x

!a n

,O

ne V

aria

ble

wher

e th

e co

effic

ient

s a 0

, a1,

a 2, …

, an

repr

esen

t rea

l num

bers

, a0

is no

t zer

o,

and

nre

pres

ents

a n

onne

gativ

e in

tege

r.

The

deg

ree

of a

pol

ynom

iali

n on

e va

riab

le is

the

gre

ates

t ex

pone

nt o

f it

s va

riab

le.T

hele

adin

g co

effi

cien

tis

the

coe

ffic

ient

of

the

term

wit

h th

e hi

ghes

t de

gree

.

Apo

lynom

ial f

unct

ion

of d

egre

e n

can

be d

escr

ibed

by

an e

quat

ion

of th

e fo

rmPo

lyno

mia

lP(

x) #

a 0xn

!a 1

xn"

1!

… !

a n"

2x2

!a n

"1x

!a n

,Fu

nctio

nwh

ere

the

coef

ficie

nts

a 0, a

1, a 2

, …, a

nre

pres

ent r

eal n

umbe

rs, a

0is

not z

ero,

an

d n

repr

esen

ts a

non

nega

tive

inte

ger.

Wh

at a

re t

he

deg

ree

and

lea

din

g co

effi

cien

t of

3x2

!2x

4!

7 "

x3?

Rew

rite

the

exp

ress

ion

so t

he p

ower

s of

xar

e in

dec

reas

ing

orde

r."

2x4

!x3

!3x

2"

7T

his

is a

pol

ynom

ial i

n on

e va

riab

le.T

he d

egre

e is

4,a

nd t

he le

adin

g co

effi

cien

t is

"2.

Fin

d f

(!5)

if

f(x)

#x3

"2x

2!

10x

"20

.f(

x) #

x3!

2x2

"10

x!

20O

rigin

al fu

nctio

nf(

"5)

#("

5)3

!2(

"5)

2"

10("

5) !

20Re

plac

e x

with

"5.

#"

125

!50

!50

!20

Eval

uate

.#

"5

Sim

plify

.

Fin

d g

(a2

!1)

if

g(x)

#x2

"3x

!4.

g(x)

#x2

!3x

"4

Orig

inal

func

tion

g(a2

"1)

#(a

2"

1)2

!3(

a2"

1) "

4Re

plac

e x

with

a2

"1.

#a4

"2a

2!

1 !

3a2

"3

"4

Eval

uate

.#

a4!

a2"

6Si

mpl

ify.

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."

!"

not a

pol

ynom

ial i

n 5;

!9

one

varia

ble;

cont

ains

6;

!tw

o va

riabl

esF

ind

f(2

) an

d f

(!5)

for

eac

h f

un

ctio

n.

7.f(

x) #

x2"

98.

f(x)

#4x

3"

3x2

!2x

"1

9.f(

x) #

9x3

"4x

2!

5x!

7!

5;16

23;!

586

73;!

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

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6G

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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,

then

f(x) →

!%

as x

→"

%

f(x) →

!%

as x

→!

%

If th

e de

gree

is e

ven

and

the

lead

ing

coef

ficie

nt is

neg

ative

, the

n

End

Beha

vior

f(x) →

"%

as x

→"

%

of P

olyn

omia

lf(x

) →"

%as

x→

!%

Func

tions

If th

e de

gree

is o

dd a

nd th

e le

adin

g co

effic

ient

is p

ositiv

e, th

enf(x

) →"

%as

x→

"%

f(x) →

!%

as x

→!

%

If th

e de

gree

is o

dd a

nd th

e le

adin

g co

effic

ient

is n

egat

ive, t

hen

f(x) →

!%

as x

→"

%

f(x) →

"%

as x

→!

%

Real

Zer

os o

fTh

e m

axim

um n

umbe

r of z

eros

of a

pol

ynom

ial f

unct

ion

is eq

ual t

o th

e de

gree

of t

he p

olyn

omia

l.

a Po

lyno

mia

lA

zero

of a

func

tion

is a

poin

t at w

hich

the

grap

h in

ters

ects

the

x-ax

is.

Func

tion

On

a gr

aph,

cou

nt th

e nu

mbe

r of r

eal z

eros

of t

he fu

nctio

n by

cou

ntin

g th

e nu

mbe

r of t

imes

the

grap

h cr

osse

s or

touc

hes

the

x-ax

is.

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 r

eal

zero

s.A

s x→

"%

,f(x

) →"

%an

d as

x→

!%

,f(x

) →!

%,

so it

is a

n od

d-de

gree

pol

ynom

ial f

unct

ion.

The

gra

ph in

ters

ects

the

x-a

xis

at 1

poi

nt,

so t

he f

unct

ion

has

1 re

al z

ero.

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 r

eal

zero

s.

1.2.

3.

even

;6ev

en;1

dou

ble

zero

odd;

3

x

f (x)

Ox

f (x)

Ox

f (x)

O

x

f (x)

O

Stu

dy

Gu

ide

and I

nte

rven

tion

(c

onti

nued

)

Poly

nom

ial F

unct

ions

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-1

7-1

Exam

ple

Exam

ple

Exer

cises

Exer

cises


An

swer

s


Skil

ls P

ract

ice

Poly

nom

ial F

unct

ions

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-1

7-1

©G

lenc

oe/M

cGra

w-Hi

ll37

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.u3

!4u

2 v2

!v4

6.2r

"r2

!

No,t

his

poly

nom

ialc

onta

ins

two

No,t

his

is n

ot a

pol

ynom

ial b

ecau

seva

riabl

es,u

and

v.$ r1 2$

cann

ot b

e w

ritte

n in

the

form

rn ,

whe

re n

is a

non

nega

tive

inte

ger.

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) #

"2x

3!

5x!

30;

!3

11.p

(x) #

x4!

8x2

"10

!1;

3812

.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

mbe

r of

rea

l ze

roes

.

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

→!

%;

odd;

1ev

en;4

odd;

3

x

f (x)

Ox

f (x)

Ox

f (x)

O

1 $ r2

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lenc

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w-Hi

ll37

8G

lenc

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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"

12No

t a p

olyn

omia

l;4.

27 !

3xy3

"12

x2y2

"10

y

$ m2 2$ca

nnot

be

writ

ten

in th

e fo

rm

No,t

his

poly

nom

ial c

onta

ins

two

m

nfo

r a n

onne

gativ

e in

tege

r n.

varia

bles

,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;!

216

!29

;!39

0;!

135

8.p(

x) #

3x3

"x2

!2x

"5

9.p(

x) #

x4!

$1 2$ x3

"$1 2$ x

10.p

(x) #

$1 3$ x3

!$2 3$ x

2!

3x

!37

;73

13;9

3!

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

mbe

r of

rea

l ze

roes

.

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

;2ev

en;1

odd;

5

20.W

IND

CH

ILL

The

fun

ctio

n C

(s) #

0.01

3s2

"s

"7

esti

mat

es t

he w

ind

chill

tem

pera

ture

C(s

) at

0&F

for

win

d sp

eeds

sfr

om 5

to

30 m

iles

per

hour

.Est

imat

e th

e w

ind

chill

tem

pera

ture

at

0&F

if t

he w

ind

spee

d is

20

mile

s pe

r ho

ur.

abou

t !22

&F

x

f (x)

Ox

f (x)

Ox

f (x)

O

2 $ m2

Pra

ctic

e (A

vera

ge)

Poly

nom

ial F

unct

ions

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-1

7-1



Rea

din

g t

o L

earn

Math

emati

csPo

lyno

mia

l Fun

ctio

ns

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-1

7-1

©G

lenc

oe/M

cGra

w-Hi

ll37

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

ucti

on t

o L

esso

n 7-

1 at

the

top

of

page

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 is

1 h

exag

onin

the

cen

ter,

6 he

xago

ns in

the

sec

ond

ring

,and

12

hexa

gons

in t

he t

hird

ring

.How

man

y he

xago

ns w

ill t

here

be

in t

he fo

urth

,fift

h,an

d si

xth

ring

s?18

;24;

30•

The

re is

1 h

exag

on in

a h

oney

com

b w

ith

1 ri

ng.T

here

are

7 h

exag

ons

ina

hone

ycom

b w

ith

2 ri

ngs.

How

man

y he

xago

ns a

re t

here

in h

oney

com

bsw

ith

3 ri

ngs,

4 ri

ngs,

5 ri

ngs,

and

6 ri

ngs?

19;3

7;61

;91

Rea

din

g t

he

Less

on

1.G

ive

the

degr

ee a

nd le

adin

g co

effi

cien

t of

eac

h po

lyno

mia

l in

one

vari

able

.

degr

eele

adin

g co

effi

cien

t

a.10

x3!

3x2

"x

!7

b.7y

2"

2y5

!y

"4y

3

c.10

0

2.M

atch

eac

h de

scri

ptio

n of

a p

olyn

omia

l fun

ctio

n fr

om t

he li

st o

n th

e le

ft w

ith

the

corr

espo

ndin

g en

d be

havi

or f

rom

the

list

on

the

righ

t.

a.ev

en d

egre

e,ne

gati

ve le

adin

g co

effi

cien

tiii

i.f(

x) →

!%

as x

→!

%;

f(x)

→!

%as

x→

"%

b.od

d de

gree

,pos

itiv

e le

adin

g co

effi

cien

tiv

ii.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.ev

en d

egre

e,po

siti

ve le

adin

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 a

n ea

sy w

ay t

o re

mem

ber

the

diff

eren

ce b

etw

een

the

end

beha

vior

of

the

grap

hsof

eve

n-de

gree

and

odd

-deg

ree

poly

nom

ial f

unct

ions

?

Sam

ple

answ

er:B

oth

ends

of t

he g

raph

of a

n ev

en-d

egre

e fu

nctio

nev

entu

ally

kee

p go

ing

in th

e sa

me

dire

ctio

n.Fo

r odd

-deg

ree

func

tions

,th

e tw

o en

ds e

vent

ually

hea

d in

opp

osite

dire

ctio

ns,o

ne u

pwar

d,th

eot

her d

ownw

ard.

100

0!

25

103

©G

lenc

oe/M

cGra

w-Hi

ll38

0G

lenc

oe A

lgeb

ra 2

Appr

oxim

atio

n by

Mea

ns o

f Pol

ynom

ials

Man

y sc

ient

ific

exp

erim

ents

pro

duce

pai

rs o

f nu

mbe

rs [

x,f(

x)]

that

can

be

rel

ated

by

a fo

rmul

a.If

the

pai

rs f

orm

a f

unct

ion,

you

can

fit

a po

lyno

mia

l to

the

pair

s in

exa

ctly

one

way

.Con

side

r th

e pa

irs

give

n by

th

e fo

llow

ing

tabl

e.

We

will

ass

ume

the

poly

nom

ial i

s of

deg

ree

thre

e.Su

bsti

tute

the

giv

en

valu

es in

to t

his

expr

essi

on.

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

will

get

the

sys

tem

of e

quat

ions

sho

wn

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 fi

nd t

he d

esir

ed p

olyn

omia

l.

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.So

lve

the

syst

em o

f equ

atio

ns 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 t

hat

repr

esen

ts t

he fo

ur o

rder

ed p

airs

.Wri

te y

our

answ

er in

the

form

y#

a!

bx!

cx2

!dx

3 .y

#!

2x3

"17

x2!

32x

"23

3.F

ind

the

poly

nom

ial t

hat

give

s th

e fo

llow

ing

valu

es.

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 c

arbo

n di

oxid

e ga

s th

at c

an b

e ab

sorb

ed b

y on

e cu

bic

cent

imet

er o

f cha

rcoa

l at

pres

sure

x.F

ind

the

valu

es fo

r 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

8f(

x)3.

15.

57.

18.

3

x8

1215

20f(

x)"

207

169

976

3801

x1

24

7f(

x)6

1139

"54

En

rich

men

t

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-1

7-1


An

swer

s


Stu

dy

Gu

ide

and I

nte

rven

tion

Gra

phin

g Po

lyno

mia

l Fun

ctio

ns

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-2

7-2

©G

lenc

oe/M

cGra

w-Hi

ll38

1G

lenc

oe A

lgeb

ra 2

Lesson 7-2

Gra

ph

Po

lyn

om

ial F

un

ctio

ns

Loca

tion

Prin

cipl

eSu

ppos

e y

#f(x

) rep

rese

nts

a po

lynom

ial f

unct

ion

and

aan

d b

are

two

num

bers

suc

h th

atf(a

) '0

and

f(b) (

0. T

hen

the

func

tion

has

at le

ast o

ne re

al z

ero

betw

een

aan

d b.

Det

erm

ine

the

valu

es o

f x

betw

een

wh

ich

eac

h r

eal

zero

of

the

fun

ctio

n f

(x)

#2x

4!

x3!

5 is

loc

ated

.Th

en d

raw

th

e gr

aph

.M

ake

a ta

ble

of v

alue

s.L

ook

at t

he v

alue

s of

f(x

) to

loca

te t

he z

eros

.The

n us

e th

e po

ints

to

sket

ch a

gra

ph o

f th

e fu

ncti

on.

The

cha

nges

in s

ign

indi

cate

tha

t 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

tabl

e of

val

ues

.Det

erm

ine

the

valu

es o

f x

atw

hic

h o

r be

twee

n w

hic

h e

ach

rea

l ze

ro i

s lo

cate

d.

1.f(

x) #

x3"

2x2

!1

2.f(

x) #

x4!

2x3

"5

3.f(

x) #

"x4

!2x

2"

1

betw

een

0 an

d !

1;be

twee

n !

2 an

d !

3;at

'1

at 1

;bet

wee

n 1

and

2be

twee

n 1

and

24.

f(x)

#x3

"3x

2!

45.

f(x)

#3x

3!

2x"

16.

f(x)

#x4

"3x

3!

1

at !

1,2

betw

een

0 an

d 1

betw

een

0 an

d 1;

betw

een

2 an

d 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-Hi

ll38

2G

lenc

oe A

lgeb

ra 2

Max

imu

m a

nd

Min

imu

m P

oin

tsA

qua

drat

ic f

unct

ion

has

eith

er a

max

imum

or

am

inim

um p

oint

on

its

grap

h.Fo

r hi

gher

deg

ree

poly

nom

ial f

unct

ions

,you

can

fin

d tu

rnin

gpo

ints

,whi

ch r

epre

sent

rel

ativ

e 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

and

grap

h th

e fu

ncti

on.

