chapter 6 polynomial functions - higher education | … · homework 6.1 ssm: intermediate algebra...

150

Chapter 6

Polynomial Functions

Homework 6.1

1. Vertex: (0, 0)

3. Vertex: (0, 0)

5. Vertex: (0, -1)

7. Vertex: (0, 5)

9. Vertex: (-5, 0)

11. Vertex: (1, 0)

13. Vertex: (4, -6)

15. Vertex: (6, 0)

SSM: Intermediate Algebra Homework 6.1

151

17. Vertex: (-5, -7)

19. Vertex: (-7, -3)

21. Vertex: (-4, -1)

23. Vertex: (-6, 3)

25.

The Domain is the set of all real numbers. Since (0, -4) is the minimum point, the range is the set of numbers where 4y ≥ − .

27.

The Domain is the set of all real numbers. Since (0, -4) is the minimum point, the range is the set of numbers where 4y ≥ .

29.

The Domain is the set of all real numbers. Since (0, -4) is the minimum point, the range is the set of numbers where 4y ≥ .

31. a. Since the function has a minimum point in quadrant III, then 0a > , 0h < , and 0k <

b. Since the function has a maximum point in quadrant II, then 0a < , 0h < , and 0k >

Homework 6.1 SSM: Intermediate Algebra

152

c. Since the function has a minimum point that lies on the x-axis when x > 0, then 0a > ,

0h > , and 0k =

d. Since the function has a minimum point that lies on the y-axis when y > 0, then 0a < ,

0h = , and 0k <

33. Answers may vary: Example:

2( 5) 3y a x= + +

Where a = -3, -2, -1, -½, ½, 1, 2, 3

35. From the vertex (5, -6), we know 2( ) ( 5) 6f x a x= − − . Since (1, 4) lies on the

parabola, substitute 1 for x and 4 for f(x) to solve for a:

2

2

4 (1 5) 6

10 ( 4)10 (16)

580.625

a

a

a

a

a

= − −= −=

=

=

The equation of the function is:

2( ) 0.625( 5) 6f x x= − −

37. The value of a for the function f is the opposite of the value of a for the function g since g has a maximum point and f has a minimum point and we can assume that the graphs of f and g have the same “shape”. Since the vertex ( ),h k of g is

( )7,3.71− and a = -2.1, an equation for g is:

2( ) 2.1( 7) 3.71g x x= − + +

39. It is possible. Example: 2 2y x= +

41. It is possible. Example: 2 2y x= −

43. a.

Quadratic Function because the points bend in a parabola shape.

b.

Yes, the parabola comes close to the points in the scattergram.

c. ( )8.86, 47.49 The percentage of Americans who were pro-choice was at a minimum of 48% in 1999, according to the model.

d. 1995, 2003

e. 94%

45. Both equations have the same vertex (2, 5). From the graph notice this is the only point that lies on both graphs.


153

47.

a. Answers may vary. Example: ( )1,2− , ( )0,1 ,

and ( )1,2 .

(-1, 2) is a solution:

2

?2

?

( ) 1

2 ( 1) 1

2 2 TRUE

f x x= +

= − +

=

(0, 1) is a solution:

2

?2

?

( ) 1

1 (0) 1

1 1 TRUE

f x x= +

= +

=

(1, 2) is a solution:

2

?2

?

( ) 1

2 (1) 1

2 2 TRUE

f x x= +

= +

=

b. Answers may vary. Example: ( )0,0 , ( )1,0 ,

and ( )1,0− .

(0, 0) is NOT a solution:

2

?2

?

( ) 1

0 (0) 1

0 1 FALSE

f x x= +

= +

=

(1, 0) is NOT a solution:

2

?2

?

( ) 1

0 (1) 1

0 2 FALSE

f x x= +

= +

=

(-1, 0) is NOT a solution:

2

?2

?

( ) 1

0 ( 1) 1

0 2 FALSE

f x x= +

= − +

=

49. a. Answers may vary. Example:

2 2( 1) and ( 1)y x y x= − = − +

b. Answers may vary. Example:

2 2 and y x y x= = −

c. Answers may vary. Example:

2 21 and 1y x y x= − = − +

d. No, it is not possible.

51. a.

b.


154

c.

d. Explanations may vary. In part b, the y-axis is stretched vertically so that the parabola is stretched vertically compared to a. Also, in part c, the x-axis is stretched horizontally so that the parabola is stretched horizontally compared to part a.

53. Adjust the WINDOW settings. Make your x-min and x-max much larger.

55. a.

b.

c.

d.

57. Step 1: Sketch a graph of 2y ax= .

Step 2: Translate the graph of 2y ax= right h units if h > 0 and left |h| units if h < 0.

Step 3: Lastly, translate the graph from step 2 up k units if k > 0 and down |k| units if k < 0.

Homework 6.2

1. 2( 1) ( ) (1)x x x x x x x− = − = −

3. 7 ( 2) 7 ( ) ( 7 )(2)x x x x x− − = − − −

27 14x x= − +

5. 3 (8 5) 3 (8 ) ( 3 )(5)x x x x x− + = − + −

224 15x x= − −

7. 2 4 ( ) (4) ( 4)x x x x x x x+ = + = +

9. 227 36 9 (3 ) ( 9 )(4)x x x x x− + = − − −

9 (3 4)x x= − −

11. 225 35 5 (5 ) ( 5 )(7)x x x x x− + = − − −


155

5 (5 7)x x= − −

13. 24 8 4 ( ) ( 4 )(2)x x x x x− − = − + −

4 ( 2)x x= − +

15. 23.7 4.7 (3.8 ) (4.7)x x x x x+ = +

(3.8 4.7)x x= +

17. ( ) ( 7)f x x x= −

19. ( ) 6 (4 5)h x x x= −

21. ( ) 2 ( 4)f x x x= − +

23. ( ) (2.5 6.2)h x x x= −

25. Yes, expanding 2 (6 9)x x + , find that

22 (6 9) 2 (6 ) 2 (9) 12 18x x x x x x x+ = + = +

27. All three students did the problem correctly, although we usually write the result as Student 2 and Student 3 did.

29. 22 (7 4) 14 8x x x x− = −

31. 2 2( 2)( 4) 4 2 8 6 8x x x x x x x+ + = + + + = + +

33. 2 2( 3)( 6) 6 3 18 3 18x x x x x x x− + = + − − = + −

35. 2 2( 8)( 8) 8 8 64 16 64x x x x x x x+ + = + + + = + +

37. 2( 7) ( 7)( 7)x x x− = − −

2

2

7 7 49

14 49

x x x

x x

= − − += − +

39. 2( 5)( 5) 5 5 25x x x x x+ − = + − −

2 25x= −

41. 2(3 5)(4 1) 12 3 20 5x x x x x+ + = + + +

212 23 5x x= + +

43. 2(4 1)(6 5) 24 20 6 5x x x x x− − = − − +

224 26 5x x= − +

45. 2 3 2( 4)( 20) 2 4 8x x x x x− + = + − −

47. 3 2 5 3 2( 6)( 3) 3 6 18x x x x x+ + = + + +

49. 2(2 5) (2 5)(2 5)x x x+ = + +

2

2

4 10 10 25

4 20 25

x x x

x x

= + + += + +

51. 2(3 2)(3 2) (3 )(3 ) 6 6 4 9 4x x x x x x x+ − = − + − = −

53. 2 2(1 )x x x x x x− − = − + = −

55. 2( 1) [( 1)( 1)]x x x− − = − − −

2

2

2

( 1)

( 2 1)

2 1

x x x

x x

x x

= − − − +

= − − += − + −

57. 23 (5 1) 3 (5 1)(5 1)x x x x x− = − −

2

2

3 2

3 (25 5 5 1)

3 (25 10 1)

63 30 3

x x x x

x x x

x x x

= − − +

= − += − +

59. 22 (4 5) 2 (4 5)(4 5)x x x x x− + = − + +

2

2

3 2

2 (16 20 20 25)

2 (16 40 25)

32 80 50

x x x x

x x x

x x x

= − + + +

= − + += − − −

61. (2.1 3.8)(1.8 5.6)x x− +

2

2

3.78 11.76 6.84 21.28

3.78 4.92 21.28

x x x

x x

= + − −= + −

63. 2( 5)( 4 1)x x x+ + +

3 2 2

3 2

4 5 20 5

9 21 5

x x x x x

x x x

= + + + + += + + +

65. 2(2 3)(3 4)x x x− + −


156

3 2 2

3 2

6 2 8 9 3 12

6 7 11 12

x x x x x

x x x

= + − − − += − − +

67. 2( 4)( 4 16)x x x+ − +

3 2 2

3

4 16 4 16 64

64

x x x x x

x

= − + + − += +

69. 2 2( 2 3)( 2)x x x x+ + + +

4 3 2 3 2 2

4 3 2

2 2 2 4 3 3 6

3 7 7 6

x x x x x x x x

x x x x

= + + + + + + + += + + + +

71. 2 2(2 3)( 2 1)x x x x+ − − +

4 3 2 3 2 2

4 3 2

2 4 2 2 3 6 3

2 3 3 7 3

x x x x x x x x

x x x x

= − + + − + − + −= − − + −

73. ( 1)( 2)( 3)x x x+ + +

2

2

3 2 2

3 2

( 1)( 3 2 6)

( 1)( 5 6)

