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8/20/2019 Lehmann IA SSM Ch6
1/34
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)
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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 4 y ≥ − .
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 4 y ≥ .
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 4 y ≥ .
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 >
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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) 3 y a x= + +
Where a = -3, -2, -1, -½, ½, 1, 2, 3
35. From the vertex (5, -6), we know2( ) ( 5) 6 f 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)
5
8
0.625
a
a
a
a
a
= − −
= −
=
=
=
The equation of the function is:
2( ) 0.625( 5) 6 f 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 2 y x= +
41. It is possible. Example:2
2 y 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.
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SSM: Intermediate Algebra Homework 6.1
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 1 y x y x= − = − +
d. No, it is not possible.
51. a.
b.
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Homework 6.2 SSM : Intermediate Algebra
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-minand x-max much larger.
55. a.
b.
c.
d.
57. Step 1: Sketch a graph of 2 y ax= .
Step 2: Translate the graph of 2 y 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 14 x x= − +
5. 3 (8 5) 3 (8 ) ( 3 )(5) x x x x x− + = − + −
224 15 x 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− + = − − −
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SSM: Intermediate Algebra Homework 6.2
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 18 x 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 8 x x x x− = −
31.2 2( 2)( 4) 4 2 8 6 8 x x x x x x x+ + = + + + = + +
33.2 2( 3)( 6) 6 3 18 3 18 x x x x x x x− + = + − − = + −
35. 2 2( 8)( 8) 8 8 64 16 64 x 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 25 x x x x x+ − = + − −
2 25 x= −
41.2(3 5)(4 1) 12 3 20 5 x x x x x+ + = + + +
212 23 5 x x= + +
43. 2(4 1)(6 5) 24 20 6 5 x x x x x− − = − − +
224 26 5 x x= − +
45. 2 3 2( 4)( 20) 2 4 8 x x x x x− + = + − −
47.3 2 5 3 2
( 6)( 3) 3 6 18 x 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 4 x 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− + −
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Homework 6.2 SSM : Intermediate Algebra
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 1
2 25 5 x x
− +
2
2
1 2 24
5 5 5
14
25
x x x
x
= + − −
= −
77.2 2( 5) ( 5) x x− − +
2 2
2 2
( 10 25) ( 10 25)
10 25 10 25
20
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 36 f x x x x x x= + + = + + +
2 12 36 x x= + +
83.2( ) 2( 3) 1 2( 3)( 3) 1h x x x x= + + = + + +
2
2
2
2
2( 3 3 9) 12( 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) 2 p 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 10 f 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 36
24
x x x x
x x x x
x
= + + − − +
= + + − + −=
This function is linear.
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SSM: Intermediate Algebra Homework 6.3
157
97.2(4) 4 3(4) 16 12 4 f = − = − =
99.2(2.8) (2.8) 3(2.8) 7.84 8.4 0.56 f = − = − = −
101.
2 2
( ) 3( ) 3 f 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 1 f = − + − = −
107.
21 1 1
2 5 12 2 2
f = − + −
1 52 1
4 2
1 51
2 2
41
2
2 1 1
= − + −
= − + −
= −
= − =
109.2 2( ) 2( ) 5( ) 1 2 5 1 f a a a a a= − + − = − + −
111.
2
( 3) 2( 3) 5( 3) 1 f 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 2
5 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 , 0 x k k + ≠ , is always prime and notfactorable.
25.2 2 264 49 (8 ) 7 (8 7)(8 7) x x x x− = + = − +
27.
2
2 21 1 1 1
9 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+ + = + +
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Homework 6.4 SSM : Intermediate Algebra
158
43. 2 3 2 ( 2)( 1) x x x x− + = − −
45. 215 27 3 (5 9) x x x x− = −
47.
2
10 23 12 (2 3)(5 4) x x x x+ + = + +
49. This expression is prime.
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 2
7 14 21 7( 2 3) x x x x+ + = + +
61. 281 16 (9 4)(9 4) x x x− = − +
63. This expression is prime.
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 25 x + is prime so2 24 100 4( 25) x x+ = + is the final step.