A r

elat

ive

max

imum

occ

urs

at x

#"

4 an

d a

rela

tive

min

imum

occ

urs

at x

#0.

Gra

ph

eac

h f

un

ctio

n b

y m

akin

g a

tabl

e of

val

ues

.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

,min

.at 2

max

.abo

ut !

1,m

ax.a

bout

!1,

min

.abo

ut 0

.5m

in.a

bout

14.

f(x)

#x4

"7x

"3

5.f(

x) #

x5"

2x2

!2

6.f(

x) #

x3!

2x2

"3

min

.abo

ut 1

max

.at 0

,m

ax.a

bout

!1,

min

.abo

ut 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

Gu

ide

and I

nte

rven

tion

(c

onti

nued

)

Gra

phin

g Po

lyno

mia

l Fun

ctio

ns

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-2

7-2

Exam

ple

Exam

ple

Exer

cises

Exer

cises



Skil

ls P

ract

ice

Gra

phin

g Po

lyno

mia

l Fun

ctio

ns

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-2

7-2

©G

lenc

oe/M

cGra

w-Hi

ll38

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

tabl

e of

val

ues

.b.

Det

erm

ine

con

secu

tive

val

ues

of

xbe

twee

n w

hic

h e

ach

rea

l ze

ro i

s lo

cate

d.

c.E

stim

ate

the

x-co

ord

inat

es a

t w

hic

h t

he

rela

tive

max

ima

and

min

ima

occu

r.

1.f(

x) #

x3"

3x2

!1

2.f(

x) #

x3"

3x!

1

zero

s be

twee

n !

1 an

d 0,

0 an

d 1,

zero

s be

twee

n !

2 an

d !

1,0

and

1,an

d 2

and

3;re

l.m

ax.a

t x#

0,an

d 1

and

2;re

l.m

ax.a

t x#

!1,

rel.

min

.at 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;

zero

bet

wee

n !

1 an

d 0;

rel.

max

.at x

#!

2,re

l.m

in.a

t x#

1,re

l.m

ax.a

t x#

0re

l.m

in.a

t x#

!1

5.f(

x) #

x4"

2x2

"2

6.f(

x) #

0.5x

4"

4x2

!4

zero

s be

twee

n !

2 an

d !

1,an

d ze

ros

betw

een

!1

and

!2,

!2

and

1 an

d 2;

rel.

max

.at x

#0,

!3,

1 an

d 2,

and

2 an

d 3;

rel.

max

.at

rel.

min

.at x

#!

1 an

d x

#1

x#

0,re

l.m

in.a

t 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-Hi

ll38

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

tabl

e of

val

ues

.b.

Det

erm

ine

con

secu

tive

val

ues

of

xbe

twee

n w

hic

h e

ach

rea

l ze

ro i

s lo

cate

d.

c.E

stim

ate

the

x-co

ord

inat

es a

t w

hic

h t

he

rela

tive

an

d r

elat

ive

min

ima

occu

r.

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 (A

vera

ge)

Gra

phin

g Po

lyno

mia

l Fun

ctio

ns

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-2

7-2

zero

s be

twee

n !

1ze

ros

betw

een

!2

and

0,1

and

2,an

d !

1,0

and

1,an

d 2

and

3;re

l.m

ax.a

t x#

2,an

d 3

and

4;re

l.m

ax.a

t x#

!1,

rel.

min

.at 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 be

twee

n !

3 an

d !

2,an

d ze

ros

betw

een

!3

and

!2,

!2

and

!1;

rel.

max

.at x

#0,

and

0 an

d 1;

rel.

min

.at x

#!

1re

l.m

in.a

t 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.

The

Con

sum

er P

rice

Ind

ex (

CP

I) g

ives

the

rel

ativ

e pr

ice

for

a fi

xed

set

of g

oods

and

ser

vice

s.T

he C

PI

from

Se

ptem

ber,

2000

to

July

,200

1 is

sho

wn

in t

he g

raph

.So

urce

: U. S

. Bur

eau

of L

abor

Stat

istics

5.D

escr

ibe

the

turn

ing

poin

ts o

f th

e gr

aph.

rel m

ax.i

n No

v.an

d Ju

ne;r

el.m

in in

Dec

.6.

If t

he g

raph

wer

e m

odel

ed b

y a

poly

nom

ial e

quat

ion,

wha

t is

the

leas

t de

gree

the

equ

atio

n co

uld

have

?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 g

raph

of

a po

lyno

mia

lfu

ncti

on t

hrou

gh t

hese

poi

nts

have

? D

escr

ibe

them

.4:

2 re

l.m

ax.a

nd 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


Rea

din

g t

o L

earn

Math

emati

csG

raph

ing

Poly

nom

ial F

unct

ions

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-2

7-2

©G

lenc

oe/M

cGra

w-Hi

ll38

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

ucti

on t

o L

esso

n 7-

2 at

the

top

of

page

353

in y

our

text

book

.

Thr

ee p

oint

s on

the

gra

ph s

how

n in

you

r te

xtbo

ok a

re (

0,14

),(7

0,3.

78),

and

(100

,9).

Giv

e th

e re

al-w

orld

mea

ning

of

the

coor

dina

tes

of t

hese

poi

nts.

Sam

ple

answ

er:I

n 19

00,1

4% o

f the

U.S

.pop

ulat

ion

was

fore

ign

born

.In

1970

,3.7

8% o

f the

pop

ulat

ion

was

fore

ign

born

.In

2000

,9%

of t

he p

opul

atio

n w

as fo

reig

n bo

rn.

Rea

din

g t

he

Less

on

1.Su

ppos

e th

at f

(x)

is a

thi

rd-d

egre

e po

lyno

mia

l fun

ctio

n an

d th

at c

and

dar

e re

alnu

mbe

rs,w

ith

d(

c.In

dica

te w

heth

er e

ach

stat

emen

t is

tru

eor

fal

se.(

Rem

embe

r th

attr

uem

eans

alw

ays

true

.)

a.If

f(c

) (0

and

f(d)

'0,

ther

e is

exa

ctly

one

rea

l zer

o be

twee

n c

and

d.fa

lse

b.If

f(c

) #f(

d) )

0,th

ere

are

no r

eal z

eros

bet

wee

n c

and

d.fa

lse

c.If

f(c

) '0

and

f(d)

(0,

ther

e is

at

leas

t on

e re

al z

ero

betw

een

can

d d.

true

2.M

atch

eac

h gr

aph

wit

h it

s de

scri

ptio

n.

a.th

ird-

degr

ee p

olyn

omia

l wit

h on

e re

lati

ve m

axim

um a

nd o

ne r

elat

ive

min

imum

;le

adin

g co

effi

cien

t ne

gati

veiii

b.fo

urth

-deg

ree

poly

nom

ial w

ith

two

rela

tive

min

ima

and

one

rela

tive

max

imum

ic.

thir

d-de

gree

pol

ynom

ial w

ith

one

rela

tive

max

imum

and

one

rel

ativ

e m

inim

um;

lead

ing

coef

fici

ent

posi

tive

ivd.

four

th-d

egre

e po

lyno

mia

l wit

h tw

o re

lati

ve m

axim

a an

d on

e re

lati

ve m

inim

umii

i.ii

.ii

i.iv

.

Hel

pin

g Y

ou

Rem

emb

er

3.T

he o

rigi

ns o

f w

ords

can

hel

p yo

u to

rem

embe

r th

eir

mea

ning

and

to

dist

ingu

ish

betw

een

sim

ilar

wor

ds.L

ook

up m

axim

uman

d m

inim

umin

a d

icti

onar

y an

d de

scri

beth

eir

orig

ins

(ori

gina

l lan

guag

e an

d m

eani

ng).

Sam

ple

answ

er:M

axim

umco

mes

from

the

Latin

wor

d m

axim

us,m

eani

ng g

reat

est.

Min

imum

com

es fr

omth

e La

tin w

ord

min

imus

,mea

ning

leas

t.

x

f (x)

Ox

f (x)

Ox

f (x)

Ox

f (x)

O

©G

lenc

oe/M

cGra

w-Hi

ll38

6G

lenc

oe A

lgeb

ra 2

Gol

den

Rect

angl

esU

se a

str

aigh

ted

ge,a

com

pas

s,an

d t

he

inst

ruct

ion

s be

low

to

con

stru

ct

a go

lden

rec

tan

gle.

1.C

onst

ruct

squ

are

AB

CD

wit

h si

des

of

2 ce

ntim

eter

s.

2.C

onst

ruct

the

mid

poin

t of

A !B!

.Cal

l the

m

idpo

int

M.

3.U

sing

Mas

the

cen

ter,

set

your

com

pass

op

enin

g at

MC

.Con

stru

ct a

n ar

c w

ith

cent

er M

that

inte

rsec

ts A !

B!.C

all t

he p

oint

of

inte

rsec

tion

P.

4.C

onst

ruct

a li

ne t

hrou

gh P

that

is

perp

endi

cula

r to

A !B!

.

5.E

xten

d D!

C!so

tha

t it

inte

rsec

ts t

he

perp

endi

cula

r.C

all t

he in

ters

ecti

on p

oint

Q.

AP

QD

is a

gol

den

rect

angl

e.C

heck

thi

s

conc

lusi

on b

y fi

ndin

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 be

low

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

bere

d n

auti

lus.

6.U

sing

Aas

a c

ente

r,dr

aw

an a

rc t

hat

pass

es t

hrou

gh

Ban

d C

.

7.U

sing

Das

a c

ente

r,dr

aw

an a

rc t

hat

pass

es t

hrou

gh

Can

d E

.

8.U

sing

Fas

a c

ente

r,dr

aw

an a

rc t

hat

pass

es t

hrou

gh

Ean

d G

.

9.C

onti

nue

draw

ing

arcs

,us

ing

H,K

,and

Mas

th

e ce

nter

s.

C

BA

GHJD

E

KM

LFD A

M

QC

PB

En

rich

men

t

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-2

7-2



Stu

dy

Gu

ide

and I

nte

rven

tion

Solv

ing

Equa

tions

Usi

ng Q

uadr

atic

Tec

hniq

ues

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-3

7-3

©G

lenc

oe/M

cGra

w-Hi

ll38

7G

lenc

oe A

lgeb

ra 2

Lesson 7-3

Qu

adra

tic

Form

Cer

tain

pol

ynom

ial e

xpre

ssio

ns in

xca

n be

wri

tten

in t

he q

uadr

atic

form

au2

!bu

!c

for

any

num

bers

a,b

,and

c,a

)0,

whe

re u

is a

n ex

pres

sion

in 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.10

1b!

49!

b""

42L

et u

#"

b!.10

1b"

49"

b!!

42 #

101(

"b!)

2"

49("

b!)!

42

c.24

a5

"12

a3

"18

Thi

s ex

pres

sion

can

not

be w

ritt

en in

qua

drat

ic f

orm

,sin

ce a

5)

(a3 )

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

!6p

2!

8

(x2 )

2"

6(x2

) !8

4(p2

)2"

6(p2

) "8

3.x8

!2x

4!

14.

x$1 8$!

2x$ 11 6$

!1

(x4 )

2"

2(x4

) "1

# x$ 11 6$$2

"2# x

$ 11 6$$ "

1

5.6x

4!

3x3

!18

6.12

x4!

10x2

"4

not p

ossi

ble

12(x

2 )2

"10

(x2 )

!4

7.24

x8!

x4!

48.

18x6

"2x

3!

12

24(x

4 )2

"x4

"4

18(x

3 )2

!2(

x3) "

12

9.10

0x4

"9x

2"

1510

.25x

8!

36x6

"49

100(

x2)2

!9(

x2) !

15no

t pos

sibl

e

11.4

8x6

"32

x3!

2012

.63x

8!

5x4

"29

48(x

3 )2

!32

(x3 )

"20

63(x

4 )2

"5(

x4) !

29

13.3

2x10

!14

x5"

143

14.5

0x3

"15

x"x!

"18

32(x

5 )2

"14

(x5 )

!14

350

# x$3 2$ $2!

15# x$3 2$ $ !

18

15.6

0x6

"7x

3!

316

.10x

10"

7x5

"7

60(x

3 )2

!7(

x3) "

310

(x5 )

2!

7(x5

) !7

Exam

ple

Exam

ple

Exer

cises

Exer

cises

©G

lenc

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cGra

w-Hi

ll38

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 exp

ress

ion

can

be w

ritt

enin

qua

drat

ic f

orm

,the

n yo

u ca

n us

e w

hat

you

know

abo

ut s

olvi

ng q

uadr

atic

equ

atio

ns t

oso

lve

the

rela

ted

poly

nom

ial e

quat

ion.

Sol

ve x

4!

40x2

"14

4 #

0.x4

"40

x2!

144

#0

Orig

inal

equ

atio

n

(x2 )

2"

40(x

2 ) !

144

#0

Writ

e th

e ex

pres

sion

on th

e le

ft in

qua

drat

ic fo

rm.

(x2

"4)

(x2

"36

) #

0Fa

ctor

.x2

"4

#0

orx2

"36

#0

Zero

Pro

duct

Pro

perty

(x"

2)(x

!2)

#0

or(x

"6)

(x!

6) #

0Fa

ctor

.

x"

2 #

0or

x!

2 #

0or

x"

6 #

0or

x!

6 #

0Ze

ro P

rodu

ct P

rope

rty

x#

2or

x#

"2

orx

#6

orx

#"

6Si

mpl

ify.

The

sol

utio

ns a

re *

2 an

d *

6.

Sol

ve 2

x"

!x"

!15

#0.

2x!

"x!

"15

#0

Orig

inal

equ

atio

n

2("

x!)2

!"

x!"

15 #

0W

rite

the

expr

essio

n on

the

left

in q

uadr

atic

form

.

(2"

x!"

5)("

x!!

3) #

0Fa

ctor

.

2"x!

"5

#0

or"

x!!