5 6 5 6

6 11 6

x x x x

x x x

x x x x x

x x x

= + + + += + + +

= + + + + += + + +

75. 1 12 2

5 5x x − +

2

2

1 2 24

5 5 51

425

x x x

x

= + − −

= −

77. 2 2( 5) ( 5)x x− − +

2 2

2 2

( 10 25) ( 10 25)

10 25 10 2520

x x x x

x x x x

x

= − + − + += − + − − −= −

79. 2 2( 4) ( 4)x x− + −

2 2

2 2

2

( 8 16) ( 8 16)

8 16 8 16

2 16 32

x x x x

x x x x

x x

= − + + − +

= − + + − += − +

81. 2( ) ( 6)( 6) 6 6 36f x x x x x x= + + = + + +

2 12 36x x= + +

83. 2( ) 2( 3) 1 2( 3)( 3) 1h x x x x= + + = + + +

2

2

2

2

2( 3 3 9) 1

2( 6 9) 1

2 12 18 1

2 12 19

x x x

x x

x x

x x

= + + + += + + +

= + + += + +

85. 2( ) 4( 5) 2 4( 5)( 5) 2p x x x x= − + = − − +

2

2

2

2

4( 5 5 25) 2

4( 10 25) 2

4 40 100 2

4 40 102

x x x

x x

x x

x x

= − − + += − + +

= − + += − +

87. 2( ) 4( 1) 1 4( 1)( 1) 1g x x x x= − − − = − − − −

2

2

2

2

4( 1) 1

4( 2 1) 1

4 8 4 1

4 8 5

x x x

x x

x x

x x

= − − − + −= − − + −

= − + − −= − + −

89. 2( ) 1.5( 2.8) 3.7k x x= + −

2

2

2

2

1.5( 2.8)( 2.8) 3.7

1.5( 2.8 2.8 7.84) 3.7

1.5( 5.6 7.84) 3.7

1.5 8.4 11.76 3.7

1.5 8.4 8.06

x x

x x x

x x

x x

x x

= + + −

= + + + −= + + −

= + + −= + +

91. 2( ) 5 ( 2) 5 10f x x x x x= − = − This function is quadratic.

93. 2 2( ) ( 6)( 6) 6 6 36 36h x x x x x x x= + + = + + − = −

This function is quadratic.

95. 2 2( ) ( 6) ( 6)h x x x= + − −

2 2

2 2

( 12 36) ( 12 36)

12 36 12 3624

x x x x

x x x x

x

= + + − − += + + − + −=

This function is linear.


157

97. 2(4) 4 3(4) 16 12 4f = − = − =

99. 2(2.8) (2.8) 3(2.8) 7.84 8.4 0.56f = − = − = −

101. 2 2( ) 3( ) 3f a a a a a= − = −

103. 2( 1) ( 1) 3( 1)f a a a+ = + − +

2

2

( 1)( 1) 3( 1)

1 3 3

2

a a a

a a a a

a a

= + + − +

= + + + − −= − −

105. 2(0) 2(0) 5(0) 1 1f = − + − = −

107. 2

1 1 12 5 1

2 2 2f = − + −

1 52 1

4 21 5

12 2

41

22 1 1

= − + −

= − + −

= −

= − =

109. 2 2( ) 2( ) 5( ) 1 2 5 1f a a a a a= − + − = − + −

111. 2( 3) 2( 3) 5( 3) 1f a a a+ = − + + + −

2

2

2

2( 6 9) 5 15 1

2 12 18 5 15 1

2 7 4

a a a

a a a

a a

= − + + + + −= − − − + + −

= − − −

113. Answers may vary. Your discussion should include square of a sum, square of a difference, and difference of two squares

Homework 6.3

1. 2 7 12 ( 3)( 4)x x x x− + = + +

3. 2 21 20 ( 20)( 1)x x x x− + = − −

5. This expression is prime since there are not two integers with product -10 and sum 2.

7. 2 23 3 18 3( 6)x x x x− − = − −

3( 3)( 2)x x= − +

9. 2 25 20 60 5( 4 12)x x x x− + + = − − −

5( 6)( 2)x x= − − +

11. 2 26 36 54 6( 6 9)x x x x+ + = + +

2

6( 3)( 3)

6( 3)

x x

x

= + += +

13. 2 2 216 4 ( 4)( 4)x x x x− = − = − +

15. 2 2 21 1 ( 1)( 1)x x x x− = − = − +

17. 2 2 225 1( 5 ) ( 5)( 5)x x x x− = − − = − − +

19. 2 2 2100 1 (10 ) 1 (10 1)(10 1)x x x x− = − = − +

21. 2 249 1( 49) ( 7)( 7)x x x x− + = − − = − − +

23. This expression is prime since an expression of the form 2 2 , 0x k k+ ≠ , is always prime and not factorable.

25. 2 2 264 49 (8 ) 7 (8 7)(8 7)x x x x− = + = − +

27. 2

2 21 1 1 19 3 3 3

x x x x − = − = − +

29. 23 10 8 (3 4)( 2)x x x x+ + = + +

31. 22 13 15 ( 5)(2 3)x x x x− + = − −

33. 29 9 2 (3 4)(3 1)x x x x+ + = − −

35. This expression is prime.

37. 29 15 4 (3 4)(3 1)x x x x− + = − −

39. 2 220 30 140 10(2 3 14)x x x x+ − = + −

10( 2)(2 7)x x= − +

41. 2 11 30 ( 5)( 6)x x x x+ + = + +


158

43. 2 3 2 ( 2)( 1)x x x x− + = − −

45. 215 27 3 (5 9)x x x x− = −

47. 210 23 12 (2 3)(5 4)x x x x+ + = + +


51. 2 15 100 ( 20)( 5)x x x x− − = − +

53. 2 23 75 3( 25)x x− − = − +

55. 2 2 48 ( 6)( 8)x x x x+ − = − +

57. 212 7 5 ( 1)(12 5)x x x x− − = − +

59. 2 27 14 21 7( 2 3)x x x x+ + = + +

61. 281 16 (9 4)(9 4)x x x− = − +


65. 28 2 15 (2 3)(4 5)x x x x− − = − +

68. 2 2 2100 10 ( 10)( 10)x x x x− = − = − +

69. 2 214 49 ( 7)( 7) ( 7)x x x x x− + = − − = −

71. 29 18 8 (3 4)(3 2)x x x x− + = − −

73. 24 79 20 ( 20)(4 1)x x x x+ − = + −

75. 2 22 22 48 2( 11 24) 2( 8)( 3)x x x x x x− + = − + = − −

77. 210 24 (2 3)(5 8)x x x x+ − = − +

79. 2 236 48 16 4(9 12 4)x x x x+ + = + +

2

4(3 2)(3 2)

4(3 2)

x x

x

= + += +

81. The student did not factor the expression correctly. It should be:

24 8 3 (2 3)(2 1)x x x x+ + = + +

83. The student did not factor the expression correctly. The expression 2 25x + is prime so

2 24 100 4( 25)x x+ = + is the final step.

85. Answers may vary. Your discussion should be include factoring techniques for factoring

2ax bx c+ + when a = 1 and when a • 1.

Homework 6.4

1. 3 23 75 3 ( 25)x x x x− = −

3 ( 5)( 5)x x x= − +

3. 3 263 7 7 (9 1)x x x x− = −

7 (3 1)(3 1)x x x= − +

5. 5 3 3 28 72 8 ( 9)x x x x− = −

38 ( 3)( 3)x x x= − +

7. 4 2 2 275 27 3 (25 9)x x x x− + = − −

23 (5 3)(5 3)x x x= − − +

9. 3 2 22 8 6 2 ( 4 3)x x x x x x+ + = + +

2 ( 1)( 3)x x x= + +

11. 4 3 2 2 24 12 40 4 ( 3 10)x x x x x x+ − = − −

24 ( 5)( 2)x x x= − +

13. 3 2 22 20 50 2 ( 10 25)x x x x x x− + − = − − +

2

2 ( 5)( 5)

2 ( 5)

x x x

x x

= − − −= − −

15. 5 4 3 3 26 33 45 3 (2 11 15)x x x x x x+ + = + +

33 (2 5)( 3)x x x= + +

17. 3 2 280 140 150 10 (8 14 15)x x x x x x+ − = + −

10 (2 5)(4 3)x x x= + −

19. 6 5 4 4 230 5 5 5 (6 1)x x x x x x− − + = − + −

4 2

4

5 (6 1)

5 (3 1)(2 1)

x x x

x x x

= − + −

= − − +


159

21. 3 2 23 4 12 ( 3) 4( 3)x x x x x x+ + + = + + +

2( 3)( 4)x x= + +

23. 3 2 24 3 12 ( 4) 3( 4)x x x x x x− + − = − + −

2( 4)( 3)x x= − +

25. 3 2 23 15 4 20 3 ( 5) 4( 5)x x x x x x− − + = − − −

2( 5)(3 4)x x= − −

27. 3 2 22 32 16 (2 1) 8(2 1)x x x x x x+ − − = + − +

2(2 1)( 8)x x= + −

29. 3 2 25 4 20 ( 5) 4( 5)x x x x x x− − + = − − −

2( 5)( 4)

( 5)( 2)( 2)x x

x x x

= − −= − − +

31. 3 2 24 12 9 27 4 ( 3) 9( 3)x x x x x x− − + = − − −

2( 3)(4 9)