85. Answers may vary. Your discussion should be
include factoring techniques for factoring2
ax 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
= − + −
= − − +
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SSM: Intermediate Algebra Homework 6.4
159
21. 3 2 23 4 12 ( 3) 4( 3) x x x x x x + + + = + + +
2( 3)( 4) x x = + +
23. 3 2 2
4 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 49 x 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 15 x 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 12 x 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 − = − −
71. This expression is prime.
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 = − +
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Homework 6.5 SSM : Intermediate Algebra
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 0 x x − =
( 3) 0
0 or 3 0
0 or 3
x x
x x
x x
− =
= − =
= =
3. 24 6 0 x x − =
2 (2 3) 0
2 0 or 2 3 0
0 or 2 33
0 or2
x x
x x
x x
x x
− =
= − =
= =
= =
5.2 8 15 0 x x + + =
( 3)( 5) 0
3 0 or 5 0
3 or 5
x x
x x
x x
+ + =
+ = + =
= − = −
7.2 4 0 x − =
( )( )
2 2
2 02 2 0
2 0 or 2 0
2 or 2
x
x x
x x
x x
− =
+ − =
+ = − =
= − =
9. 27 6 1 1 x x + + =
( )
( )
27 6 0
7 6 0
7 6 0
7 0 or 6 0
0 or 6
x x
x x
x x
x x
x x
+ =
+ =
+ =
= + =
= = −
11. 23 6 x x =
( )
( )
23 6 0
3 2 0
3 2 0
3 0 or 2 0
0 or 2
x x
x x
x x
x x
x x
− =
− =
− =
= − =
= =
13. 29 2 5 x x = − +
( )( )
22 9 5 0
2 1 5 0
2 1 0 or 5 0
2 1 or 5
1 or 5
2
x x
x x
x x
x x
x x
+ − =
− + =
− = + =
= = −
= = −
15. 29 12 4 0 x x − + =
( )( )3 2 3 2 0
3 2 0
3 2
2
3
x x
x
x
x
− − =
− =
=
=
17.23 12 x =
( )
( )( )
( )( )
2
2
3 12 0
3 4 0
3 2 2 0
2 2 0
2 0 or 2 0
2 or 2
x
x
x x
x x
x x
x x
− =
− =
− + =
− + =
− = + =
= = −
19. 216 25 x =
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161
( )( )
2
2 2
16 25 0
(4 ) 5 0
4 5 4 5 0
4 5 0 or 4 5 0
4 5 or 4 5
5 5 or
4 4
x
x
x x
x x
x x
x x
− =
− =
− + =
− = + == = −
= = −
21. (5 3) 4 4 x x + − = −
( )
2
2
5 3 4 4
5 3 0
5 3 0
0 or 5 3 0
0 or 5 33
0 or5
x x
x x
x x
x x
x x
x x
+ − = −
+ =
+ =
= + =
= = −= = −
23. 3 ( 2) 4 x x x − =
( )
2
2
3 6 4
3 10 0
3 10 0
0 or 3 10 0
0 or 3 10
100 or
3
x x x
x x
x x
x x
x x
x x
− =
− =
− =
= − == =
= =
25.2
9 49 0 x − =
( )( )
2 2(3 ) 7 0
3 7 3 7 0
3 7 0 or 3 7 0
3 7 or 3 7
7 7 or
3 3
x
x x
x x
x x
x x
− =
− + =
− = + == = −
= = −
27. 2
1 x =
( )( )
2
2 2
1 0
1 0
1 1 0
1 0 or 1 0
1 or 1
x
x
x x
x x
x x
− =
− =
− + =
− = + == = −
29.2
64 28 9 x x = −
( )( )
29 28 64 0
3 8 3 8 0
3 8 0
3 8
8
3
x x
x x
x
x
x
− + =
− − =
− ==
=
31.2
4 8 32 x x − =
( )
( )( )
2
2
2
4 8 32 0