3 #

0Ze

ro P

rodu

ct P

rope

rty

"x!

#or

"x!

#"

3Si

mpl

ify.

Sinc

e th

e pr

inci

pal s

quar

e ro

ot o

f a

num

ber

cann

ot b

e ne

gati

ve,"

x!#

"3

has

no s

olut

ion.

The

sol

utio

n is

or

6.

Sol

ve e

ach

equ

atio

n.

1.x4

#49

2.x4

"6x

2#

"8

3.x4

"3x

2#

54

'!

7",'

i!7"

'2,

'!

2"'

3,'

i!6"

4.3t

6"

48t2

#0

5.m

6"

16m

3!

64 #

06.

y4"

5y2

!4

#0

0,'

2,'

2i2,

!1

'i!

3"'

1,'

2

7.x4

"29

x2!

100

#0

8.4x

4"

73x2

!14

4 #

09.

"!

12 #

0

'5,

'2

'4,

',

10.x

"5"

x!!

6 #

011

.x"

10"

x!!

21 #

012

.x$2 3$

"5x

$1 3$!

6 #

0

4,9

9,49

27,81 $ 4

1 $ 33 $ 2

7 $ x1 $ x2

1 $ 425 $ 4

5 $ 2Stu

dy

Gu

ide

and I

nte

rven

tion

(c

onti

nued

)

Solv

ing

Equa

tions

Usi

ng Q

uadr

atic

Tec

hniq

ues

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-3

7-3

Exam

ple1

Exam

ple1

Exam

ple2

Exam

ple2

Exer

cises

Exer

cises


An

swer

s


Skil

ls P

ract

ice

Solv

ing

Equa

tions

Usi

ng Q

uadr

atic

Tec

hniq

ues

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-3

7-3

©G

lenc

oe/M

cGra

w-Hi

ll38

9G

lenc

oe A

lgeb

ra 2

Lesson 7-3

Wri

te e

ach

exp

ress

ion

in

qu

adra

tic

form

,if

pos

sibl

e.

1.5x

4!

2x2

"8

5(x2

)2"

2(x2

) !8

2.3y

8"

4y2

!3

not p

ossi

ble

3.10

0a6

!a3

100(

a3)2

"a3

4.x8

!4x

4!

9(x

4 )2

"4(

x4) "

9

5.12

x4"

7x2

12(x

2 )2

!7(

x2)

6.6b

5!

3b3

"1

not p

ossi

ble

7.15

v6"

8v3

!9

15(v

3 )2

!8(

v3) "

98.

a9"

5a5

!7a

a[(a

4 )2

!5(

a4) "

7]

Sol

ve e

ach

equ

atio

n.

9.a3

"9a

2!

14a

#0

0,7,

210

.x3

#3x

20,

3

11.t

4"

3t3

"40

t2#

00,

!5,

812

.b3

"8b

2!

16b

#0

0,4

13.m

4#

4!

!2",

!2",

!i!

2",i!

2"14

.w3

"6w

#0

0,!

6",!

!6"

15.m

4"

18m

2#

"81

!3,

316

.x5

"81

x#

00,

!3,

3,!

3i,3

i

17.h

4"

10h2

#"

9!

1,1,

!3,

318

.a4

"9a

2!

20 #

0!

2,2,

!5",

!!

5"

19.y

4"

7y2

!12

#0

20.v

4"

12v2

!35

#0

2,!

2,!

3",!

!3"

!5",

!!

5",!

7",!

!7"

21.x

5"

7x3

!6x

#0

22.c

$2 3$!

7c$1 3$

!12

#0

0,!

1,1,

!6",

!!

6"!

64,!

27

23.z

"5"

z!#

"6

4,9

24.x

"30

"x!

!20

0 #

010

0,40

0

©G

lenc

oe/M

cGra

w-Hi

ll39

0G

lenc

oe A

lgeb

ra 2

Wri

te e

ach

exp

ress

ion

in

qu

adra

tic

form

,if

pos

sibl

e.

1.10

b4!

3b2

"11

2."

5x8

!x2

!6

3.28

d6!

25d3

10(b

2 )2

"3(

b2) !

11no

t pos

sibl

e28

(d3 )

2"

25(d

3 )

4.4s

8!

4s4

!7

5.50

0x4

"x2

6.8b

5"

8b3

"1

4(s4

)2"

4(s4

) "7

500(

x2)2

!x2

not p

ossi

ble

7.32

w5

"56

w3

!8w

8.e$2 3$

!7e

$1 3$"

109.

x$1 5$!

29x$ 11 0$

! 2

8w[4

(w2 )

2!

7(w

2 ) "

1]( e$1 3$ )2

"7( e

$1 3$ ) !10

( x$ 11 0$)2

"29

( x$ 11 0$) "

2

Sol

ve e

ach

equ

atio

n.

10.y

4"

7y3

"18

y2#

0!

2,0,

911

.s5

!4s

4"

32s3

#0

!8,

0,4

12.m

4"

625

#0

!5,

5,!

5i,5

i13

.n4

"49

n2#

00,

!7,

7

14.x

4"

50x2

!49

#0

!1,

1,!

7,7

15.t

4"

21t2

!80

#0

!4,

4,!

5",!

!5"

16.4

r6"

9r4

#0

0,$3 2$ ,

!$3 2$

17.x

4"

24 #

"2x

2!

2,2,

!i!

6",i!

6"

18.d

4#

16d

2"

48 !

2,2,

!2!

3",2!

3"19

.t3

"34

3 #

07,

,

20.x

$1 2$"

5x$1 4$

!6

#0

16,8

121

.x$4 3$

"29

x$2 3$!

100

#0

8,12

5

22.y

3"

28y$3 2$

!27

#0

1,9

23.n

"10

"n!

!25

#0

25

24.w

"12

"w!

!27

#0

9,81

25.x

"2"

x!"

80 #

010

0

26.P

HY

SIC

SA

pro

ton

in a

mag

neti

c fi

eld

follo

ws

a pa

th o

n a

coor

dina

te g

rid

mod

eled

by

the

func

tion

f(x

) #x4

"2x

2"

15.W

hat

are

the

x-co

ordi

nate

s of

the

poi

nts

on t

he g

rid

whe

re t

he p

roto

n cr

osse

s th

e x-

axis

?!

!5" ,

!5"

27.S

URV

EYIN

GV

ista

cou

nty

is s

etti

ng a

side

a la

rge

parc

el o

f la

nd t

o pr

eser

ve it

as

open

spac

e.T

he c

ount

y ha

s hi

red

Meg

han’

s su

rvey

ing

firm

to

surv

ey t

he p

arce

l,w

hich

is in

the

shap

e of

a r

ight

tri

angl

e.T

he lo

nger

leg

of t

he t

rian

gle

mea

sure

s 5

mile

s le

ss t

han

the

squa

re o

f th

e sh

orte

r le

g,an

d th

e hy

pote

nuse

of

the

tria

ngle

mea

sure

s 13

mile

s le

ssth

an t

wic

e th

e sq

uare

of

the

shor

ter

leg.

The

leng

th o

f ea

ch b

ound

ary

is a

who

le n

umbe

r.F

ind

the

leng

th o

f ea

ch b

ound

ary.

3 m

i,4

mi,

5 m

i

!7

"7i

!3"

$$ 2

!7

!7i

!3"

$$ 2

Pra

ctic

e (A

vera

ge)

Solv

ing

Equa

tions

Usi

ng Q

uadr

atic

Tec

hniq

ues

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-3

7-3



Rea

din

g t

o L

earn

Math

emati

csSo

lvin

g Eq

uatio

ns U

sing

Qua

drat

ic T

echn

ique

s

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-3

7-3

©G

lenc

oe/M

cGra

w-Hi

ll39

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

ucti

on t

o L

esso

n 7-

3 at

the

top

of

page

360

in y

our

text

book

.

Exp

lain

how

the

for

mul

a gi

ven

for

the

volu

me

of t

he b

ox c

an b

e ob

tain

edfr

om t

he d

imen

sion

s sh

own

in t

he f

igur

e.

Sam

ple

answ

er:T

he v

olum

e of

a re

ctan

gula

r box

is g

iven

by

the

form

ula

V#

!wh.

Subs

titut

e 50

!2x

for !

,32

!2x

for w

,and

xfo

r hto

get

V(

x) #

(50

!2x

)(32

!2x

)(x) #

4x3

!16

4x2

"16

00x.

Rea

din

g t

he

Less

on

1.W

hich

of

the

follo

win

g ex

pres

sion

s ca

n be

wri

tten

in q

uadr

atic

for

m?

b,c,

d,f,

g,h,

ia.

x3!

6x2

!9

b.x4

"7x

2!

6c.

m6

!4m

3!

4

d.y

"2y

$1 2$"

15e.

x5!

x3!

1f.

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 ex

pres

sion

fro

m t

he li

st o

n th

e le

ft w

ith

its

fact

oriz

atio

n fr

om t

he li

st o

n th

e ri

ght.

a.x4

"3x

2"

40vi

i.(x

3!

3)(x

3"

3)

b.x4

"10

x2!

25v

ii.

("x!

!3)

("x!

"3)

c.x6

"9

iii

i.("

x!!

3)2

d.x

"9

iiiv

.(x

2!

1)(x

4"

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 a

n ea

sy w

ay t

o te

ll w

heth

er a

tri

nom

ial i

n on

e va

riab

le c

onta

inin

g on

e co

nsta

ntte

rm c

an b

e w

ritt

en in

qua

drat

ic f

orm

?

Sam

ple

answ

er:L

ook

at th

e tw

o te

rms

that

are

not

con

stan

ts a

ndco

mpa

re th

e ex

pone

nts

on th

e va

riabl

e.If

one

of th

e ex

pone

nts

is tw

ice

the

othe

r,th

e tri

nom

ial c

an b

e w

ritte

n in

qua

drat

ic fo

rm.

©G

lenc

oe/M

cGra

w-Hi

ll39

2G

lenc

oe A

lgeb

ra 2

Odd

and

Eve

n Po

lyno

mia

l Fun

ctio

ns

En

rich

men

t

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-3

7-3

Fun

ctio

ns w

hose

gra

phs

are

sym

met

ric

wit

hre

spec

t to

the

ori

gin

are

calle

d od

dfu

ncti

ons.

If f

("x)

#"

f(x)

for

all x

in t

he d

omai

n of

f(x

),th

en f

(x)

is o

dd.

Fun

ctio

ns w

hose

gra

phs

are

sym

met

ric

wit

hre

spec

t to

the

y-a

xis

are

calle

d ev

enfu

ncti

ons.

If f

("x)

#f(

x) f

or a

ll x

in t

he d

omai

n of

f(x

),th

en f

(x)

is e

ven.

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)Re

plac

e x

with

"x.

#"

x3!

3xSi

mpl

ify.

#"

(x3

"3x

)Fa

ctor

out

"1.

#"

f(x)

Subs

tutu

te.

The

refo

re,f

(x)

is o

dd.

The

gra

ph a

t th

e ri

ght

veri

fies

tha

t f(

x) is

odd

.T

he g

raph

of

the

func

tion

is s

ymm

etri

c w

ith

resp

ect

to t

he o

rigi

n.

Det

erm

ine

wh

eth

er e

ach

fu

nct

ion

is

odd

,eve

n,o

r n

eith

erby

gra

ph

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) #

x7od

d4.

f(x)

#x3

"x2

neith

er

5.f(

x) #

3x3

!1

neith

er6.

f(x)

#x8

"x5

"6

neith

er

7.f(

x) #

"8x

5"

2x3

!6x

odd

8.f(

x) #

x4"

3x3

!2x

2"

6x!

1ne

ither

9.f(

x) #

x4!

3x2

!11

even

10.f

(x) #

x7"

6x5

!2x

3!

xod

d

11.C

ompl

ete

the

follo

win

g de

fini

tion

s:A

pol

ynom

ial f

unct

ion

is o

dd if

and

onl

y

if a

ll th

e te

rms

are

of

degr

ees.

A p

olyn

omia

l fun

ctio

n is

eve

n

if a

nd o

nly

if a

ll th

e te

rms

are

of

degr

ees.

even

odd

x

f (x)

O1

2–2

–1

4 2 –2 –4

f(x) #

x3 !

3x


An

swer

s


Stu

dy

Gu

ide

and I

nte

rven

tion

The

Rem

aind

er a

nd F

acto

r The

orem

s

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-4

7-4

©G

lenc

oe/M

cGra

w-Hi

ll39

3G

lenc

oe A

lgeb

ra 2

Lesson 7-4

Syn

thet

ic S

ub

stit

uti

on

Rem

aind

erTh

e re

mai

nder

, whe

n yo

u di

vide

the

polyn

omia

l f(x

) by

(x"

a), i

s th

e co

nsta

nt f(

a).

Theo

rem

f(x) #

q(x)

+(x

"a)

!f(a

), wh

ere

q(x)

is a

pol

ynom

ial w

ith d

egre

e on

e le

ss th

an th

e de

gree

of 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

1Sy

nthe

tic

Subs

titu

tion

By

the

Rem

aind

er T

heor

em,f

("2)

sho

uld

be t

he r

emai

nder

whe

n yo

u di

vide

the

poly

nom

ial b

y x

!2.

"2

32

"5

1"

2"

68

"6

103

"4

3"

58

The

rem

aind

er is

8,s

o f(

"2)

#8.

Met

hod

2D

irec

t Su

bsti

tuti

onR

epla

ce x

wit

h "

2.f(

x) #

3x4

!2x

3"

5x2

!x

"2

f("

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)

#5x

3"

2x!

1,fi

nd

f(3

).A

gain

,by

the

Rem

aind

er T

heor

em,f

(3)

shou

ld b

e th

e re

mai

nder

whe

n yo

u di

vide

the

poly

nom

ial b

y x

"3.

35

02

"1

1545

141

515

4714

0T

he r

emai

nder

is 1

40,s

o f(

3) #

140.

Use

syn

thet

ic s

ubs

titu

tion

to

fin

d f

(!5)

an

d f

#$fo

r ea

ch f

un

ctio

n.

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

ubs

titu

tion

to

fin

d f

(4)

and

f(!