( 3)(2 3)(2 3)x x

x x x

= − −= − − +

33. 3 2 212 4 3 1 4 (3 1) 1(3 1)x x x x x x+ − − = + − +

2(3 1)(4 1)

(3 1)(2 1)(2 1)x x

x x x

= + −= + − +

35. 3 22 50 2 ( 25)x x x x− = −

2 ( 5)( 5)x x x= − +

37. 3 2 24 7 28 ( 4) 7( 4)x x x x x x+ − − = + − +

2( 4)( 7)x x= + −

39. 2 81 ( 9)( 9)x x x− = + −

41. 28 10 3 (4 3)(2 1)x x x x− + = − −

43. 29 15 3 (3 5)x x x x− = −

45. 4 3 2 2 23 18 24 3 ( 6 8)x x x x x x+ + = + +

23 ( 4)( 2)x x x= + +

47. 2 214 49 14 49x x x x+ + = + +

2

( 7)( 7)

( 7)

x x

x

= + += +

51. 2 ( 1)x x x x+ = +

53. 3 216 ( 16) ( 4)( 4)x x x x x x x− = − − = − − +

55. 3 2 24 20 5 4 ( 5) 1( 5)x x x x x x+ − − = + − +

2( 5)(4 1)

( 5)(2 1)(2 1)x x

x x x

= + −= + − +

57. 3 2100 ( 100) ( 10)( 10)x x x x x x x− = − = + −

59. 2 3 3 215 6 5 6 5 15x x x x x x− + − = − − + This expression is prime.

61. 4 3 2 2 25 35 60 5 ( 7 12)x x x x x x− + = − +

25 ( 4)( 3)x x x= − −

63. 2 3 3 226 12 12 12 26 12x x x x x x+ + = + +

22 (6 13 6)

2 (2 3)(3 2)x x x

x x x

= + += + +

65. 2 24 16 16 4( 4 4)x x x x− + = − +

2

4( 2)( 2)

4( 2)

x x

x

= − −= −

67. 2 22 1 1( 2 1)x x x x− − − = − + +

2

( 1)( 1)

( 1)

x x

x

= − + += − +

69. 4 5 48 6 2 (3 4)x x x x− = − −


73. 5 4 3 3 220 20 15 5 (4 4 3)x x x x x x+ − = + −

35 (2 1)(2 3)x x x= − +


160

75. The student hasn’t finished the problem, only done the first step. To finish the problem factor ( 5)x + out of both terms to give a final answer of:

2( 5)( 3)x x+ −

77. The student’s answer of (4 6)(4 6)x x+ − is not correct because it is not factored completely (2 can be factored out). The student should have factored out the GCF first. The answer would be:

2 216 36 4(4 9) 4(2 3)(2 3)x x x x− = − = − +

79. Answers may vary. See box titled “Five-Step Factoring Strategy” on page 281 of the text or look under the “Key Points of This Section” which precedes this homework section.

Homework 6.5

1. 2 3 0x x− =

( 3) 0

0 or 3 00 or 3

x x

x x

x x

− == − == =

3. 24 6 0x x− =

2 (2 3) 0

2 0 or 2 3 00 or 2 3

30 or

2

x x

x x

x x

x x

− == − =

= =

= =

5. 2 8 15 0x x+ + =

( 3)( 5) 0

3 0 or 5 03 or 5

x x

x x

x x

+ + =+ = + == − = −

7. 2 4 0x − =

( )( )2 22 0

2 2 0

2 0 or 2 02 or 2

x

x x

x x

x x

− =+ − =

+ = − == − =

9. 27 6 1 1x x+ + =

( )( )

27 6 0

7 6 0

7 6 0

7 0 or 6 00 or 6

x x

x x

x x

x x

x x

+ =+ =

+ == + =

= = −

11. 23 6x x=

( )( )

23 6 0

3 2 0

3 2 0

3 0 or 2 00 or 2

x x

x x

x x

x x

x x

− =− =

− == − =

= =

13. 29 2 5x x= − +

( )( )

22 9 5 0

2 1 5 0

2 1 0 or 5 02 1 or 5

1 or 5

2

x x

x x

x x

x x

x x

+ − =− + =

− = + == = −

= = −

15. 29 12 4 0x x− + =

( )( )3 2 3 2 0

3 2 03 2

23

x x

x

x

x

− − =− ==

=

17. 23 12x =

( )( )( )

( )( )

2

2

3 12 0

3 4 0

3 2 2 0

2 2 0

2 0 or 2 02 or 2

x

x

x x

x x

x x

x x

− =

− =

− + =

− + =− = + == = −

19. 216 25x =


161

( )( )

2

2 2

16 25 0

(4 ) 5 0

4 5 4 5 0

4 5 0 or 4 5 04 5 or 4 5

5 5 or

4 4

x

x

x x

x x

x x

x x

− =

− =− + =

− = + == = −

= = −

21. (5 3) 4 4x x + − = −

( )

2

2

5 3 4 4

5 3 0

5 3 0

0 or 5 3 00 or 5 3

30 or

5

x x

x x

x x

x x

x x

x x

+ − = −+ =

+ == + == = −

= = −

23. 3 ( 2) 4x x x− =

( )

2

2

3 6 4

3 10 0

3 10 0

0 or 3 10 00 or 3 10

100 or

3

x x x

x x

x x

x x

x x

x x

− =− =

− == − == =

= =

25. 29 49 0x − =

( )( )

2 2(3 ) 7 0

3 7 3 7 0

3 7 0 or 3 7 03 7 or 3 7

7 7 or

3 3

x

x x

x x

x x

x x

− =− + =

− = + == = −

= = −

27. 2 1x =

( )( )

2

2 2

1 0

1 0

1 1 0

1 0 or 1 01 or 1

x

x

x x

x x

x x

− =− =− + =

− = + == = −

29. 264 28 9x x= −

( )( )

29 28 64 0

3 8 3 8 0

3 8 0 3 8

83

x x

x x

x

x

x

− + =− − =

− ==

=

31. 24 8 32x x− =

( )

( )( )

2

2

2

4 8 32 0

4 2 8 0

2 8 0

4 2 0

4 0 or 2 04 or 2

x x

x x

x x

x x

x x

x x

− − =

− − =

− − =− + =

− = + == = −

33. 2 12 36 0x x− + =

( )( )6 6 0

6 0 6

x x

x

x

− − =− ==

35. 2 10

25x − =

22 1

05

1 10

5 51 1

0 or 05 51 1

or 5 5

x

x x

x x

x x

− = − + =

− = + =

= = −

37. 212 2 2 0x x− − =

( )

( )( )

2

2

2 6 1 0

6 1 0

2 1 3 1 0

2 1 0 or 3 1 02 1 or 3 1

1 1 or

2 3

x x

x x

x x

x x

x x

x x

− − =

− − =− + =

− = + == = −

= = −


162

39. 21 16

4 2x x− =

( )

( )( )

2

2

2

1 14 4 6

4 2

2 24

2 24 0

6 4 0

6 0 or 4 06 or 4

x x

x x

x x

x x

x x

x x

− = − =− − =− + =

− = + == = −

41. 36 24 0x x− =

( )( )( )( )

2

2 2

6 4 0

6 2 0

6 2 2 0

6 0 or 2 0 or 2 00 or 2 or 2

x x

x x

x x x

x x x

x x x

− =

− =

− + == − = + =

= = = −

43. 3 24 2 36 18 0x x x− − + =

( ) ( )( )( )

( )( )( )( )( )( )( )

( )( )( )

2

2

2

2 2

2 2 1 18 2 1 0

2 1 2 18 0

2 2 1 9 0

2 2 1 3 0

2 2 1 3 3 0

2 1 3 3 0

2 1 0 or 3 0 or 3 02 1 or 3 or 3

1 or 3 or 3

2

x x x

x x

x x

x x

x x x

x x x

x x x

x x x

x x x

− − − =

− − =

− − =

− − =

− − + =

− − + =− = − = + == = = −

= = = −

45. 3 24 7x x=

( )

3

2

2

4 7 0

4 7 0

0 or 4 7 00 or 4 7

70 or

4

x

x x

x x

x x

x x

− =− =

= − == =

= =

47. 3 24 9 20 45x x x− = −

( ) ( )( )( )( )( )( )( )( )

3 2

2

2

2 2

4 20 9 45 0

4 5 9 5 0

5 4 9 0

5 (2 ) 3 0

5 2 3 2 3 0

5 0 or 2 3 0 or 2 3 05 or 2 3 or 2 3

3 35 or or

2 2

x x x

x x x

x x

x x

x x x

x x x

x x x

x x x

− − + =− − − =

− − =

− − =

− − + =− = − = + == = = −

= = = −

49. 3 29 12 4 27x x x− = −

( ) ( )( )( )( )( )( )( )( )

3 2

2

2

2 2

9 27 4 12 0

9 3 4 3 0

3 9 4 0

3 (3 ) 2 0

3 3 2 3 2 0

3 0 or 3 2 0 or 3 2 03 or 3 2 or 3 2

2 25 or or

3 3

x x x

x x x

x x

x x

x x x

x x x

x x x

x x x

+ − − =

+ − + =

+ − =

+ − =

+ − + =+ = − = + == = = −

= = = −

51. 3 218 3 6x x x+ =

( )( )( )