4 2 8 0
2 8 0
4 2 0
4 0 or 2 0
4 or 2
x x
x x
x x
x x
x x
x x
− − =
− − =
− − =
− + =
− = + == = −
33.2
12 36 0 x x − + =
( )( )6 6 0
6 0
6
x x
x
x
− − =
− ==
35. 2 1
025
x − =
2
2 10
5
1 10
5 5
1 10 or 0
5 5
1 1 or
5 5
x
x x
x x
x x
− =
− + =
− = + =
= = −
37. 2
12 2 2 0 x x − − =
( )
( )( )
2
2
2 6 1 0
6 1 0
2 1 3 1 0
2 1 0 or 3 1 0
2 1 or 3 1
1 1 or
2 3
x x
x x
x x
x x
x x
x x
− − =
− − =
− + =
− = + == = −
= = −
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39. 21 1
64 2
x x − =
( )
( )( )
2
2
2
1 14 4 6
4 22 24
2 24 0
6 4 0
6 0 or 4 0
6 or 4
x x
x x
x x
x x
x x
x x
− = − =
− − =
− + =
− = + == = −
41.3
6 24 0 x x − =
( )
( )
( )( )
2
2 2
6 4 0
6 2 0
6 2 2 0
6 0 or 2 0 or 2 0
0 or 2 or 2
x x
x x
x x x
x x x
x x x
− =
− =
− + =
= − = + == = = −
43. 3 2
4 2 36 18 0 x 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 0
2 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 2
4 7 x x =
( )
3
2
2
4 7 0
4 7 0
0 or 4 7 0
0 or 4 7
70 or
4
x
x x
x x
x x
x x
− =
− == − =
= =
= =
47.3 2
4 9 20 45 x 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 0
5 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 2
9 12 4 27 x x x − = −
( ) ( )
( )( )
( )( )
( )( )( )
3 2
2
2
2 2
9 27 4 12 09 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 0
3 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 2
18 3 6 x 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 0
0 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. 2
17 28 3 x x − = −
( )( )
23 17 28 0
3 4 7 0
3 4 0 or 7 0
3 4 or 7
4 or 7
3
x x
x x
x x
x x
x x
− − =
+ − =
+ = − == − =
= − =
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55. ( )( )2 5 40 x x+ + =
( ) ( )
2
2
7 10 40
7 30 0
3 10 0
3 0 or 10 0
3 or 10
x x
x x
x x
x x
x x
+ + =
+ − =
− + =− = + == = −
57. ( ) ( )4 1 24 3 2 x x x x− − = −
( )( )
2 2
2
4 4 24 3 6
2 24 0
4 6 0
4 0 or 6 0
4 or 6
x x x x
x x
x x
x x
x x
− − = −
+ − =
− + =
− = + =
= = −
59. ( )( )2 5 6 2 3 x x x x+ + = + +
61. 2 5 6 0 x x+ + =
( )( )2 3 0
2 0 or 3 0
2 or 3
x x
x x
x x
+ + =
+ = + == − = −
63. 2 7 10 0 x x− + =
( )( )2 5 02 0 or 5 0
2 or 5
x x x x
x x
− − =− = − == =
65. ( )( )2 7 10 2 5 x 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 0
4 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 ( ) 0 f x =
( )( )
20 9 20
0 5 4
5 0 or 4 0
5 or 4
x x
x x
x x
x x
= − +
= − −
− = − == =
The x-intercepts are (5, 0) and (4, 0).
73. Solve for x when ( ) 0 f x =
( )( )
2
2 2
0 16
0 4
0 2 2
2 0 or 2 0
2 or 2
x
x
x x
x x
x x
= −
= −
= + −
+ = − =
= − =
The x-intercepts are (-2, 0) and (2, 0).