3) f

or e

ach

fu

nct

ion

.

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) #

3x4

"2x

3"

x2!

2x"

514

04;2

9862

7;27

711

.f(x

) #2x

4"

4x3

"x2

"6x

!3

12.f

(x) #

4x4

"4x

3!

3x2

"2x

"3

219;

282

805;

462

29 $ 1635 $ 83 $ 4

1 $ 2

Exer

cises

Exer

cises

©G

lenc

oe/M

cGra

w-Hi

ll39

4G

lenc

oe A

lgeb

ra 2

Fact

ors

of

Poly

no

mia

lsT

he F

acto

r T

heo

rem

can

help

you

fin

d al

l the

fac

tors

of

apo

lyno

mia

l.

Fact

or T

heor

emTh

e bi

nom

ial x

"a

is a

fact

or o

f the

pol

ynom

ial f

(x) i

f and

onl

y 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

The

orem

,the

bin

omia

l x!

5 is

a f

acto

r of

the

pol

ynom

ial i

f "5

is a

zer

o of

the

poly

nom

ial f

unct

ion.

To c

heck

thi

s,us

e sy

nthe

tic

subs

titu

tion

.

"5

12

"13

10"

515

"10

1"

32

0

Sinc

e th

e re

mai

nder

is 0

,x!

5 is

a f

acto

r of

the

pol

ynom

ial.

The

pol

ynom

ial

x3!

2x2

"13

x!

10 c

an b

e fa

ctor

ed a

s (x

!5)

(x2

"3x

!2)

.The

dep

ress

ed p

olyn

omia

l x2

"3x

!2

can

be f

acto

red

as (x

"2)

(x"

1).

So x

3!

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 bi

nom

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

Gu

ide

and I

nte

rven

tion

(c

onti

nued

)

The

Rem

aind

er a

nd F

acto

r The

orem

s

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-4

7-4

Exam

ple

Exam

ple

Exer

cises

Exer

cises



Skil

ls P

ract

ice

The

Rem

aind

er a

nd F

acto

r The

orem

s

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-4

7-4

©G

lenc

oe/M

cGra

w-Hi

ll39

5G

lenc

oe A

lgeb

ra 2

Lesson 7-4

Use

syn

thet

ic s

ubs

titu

tion

to

fin

d f

(2)

and

f(!

1) f

or e

ach

fu

nct

ion

.

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

"3x

3!

2x2

"2x

!6

2,14

11.f

(x) #

x5"

7x3

"4x

!10

12.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 bi

nom

ials

.

13.x

3!

2x2

"x

"2;

x!

114

.x3

!x2

"5x

!3;

x"

1x

!1,

x"

2x

!1,

x"

3

15.x

3!

3x2

"4x

"12

;x!

316

.x3

"6x

2!

11x

"6;

x"

3x

!2,

x"

2x

!1,

x!

2

17.x

3!

2x2

"33

x"

90;x

!5

18.x

3"

6x2

!32

;x"

4x

"3,

x!

6x

!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"

2x

!1,

x"

2x2

"x

"1

©G

lenc

oe/M

cGra

w-Hi

ll39

6G

lenc

oe A

lgeb

ra 2

Use

syn

thet

ic s

ubs

titu

tion

to

fin

d f

(!3)

an

d f

(4)

for

each

fu

nct

ion

.

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

"2x

!8

!31

,32

8.f(

x) #

x3"

x2!

4x"

4!

52,6

0

9.f(

x) #

x3!

3x2

!2x

"50

!56

,70

10.f

(x) #

x4!

x3"

3x2

"x

!12

42,2

80

11.f

(x) #

x4"

2x2

"x

!7

73,2

2712

.f(x

) #2x

4"

3x3

!4x

2"

2x!

128

6,37

7

13.f

(x) #

2x4

"x3

!2x

2"

2618

1,45

414

.f(x

) #3x

4"

4x3

!3x

2"

5x"

339

0,53

7

15.f

(x) #

x5!

7x3

"4x

"10

16.f

(x) #

x6!

2x5

"x4

!x3

"9x

2!

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 bi

nom

ials

.

17.x

3!

3x2

"6x

"8;

x"

218

.x3

!7x

2!

7x"

15;x

"1

x"

1,x

"4

x"

3,x

"5

19.x

3"

9x2

!27

x"

27;x

"3

20.x

3"

x2"

8x!

12;x

!3

x!

3,x

!3

x!

2,x

!2

21.x

3!

5x2

"2x

"24

;x"

222

.x3

"x2

"14

x!

24;x

!4

x"

3,x

"4

x!

3,x

!2

23.3

x3"

4x2

"17

x!

6;x

!2

24.4

x3"

12x2

"x

!3;

x"

3x

!3,

3x!

12x

!1,

2x"

1

25.1

8x3

!9x

2"

2x"

1;2x

!1

26.6

x3!

5x2

"3x

"2;

3x"

23x

"1,

3x!

12x

"1,

x"

1

27.x

5!

x4"

5x3

"5x

2!

4x!

4;x

!1

28.x

5"

2x4

!4x

3"

8x2

"5x

!10

;x"

2x

!1,

x"

1,x

!2,

x"

2x

!1,

x"

1,x2

"5

29.P

OPU

LATI

ON

The

pro

ject

ed p

opul

atio

n in

tho

usan

ds f

or a

cit

y ov

er t

he n

ext

seve

ral

year

s ca

n be

est

imat

ed b

y th

e fu

ncti

on P

(x) #

x3!

2x2

"8x

!52

0,w

here

xis

the

num

ber

of y

ears

sin

ce 2

000.

Use

syn

thet

ic s

ubst

itut

ion

to e

stim

ate

the

popu

lati

on

for

2005

.65

5,00

0

30.V

OLU

ME

The

vol

ume

of w

ater

in a

rec

tang

ular

sw

imm

ing

pool

can

be

mod

eled

by

the

poly

nom

ial 2

x3"

9x2

!7x

!6.

If t

he d

epth

of

the

pool

is g

iven

by

the

poly

nom

ial

2x!

1,w

hat

poly

nom

ials

exp

ress

the

leng

th a

nd w

idth

of

the

pool

?x

!3

and

x!

2

Pra

ctic

e (A

vera

ge)

The

Rem

aind

er a

nd F

acto

r The

orem

s

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-4

7-4


An

swer

s


Rea

din

g t

o L

earn

Math

emati

csTh

e Re

mai

nder

and

Fac

tor T

heor

ems

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-4

7-4

©G

lenc

oe/M

cGra

w-Hi

ll39

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

ucti

on t

o L

esso

n 7-

4 at

the

top

of

page

365

in y

our

text

book

.

Show

how

you

wou

ld u

se t

he m

odel

in t

he in

trod

ucti

on t

o es

tim

ate

the

num

ber

of in

tern

atio

nal t

rave

lers

(in

mill

ions

) to

the

Uni

ted

Stat

es in

the

year

200

0.(S

how

how

you

wou

ld s

ubst

itut

e nu

mbe

rs,b

ut d

o no

t ac

tual

lyca

lcul

ate

the

resu

lt.)

Sam

ple

answ

er:0

.02(

14)3

!0.

6(14

)2"

6(14

) "25

.9

Rea

din

g t

he

Less

on

1.C

onsi

der

the

follo

win

g sy

nthe

tic

divi

sion

.1

32

"6

43

5"

13

5"

13

a.U

sing

the

div

isio

n sy

mbo

l ,,w

rite

the

div

isio

n pr

oble

m t

hat

is r

epre

sent

ed b

y th

issy

nthe

tic

divi

sion

.(D

o no

t in

clud

e th

e an

swer

.)(3

x3"

2x2

!6x

"4)

((x

!1)

b.Id

enti

fy e

ach

of t

he f

ollo

win

g fo

r th

is d

ivis

ion.

divi

dend

divi

sor

quot

ient

rem

aind

er

c.If

f(x

) #3x

3!

2x2

"6x

!4,

wha

t is

f(1

)?3

2.C

onsi

der

the

follo

win

g sy

nthe

tic

divi

sion

."

31

00

27"

39

"27

1"

39

0

a.T

his

divi

sion

sho

ws

that

is

a f

acto

r of

.

b.T

he d

ivis

ion

show

s th

at

is a

zer

o of

the

pol

ynom

ial f

unct

ion

f(x)

#.

c.T

he d

ivis

ion

show

s th

at t

he p

oint

is

on

the

grap

h of

the

pol

ynom

ial

func

tion

f(x

) #.

Hel

pin

g Y

ou

Rem

emb

er

3.T

hink

of

a m

nem

onic

for

rem

embe

ring

the

sen

tenc

e,“D

ivid

end

equa

ls q

uoti

ent

tim

esdi

viso

r pl

us r

emai

nder

.”Sa

mpl

e an

swer

:Def

inite

ly e

very

qui

et te

ache

r des

erve

s pr

oper

rew

ards

.

x3"

27(!

3,0)

x3"

27!

3x3

"27

x "

3

33x

3"

5x!

1x

! 1

3x3

"2x

2!

6x"

4

©G

lenc

oe/M

cGra

w-Hi

ll39

8G

lenc

oe A

lgeb

ra 2

Usin

g M

axim

um V

alue

sM

any

tim

es m

axim

um s

olut

ions

are

nee

ded

for

diff

eren

t si

tuat

ions

.For

in

stan

ce,w

hat

is t

he a

rea

of t

he la

rges

t re

ctan

gula

r fi

eld

that

can

be

encl

osed

w

ith

2000

fee

t of

fen

cing

?

Let

xan

d y

deno

te t

he le

ngth

and

wid

th

of t

he f

ield

,res

pect

ivel

y.

Peri

met

er:2

x!

2y#

2000

→y

#10

00 "

xA

rea:

A#

xy#

x(10

00 "

x) #

"x2

!10

00x

Thi

s pr

oble

m is

equ

ival

ent

to f

indi

ng

the

high

est

poin

t on

the

gra

ph o

f A

(x) #

"x2

!10

00x

show

n on

the

rig

ht.

Com

plet

e th

e sq

uare

for

"x2

!10

00x.

A#

"(x

2"

1000

x!

5002

) !50

02

#"

(x"

500)

2!

5002

Bec

ause

the

ter

m "

(x"

500)

2is

eit

her

nega

tive

or

0,th

e gr

eate

st v

alue

of A

is 5

002 .

The

max

imum

are

a en

clos

ed is

50

02or

250

,000

squ

are

feet

.

Sol

ve e

ach

pro

blem

.

1.F

ind

the

area

of

the

larg

est

rect

angu

lar

gard

en t

hat

can

be e

nclo

sed

by

300

feet

of

fenc

e.

5625

ft2

2.A

far

mer

will

mak

e a

rect

angu

lar

pen

wit

h 10

0 fe

et o

f fe

nce

usin

g pa

rt

of h

is b

arn

for

one

side

of

the

pen.

Wha

t is

the

larg

est

area

he

can

encl

ose?

1250

ft2

3.A

n ar

ea a

long

a s

trai

ght

ston

e w

all i

s to

be

fenc

ed.T

here

are

600

met

ers

of f

enci

ng a

vaila

ble.

Wha

t is

the

gre

ates

t re

ctan

gula

r ar

ea t

hat

can

be

encl

osed

?

45,0

00 m

2

A

xO

1000

x

y

En

rich

men

t

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-4

7-4



Stu

dy

Gu

ide

and I

nte

rven

tion

Root

s an

d Ze

ros

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-5

7-5

©G

lenc

oe/M

cGra

w-Hi

ll39

9G

lenc

oe A

lgeb

ra 2

Lesson 7-5

Type

s of

Roo

tsT

he fo

llow

ing

stat

emen

ts a

re e

quiv

alen

t fo

r an

y po

lyno

mia

l fun

ctio

n f(

x).

•c

is a

zer

o of

the

pol

ynom

ial f

unct

ion

f(x)

.•

(x"

c) is

a f

acto

r of

the

pol

ynom

ial f

(x).

•c

is a

roo

t or

sol

utio

n of

the

pol

ynom

ial e

quat

ion

f(x)

#0.

If c

is r

eal,

then

(c,0

) is

an

inte

rcep

t of

the

gra

ph o

f f(x

).

Fund

amen

tal

Ever

y po

lynom

ial e

quat

ion

with

deg

ree

grea

ter t

han

zero

has

at l

east

one

root

in th

e se

tTh

eore

m o

f Alg

ebra

of c

ompl

ex n

umbe

rs.

Coro

llary

to th

e A

polyn

omia

l equ

atio

n of

the

form

P(x

) #0

of d

egre

e n

with

com

plex

coe

fficie

nts

has

Fund

amen

tal

exac

tly n

root

s in

the

set o

f com

plex

num

bers

.Th

eore

m o

f Alg

ebra

s

If P

(x) i

s a

polyn

omia

l with

real

coe

fficie

nts

whos

e te

rms

are

arra

nged

in d

esce

ndin

gpo

wers

of t

he v

aria

ble,

Desc

arte

s’Ru

le•

the

num

ber o

f pos

itive

real

zer

os o

f y#

P(x

) is

the

sam

e as

the

num

ber o

f cha

nges

in

of S

igns

sign

of th

e co

effic

ient

s of

the

term

s, o

r is

less

than

this

by a

n ev

en n

umbe

r, an

d•

the

num

ber o

f neg

ative

real

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os o

f y#

P(x

) is

the

sam

eas

the

num

ber o

f cha

nges

in

sign

of th

e co

effic

ient

s of

the

term

s of

P("

x), o

r is

less

than

this

num

ber b

y an

eve

nnu

mbe

r.

Sol

ve t

he

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atio

n

6x3

"3x

#0

and

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te t

he

nu

mbe

r an

d t

ype

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the

Zer

o P

rodu

ct P

rope

rty.

3x#

0or

2x2

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x#

0or

2x2

#"

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equ

atio

n ha

s on

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al r

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0,

and

two

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inar

y ro

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*.

i"2!

$2

i"2!

$2

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te t

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nu

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r of

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itiv

ere

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,neg

ativ

e re

al z

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mag

inar

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ros

for

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x) h

as d

egre

e 4,

it h

as 4

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os.