3 2

2

18 3 6 0

3 6 2 0

3 3 2 1 0

3 0 or 3 2 0 or 1 00 or 3 2 or 1

20 or or 1

3

x x x

x x x

x x x

x x x

x x x

x x x

+ − =

− − =

+ − == + = − =

= = − =

= = − =

53. 217 28 3x x− = −

( )( )

23 17 28 0

3 4 7 0

3 4 0 or 7 03 4 or 7

4 or 7

3

x x

x x

x x

x x

x x

− − =+ − =

+ = − == − =

= − =


163

55. ( )( )2 5 40x x+ + =

( )( )

2

2

7 10 40

7 30 0

3 10 0

3 0 or 10 03 or 10

x x

x x

x x

x x

x x

+ + =+ − =− + =

− = + == = −

57. ( ) ( )4 1 24 3 2x x x x− − = −

( )( )

2 2

2

4 4 24 3 6

2 24 0

4 6 0

4 0 or 6 04 or 6

x x x x

x x

x x

x x

x x

− − = −+ − =− + =

− = + == = −

59. ( )( )2 5 6 2 3x x x x+ + = + +

61. 2 5 6 0x x+ + =

( )( )2 3 0

2 0 or 3 02 or 3

x x

x x

x x

+ + =+ = + == − = −

63. 2 7 10 0x x− + =

( )( )2 5 0

2 0 or 5 02 or 5

x x

x x

x x

− − =− = − == =

65. ( )( )2 7 10 2 5x x x x− + = − −

67. False, a and b can have values other than a = 2 and b = 10. (For example a = 4 and b = 5)

69. The student did not do factor by grouping correctly (step 3 has the error). The correct solution is:

( ) ( )( )( )( )( )( )( )( )

3 2

2

2

2 2

4 9 36 0

4 9 4 0

4 9 0

4 ( ) 3 0

4 3 3 0

4 0 or 3 0 or 3 04 or 3 or 3

x x x

x x x

x x

x x

x x x

x x x

x x x

+ − − =+ − + =

+ − =

+ − =

+ − + =+ = − = + == − = = −

71. Solve for x when ( ) 0f x =

( )( )20 9 20

0 5 4

5 0 or 4 05 or 4

x x

x x

x x

x x

= − += − −− = − == =

The x-intercepts are (5, 0) and (4, 0).


( )( )

2

2 2

0 16

0 4

0 2 2

2 0 or 2 02 or 2

x

x

x x

x x

x x

= −= −= + −+ = − == − =

The x-intercepts are (-2, 0) and (2, 0).


( )

( )( )

2

2

2

0 2 20 32

0 2 10 16

0 10 16

0 8 2

8 0 or 2 08 or 2

x x

x x

x x

x x

x x

x x

= − +

= − +

= − += − −− = − == =



( )( )

20 36 25

0 6 5 6 5

6 5 0 or 6 5 06 5 or 6 5

5 5 or

6 6

x

x x

x x

x x

x x

= −= − +− = + == = −

= = −

The x-intercepts are 5

,06

and 5

,06

− .



164

( )( )

20 12 7 10

0 3 2 4 5

3 2 0 or 4 5 03 2 or 4 5

2 5 or

3 4

x x

x x

x x

x x

x x

= − −= + −+ = − == − =

= − =

The x-intercepts are 2

,03

− and

5,0

4

.


( )20 3 30

0 3 10

3 0 or 10 00 or 10

x x

x x

x x

x x

= −= −= − =

= =



( )( )( )

3 2

2

0 2 2 84

0 2 42

0 2 7 6

2 0 or 7 0 or 6 00 or 7 or 6

x x x

x x x

x x x

x x x

x x x

= − −

= − −

= − += − = + =

= = = −

The x-intercepts are (0, 0), (7, 0) and (-6, 0).


( ) ( )( )( )( )( )( )( )( )

3 2

2

2

2 2

0 2 2

0 2 1 2

0 2 1

0 2 1

0 2 1 1

2 0 or 1 0 or 1 02 or 1 or 1

x x x

x x x

x x

x x

x x x

x x x

x x x

= + − −= + − +

= + −

= + −

= + − ++ = − = + == − = = −

The x-intercepts are (-2, 0), (1, 0) and (-1, 0).

87. a.

The scattergram suggests that the data can be best modeled by a quadratic function since the points are making a parabola shape.

b.

It appears that f models the data quite well.

c. Solve for t when f(t) = 109

( )( )

2

2

109 22 130

0 22 21

0 21 1

21 0 or 1 021 or 1

t t

t t

t t

t t

t t

= − += − += − −

− = − == =

Sales will be 109 million cases in the years 1981 or 2001.

d. Solve for f(t) when t = 23 (year 2003)

2(23) (23) 22(23) 130 153f = − + =

In 2003, the sales will be 153 million cases.


165

89. a.

It appears that f models the data quite well.

b. Solve for t when f(t) = 63

( )

( )

( )( )

2

2

2

2

2

2

463 10 91

74

7 63 10 91 77

441 4 70 637

0 4 70 196

0 2 2 35 98

0 2 35 98

0 2 7 14

2 7 0 or 14 02 7 or 14

7 or 14

23.5 or 14

t t

t t

t t

t t

t t

t t

t t

t t

t t

t t

t t

= − +

= − + = − +

= − +

= − +

= − += − −− = − == =

= =

= =

The model predicts that 63% of Americans will be pro-choice in 2004 and in 1994.

c. Solve for t when f(t) = 91

( )

( )

2

2

2

2

491 10 91

74

7 91 10 91 77

637 4 70 637

0 4 70

0 2 2 35

2 0 or 2 35 00 or 2 35

350 or

20 or 17.5

t t

t t

t t

t t

t t

t t

t t

t t

t t

= − +

= − + = − +

= −= −= − =

= =

= =

= ≈

91% of Americans will be pro-choice in 2008 and in 1990. It is not likely that these predictions will be correct since public opinion has been between 48% and 58% for the past 30 years.

91. ( ) ( ) ( )29 9 9 6 66f = − − =

93. ( ) ( ) ( )27 7 7 6 50f − = − − − − =

95. ( ) 2 6f x x x= − −

( )( )

2

2

14 6

0 20

0 5 4

5 0 or 4 05 or 4

x x

x x

x x

x x

x x

= − −= − −= − +− = + == = −

97. ( ) 2 6f x x x= − −

( )

2

2

6 6

0

0 1

0 or 1 00 or 1

x x

x x

x x

x x

x x

− = − −= −= −= − == =

99. Since ( )2,5− lies on the parabola, ( )2 5f − = .

101. Since ( )1, 1− lies on the parabola, ( )1 1f = − .

103. Since ( )1.7, 4− and ( )3.7,4 lie on the parabola,

( )1.7 and 3.7 when 4a a f a= − = = .

105. Since ( )1, 1− lies on the parabola,

( )1 when 1a f a= = − .

107. Since ( )0,19 is in the table, ( )0 19f = .

109. Since ( )4,3 is in the table, ( )4 3f = .

111. Since ( )0,19 and ( )6,19 are in the table,

( )0 and 6 when 19x x f x= = = .

113. Since ( )3,1 is in the table, ( )3 when 1x f x= = .


166

115. a. Since ( )1,9 and ( )5,9 are in the table,

( )1 and 5 when 9x x f x= = = .

b. The function f does not have an inverse function since the output 9 corresponds to not one input, but two (1 and 5).

117. Answers may vary. Example:

2( ) ( 5)( 1) 4 5f x x x x x= + − = + −


2( 3)( 1) 2 3y x x x x= + − = + −

121. Answers may vary, but must be of the form:

( ) ( 6)( 3)h x a x x= + + , where a > 0

because the graph has a minimum point and x-intercepts ( )6,0− and ( )3,0− .

Example: 2( ) ( 6)( 3) 9 18h x x x x x= + + = + +

123. Student 1’s work is correct. Student 2’s work is incorrect because we cannot divide by x since x is possibly 0.

125. Answers may vary. See box titled “Solving Quadratic Equations” on page 28 8 of the text or look under the “Key Points of This Section” which precedes this homework section.

Homework 6.6

1. Since 0 10

52+ = the x-coordinate of the vertex

must be 5.

3. Since 0 6

32+ = the x-coordinate of the vertex must

be 3.

5. Since 0 ( 4)

22

+ − = − the x-coordinate of the vertex

must be -2.

7. Since 0 7

3.52+ = the x-coordinate of the vertex

must be 3.5.

9. Since 2 8

52+ = the x-coordinate of the vertex must

be 5.

11. Since 4 6

12

− + = the x-coordinate of the vertex

must be 1.

13. A symmetric point to the y-intercept has a value of x that is 2 units to the right of 2x = (value of x at the vertex). The value of y is the same as that of the y-intercept, so another point on the parabola is ( )4,9 .

15. A symmetric point to the given point ( )1,1 has a

value of x that is 2 units to the right of 3x = (value of x at the vertex). The value of y is the same as that of the given point, so symmetric point to ( )1,1

on the parabola is ( )5,1 .