75. Solve for x when ( ) 0 f x =
( )
( )( )
2
2
2
0 2 20 32
0 2 10 16
0 10 16
0 8 2
8 0 or 2 0
8 or 2
x x
x x
x x
x x
x x
x x
= − +
= − +
= − +
= − −
− = − =
= =
The x-intercepts are (8, 0) and (2, 0).
77. Solve for x when ( ) 0 f x =
( )( )
20 36 25
0 6 5 6 5
6 5 0 or 6 5 0
6 5 or 6 5
5 5 or
6 6
x
x x
x x
x x
x x
= −
= − +
− = + == = −
= = −
The x-intercepts are5
,06
and5
,06
−
.
79. Solve for x when ( ) 0 f x =
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164
( )( )
20 12 7 10
0 3 2 4 5
3 2 0 or 4 5 0
3 2 or 4 5
2 5 or
3 4
x x
x x
x x
x x
x x
= − −
= + −
+ = − == − =
= − =
The x-intercepts are2
,03
−
and5
,04
.
81. Solve for x when ( ) 0 f x =
( )
20 3 30
0 3 10
3 0 or 10 0
0 or 10
x x
x x
x x
x x
= −
= −
= − =
= =
The x-intercepts are (0, 0) and (10, 0).
83. Solve for x when ( ) 0 f x =
( )
( )( )
3 2
2
0 2 2 84
0 2 42
0 2 7 6
2 0 or 7 0 or 6 0
0 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).
85. Solve for x when ( ) 0 f x =
( ) ( )
( )( )
( )( )
( )( )( )
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 0
2 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 0
21 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) 130 153 f = − + =
In 2003, the sales will be 153 million cases.
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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
7
4
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 0
2 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
7
47 91 10 91 7
7
637 4 70 637
0 4 70
0 2 2 35
2 0 or 2 35 0
0 or 2 35
350 or
2
0 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. ( ) ( ) ( )2
9 9 9 6 66 f = − − =
93. ( ) ( ) ( )2
7 7 7 6 50 f − = − − − − =
95. ( ) 2 6 f x x x= − −
( )( )
2
2
14 6
0 20
0 5 4
5 0 or 4 0
5 or 4
x x
x x
x x
x x
x x
= − −
= − −
= − +
− = + == = −
97. ( ) 2 6 f x x x= − −
( )
2
2
6 6
0
0 1
0 or 1 0
0 or 1
x x
x x
x x
x x
x x
− = − −
= −
= −
= − == =
99. Since ( )2,5− lies on the parabola, ( )2 5 f − = .
101. Since ( )1, 1− lies on the parabola, ( )1 1 f = − .
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 19 f = .
109. Since ( )4,3 is in the table, ( )4 3 f = .
111. Since ( )0,19 and ( )6,19 are in the table,
( )0 and 6 when 19 x x f x= = = .
113. Since ( )3,1 is in the table, ( )3 when 1 x f x= = .
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166
115. a. Since ( )1,9 and ( )5,9 are in the table,
( )1 and 5 when 9 x x f x= = = .
b. The function f does not have an inverse
function since the output 9 corresponds to notone input, but two (1 and 5).
117. Answers may vary. Example:
2( ) ( 5)( 1) 4 5 f x x x x x= + − = + −
119. Answers may vary. Example:
2( 3)( 1) 2 3 y 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” whichprecedes this homework section.
Homework 6.6
1. Since0 10
52
+= the x-coordinate of the vertex
must be 5.
3. Since0 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. Since0 7
3.52
+= the x-coordinate of the vertex
must be 3.5.
9. Since2 8
52
+= the x-coordinate of the vertex must
be 5.
11. Since4 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 2 x = (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 3 x = (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 7 y = − + =
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 0
0 or 6
x x
x x
x x
x x
x x
= − +
= −
= −
= − =
= =
Therefore the symmetric points are ( )0,7 and
( )6,7 . Since0 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 2 y = − + = − . So the vertex is ( )3, 2− .