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es’ R

ule

of S

igns

to

dete

rmin

e th

enu

mbe

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d ty

pe o

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l zer

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ince

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,the

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ind

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ere

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ne s

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chan

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ly 1

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tive

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l zer

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ple1

Exam

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Exer

cises

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ve e

ach

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atio

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al z

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ativ

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mag

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.

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x) #

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juga

teSu

ppos

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al n

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rs w

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)0.

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!bi

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zero

of a

pol

ynom

ial

Theo

rem

func

tion

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fficie

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a"

biis

also

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ero

of th

e fu

nctio

n.

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d a

ll o

f th

e ze

ros

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(x)

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ree

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os.

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ther

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s 3

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itiv

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alze

ros.

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e th

ere

is 1

sig

n ch

ange

for

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f("

x),t

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unct

ion

has

1 ne

gati

vere

al z

ero.

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syn

thet

ic s

ubst

itut

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to t

est

som

e po

ssib

le z

eros

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21

0"

1538

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24

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31

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39

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he p

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l fun

ctio

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ow t

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ynth

etic

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stit

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n ag

ain

to f

ind

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roof

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dep

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ed p

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l.

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13

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0

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5 is

ano

ther

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o.U

se t

he Q

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atic

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mul

a on

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l x2

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to f

ind

the

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r 2

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s,1

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os a

t 3

and

"5

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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!

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9!

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f(x)

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dy

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ide

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nte

rven

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onti

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)

Root

s an

d Ze

ros

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-5

7-5

Exam

ple

Exam

ple

Exer

cises

Exer

cises


An

swer

s


Skil

ls P

ract

ice

Root

s an

d Ze

ros

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-5

7-5

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lenc

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Lesson 7-5

Sol

ve e

ach

equ

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n.S

tate

th

e n

um

ber

and

typ

e of

roo

ts.

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inar

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real

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agin

ary

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te t

he

pos

sibl

e n

um

ber

of p

osit

ive

real

zer

os,n

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ive

real

zer

os,a

nd

imag

inar

y ze

ros

of e

ach

fu

nct

ion

.

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x) #

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x) #

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x!

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2 or

0;1

;2 o

r 0

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x) #

x3"

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(x) #

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;0;2

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3 or

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r 0

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1;1;

22

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;2 o

r 0;4

or 2

or 0

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d a

ll t

he

zero

s of

eac

h f

un

ctio

n.

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(x) #

x3"

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fun

ctio

n o

f le

ast

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ree

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1,1,

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e n

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ber

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os,a

nd

imag

inar

y ze

ros

of e

ach

fu

nct

ion

.

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x) #

4x3

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x!

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r 0;4

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r 0;4

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r 0

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d a

ll t

he

zero

s of

eac

h f

un

ctio

n.

11.h

(x) #

2x3

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2"

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(x) #

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(x) #

x3"

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) #x4

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x2!

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2 !

i!

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(x) #

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2"

14x

"8

16.f

(x) #

x4"

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!6x

2!

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"40

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2,!

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3 !

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"i

Wri

te a

pol

ynom

ial

fun

ctio

n o

f le

ast

deg

ree

wit

h i

nte

gral

coe

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ien

ts t

hat

has

th

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ven

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os.

17."

5,3i

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2,3

!i

f(x)

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2"

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x) #

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19."

1,4,

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x) #

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20

21.C

RA

FTS

Step

han

has

a se

t of

pla

ns t

o bu

ild a

woo

den

box.

He

wan

ts t

o re

duce

the

volu

me

of t

he b

ox t

o 10

5 cu

bic

inch

es.H

e w

ould

like

to

redu

ce t

he le

ngth

of

each

dim

ensi

on in

the

pla

n by

the

sam

e am

ount

.The

pla

ns c

all f

or t

he b

ox t

o be

10

inch

es b

y8

inch

es b

y 6

inch

es.W

rite

and

sol

ve a

pol

ynom

ial e

quat

ion

to f

ind

out

how

muc

hSt

ephe

n sh

ould

tak

e fr

om e

ach

dim

ensi

on.

(10

!x)

(8 !

x)(6

!x)

#10

5;3

in.

Pra

ctic

e (A

vera

ge)

Root

s an

d Ze

ros

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-5

7-5



Rea

din

g t

o L

earn

Math

emati

csRo

ots

and

Zero

s

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-5

7-5

©G

lenc

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w-Hi

ll40

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ra 2

Lesson 7-5

Pre-

Act

ivit

yH

ow c

an t

he

root

s of

an

equ

atio

n b

e u

sed

in

ph

arm

acol

ogy?

Rea

d th

e in

trod

ucti

on t

o L

esso

n 7-

5 at

the

top

of

page

371

in y

our

text

book

.

Usi

ng t

he m

odel

giv

en in

the

intr

oduc

tion

,wri

te a

pol

ynom

ial e

quat

ion

wit

h 0

on o

ne s

ide

that

can

be

solv

ed t

o fi

nd t

he t

ime

or t

imes

at

whi

chth

ere

is 1

00 m

illig

ram

s of

med

icat

ion

in a

pat

ient

’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

heth

er e

ach

stat

emen

t is

tru

eor

fal

se.

a.E

very

pol

ynom

ial e

quat

ion

of d

egre

e gr

eate

r th

an o

ne h

as a

t le

ast

one

root

in t

he s

etof

rea

l num

bers

.fa

lse

b.If

cis

a r

oot

of t

he p

olyn

omia

l equ

atio

n f(

x) #

0,th

en (

x"

c) is

a f

acto

r of

the

poly

nom

ial f

(x).

true

c.If

(x!

c) is

a f

acto

r of

the

pol

ynom

ial f

(x),

then

cis

a z

ero

of t

he p

olyn

omia

l fu

ncti

on f

.fa

lse

d.A

pol

ynom

ial f

unct

ion

fof

deg

ree

nha

s ex

actl

y (n

"1)

com

plex

zer

os.

fals

e

2.L

et f

(x) #

x6"

2x5

!3x

4"

4x3

!5x

2!

6x"

7.

a.W

hat

are

the

poss

ible

num

bers

of

posi

tive

rea

l zer

os o

f f?

5,3,

or 1

b.W

rite

f("

x) in

sim

plif

ied

form

(w

ith

no p

aren

thes

es).

x6"

2x5

"3x

4"

4x3

"5x

2!

6x!

7W

hat

are

the

poss

ible

num

bers

of

nega

tive

rea

l zer

os o

f f?

1c.

Com

plet

e th

e fo

llow

ing

char

t to

sho

w t

he p

ossi

ble

com

bina

tion

s of

pos

itiv

e re

al z

eros

,ne

gati

ve r

eal z

eros

,and

imag

inar

y ze

ros

of t

he p

olyn

omia

l fun

ctio

n f.

Num

ber o

fNu

mbe

r of

Num

ber o

f To

tal N

umbe

r Po

sitiv

e Re

al Z

eros

Nega

tive

Real

Zer

osIm

agin

ary

Zero

sof

Zer

os

51

06

31

26

11

46

Hel

pin

g Y

ou

Rem

emb

er

3.It

is e

asie

r to

rem

embe

r m

athe

mat

ical

con

cept

s an

d re

sult

s if

you

rel

ate

them

to

each

othe

r.H

ow c

an t

he C

ompl

ex C

onju

gate

s T

heor

em h

elp

you

rem

embe

r th

e pa

rt o

fD

esca

rtes

’ Rul

e of

Sig

ns t

hat

says

,“or

is le

ss t

han

this

num

ber

by a

n ev

en n

umbe

r.”Sa

mpl

e an

swer

:For

a p

olyn

omia

l fun

ctio

n in

whi

ch th

e po

lyno

mia

l has

real

coe

ffici

ents

,im

agin

ary

zero

s co

me

in c

onju

gate

pai

rs.T

here

fore

,the

rem

ust b

e an

eve

n nu

mbe

r of i

mag

inar

y ze

ros.

For e

ach

pair

of im

agin

ary

zero

s,th

e nu

mbe

r of p

ositi

ve o

r neg

ativ

e ze

ros

decr

ease

s by

2.

©G

lenc

oe/M

cGra

w-Hi

ll40

4G

lenc

oe A

lgeb

ra 2

The

Bise

ctio

n M

etho

d fo

r App

roxi

mat

ing

Real

Zer

osT

he b

isec

tion

met

hod

can

be u

sed

to a

ppro

xim

ate

zero

s of

pol

ynom

ial

func

tion

s lik

e f(

x) #

x3!

x2"

3x"

3.

Sinc

e f(

1) #

"4

and

f(2)

#3,

ther

e is

at

leas

t on

e re

al z

ero

betw

een

1 an

d 2.

The

mid

poin

t of

thi

s in

terv

al is

$1! 2

2$

#1.

5.Si

nce

f(1.

5) #

"1.

875,

the

zero

is

betw

een

1.5

and

2.T

he m

idpo

int

of t

his

inte

rval

is $1.

5 2!2

$#

1.75

.Sin

ce

f(1.

75) i

s ab

out

0.17

2,th

e ze

ro is

bet

wee

n 1.

5 an

d 1.

75.T

he m

idpo

int

of t

his

inte

rval

is $1.

5! 2

1.75

$#

1.62

5 an

d f(

1.62

5) is

abo

ut "

0.94

.The

zer

o is

bet

wee

n

1.62

5 an

d 1.

75.T

he m

idpo

int

of t

his

inte

rval

is $1.

625

2!1.

75$

#1.

6875

.Sin

ce

f(1.

6875

) is

abo

ut "

0.41

,the

zer

o is

bet

wee

n 1.

6875

and

1.7

5.T

here

fore

,the

ze

ro is

1.7

to

the

near

est

tent

h.

The

dia

gram

bel

ow s

umm

ariz

es t

he r

esul

ts o

btai

ned

by t

he b

isec

tion

met

hod.

Usi

ng

the

bise

ctio

n 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

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-5

7-5


An

swer

s


Stu

dy

Gu

ide

and I

nte

rven

tion

Ratio

nal Z

ero

Theo

rem

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-6

7-6

©G

lenc

oe/M

cGra

w-Hi

ll40

5G

lenc

oe A

lgeb

ra 2

Lesson 7-6

Iden

tify

Rat

ion

al Z

ero

s

Ratio

nal Z

ero

Let f

(x) #

a 0xn

!a 1

xn"

1!

… !

a n"

2x2

!a n

"1x

!an

repr

esen

t a p

olyn

omia

l fun

ctio

n Th

eore

mwi

th in

tegr

al c

oeffi

cient

s. If

$p q$is

a ra

tiona

l num

ber i

n sim

ples

t for

m a

nd is

a z

ero

of y

#f(x

),th

en p

is a

fact

or o

f an

and

qis

a fa

ctor

of a

0.

Coro

llary

(Int

egra

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 ra

tiona

l Ze

ro T

heor

em)

zero

s of

the

func

tion

mus

t be

fact

ors

of a

n.

Lis

t al

l of

th

e p

ossi

ble

rati

onal

zer

os o

f ea

ch f

un

ctio

n.

a.f(

x) #

3x4

!2x

2"

6x!

10

If $p q$

is a

rat

iona

l roo

t,th

en p

is a

fac

tor

of "

10 a

nd q

is a

fac

tor

of 3

.The

pos

sibl

e va

lues

fo

r p

are

*1,

*2,

*5,

and

*10

.The

pos

sibl

e va

lues

for

qar

e *

1 an

d *

3.So

all

of t

he

poss

ible

rat

iona

l 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

Sinc

e th

e co

effi

cien

t of

x3

is 1

,the

pos

sibl

e ra

tion

al z

eros

mus

t be

the

fac

tors

of

the

cons

tant

ter

m "

36.S

o th

e po

ssib

le r

atio

nal z

eros

are

*1,

*2,

*3,

*4,

*6,

*9,

*12

,*18

,an

d *

36.

Lis

t al

l of

th

e p

ossi

ble

rati

onal

zer

os o

f ea

ch f

un

ctio

n.

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"

7x3

"4x

2!

x"

494.

p(x)

#2x

4"

5x3

!8x

2!

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) #

6x5

"3x

4!

12x3

!18

x2"

9x!

2110

.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-Hi

ll40

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 c

orol

lary

to

the

Fun

dam

enta

l The

orem

of A

lgeb

ra,w

e kn

ow t

hat

ther

e ar

e ex

actl

y 3

com

plex

roo

ts.A

ccor

ding

to

Des

cart

es’ R

ule

of S

igns

the

re a

re 2

or

0 po

siti

ve

real

roo

ts a

nd 1

neg

ativ

e re

al r

oot.

The

pos

sibl

e ra

tion

al z

eros

are

*1,

*2,

*3,

*4,

*6,

*12

,*

,*,*

,*,*

,*.M

ake

a ta

ble

and

test

som

e po

ssib

le r

atio

nal z

eros

.

Sinc

e f(

1) #

0,yo

u kn

ow t

hat

x#

1 is

a z

ero.

The

dep

ress

ed p

olyn

omia

l is

5x2

!17

x"

12,w

hich

can

be

fact

ored

as

(5x

"3)

(x!

4).

By

the

Zero

Pro

duct

Pro

pert

y,th

is e

xpre

ssio

n eq

uals

0 w

hen

x#

or x

#"

4.T

he r

atio

nal z

eros

of

this

fun

ctio

n ar

e 1,

,and

"4.

Fin

d a

ll o

f th

e ze

ros

of f

(x)

#8x

4"

2x3

"5x

2"

2x!

3.T

here

are

4 c

ompl

ex r

oots

,wit

h 1

posi

tive

rea

l roo

t an

d 3

or 1

neg

ativ

e re

al r

oots

.The

po

ssib

le r

atio

nal z

eros

are

*1,

*3,

*,*

,*,*

,*,a

nd *

.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

Gu

ide

and I

nte

rven

tion

(c

onti

nued

)

Ratio

nal Z

ero

Theo

rem

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

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.

Sinc

e 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

sub

stit

utio

n ag

ain.

Any

rem

aini

ngra

tion

al r

oots

mus

t be

neg

ativ

e.