17. First, find the y-intercept by substituting 0 for x in the function:

( )20 6 0 7 7y = − + =

The y-intercept is ( )0,7 . Next find the symmetric

point to ( )0,7 . Substitute 7 for y in the function

and solve for x:

( )

2

2

7 6 7

0 6

0 6

0 or 6 00 or 6

x x

x x

x x

x x

x x

= − += −= −= − == =

Therefore the symmetric points are ( )0,7 and

( )6,7 . Since 0 6

32+ = , the x-coordinate of the

vertex is 3. To find the y-coordinate of the vertex, substitute 3 for x and solve for y:

( )23 6 3 7 2y = − + = − . So the vertex is ( )3, 2− .


167


( )20 8 0 9 9y = + + =



and solve for x:

( )

2

2

9 8 9

0 8

0 8

0 or 8 00 or 8

x x

x x

x x

x x

x x

= + += += += + == = −


( )8,9− . Since 0 ( 8)

42

+ − = − , the x-coordinate of

the vertex is -4. To find the y-coordinate of the vertex, substitute -4 for x and solve for y:

( ) ( )24 8 4 9 7y = − + − + = − . So the vertex is

( )4, 7− − .


( ) ( )20 8 0 10 10y = − + − = −

The y-intercept is ( )0, 10− . Next find the

symmetric point to ( )0, 10− . Substitute -10 for y in the function and solve for x:

( )

2

2

10 8 10

0 8

0 8

0 or 8 00 or 8

x x

x x

x x

x x

x x

− = − + −= − += − −

− = − == =

Therefore the symmetric points are ( )0, 10− and

( )8, 10− . Since 0 8



( ) ( )24 8 4 10 6y = − + − = . So the vertex is ( )4,6 .


( ) ( )23 0 6 0 4 4y = + − = −

The y-intercept is ( )0, 4− . Next find the symmetric

point to ( )0, 4− . Substitute -4 for y in the function

and solve for x:

( )

2

2

4 3 6 4

0 3 6

0 3 2

3 0 or 2 00 or 2

x x

x x

x x

x x

x x

− = + −= += += + =

= = −


( )2, 4− − . Since 0 ( 2)

12

+ − = − , the x-coordinate of

the vertex is -1. To find the y-coordinate of the vertex, substitute -1 for x and solve for y:


168

( ) ( )23 1 6 1 4 7y = − + − − = − . So the vertex is

( )1, 7− − .


( ) ( )23 0 12 0 5 5y = − + − = −



and solve for x:

( )

2

2

5 3 12 5

0 3 12

0 3 4

3 0 or 4 00 or 4

x x

x x

x x

x x

x x

− = − + −= − += − −

− = − == =


( )4, 5− . Since 0 4



( ) ( )23 2 12 2 5 7y = − + − = . So the vertex is ( )2,7 .


( ) ( )24 0 9 0 5 5y = − − − = −



and solve for x:

( )

2

2

5 4 9 5

0 4 9

0 4 9

0 or 4 9 00 or 4 9

90 or 2.25

4

x x

x x

x x

x x

x x

x x

− = − − −

= − −= − +

− = + == = −

= = − ≈ −


( )2.25, 5− − . Since 0 ( 2.25)

1.1252

+ − = − , the x-

coordinate of the vertex is -1.125. To find the y-coordinate of the vertex, substitute -1.125 for x and solve for y:

( ) ( )23 1.125 12 1.125 5 0.0625y = − − + − − = . So

the vertex is ( )1.125,0.0625− .


( ) ( )22 0 7 0 7 7y = − + =


point to ( )0,7 . Substitute 7 for y in the function and solve for x:


169

( )

2

2

7 2 7 7

0 2 7

0 2 7

0 or 2 7 00 or 2 7

70 or 3.5

2

x x

x x

x x

x x

x x

x x

= − +

= −= −= − == =

= = =


( )3.5,7 . Since 0 3.5

1.752

+ = , the x-coordinate of

the vertex is 1.75. To find the y-coordinate of the vertex, substitute 1.75 for x and solve for y:

( ) ( )22 1.75 7 1.75 7 0.875y = − + = . So the vertex is

( )1.75,0.875 .

31. First, change the equation to standard form: 2

2

4 6 8

4 8 6

x y x

y x x

− + == − +

Next, find the y-intercept by substituting 0 for x in the function:

( ) ( )24 0 8 0 6 6y = − + =


point to ( )0,6 . Substitute 6 for y in the function and solve for x:

( )

2

2

6 4 8 6

0 4 8

0 4 2

4 0 or 2 00 or 2

x x

x x

x x

x x

x x

= − += −= −

= − == =


( )2,6 . Since 0 2



( ) ( )24 1 8 1 6 2y = − + = . So the vertex is ( )1,2 .

33. First, change the equation to standard form:

( )( )

2

2

2

2

2 3 15

2 6 9 15

2 12 18 15

2 12 33

y x

y x x

y x x

y x x

= − +

= − + +

= − + +

= − +


( ) ( )22 0 12 0 33 33y = − + =



and solve for x:

( )

2

2

33 2 12 33

0 2 12

0 2 6

2 0 or 6 00 or 6

x x

x x

x x

x x

x x

= − += −= −

= − == =


( )6,33 . Since 0 6



( ) ( )22 3 12 3 33 15y = − + = . So the vertex is

( )3,15 .


170

35. The equation is in vertex form. To sketch the graph of 2 6y x= − , which has a vertex of ( )0, 6− ,

translate the graph of 2y x= down 6 units.


( ) ( )22.8 0 8.7 0 4 4y = − + =

The y-intercept is ( )0, 4 . Next find the symmetric

point to ( )0, 4 . Substitute the 4 for y in the

function and solve for x:

( )

2

2

4 2.8 8.7 4

0 2.8 8.7

0 2.8 8.7

0 or 2.8 8.7 00 or 2.8 8.7

8.70 or 3.11

2.8

x x

x x

x x

x x

x x

x x

= − += −= −= − == =

= = ≈

Therefore the symmetric points are ( )0, 4 and

( )3.11,4 . Since 0 3.11

1.562

+ ≈ , the x-coordinate

of the vertex is 1.56. To find the y-coordinate of the vertex, substitute 1.56 for x and solve for y:

( ) ( )22.8 1.56 8.7 1.56 4 2.76y = − + ≈ − . So the

vertex is ( )1.56, 2.76− .


( ) ( )23.9 0 6.9 0 3.4 3.4y = + − = −

The y-intercept is ( )0, 3.4− . Next find the

symmetric point to ( )0, 3.4− . Substitute -3.4 for y

in the function and solve for x:

( )

2

2

3.4 3.9 6.9 3.4

0 3.9 6.9

0 3.9 6.9

0 or 3.9 6.9 00 or 3.9 6.9

6.90 or 1.77

3.9

x x

x x

x x

x x

x x

x x

− = + −

= += += + == = −

= = − ≈ −

Therefore the symmetric points are ( )0, 3.4− and

( )1.77, 3.4− − . Since 0 ( 1.77)

0.882

+ − ≈ − , the x-

coordinate of the vertex is -0.88. To find the y-coordinate of the vertex, substitute -0.88 for x and solve for y:

( ) ( )23.9 1.77 6.9 1.77 3.4 6.45y = − + − − = − . So the

vertex is ( )1.77, 6.45− − .


171

41. First, change the equation to standard form: 2

2

2

2

3.6 2.63 8.3 7.1

3.6 8.3 2.63 7.1

8.3 2.63 7.13.6

2.31 7.31 1.97

y x x

y x x

x xy

y x x

− = −= + −

+ −=

= + −


( ) ( )22.31 0 7.31 0 1.97 1.97y = + − = −

The y-intercept is ( )0, 1.97− . Next find the

symmetric point to ( )0, 1.97− . Substitute -1.97 for

y in the function and solve for x:

( )

2

2

1.97 2.31 7.31 1.97

0 2.31 7.31

0 2.31 7.31

0 or 2.31 7.31 00 or 2.31 7.31

7.310 or 3.16

2.31

x x

x x

x x

x x

x x

x x

− = + −

= += += + == = −

= = − ≈ −

Therefore the symmetric points are ( )0, 1.97− and

( )3.16, 1.97− − . Since ( )0 3.16

1.582

+ −= − , the x-

coordinate of the vertex is 1.58− . To find the y-coordinate of the vertex, substitute 1.58− for x and solve for y:

( ) ( )22.31 1.58 7.31 1.58 1.97 7.75y = − + − − = − .

So the vertex is ( )1.58, 7.75− − .

43. Since the x-intercepts are symmetric points and 2 6

42+ = , the x-coordinate of the vertex is 4.

45. Since the x-intercepts are symmetric points and 9 4 52 2

− + = − , the x-coordinate of the vertex is

52

− .

47. To find the x-intercepts, let 0y = and solve for x:

( )20 5 10

0 5 2

5 0 or 2 00 or 2

x x

x x

x x

x x

= −= −

= − == =

The x-intercepts are ( )0,0 and ( )2,0 . The y-

intercept is, therefore, ( )0,0 . Since the x-intercepts

are symmetric points and 0 2

12+ = , the x-

coordinate of the vertex is 1. Substitute 1 for x in the function and solve for y:

( ) ( )25 1 10 1 5y = − = − . So, the vertex is ( )1, 5− .


( )20 4

0 4

0 or 4 00 or 4

x x

x x

x x

x x

= −= −= − == =




22+ = , the x-

coordinate of the vertex is 2. Substitute 2 for x in


172

the function and solve for y:

( ) ( )25 2 10 2 4y = − = − . So, the vertex is ( )2, 4− .


( )( )20 10 24

0 6 4

6 0 or 4 06 or 4

x x

x x

x x

x x

= − += − −− = − == =

The x-intercepts are ( )6,0 and ( )4,0 . To find the y-intercept, let x = 0 and solve for y:

( ) ( )20 10 0 24 24y = − + = . The y-intercept is

( )0, 24 . Since the x-intercepts are symmetric

points and 6 4

52+ = , the x-coordinate of the vertex

is 5. Substitute 5 for x in the function and solve for

y: ( ) ( )25 10 5 24 1y = − + = − . So, the vertex is

( )5, 1− .


( )( )20 8 7

0 7 1

7 0 or 1 07 or 1

x x

x x

x x

x x

= − += − −− = − == =

The x-intercepts are ( )7,0 and ( )1,0 . To find the y-intercept, let x = 0 and solve for y:

( ) ( )20 8 0 7 7y = − + = . The y-intercept is ( )0,7 .

Since the x-intercepts are symmetric points and 7 1

42+ = , the x-coordinate of the vertex is 4.

Substitute 4 for x in the function and solve for y:

( ) ( )24 8 4 7 9y = − + = − . So, the vertex is ( )4, 9− .


( )( )20 9

0 3 3

3 0 or 3 03 or 3

x

x x

x x

x x

= −= − +− = + == = −

The x-intercepts are ( )3,0 and ( )3,0− . To find the y-intercept, let x = 0 and solve for y:

( )20 9 9y = − = − . The y-intercept is ( )0, 9− .

Since the x-intercepts are symmetric points and ( )3 3

02

+ −= , the x-coordinate of the vertex is 0.

So the vertex is ( )0, 9− .


173


( )( )

20 2 11 21

0 2 3 7

2 3 0 or 7 02 3 or 7

3= 1.5 or 7

2

x x

x x

x x

x x

x x

= − −= + −+ = − == =

= − =

The x-intercepts are ( )1.5,0− and ( )7,0 . To find the y-intercept, let x = 0 and solve for y:

( ) ( )22 0 11 0 21 21y = − − = − . The y-intercept is

( )0, 21− . Since the x-intercepts are symmetric

points and ( )7 1.5

2.752

+ −= , the x-coordinate of

the vertex is 2.75. Substitute 2.75 for x in the function and solve for y:

( ) ( )22 2.75 11 2.75 21 36.125y = − − = − . So, the

vertex is ( )2.75, 36.125− .

59. a. Begin by finding the h-coordinate of the h-intercept:

( ) ( ) ( )20 16 0 140 0 3 3h = − + + =

This means the height of the ball is 3 feet when the matter makes contact (t = 0).

b. Since the function is in the form 2( )h t at bt c= + + and a = -16 < 0, the vertex

is the maximum point. Find the h(t)-coordinate of the vertex. Since the h-intercept is (0, 3) from part a, find the symmetric point by substituting 3 for h(t) in the function and solve for t:

( )

2

2

3 16 140 3

0 16 140

0 4 4 35

4 0 or 4 35 00 or 4 35

350 or 8.75

4

t t

t t

t t

t t

t t

t t

= − + +

= − += − −

− = − == =

= = =

With the same h-coordinate as the h-intercept, the symmetric point is (8.75, 3). Since the average of the t-coordinates of the symmetric points is approximately 4.37, the t-coordinate of the vertex is approximately 4.37. Compute h(4.37) to find the h-coordinate of the vertex:

( ) ( ) ( )24.37 16 4.37 140 4.37 3

309.25

h = − + +≈

So the vertex is (4.37, 309.25), which means that the maximum height of the ball is 309.25 feet and it is reached at 4.37 seconds.

c.

61. a. The function P is in the form 2( )P t at bt c= + + and a = 0.99 > 0, so the

vertex is the minimum point where the profit is the least. Find the vertex. Begin by finding the P-intercept of the function: let t = 0,

( ) ( ) ( )20 0.09 0 1.65 0 9.72 9.72f = − + =

The P-intercept is (0, 9.72), Next, find the symmetric point to (0, 9.72). Substitute 9.72 for P(t) in the function and solve for t:


174

( )

2

2

9.72 0.09 1.65 9.72

0 0.09 1.65

0 0.09 1.65

0 or 0.09 1.65 00 or 0.09 1.65

1.650 or 18.3

0.09

x x

x x

x x

x x

x x

x x

= − += −= −= − == =

= = ≈

With the same P-coordinate as the P-intercept, the symmetric point is (18.3, 97.2). Since the average of the t-coordinates is approximately 9.17, the t-coordinate of the vertex is 9.17. Computer P(9.17) to find the P-coordinate of the vertex:

( ) ( ) ( )29.17 0.09 9.17 1.65 9.17 9.72

2.16

P = − +≈

So the vertex is (9.17, 2.16), which means that in 1999 ( )9.17t ≈ , the profit was the least at 2.16 million dollars.

b.

63. The function f is in the form 2( )f t at bt c= + + where a = 0.0038 > 0, so the vertex is the minimum point of the parabola where the median age is at a minimum. Find the vertex. Begin by finding the f-intercept: let t = 0,

( ) ( ) ( )20 0.00248 0 0.29 0 31.48 31.48f = − + = .

The f-intercept is (0, 31.48). Next, find the symmetric point to (0, 31.48). Substitute 31.48 for f(t) in the function and solve for t:

( )

2

2

31.48 0.00248 0.29 31.48

0 0.00248 0.29

0 0.00248 0.29

0 or 0.00248 0.29 00 or 0.00248 0.29

0.290 or 116.94

0.00248

t t

t t

t t

t t

t t

t t

= − += −= −= − == =

= = ≈

With the same f-coordinate as the f-intercept, the symmetric point is (116.94, 31.48). Since the average of the t-coordinates of the symmetric points is approximately 58.47, the t-coordinate of the vertex is 58.47. Compute f(58.47) to find the f-coordinate of the vertex:

( ) ( ) ( )258.47 0.00248 58.47 0.29 58.47 31.48

23.00

f = − +≈

So the vertex is (58.47, 23) which means that the minimum median age for men was 23 years old in 1958 (t • 58.47).

65. Since the function is in the form 2( )f t at bt c= + + and a = -5.07 < 0, the vertex is the maximum. Find the vertex. Begin by finding the f-intercept: let t = 0,

( ) ( ) ( )20 5.07 0 102.93 0 460.40 460.40f = − + − = − .

The f-intercept is (0, -460.4). Next, find the symmetric point to (0, -460.4). Substitute -460.40 for f(t) in the function and solve for t:

( )

2

2

460.4 5.07 102.93 460.40

0 5.07 102.93

0 5.07 102.93

0 or 5.07 102.93 00 or 5.07 102.93

102.930 or 20.3

5.07

t t

t t

t t

t t

t t

t t

− = − + −

= − += − −

− = − == =

= = ≈

With the same f-coordinate as the f-intercept, the symmetric point is (20.3, -460.4). Since the average of the t-coordinates of the symmetric points is approximately 10.15, the t-coordinate of the vertex is approximately 10.15. Compute f(10.15) to find the f-coordinate of the vertex:

( ) ( ) ( )210.15 5.07 10.15 102.93 10.15 460.4

62.02

f = − + −≈

SSM: Intermediate Algebra Chapter 6 Review Exercises

175

So the vertex is (10.15, 62.02), which means that the sales of premium sports cars was at a maximum of 62.02 thousand cars in 2000 (t = 10.15).

67. a. First find the y-intercept:

2(0) 0 4(0) 12 12f = + − = − .

The y-intercept is (0, -12). Next, find the symmetric point to (0, -12) by substituting -12 for f(x) in the function and solve for x:

( )

2

2

12 4 12

0 4

0 4

0 or 4 00 or 4

x x

x x

x x

x x

x x

− = + −= += += + == = −

The symmetric points are (0, -12) and ( )4, 12− − . Since the average of the x-

coordinates is 0 ( 4)

22

+ − = − , the x-coordinates

of the vertex is -2.

b. First find the x-intercepts:

( )( )20 4 12

0 2 6

2 0 or 6 02 or 6

x x

x x

x x

x x

= + −= − +− = + == = −

The x-intercepts are (2, 0) and (-6, 0). These points are symmetric and since the average of

the x-coordinates is 2 ( 6)

22

+ − = − , the x-

coordinate of the vertex is -2.

c. The results for part a and b are the same.

d. Finding the x-coordinate of the vertex by averaging the x-coordinates of the y-intercept and its symmetric point is the easier method.

e. Finding the x-coordinate of the vertex by averaging the x-coordinates of the y-intercept and its symmetric point is the easier method since this function is prime.

f. It is easier to find the x-coordinate of the vertex for a prime function by averaging the x

coordinates of the y-intercept and its symmetric point.

69. (3, 2) is the vertex for both f and k. The vertex of g is approximately (2.7, 1.8). The vertex of h is approximately (3.3, 1.7).

71. Answers may vary. See page 295 of the text and the description of figure 54.

Chapter 6 Review Exercises

1.

2.

3.

Chapter 6 Review Exercises SSM: Intermediate Algebra

176

4.

5. Since the parabola has a maximum point, a < 0. Since the vertex is in quadrant II, h < 0 and k > 0.

6. This is the graph of 2y x= − , translated up 4 units

and to the right 4 units, so ( )24 4y x= − − + . (The

vertex is (4, 4)).