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

The

zer

os o

f th

is f

unct

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

Ratio

nal Z

ero

Theo

rem

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-6

7-6

©G

lenc

oe/M

cGra

w-Hi

ll40

7G

lenc

oe A

lgeb

ra 2

Lesson 7-6

Lis

t al

l of

th

e p

ossi

ble

rati

onal

zer

os o

f ea

ch f

un

ctio

n.

1.n(

x) #

x2!

5x!

32.

h(x)

#x2

"2x

"5

'1,

'3

'1,

'5

3.w

(x) #

x2"

5x!

124.

f(x)

#2x

2!

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,

39.

p(x)

#x3

"x2

!x

"1

10.z

(x) #

x3"

4x2

!6x

"4

12

11.h

(x) #

x3"

x2!

4x"

412

.g(x

) #3x

3"

9x2

"10

x"

8

14

13.g

(x) #

2x3

!7x

2"

7x"

1214

.h(x

) #2x

3"

5x2

"4x

!3

!4,

!1,

$3 2$!

1,$1 2$ ,

315

.p(x

) #3x

3"

5x2

"14

x"

4 #

016

.q(x

) #3x

3!

2x2

!27

x!

18

!$1 3$

!$2 3$

17.q

(x) #

3x3

"7x

2!

418

.f(x

) #x4

"2x

3"

13x2

!14

x!

24

!$2 3$ ,

1,2

!3,

!1,

2,4

19.p

(x) #

x4"

5x3

"9x

2"

25x

"70

20.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!

5x2

!11

x!

1522

.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

!4x

3!

5x2

!4x

!4

$1 3$ ,$3 2$ ,

,!

2,!

2,!

i,i

1 !

!5"

$2

1 "

!5"

$2

©G

lenc

oe/M

cGra

w-Hi

ll40

8G

lenc

oe A

lgeb

ra 2

Lis

t al

l of

th

e p

ossi

ble

rati

onal

zer

os o

f ea

ch f

un

ctio

n.

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

"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"

49x2

0,!

7,7

11.h

(x) #

x3"

7x2

!17

x"

153

12.b

(x) #

x3!

6x!

20!

2

13.f

(x) #

x3"

6x2

!4x

"24

614

.g(x

) #2x

3!

3x2

"4x

"4

!2

15.h

(x) #

2x3

"7x

2"

21x

!54

#0!3,

2,$9 2$

16.z

(x) #

x4"

3x3

!5x

2"

27x

"36

!1,

4

17.d

(x) #

x4!

x3!

16no

ratio

nal z

eros

18.n

(x) #

x4"

2x3

"3

!1

19.p

(x) #

2x4

"7x

3!

4x2

!7x

"6

20.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) #

2x4

!7x

3"

2x2

"19

x"

1222

.q(x

) #x4

"4x

3!

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.T

RA

VEL

The

hei

ght

of a

box

tha

t Jo

an is

shi

ppin

g is

3 in

ches

less

tha

n th

e w

idth

of

the

box.

The

leng

th is

2 in

ches

mor

e th

an t

wic

e th

e w

idth

.The

vol

ume

of t

he b

ox is

154

0 in

3 .W

hat

are

the

dim

ensi

ons

of t

he b

ox?

22 in

.by

10 in

.by

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

the

vol

ume

of t

he p

yram

id is

432

m3 ,

how

tal

l is

it?

Use

the

for

mul

a V

#$1 3$ B

h.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 (A

vera

ge)

Ratio

nal Z

ero

Theo

rem

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-6

7-6


An

swer

s


Rea

din

g t

o L

earn

Math

emati

csRa

tiona

l Zer

o Th

eore

m

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-6

7-6

©G

lenc

oe/M

cGra

w-Hi

ll40

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

blem

s in

volv

ing

larg

en

um

bers

?

Rea

d th

e in

trod

ucti

on t

o L

esso

n 7-

6 at

the

top

of

page

378

in y

our

text

book

.

Rew

rite

the

pol

ynom

ial e

quat

ion

w(w

!8)

(w"

5) #

2772

in t

he f

orm

f(

x) #

0,w

here

f(x

) is

a p

olyn

omia

l wri

tten

in d

esce

ndin

g po

wer

s of

x.

w3

"3w

2!

40w

!27

72 #

0

Rea

din

g t

he

Less

on

1.Fo

r ea

ch o

f th

e fo

llow

ing

poly

nom

ial f

unct

ions

,lis

t al

l the

pos

sibl

e va

lues

of p

,all

the

poss

ible

val

ues

of q

,and

all

the

poss

ible

rat

iona

l zer

os $p q$.

a.f(

x) #

x3"

2x2

"11

x!

12

poss

ible

val

ues

of p

:'

1,'

2,'

3,'

4,'

6,'

12po

ssib

le v

alue

s of

q:

'1

poss

ible

val

ues

of $p q$:

'1,

'2,

'3,

'4,

'6,

'12

b.f(

x) #

2x4

!9x

3"

23x2

"81

x!

45

poss

ible

val

ues

of p

:'

1,'

3,'

5,'

9,'

15,'

45po

ssib

le v

alue

s 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 in

you

r ow

n w

ords

how

Des

cart

es’ R

ule

of S

igns

,the

Rat

iona

l Zer

o T

heor

em,a

ndsy

nthe

tic

divi

sion

can

be

used

tog

ethe

r to

fin

d al

l of

the

rati

onal

zer

os o

f a

poly

nom

ial

func

tion

wit

h in

tege

r co

effi

cien

ts.

Sam

ple

answ

er:U

se D

esca

rtes

’Rul

e to

find

the

poss

ible

num

bers

of

posi

tive

and

nega

tive

real

zer

os.U

se th

e Ra

tiona

l Zer

o Th

eore

m to

list

all

poss

ible

ratio

nal z

eros

.Use

syn

thet

ic d

ivis

ion

to te

st w

hich

of t

henu

mbe

rs o

n th

e lis

t of p

ossi

ble

ratio

nal z

eros

are

act

ually

zer

os o

f the

poly

nom

ial f

unct

ion.

(Des

cart

es’R

ule

may

hel

p yo

u to

lim

it th

epo

ssib

ilitie

s.)

Hel

pin

g Y

ou

Rem

emb

er

3.So

me

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 in

the

den

omin

ator

s w

hen

form

ing

a lis

t of

pos

sibl

e ra

tion

al z

eros

of

a po

lyno

mia

lfu

ncti

on.H

ow c

an y

ou u

se t

he li

near

pol

ynom

ial e

quat

ion

ax!

b#

0,w

here

aan

d b

are

nonz

ero

inte

gers

,to

rem

embe

r th

is?

Sam

ple

answ

er:T

he s

olut

ion

of th

e eq

uatio

n is

!$b a$ .

The

num

erat

or

bis

a fa

ctor

of t

he c

onst

ant t

erm

in a

x"

b.Th

e de

nom

inat

or a

is a

fact

orof

the

lead

ing

coef

ficie

nt in

ax

"b.

©G

lenc

oe/M

cGra

w-Hi

ll41

0G

lenc

oe A

lgeb

ra 2

Infin

ite C

ontin

ued

Frac

tions

Som

e in

fini

te e

xpre

ssio

ns a

re a

ctua

lly e

qual

to

real

num

bers

! The

infi

nite

con

tinu

ed f

ract

ion

at t

he r

ight

ison

e ex

ampl

e.

If y

ou u

se x

to s

tand

for

the

infi

nite

fra

ctio

n,th

en t

heen

tire

den

omin

ator

of

the

firs

t fr

acti

on o

n th

e ri

ght

isal

so e

qual

to

x.T

his

obse

rvat

ion

lead

s to

the

fol

low

ing

equa

tion

:

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 m

ore

term

s yo

u ad

d to

the

fra

ctio

ns a

bove

,the

clo

ser

thei

r va

lue

appr

oach

es t

he v

alue

of

the

infi

nite

con

tinu

ed f

ract

ion.

Wha

t va

lue

do t

he

frac

tion

s se

em t

o be

app

roac

hing

?ab

out 1

.6

7.R

ewri

te x

#1

!$1 x$

as a

qua

drat

ic e

quat

ion

and

solv

e fo

r x.

x2!

x!

1 #

0;x

#;x

%1.

618

or !

0.61

8 (T

he p

ositi

ve ro

ot is

the

valu

e of

the

infin

ite fr

actio

n,be

caus

e th

e or

igin

al fr

actio

n is

cle

arly

not

neg

ativ

e.)

8.F

ind

the

valu

e of

the

fol

low

ing

infi

nite

con

tinu

ed f

ract

ion.

3 !

x#

3 "

$1 x$ ;x

#or

abo

ut 3

.30

3 "

!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

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-6

7-6

x#

1 !

1

1 !

1

1 !

1

1 !

11

!…



Stu

dy

Gu

ide

and I

nte

rven

tion

Ope

ratio

ns o

n Fu

nctio

ns

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-7

7-7

©G

lenc

oe/M

cGra

w-Hi

ll41

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)Di

ffere

nce

(f"

g)(x

) #f(x

) "g(

x)O

pera

tions

with

Fun

ctio

nsPr

oduc

t(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)Ad

ditio

n 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)

Subt

ract

ion

of fu

nctio

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 fu

nctio

ns

#(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)Di

strib

utive

Pro

perty

#3x

3"

2x2

!9x

2"

6x"

12x

!8

Dist

ribut

ive P

rope

rty

#3x

3!

7x2

"18

x!

8Si

mpl

ify.

#$(x

)#

Divis

ion

of fu

nctio

ns

#,x

)$2 3$

f(x) #

x2!

3x"

4 an

d 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"

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

!11

x"

43x

2"

x"

2;3x

2!

3x"

8;3x

2"

13x

!5;

!3x

2!

9x"

3;6x

3!

11x2

"13

x!

15;

6x3

"19

x2!

19x

"4;

,x*

,x*

,!4

5.f(

x) #

x2"

1;g(

x) #

x2!

1 "

;x2

!1

!;x

!1;

x3"

x2!

x!

1,x

*!

11

$ x"

11

$ 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-Hi

ll41

2G

lenc

oe A

lgeb

ra 2

Co

mp

osi

tio

n o

f Fu

nct

ion

s

Com

posi

tion

Supp

ose

fand

gar

e fu

nctio

ns s

uch

that

the

rang

e of

gis

a su

bset

of t

he d

omai

n of

f.of

Fun

ctio

nsTh

en th

e co

mpo

site

func

tion

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

#(3

x"

4)2

"1

#3x

2"

7#

9x2

"24

x!

16 "

1#

9x2

"24

x!

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!g

does

not

exi

st;

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) #

!4x

4"

8x2

!4

5.f(

x) #

x2!

2x;g

(x) #

x"

96.

f(x)

#5x

!4;

g(x)

#3

"x

[f!

g](x

) #x2

!16

x"

63,

[f!

g](x

) #19

!5x

,[g

!f]

(x) #

x2"

2x!

9[g

!f]

(x) #

!1

!5x

Stu

dy

Gu

ide

and I

nte

rven

tion

(c

onti

nued

)

Ope

ratio

ns o

n Fu

nctio

ns

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-7

7-7

Exam

ple1

Exam

ple1

Exam

ple2

Exam

ple2

Exer

cises

Exer

cises


An

swer

s


Skil

ls P

ract

ice

Ope

ratio

ns o

n Fu

nctio

ns

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-7

7-7

©G

lenc

oe/M

cGra

w-Hi

ll41

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) #

x2x2

!x

"4;

x2"

x!

4;4.

f(x)

#3x

2$3x

3 x"5

$,x

*0;

$3x3 x!

5$

,x*

0;g(

x) #

4 "

x4x

2!

x3;

,x*

4g(

x) #

$5 x$15

x,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!

12x

"3;

!12

x!

1h(

x) #

x!

2h(

x) #

4x"

1

11.g

(x) #

x"

6x;

x12

.g(x

) #x

"3

x2!

3;x2

!6x

"9

h(x)

#x

!6

h(x)

#x2

13.g

(x) #

5x5x

2"

5x!

5;14

.g(x

) #x

!2

2x2

!1;

2x2

"8x

"5

h(x)

#x2

!x

"1

25x2

"5x

!1

h(x)

#2x

2"

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-Hi

ll41

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" 2

1$

,x*

0;2x

2"

7x"

3;7x

"21

;

2x2

!5x

!3;

$8x4 x2!

1$

,x*

0;x4

"7x

3"

3x2

!63

x!

108;

$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 e

xist

;{(!

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

!6

h(x)

#x

"4

h(x)

#2x

!3

h(x)

#3x

23x

2"

6;3x

!12

;3x

!4

!16

x!

24;!

16x

"3

3x2

"36

x"

108

11.g

(x) #

x!

312

.g(x

) #"

2x13

.g(x

) #x

"2

h(x)

#2x

2h(

x) #

x2!

3x!

2h(

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)]

2017

.f[h

("9)

]25

18.h

[g("

3)]

!11

19.g

[f(8

)]32

020

.h[f

(20)

]40

421

.[f

!(h

!g)

]("

1)1

22.[

f!

(g!

h)](

4)16

0023

.BU

SIN

ESS

The

fun

ctio

n f(

x) #

1000

"0.

01x2

mod

els

the

man

ufac

turi

ng c

ost

per

item

whe

n x

item

s ar

e pr

oduc

ed,a

nd g

(x) #

150

"0.

001x

2m

odel

s th

e se

rvic

e co

st p

er it

em.

Wri

te a

fun

ctio

n C

(x)

for

the

tota

l man

ufac

turi

ng a

nd s

ervi

ce c

ost

per

item

.C

(x) #

1150

!0.

011x

2

24.M

EASU

REM

ENT

The

for

mul

a f

#$ 1n 2$

conv

erts

inch

es n

to f

eet

f,an

d m

#$ 52

f 80$co

nver

ts

feet

to

mile

s m

.Wri

te a

com

posi

tion

of

func

tion

s th

at c

onve

rts

inch

es t

o m

iles.