7. 23 ( 2) ( 3 ) ( 3 )2 3 6x x x x x x x− + = − + − = − −

8. 2( 4) ( 4)( 4)x x x+ = + +

2

2

4 4 16

8 16

x x x

x x

= + + += + +

9. 2 2( 2)( 5) 5 2 10 3 10x x x x x x x+ − = − + − = − −

10. 23 (2 5) 3 (2 5)(2 5)x x x x x− = − −

2 2

2

3 2

3 [(2 ) 10 10 5 ]

3 (4 20 25)

12 60 75

x x x x

x x x

x x x

= − − += − +

= − +

11. 2(2 3)(5 1) (10 2 15 3)x x x x x− − + = − + − −

2

2

(10 13 3)

10 13 3

x x

x x

= − − −= − + +

12. 22 (4 7)(4 7) 2 (16 28 28 49)x x x x x x x− − + = − − + −

2

3

2 (16 49)

32 98

x x

x x

= − −= − +

13. 2 2

1 1 1 1 1 15 8 5 8 5 8

x x x − + = −

21 125 64

x= −

14. 2 2 4 2 2( 3)( 7) 7 3 21x x x x x− − = − − +

4 210 21x x= − +

15. 2 3 2 2( 2)( 5 4) 5 4 2 10 8x x x x x x x x+ − + = − + + − +

3 23 6 8x x x= − − +

16. 2 2(3 2 1)( 3 2)x x x x− + + −

4 3 2 3 2 2

4 3 2

3 9 6 2 6 4 3 2

3 7 11 7 2

x x x x x x x x

x x x x

= + − − − + + + −= + − + −

17. The following functions factor down to 3( 1)( 4)y x x= − + :

23 9 12; 3( 4)( 1)

3 ( 3) 12; ( 4)(3 3)y x x y x x

y x x y x x

= + − = + −= + − = + −

18. 2( ) 2( 1) 3 2( 1)( 1) 3f x x x x= − + − = − + + −

2

2

2

2

2( 1) 3

2( 2 1) 3

2 4 2 3

2 4 5

x x x

x x

x x

x x

= − + + + −= − + + −

= − − − −= − − −

19. 24 12 ( 4 ) ( 4 )(3) 4 ( 3)x x x x x x x− + = − − − = − +

20. 2 2 24 ( 6)( 4)x x x x− − = − +

21. 2 2 2 27 63 7( 9) 7( 3 ) 7( 3)( 3)x x x x x− = − = − = − +

22. 2

2 21 1 1 14 2 2 2

x x x x − = − = − +

23. 2 23 6 45 3( 2 15) 3( 5)( 2)x x x x x x− − = − − = − +

24. 22.4 7.9 (2.4 7.9)x x x x− = −

25. 2 22 16 32 2( 8 16)x x x x− + = − +

2

2( 4)( 4)

2( 4)

x x

x

= − −

= −

26. 28 14 15 (2 5)(4 3)x x x x+ − = + −


177

27. 2 26 33 36 3(2 11 12)x x x x− + = − +

3(2 3)( 4)x x= − −

28. 3 25 45 5 ( 9)x x x x− = −

2 25 ( 3 )

5 ( 3)( 3)x x

x x x

= −= + −

29. 3 2 23 27 42 3 ( 9 14)x x x x x x− + = − +

3 ( 2)( 7)x x x= − −

30. 3 2 24 8 9 18 4 ( 2) 9( 2)x x x x x x+ − − = + − +

2

2 2

( 2)(4 9)

( 2)[(2 ) 3 ]( 2)(2 3)(2 3)

x x

x x

x x x

= + −= + −= + − +

31. 2 2 24 0x x− − =

( 6)( 4) 0

6 0 or 4 06 or 4

x x

x x

x x

− + =− = + == = −

32. 2 25 0x − =

2 25 0( 5)( 5) 0

5 0 or 5 05 or 5

x

x x

x x

x x

− =− + =

− = + == = −

33. 2 10

81x − =

22 1

09

1 10

9 91 1

0 or 09 91 1

or 9 9

x

x x

x x

x x

− = − + =

− = + =

= = −

34. 23 7 2 0x x− + =

(3 1)( 2) 03 1 0 or 2 03 1 or 2

1 or 2

3

x x

x x

x x

x x

− − =− = − == =

= =

35. 210 100x x=

210 100 010 ( 10) 010 0 or 10 0

0 or 10

x x

x x

x x

x x

− =− =

= − == =

36. 2 23 10 5 5x x x x− + = − + +

22 9 5 0(2 1)( 5) 02 1 0 or 5 02 1 or 5

1 or 5

2

x x

x x

x x

x x

x x

+ − =− + =

− = + == = −

= = −

37. 2 33 30 6x x x+ =

3 2

2

6 3 30 0

3 (2 10) 03 (2 5)( 2) 03 0 or 2 5 0 or 2 0

0 or 2 5 or 25

0 or or 22

x x x

x x x

x x x

x x x

x x x

x x x

− − =− − =

− + == − = + =

= = = −

= = = −

38. 3 24 12 3x x x− = −

3 2

2

2

2 2

3 4 12 0

( 3) 4( 3) 0

( 3)( 4) 0

( 3)( 2 ) 0( 3)( 2)( 2) 0

3 0 or 2 0 or 2 03 or 2 or 2

x x x

x x x

x x

x x

x x x

x x x

x x x

+ − + =+ − + =

+ − =+ − =+ − + =

+ = − = + == − = = −


178

39. 23 ( 7) 13 2 7x x x− + = −

2 2

2

3 21 13 2 7

21 20 0( 20)( 1) 0

20 0 or 1 020 or 1

x x x

x x

x x

x x

x x

− + = −− + =− − =

− = − == =

40. 2 220 24 4 9x x x− = −

216 24 9 0(4 3)(4 3) 04 3 04 3

34

x x

x x

x

x

x

− + =− − =

− ==

=

41. To find the x-intercepts, of the function f substitute 0 for f(x) and solve for x:

2

2

( ) 6 7 5

0 6 7 50 (3 5)(2 1)3 5 0 or 2 1 03 5 or 2 1

5 1 or

3 2

f x x x

x x

x x

x x

x x

x x

= − −= − −= − +− = + == = −

= = −

So, the x-intercepts are 5

,03

and 1

,02

− .

42. To find the x-intercepts, of the function f substitute 0 for f(x) and solve for x:

2

2 2

( ) 64 49

0 (8 ) 70 (8 7)(8 7)8 7 0 or 8 7 08 7 or 8 7

7 7 or

8 8

f x x

x

x x

x x

x x

x x

= −= −= − +− = + == = −

= = −

So, the x-intercepts are 7

,08

and 7

,08

− .

43. 2(0) 5(0) (0) 4 4f = − − = −

44. 2

1 1 15 4

5 5 5f = − −

1 15 4

25 51 1

45 5

4

= − −

= − −

= −

45. ( ) ( ) ( )22 5 2 2 4f a a a+ = + − + −

( )( ) ( )( )2

2

2

5 2 2 2 4

5 4 4 2 4

5 20 20 6

5 19 14

a a a

a a a

a a a

a a

= + + − + −

= + + − − −

= + + − −

= + +

46. ( ) 25 4f x x x= − −

25 4 0(5 4)( 1) 05 4 0 or 1 05 4 or 1

4 or 1

5

x x

x x

x x

x x

x x

− − =+ − =

+ = − == − =

= − =

47. ( ) 25 4f x x x= − −

2

2

5 4 2

5 6 0(5 6)( 1) 05 6 0 or 1 05 6 or 1

6 or 1

5

x x

x x

x x

x x

x x

x x

− − =− − =− + =

− = + == = −

= = −

48. ( ) 25 4f x x x= − −

2

2

5 4 4

5 0(5 1) 0

0 or 5 1 00 or 5 1

10 or

5

x x

x x

x x

x x

x x

x x

− − = −− =− =

= − == =

= =


179

49. Since 2 9 72 2

− + = the x-coordinate of the vertex

must be 72

.


( )22(0) 8 0 5 5y = − + + =



and solve for x:

( )

2

2

5 2 8 5

0 2 8

0 2 4

2 0 or 4 00 or 4

x x

x x

x x

x x

x x

= − + += − += − −

− = − == =


( )4,5 . Since 0 4



( )22(2) 8 2 5 13y = − + + = . So the vertex is

( )2,13 .

51. The equation is in vertex form. To sketch the graph of

2 29 9y x x= − = − + , which has a vertex of

( )0,9 , translate the graph of 2y x= − up 9 units.


( )20 3 6

0 3 2

3 0 or 2 00 or 2

x x

x x

x x

x x

= −= −= − =

= =




12+ = , the x-

coordinate of the vertex is 1. Substitute 1 for x in

the function and solve for y: ( ) ( )23 1 6 1 3y = − = − .

So, the vertex is ( )1, 3− .

53. First, change the equation to standard form: 2 2

2

1.7 2.6 6.7 10 2.1

4.1 11.7 2.1

x x y x x

y x x

+ + = − += − +


( ) ( )24.1 0 11.7 0 2.1 2.1y = − + =

The y-intercept is ( )0,2.1 . Next find the symmetric

point to ( )0,2.1 . Substitute 2.1 for y in the

function and solve for x:


180

( )

2

2

2.1 4.1 11.7 2.1

0 4.1 11.7

0 4.1 11.7

0 or 4.1 11.7 00 or 4.1 11.7

11.70 or 2.85

4.1

x x

x x

x x

x x

x x

x x

= − += −= −= − == =

= = ≈

Therefore the symmetric points are ( )0,2.1 and

( )2.85, 2.1 . Since 0 2.85

1.432

+ ≈ , the x-coordinate

of the vertex is 1.43 . To find the y-coordinate of the vertex, substitute 1.43 for x and solve for y:

( ) ( )24.1 1.43 11.7 1.43 2.1 6.25y = − + ≈ − . So the

vertex is ( )1.43, 6.25− .

54. Answers may vary. Example: 2( 5)y x= −


2( 5)( 6) 8 12y x x x x= − − = − +

56. a. Since the function is in the form 2( )h t at bt c= + + and a = -16 < 0, the vertex

is the maximum point. Find the h(t)-coordinate of the vertex.

( ) ( ) ( )20 16 0 100 0 3 3h = − + + =

So, the h-intercept is (0, 3). Next, find the symmetric point by substituting 3 for h(t) in the function and solve for t:

( )

2

2

3 16 100 3

0 16 100

0 4 4 25

4 0 or 4 25 00 or 4 25

250 or 6.25

4

t t

t t

t t

t t

t t

t t

= − + +

= − += − −

− = − == =

= = =

The symmetric points are (0, 3) and (6.25, 3). Since the average of the t-coordinates is 0 6.25

3.1252

+ = , the t-coordinate of the

vertex is 3.125. Substitute 3.125 for t in the function to find the h-coordinate of the vertex:

( ) ( ) ( )23.125 16 3.125 100 3.125 3

159.25

h = − + +=

So the vertex is (3.125, 159.25), which means that the maximum height of the ball is 159.25 feet and it is reached in 3.125 seconds.

b. Solve for t when h(t) = 3. From part a, we see that when h(t) = 3, t is either 0 or 6.25. In this case, the fielder had 6.25 seconds to get into position.

c.

61. The function f is in the form 2( )f t at bt c= + + where a = 0.0027 > 0, so the vertex is the minimum point of the parabola where the median age is at a minimum. Find the vertex. Begin by finding the f-intercept: let t = 0,

( ) ( ) ( )20 0.0027 0 0.31 0 29.29 29.29f = − + = .

SSM: Intermediate Algebra Chapter 6 Test

181

The f-intercept is (0, 29.29). Next, find the symmetric point to (0, 29.29). Substitute 29.29 for f(t) in the function and solve for t:

( )

2

2

29.29 0.0027 0.31 29.29

0 0.0027 0.31

0 0.0027 0.31

0 or 0.0027 0.31 00 or 0.0027 0.31

0.310 or 114.81

0.0027

t t

t t

t t

t t

t t

t t

= − + == −= −= − == =

= = ≈

With the same f-coordinate as the f-intercept, the symmetric point is (114.81, 29.29). Since the average of the t-coordinates of the symmetric points is approximately 57.41, the t-coordinate of the vertex is 57.41. Compute f(57.41) to find the f-coordinate of the vertex:

( ) ( ) ( )257.41 0.0027 57.41 0.31 57.41 29.29

20.39

f = − +=

So the vertex is (57.41, 20.39) which means that the median age for women at their first marriage was at a minimum of 20.4 years old in 1957 (t • 57.41).

Chapter 6 Test

1.

2. Since the vertex lies on the x-axis when x > 0, h > 0 and k = 0. Since the parabola is turned upward (has a minimum point), a > 0.

3. Answers may vary. Example: 2( 2) 7y x= − +

4. 22( 2)( 7) 2( 7 2 14)x x x x x− − + = − + − −

2

2

2( 5 14)

2 10 28

x x

x x

= − + −= − − +

5. 2 2(3 7)(3 7) (3 ) 49 9 49x x x x− + = − = −

6. 22 (5 8) 2 (5 8)(5 8)x x x x x+ = + +

2

2

3 2

2 [(5 ) 40 40 64]

2 (25 80 64)

50 160 128

x x x x

x x x

x x x

= + + += + +

= + +

7. 2 2(2 3)( 2 1)x x x x− + + −

4 3 2 3 2 2

4 3 2

2 4 2 2 3 6 3

2 3 7 3

x x x x x x x x

x x x x

= + − − − + + + −= + − + −

8. 2( ) 2( 6) 11 2( 6)( 6) 11f x x x x= − + + = − + + +

2

2

2

2

2( 6 6 36) 11

2( 12 36) 11

2 24 72 11

2 24 61

x x x

x x

x x

x x

= − + + + += − + + +

= − − − += − − −

9. 28 13 6 ( 2)(8 3)x x x x+ − = + −

10. 3 2 22 12 18 2 ( 6 9)x x x x x x− + = − +

2

2 ( 3)( 3)

2 ( 3)

x x x

x x

= − −= −

11. 2 3 10 0x x− − =

( 5)( 2) 0

5 0 or 2 02 or 2

x x

x x

x x

− + =− = + == = −

12. (2 7)( 3) 10x x− − =

2

2

2

2 6 7 21 10

2 13 21 10

2 13 11 0(2 11)( 1) 02 11 0 or 1 02 11 or 1

11 or 1

2

x x x

x x

x x

x x

x x

x x

x x

− − + =− + =

− + =− − =

− = − == =

= =

13. 225 16 0x − =

Chapter 6 Test SSM: Intermediate Algebra

182

( )( )( )

22(5 ) 4 0

5 4 5 4 0

5 4 0 or 5 4 05 4 or 5 4

4 4 or

5 5

x

x x

x x

x x

x x

− =

− + =− = + == = −

= = −

14. 3 22 3 18 27x x x+ = +

3 2

2

2

2 2

2 3 18 27 0

(2 3) 9(2 3) 0

(2 3)( 9) 0

(2 3)( 3 ) 0(2 3)( 3)( 3) 02 3 0 or 3 0 or 3 02 3 or 3 or 3

3 or 3 or 3

2

x x x

x x x

x x

x x

x x x

x x x

x x x

x x x

+ − − =+ − + =

+ − =

+ − =+ − + =

+ = − = + == − = = −

= − = = −

15. Since the vertex of the parabola is (4, 7), the standard form of its equation is

2( 4) 7y a x= − +

The point (1, 3) satisfies this equation so we can substitute 1 for x and 3 for y to solve for a:

2

2

3 (1 4) 7

3 ( 3) 73 9 7

4 949

a

a

a

a

a

= − += − += +

− =

= −

So the equation of the parabola is

24( 4) 7

9y x= − − +

Any point that satisfies this equation will lie on the parabola. One such point is (7, 3), which is the symmetric point to (1, 3).

16. a. To find the x-intercept, solve for x when ( ) 0f x =

20 5 240 ( 8)( 3)

8 0 or 3 08 or 3

x x

x x

x x

x x

= − −= − +− = + == = −

b. Since the x-intercepts are symmetric points, the average of the x-coordinates for these points is the x-coordinate of the vertex, which

is 8 ( 3)

2.52

+ − = . Substitute 2.5 for x in the

function to find the y-coordinate of the vertex:

2(2.5) 5(2.5) 24 30.25y = − − = −

So, the vertex is (2.5, -30.25).

17. ( ) ( ) ( )20 3 0 10 0 8 8f = + − = −

18. ( ) ( ) ( )23 3 3 10 3 8 11f − = − + − − = −

19. 20 3 10 8x x= + −

0 (3 2)( 4)3 2 0 or 4 03 2 or 4

2 or 4

3

x x

x x

x x

x x

= − +− = + == = −

= = −

20. 21 3 10 8x x= + −

2

2

5 3 10 8

0 3 10 130 (3 13)( 1)3 13 0 or 1 03 13 or 1

13 or 1

3

x x

x x

x x

x x

x x

x x

= + −= + −= + −+ = − == =

= =

21. Since (0, 0.75) lies on the curve, (0) 0.75f = .

22. Since (-3, 0) lies on the curve, ( 3) 0f − = .

23. This is not true for any value of x.

24. Since (1, 1) lies on the curve, x = 1 when ( ) 1f x = .

25. Since (-3, -3) and (5, -3) lie on the curve, x = 3 and x = 5 when ( ) 3f x = −

SSM: Intermediate Algebra Chapter 6 Test

183

26. a. Solve for t when P(t) = 14.1.

2

2

14.1 0.2 2.6 14.1

0 0.2 2.60 0.2 ( 13)0.2 0 or 13 0

0 or 13

t t

t t

t t

t t

t t

= − += −= −

= − == =

So the profit was 14.1 million dollars in 1990 and in 2003.

b. The function f is in the form 2( )f t at bt c= + + where a = 0.2 > 0, so the

vertex is the minimum point of the parabola. Find the vertex. From part a, we know that the point (0, 14.1) and (13, 14.1) lie on P(t) and these are symmetric points. The average of the

t-coordinates of these points is 0 13

6.52+ = ,

which is the t-coordinate of the vertex. Substitute 6.5 for t in the function to find the P-coordinate:

( ) ( ) ( )26.5 0.2 6.5 2.6 6.5 14.1

5.65

P = − +=

So, the vertex is (6.5, 5.65), which means that profit was least in 1997 (t = 7) at 5.65 million dollars.

c.

chapter 6 polynomial functions - higher education | … · homework 6.1 ssm: intermediate algebra...

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