[m!

f]n

#$ 63

,n 360

$

1 $ x2

Pra

ctic

e (A

vera

ge)

Ope

ratio

ns o

n Fu

nctio

ns

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-7

7-7



Rea

din

g t

o L

earn

Math

emati

csO

pera

tions

on

Func

tions

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-7

7-7

©G

lenc

oe/M

cGra

w-Hi

ll41

5G

lenc

oe A

lgeb

ra 2

Lesson 7-7

Pre-

Act

ivit

yW

hy is

it im

port

ant

to c

ombi

ne f

unct

ions

in b

usin

ess?

Rea

d th

e in

trod

ucti

on t

o L

esso

n 7-

7 at

the

top

of

page

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 s

ale

of

50 b

irdh

ouse

s.(D

o no

t ac

tual

ly c

alcu

late

her

pro

fit.

)Sa

mpl

e an

swer

:1.

Find

the

reve

nue

by s

ubst

itutin

g 50

for x

in th

e ex

pres

sion

125x

.Nex

t,fin

d th

e co

st b

y su

bstit

utin

g 50

for x

in th

eex

pres

sion

65x

"54

00.F

inal

ly,su

btra

ct th

e co

st fr

om th

ere

venu

e to

find

the

prof

it.2.

Form

the

prof

it fu

nctio

n p(

x) #

r(x)

!c(

x) #

125x

!(6

5x"

5400

) #60

x!

5400

.Su

bstit

ute

50 fo

r xin

the

expr

essi

on 6

0x!

5400

.

Rea

din

g t

he

Less

on

1.D

eter

min

e w

heth

er e

ach

stat

emen

t is

tru

eor

fal

se.(

Rem

embe

r th

at t

rue

mea

ns

alw

ays

true

.)

a.If

fan

d g

are

poly

nom

ial f

unct

ions

,the

n f

!g

is a

pol

ynom

ial f

unct

ion.

true

b.If

fan

d g

are

poly

nom

ial f

unct

ions

,the

n is

a p

olyn

omia

l fun

ctio

n.fa

lse

c.If

fan

d g

are

poly

nom

ial f

unct

ions

,the

dom

ain

of t

he f

unct

ion

f+

gis

the

set

of

all

real

num

bers

.tru

ed.

If f

(x) #

3x!

2 an

d g(

x) #

x"

4,th

e do

mai

n of

the

fun

ctio

n is

the

set

of

all r

eal

num

bers

.fa

lse

e.If

fan

d g

are

poly

nom

ial f

unct

ions

,the

n (f

!g)

(x) #

(g!

f)(x

).fa

lse

f.If

fan

d g

are

poly

nom

ial f

unct

ions

,the

n (f

+g)

(x) #

(g+

f)(x

)tru

e

2.L

et f

(x) #

2x"

5 an

d g(

x) #

x2!

1.

a.E

xpla

in in

wor

ds h

ow y

ou w

ould

fin

d (f

!g)

("3)

.(D

o no

t ac

tual

ly d

o an

y ca

lcul

atio

ns.)

Sam

ple

answ

er:S

quar

e !

3 an

d ad

d 1.

Take

the

num

ber y

ou g

et,

mul

tiply

it b

y 2,

and

subt

ract

5.

b.E

xpla

in in

wor

ds h

ow y

ou w

ould

fin

d (g

!f)

("3)

.(D

o no

t ac

tual

ly d

o an

yca

lcul

atio

ns.)

Sam

ple

answ

er:M

ultip

ly !

3 by

2 a

nd s

ubtra

ct 5

.Tak

e th

enu

mbe

r you

get

,squ

are

it,an

d ad

d 1.

Hel

pin

g Y

ou

Rem

emb

er

3.So

me

stud

ents

hav

e tr

oubl

e re

mem

beri

ng t

he c

orre

ct o

rder

in w

hich

to

appl

y th

e tw

oor

igin

al f

unct

ions

whe

n ev

alua

ting

a c

ompo

site

fun

ctio

n.W

rite

thr

ee s

ente

nces

,eac

h of

whi

ch e

xpla

ins

how

to

do t

his

in a

slig

htly

dif

fere

nt w

ay.(

Hin

t:U

se t

he w

ord

clos

est

inth

e fi

rst

sent

ence

,the

wor

ds in

side

and

outs

ide

in t

he s

econ

d,an

d th

e w

ords

left

and

righ

tin

the

thi

rd.)

Sam

ple

answ

er:1

.The

func

tion

that

is w

ritte

n cl

oses

t to

the

varia

ble

is a

pplie

d fir

st.2

.Wor

k fro

m th

e in

side

to th

e ou

tsid

e.3.

Wor

k fro

m ri

ght t

o le

ft.

f $ g

f $ g

©G

lenc

oe/M

cGra

w-Hi

ll41

6G

lenc

oe A

lgeb

ra 2

Rela

tive

Max

imum

Val

ues

The

gra

ph o

f f(x

) #x3

"6x

"9

show

s a

rela

tive

max

imum

val

ue s

omew

here

be

twee

n f(

"2)

and

f("

1).Y

ou c

an o

btai

n a

clos

er a

ppro

xim

atio

n by

com

pari

ng

valu

es s

uch

as t

hose

sho

wn

in t

he t

able

.

To t

he n

eare

st t

enth

a r

elat

ive

max

imum

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

.02.

f(x)

#x3

"3x

"3

rel.

max

.of !

1.0

3.f(

x) #

x3"

9x"

2re

l.m

ax.o

f 8.4

4.f(

x) #

x3!

2x2

"12

x"

24re

l.m

ax.o

f 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.

375

"1.

4"

3.34

4"

1.3

"3.

397

"1

"4

x

f(x)

O2

–2–4 –8 –12

–16

–20

–44

f(x)

# x

3 ! 6

x !

9

En

rich

men

t

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-7

7-7


An

swer

s


Stu

dy

Gu

ide

and I

nte

rven

tion

Inve

rse

Func

tions

and

Rel

atio

ns

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-8

7-8

©G

lenc

oe/M

cGra

w-Hi

ll41

7G

lenc

oe A

lgeb

ra 2

Lesson 7-8

Fin

d In

vers

es

Inve

rse

Rela

tions

Two

rela

tions

are

inve

rse

rela

tions

if a

nd o

nly

if wh

enev

er o

ne re

latio

n co

ntai

ns th

e el

emen

t (a,

b),

the

othe

r rel

atio

n co

ntai

ns th

e el

emen

t (b,

a).

Prop

erty

of I

nver

se

Supp

ose

fand

f"1

are

inve

rse

func

tions

.Fu

nctio

nsTh

en 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.

Step

1R

epla

ce f

(x)

wit

h y

in t

he o

rigi

nal e

quat

ion.

f(x)

#$2 5$ x

"→

y#

$2 5$ x"

Step

2In

terc

hang

e x

and

y.

x#

$2 5$ y"

Step

3So

lve

for

y.

x#

$2 5$ y"

Inve

rse

5x#

2y"

1M

ultip

ly ea

ch s

ide

by 5

.

5x!

1 #

2yAd

d 1

to e

ach

side.

(5x

!1)

#y

Divid

e ea

ch s

ide

by 2

.

The

inve

rse

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) #

2x !

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-Hi

ll41

8G

lenc

oe A

lgeb

ra 2

Inve

rses

of

Rel

atio

ns

and

Fu

nct

ion

s

Inve

rse

Func

tions

Two

func

tions

fan

d g

are

inve

rse

func

tions

if a

nd o

nly

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

The

fun

ctio

ns a

re 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

$2 3$

Sinc

e [f

!g]

(x) )

x,th

e fu

ncti

ons

are

not

inve

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

nog(

x) #

$1 8$ x!

12no

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

Gu

ide

and I

nte

rven

tion

(c

onti

nued

)

Inve

rse

Func

tions

and

Rel

atio

ns

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-8

7-8

Exam

ple1

Exam

ple1

Exam

ple2

Exam

ple2

Exer

cises

Exer

cises



Skil

ls P

ract

ice

Inve

rse

Func

tions

and

Rel

atio

ns

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-8

7-8

©G

lenc

oe/M

cGra

w-Hi

ll41

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)

}{(1

2,!

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)

#$1 3$ x

f!1 (

x) #

x!

2

10.g

(x) #

2x"

111

.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"

1no

14.f

(x) #

2x!

3ye

s15

.f(x

) #5x

"5

yes

g(x)

#1

"x

g(x)

#$1 2$ (

x"

3)g(

x) #

$1 5$ x!

1

16.f

(x) #

2xye

s17

.h(x

) #6x

"2

no18

.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-Hi

ll42

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) #

$4 3$ xg!

1 (x)

#x

!3

y#

$x" 3

2$

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!

6ye

s11

.f(x

) #"

4x!

1ye

s12

.g(x

) #13

x"

13no

g(x)

#x

"6

g(x)

#$1 4$ (

1 "

x)h(

x) #

$ 11 3$x

"1

13.f

(x) #

2xno

14.f

(x) #

$6 7$ xye

s15

.g(x

) #2x

"8

yes

g(x)

#"

2xg(

x) #

$7 6$ xh(

x) #

$1 2$ x!

4

16. M

EASU

REM

ENT

The

poi

nts

(63,

121)

,(71

,180

),(6

7,14

0),(

65,1

08),

and

(72,

165)

giv

eth

e w

eigh

t in

pou

nds

as a

fun

ctio

n of

hei

ght

in in

ches

for

5 s

tude

nts

in a

cla

ss.G

ive

the

poin

ts f

or t

hese

stu

dent

s th

at r

epre

sent

hei

ght

as a

fun

ctio

n 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 C

lear

ys a

re r

epla

cing

the

flo

orin

g in

the

ir 1

5 fo

ot b

y 18

foo

t ki

tche

n.T

he n

ew f

loor

ing

cost

s $1

7.99

per

squ

are

yard

.The

for

mul

a f(

x) #

9xco

nver

ts s

quar

e ya

rds

to s

quar

e fe

et.

17.F

ind

the

inve

rse

f"1 (

x).W

hat

is t

he s

igni

fica

nce

of f"

1 (x)

for

the

Cle

arys

?f!

1 (x)

#$x 9$ ;

It w

ill a

llow

them

to c

onve

rt th

e sq

uare

foot

age

of th

eir k

itche

n flo

or to

squa

re y

ards

,so

they

can

then

cal

cula

te th

e co

st o

f the

new

floo

ring.

18.W

hat

will

the

new

flo

orin

g co

st t

he C

lear

y’s?

$539

.70

x

y

Ox

g (x)

Ox

f (x)

OPra

ctic

e (A

vera

ge)

Inve

rse

Func

tions

and

Rel

atio

ns

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-8

7-8


An

swer

s


Rea

din

g t

o L

earn

Math

emati

csIn

vers

e Fu

nctio

ns a

nd R

elat

ions

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-8

7-8

©G

lenc

oe/M

cGra

w-Hi

ll42

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

ucti

on t

o L

esso

n 7-

8 at

the

top

of

page

390

in y

our

text

book

.

A fu

ncti

on m

ulti

plie

s a

num

ber

by 3

and

the

n ad

ds 5

to

the

resu

lt.W

hat

does

the

inve

rse

func

tion

do,

and

in w

hat

orde

r?Sa

mpl

e an

swer

:It f

irst

subt

ract

s 5

from

the

num

ber a

nd th

en d

ivid

es 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

,the

dom

ain

of o

ne r

elat

ion

is t

he

ofth

e ot

her.

b.Su

ppos

e th

at g

(x)

is a

rel

atio

n an

d th

at t

he p

oint

(4,

"2)

is o

n it

s gr

aph.

The

n a

poin

t

on t

he g

raph

of g

"1 (

x) is

.

c.T

he

test

can

be

used

on

the

grap

h of

a f

unct

ion

to d

eter

min

e

whe

ther

the

fun

ctio

n ha

s an

inve

rse

func

tion

.

d.If

you

are

giv

en t

he g

raph

of

a fu

ncti

on,y

ou c

an f

ind

the

grap

h of

its

inve

rse

by

refl

ecti

ng t

he o

rigi

nal g

raph

ove

r th

e lin

e w

ith

equa

tion

.

e.If

fan

d g

are

inve

rse

func

tion

s,th

en (

f!

g)(x

) #

and

(g!

f)(x

) #.

f.A

fun

ctio

n ha

s an

inve

rse

that

is a

lso

a fu

ncti

on o

nly

if t

he g

iven

fun

ctio

n is

.

g.Su

ppos

e th

at h

(x)

is a

fun

ctio

n w

hose

inve

rse

is a

lso

a fu

ncti

on.I

f h(5

) #

12,t

hen

h"1 (

12)

#.

2.A

ssum

e th

at f(

x) is

a o

ne-t

o-on

e fu

ncti

on d

efin

ed b

y an

alg

ebra

ic e

quat

ion.

Wri

te t

he f

our

step

s yo

u w

ould

fol

low

in o

rder

to

find

the

equ

atio

n fo

r f"

1 (x)

.

1.Re

plac

e f(

x) w

ith y

in th

e or

igin

al e

quat

ion.

2.In

terc

hang

e x

and

y.3.

Solv

e fo

r y.

4.Re

plac

e y

with

f!

1 (x)

.

Hel

pin

g Y

ou

Rem

emb

er3.

A g

ood

way

to

rem

embe

r so

met

hing

new

is t

o re

late

it t

o so

met

hing

you

alr

eady

kno

w.

How

are

the

ver

tica

l and

hor

izon

tal l

ine

test

s re

late

d?Sa

mpl

e an

swer

:The

ver

tical

line

test

det

erm

ines

whe

ther

a re

latio

n is

a fu

nctio

n be

caus

e th

e or

dere

dpa

irs in

a fu

nctio

n ca

n ha

ve n

o re

peat

ed x

-val

ues.

The

horiz

onta

l lin

e te

stde

term

ines

whe

ther

a fu

nctio

n is

one

-to-o

ne b

ecau

se a

one

-to-o

nefu

nctio

n ca

nnot

hav

e an

y re

peat

ed y

-val

ues.

5

one-

to-o

ne

xx

y #

x

horiz

onta

l lin

e(!

2,4)

rang

e

©G

lenc

oe/M

cGra

w-Hi

ll42

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

the

golf

bal

l int

o th

e ho

le in

as

few

sho

ts a

s po

ssib

le.A

s in

the

dia

gram

at

the

righ

t,th

e ho

le is

oft

en p

lace

d so

tha

t a

dire

ct s

hot

is im

poss

ible

.Ref

lect

ions

can

be u

sed

to h

elp

dete

rmin

e th

e di

rect

ion

that

the

bal

l sho

uld

bero

lled

in o

rder

to

scor

e a

hole

-in-

one.

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

imag

e of

the

“ho

le”

wit

h re

spec

t to

E!F!

and

labe

l it

H-.

The

inte

rsec

tion

of B !

H!-!w

ith

wal

l E!F!

is t

he p

oint

at

whi

ch t

he

shot

sho

uld

be d

irec

ted.

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

imag

e of

Hw

ith

resp

ect

to E!

F!an

d la

bel i

t H

-.In

thi

s ca

se,B !

H!-!i

nter

sect

s J!K!

befo

re in

ters

ecti

ng E!

F!.T

hus,

this

path

can

not

be u

sed.

To f

ind

a us

able

pat

h,fi

nd t

he r

efle

ctio

n im

age

of H

-w

ith

resp

ect

to G !

F!an

d la

bel i

t H

..N

ow,t

he

inte

rsec

tion

of B !

H!.!w

ith

wal

l G!F!

is t

he p

oint

at

whi

ch t

he s

hot

shou

ld b

e di

rect

ed.

Cop

y ea

ch f

igu

re.T

hen

,use

ref

lect

ion

s to

det

erm

ine

a p

ossi

ble

pat

h f

or a

hol

e-in

-on

e.

1.2.

3.

H

B

H

B

H

B

B GF

JK

H' H"

E

H

Ball

Hole

E

H'

F

Ball

Hole

En

rich

men

t

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-8

7-8

Exam

ple1

Exam

ple1

Exam

ple2

Exam

ple2



Stu

dy

Gu

ide

and I

nte

rven

tion

Squa

re R

oot F

unct

ions

and

Ineq

ualit

ies

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-9

7-9

©G

lenc

oe/M

cGra

w-Hi

ll42

3G

lenc

oe A

lgeb

ra 2

Lesson 7-9

Squ

are

Ro

ot

Fun

ctio

ns

A f

unct

ion

that

con

tain

s th

e sq

uare

roo

t of

a v

aria

ble

expr

essi

on is

a s

quar

e ro

ot f

un

ctio

n.

Gra

ph

y#

!3x

!"

2".S

tate

its

dom

ain

an

d r

ange

.

Sinc

e th

e ra

dica

nd c

anno

t be

neg

ativ

e,3x

"2

/0

or x

/$2 3$ .

The

x-i

nter

cept

is $2 3$ .

The

ran

ge is

y/

0.

Mak

e a

tabl

e of

val

ues

and

grap

h th

e fu

ncti

on.

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-Hi

ll42

4G

lenc

oe A

lgeb

ra 2

Squ

are

Ro

ot

Ineq

ual

itie

sA

squ

are

root

in

equ

alit

yis

an

ineq

ualit

y th

at c

onta

ins

the

squa

re r

oot

of a

var

iabl

e ex

pres

sion

.Use

wha

t yo

u kn

ow a

bout

gra

phin

g sq

uare

roo

tfu

ncti

ons

and

quad

rati

c in

equa

litie

s to

gra

ph s

quar

e ro

ot in

equa

litie

s.

Gra

ph

y,

!2x

!"

1""

2.G

raph

the

rel

ated

equ

atio

n y

#"

2x"

1!

!2.

Sinc

e th

e bo

unda

ry

shou

ld b

e in

clud

ed,t

he g

raph

sho

uld

be s

olid

.

The

dom

ain

incl

udes

val

ues

for

x/

$1 2$ ,so

the

gra

ph is

to

the

righ

t

of x

#$1 2$ .

The

ran

ge in

clud

es o

nly

num

bers

gre

ater

tha

n 2,

so t

he

grap

h is

abo

ve 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

0"

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 !

2x

y

O

y # 2

!"

""

2x !

3 x

y

Oy #

!"

"x "

1 !

4

x

y

O

y # !

(""

3x !

4

x

y

O

y # 3

!("

"2x

! 1

x

y

O

y # !

""

x " 3

x

y

O

y # 2

!(x

x

y

O

x

y

O

y # !

""

"2x

! 1

" 2

Stu

dy

Gu

ide

and I

nte

rven

tion

(c

onti

nued

)

Squa

re R

oot F

unct

ions

and

Ineq

ualit

ies

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-9

7-9

Exam

ple

Exam

ple

Exer

cises

Exer

cises


An

swer

s


Skil

ls P

ract

ice

Squa

re R

oot F

unct

ions

and

Ineq

ualit

ies

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-9

7-9

©G

lenc

oe/M

cGra

w-Hi

ll42

5G

lenc

oe A

lgeb

ra 2

Lesson 7-9

Gra

ph

eac

h f

un

ctio

n.S

tate

th

e d

omai

n a

nd

ran

ge o

f ea

ch f

un

ctio

n.

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+

0D:

x+

2.5,

R:y

,0

D:x

+!

4,R:

y+

!2

Gra

ph

eac

h i

neq

ual

ity.

7.y

'"

4x!8.

y/

"x

!1

!9.

y0

"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

©G

lenc

oe/M

cGra

w-Hi

ll42

6G

lenc

oe A

lgeb

ra 2

Gra

ph

eac

h f

un

ctio

n.S

tate

th

e d

omai

n a

nd

ran

ge o

f ea

ch f

un

ctio

n.

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

+$4 3$ ,

R:y

+0

D:x

+!

7,R:

y+

!4

D:x

+!

$3 2$ ,R:

y,

1

Gra

ph

eac

h i

neq

ual

ity.

7.y

/"

"6x!

8.y

0"

x"

5!

!3

9.y

("

2"3x

!2

!

10.R

OLL

ER C

OA

STER

ST

he v

eloc

ity

of a

rol

ler

coas

ter

as it

mov

es d

own

a hi

ll is

v

#"

v 02

!!

64h

!,w

here

v0

is t

he in

itia

l vel

ocit

y an

d h

is t

he v

erti

cal d

rop

in f

eet.

If

v#

70 f

eet

per

seco

nd a

nd v

0#

8 fe

et p

er s

econ

d,fi

nd h

.ab

out 7

5.6

ft

11.W

EIG

HT

Use

the

for

mul

a d

#'(

"39

60,w

hich

rel

ates

dis

tanc

e fr

om E

arth

d

in m

iles

to w

eigh

t.If

an

astr

onau

t’s w

eigh

t on

Ear

th W

Eis

148

pou

nds

and

in s

pace

Ws

is11

5 po

unds

,how

far

fro

m E

arth

is t

he a

stro

naut

?ab

out 5

32 m

i

3960

2W

E$

$ Ws

x

y O

x

y

O

x

y

O

x

y Ox

y

O

x

y

O

x

y

O

x

y

O

x

y

O

Pra

ctic

e (A

vera

ge)

Squa

re R

oot F

unct

ions

and

Ineq

ualit

ies

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-9

7-9



Rea

din

g t

o L

earn

Math

emati

csSq

uare

Roo

t Fun

ctio

ns

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-9

7-9

©G

lenc

oe/M

cGra

w-Hi

ll42

7G

lenc

oe A

lgeb

ra 2

Lesson 7-9

Pre-

Act

ivit

yH

ow a

re s

quar

e ro

ot f

un

ctio

ns

use

d i

n b

rid

ge d

esig

n?

Rea

d th

e in

trod

ucti

on t

o L

esso

n 7-

9 at

the

top

of

page

395

in y

our

text

book

.

If t

he w

eigh

t to

be

supp

orte

d by

a s

teel

cab

le is

dou

bled

,sho

uld

the

diam

eter

of

the

supp

ort

cabl

e al

so b

e do

uble

d? I

f no

t,by

wha

t nu

mbe

rsh

ould

the

dia

met

er b

e m

ulti

plie

d?

no;!

2"

Rea

din

g t

he

Less

on

1.M

atch

eac

h sq

uare

roo

t fu

ncti

on f

rom

the

list

on

the

left

wit

h it

s do

mai

n an

d ra

nge

from

the

list

on t

he r

ight

.

a.y

#"

x!iv

i.do

mai

n:x

/0;

rang

e:y

/3

b.y

#"

x!

3!

viii

ii.

dom

ain:

x/

0;ra

nge:

y0

0

c.y

#"

x!!

3i

iii.

dom

ain:

x/

0;ra

nge:

y0

"3

d.y

#"

x"

3!

viv

.do

mai

n:x

/0;

rang

e:y

/0

e.y

#"

"x!

iiv.

dom

ain:

x/

3;ra

nge:

y/

0

f.y

#"

"x

"3

!vi

ivi

.do

mai

n:x

03;

rang

e:y

/3

g.y

#"

3 "

x!

!3

vivi

i.do

mai

n:x

/3;

rang

e:y

00

h.

y#

""

x!"

3iii

viii

.do

mai

n:x

/"

3;ra

nge:

y/

0

2.T

he g

raph

of

the

ineq

ualit

y y

0"

3x!

6!

is a

sha

ded

regi

on.W

hich

of

the

follo

win

gpo

ints

lie

insi

de t

his

regi

on?

(3,0

)(2

,4)

(5,2

)(4

,"2)

(6,6

)

(3,0

),(5

,2),

(4,!

2)

Hel

pin

g Y

ou

Rem

emb

er

3.A

goo

d w

ay t

o re

mem

ber

som

ethi

ng is

to

expl

ain

it t

o so

meo

ne e

lse.

Supp

ose

you

are

stud

ying

thi

s le

sson

wit

h a

clas

smat

e w

ho t

hink

s th

at y

ou c

anno

t ha

ve s

quar

e ro

otfu

ncti

ons

beca

use

ever

y po

siti

ve r

eal n

umbe

r ha

s tw

o sq

uare

roo

ts.H

ow w

ould

you

expl

ain

the

idea

of

squa

re r

oot

func

tion

s to

you

r cl

assm

ate?

Sam

ple

answ

er:T

o fo

rm a

squ

are

root

func

tion,

choo

se e

ither

the

posi

tive

or n

egat

ive

squa

re ro

ot.F

or e

xam

ple,

y#

!x"

and

y#

!!

x"ar

etw

o se

para

te fu

nctio

ns.

©G

lenc

oe/M

cGra

w-Hi

ll42

8G

lenc

oe A

lgeb

ra 2

Read

ing

Alge

bra

If t

wo

mat

hem

atic

al p

robl

ems

have

bas

ic s

truc

tura

l sim

ilari

ties

,th

ey a

re s

aid

to b

e an

alog

ous.

Usi

ng a

nalo

gies

is o

ne w

ay o

fdi

scov

erin

g an

d pr

ovin

g ne

w t

heor

ems.

The

fol

low

ing

num

bere

d se

nten

ces

disc

uss

a th

ree-

dim

ensi

onal

anal

ogy

to t

he P

ytha

gore

an t

heor

em.

01C

onsi

der

a te

trah

edro

n w

ith

thre

e pe

rpen

dicu

lar

face

s th

atm

eet

at v

erte

x O

.02

Supp

ose

you

wan

t to

kno

w h

ow t

he a

reas

A,B

,and

Cof

the

thre

e fa

ces

that

mee

t at

ver

tex

Oar

e re

late

d to

the

are

a D

of t

he f

ace

oppo

site

ver

tex

O.

03It

is n

atur

al t

o ex

pect

a f

orm

ula

anal

ogou

s to

the

P

ytha

gore

an t

heor

em z

2#

x2!

y2,w

hich

is t

rue

for

a si

mila

r si

tuat

ion

in t

wo

dim

ensi

ons.

04To

exp

lore

the

thr

ee-d

imen

sion

al c

ase,

you

mig

ht g

uess

a

form

ula

and

then

try

to

prov

e it

.05

Tw

o re

ason

able

gue

sses

are

D3

#A

3!

B3

!C

3an

d D

2#

A2

!B

2!

C2 .

Ref

er t

o th

e n

um

bere

d s

ente

nce

s to

an

swer

th

e qu

esti

ons.

1.U

se s

ente

nce

01 a

nd t

he t

op d

iagr

am.T

he p

refi

x te

tra-

mea

ns f

our.

Wri

te a

nin

form

al d

efin

itio

n of

tet

rahe

dron

.

a th

ree-

dim

ensi

onal

figu

re w

ith fo

ur fa

ces

2.U

se s

ente

nce

02 a

nd t

he t

op d

iagr

am.W

hat

are

the

leng

ths

of t

he s

ides

of

each

fac

e of

the

tet

rahe

dron

?a,

b,an

d c;

a,q,

and

r;b,

p,an

d r;

c,p,

and

q

3.R

ewri

te s

ente

nce

01 t

o st

ate

a tw

o-di

men

sion

al a

nalo

gue.

Cons

ider

a tr

iang

le w

ith tw

o pe

rpen

dicu

lar s

ides

that

mee

t at v

erte

x C.

4.R

efer

to

the

top

diag

ram

and

wri

te e

xpre

ssio

ns f

or t

he a

reas

A,B

,and

Cm

enti

oned

in s

ente

nce

02.

Poss

ible

ans

wer

:A#

$1 2$ pr,

B#

$1 2$ pq,

C#

$1 2$ rq

5.To

exp

lore

the

thr

ee-d

imen

sion

al c

ase,

you

mig

ht b

egin

by

expr

essi

ng a

,b,

and

cin

ter

ms

of p

,q,a

nd r.

Use

the

Pyt

hago

rean

the

orem

to

do t

his.

a2#

q2"

r2,b

2#

r2"

p2,c

2#

p2"

q2

6.W

hich

gue

ss in

sen

tenc

e 05

see

ms

mor

e lik

ely?

Jus

tify

you

r an

swer

.

See

stud

ents

’exp

lana

tions

.

y O

z

x

b

c

Op

a

qr

En

rich

men

t

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

7-9

7-9

Documents

Chapter 7 Resource Masters - Math Class · PDF file©Glencoe/McGraw-Hill iv Glencoe Algebra 2 Teacher’s Guide to Using the Chapter 7 Resource Masters The Fast File Chapter Resource