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19. First, find the y-intercept by substituting 0 for x in
the function:
( )20 8 0 9 9 y = + + =
The y-intercept is ( )0,9 . Next find the symmetric
point to ( )0,9 . Substitute 9 for y in the function
and solve for x:
( )
2
2
9 8 9
0 8
0 8
0 or 8 0
0 or 8
x x
x x
x x
x x
x x
= + +
= +
= +
= + =
= = −
Therefore the symmetric points are ( )0,9 and
( )8,9− . Since0 ( 8)
42
+ −= − , the x-coordinate of
the vertex is -4. To find the y-coordinate of thevertex, substitute -4 for x and solve for y:
( ) ( )2
4 8 4 9 7 y = − + − + = − . So the vertex is
( )4, 7− − .
21. First, find the y-intercept by substituting 0 for x in
the function:
( ) ( )2
0 8 0 10 10 y = − + − = −
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 0
0 or 8
x x
x x
x x
x x
x x
− = − + −
= − +
= − −
− = − =
= =
Therefore the symmetric points are ( )0, 10− and
( )8, 10− . Since0 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 10 6 y = − + − = . So the vertex is ( )4,6 .
23. First, find the y-intercept by substituting 0 for x in
the function:
( ) ( )2
3 0 6 0 4 4 y = + − = −
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 0
0 or 2
x x
x x
x x
x x
x x
− = + −
= +
= +
= + =
= = −
Therefore the symmetric points are ( )0, 4− and
( )2, 4− − . Since0 ( 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:
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168
( ) ( )2
3 1 6 1 4 7 y = − + − − = − . So the vertex is
( )1, 7− − .
25. First, find the y-intercept by substituting 0 for x in
the function:
( ) ( )2
3 0 12 0 5 5 y = − + − = −
The y-intercept is ( )0, 5− . Next find the symmetric
point to ( )0, 5− . Substitute -5 for y in the function
and solve for x:
( )
2
2
5 3 12 5
0 3 12
0 3 4
3 0 or 4 0
0 or 4
x x
x x
x x
x x
x x
− = − + −
= − +
= − −
− = − =
= =
Therefore the symmetric points are ( )0, 5− and
( )4, 5− . Since0 4
22
+= , the x-coordinate of the
vertex is 2. To find the y-coordinate of the vertex,
substitute 2 for x and solve for y:
( ) ( )2
3 2 12 2 5 7 y = − + − = . So the vertex is ( )2,7 .
27. First, find the y-intercept by substituting 0 for x in
the function:
( ) ( )2
4 0 9 0 5 5 y = − − − = −
The y-intercept is ( )0, 5− . Next find the symmetric
point to ( )0, 5− . Substitute -5 for y in the function
and solve for x:
( )
2
2
5 4 9 5
0 4 9
0 4 9
0 or 4 9 0
0 or 4 9
9
0 or 2.254
x x
x x
x x
x x
x x
x x
− = − − −
= − −
= − +
− = + =
= = −
= = − ≈ −
Therefore the symmetric points are ( )0, 5− and
( )2.25, 5− − . Since0 ( 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:
( ) ( )2
3 1.125 12 1.125 5 0.0625 y = − − + − − = . So
the vertex is ( )1.125,0.0625− .
29. First, find the y-intercept by substituting 0 for x in
the function:
( ) ( )2
2 0 7 0 7 7 y = − + =
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:
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( )
2
2
7 2 7 7
0 2 7
0 2 7
0 or 2 7 0
0 or 2 7
70 or 3.5
2
x x
x x
x x
x x
x x
x x
= − +
= −
= −
= − =
= =
= = =
Therefore the symmetric points are ( )0,7 and
( )3.5,7 . Since0 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:
( ) ( )2
2 1.75 7 1.75 7 0.875 y = − + = . 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:
( ) ( )2
4 0 8 0 6 6 y = − + =
The y-intercept is ( )0,6 . Next find the symmetric
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 0
0 or 2
x x
x x
x x
x x
x x
= − +
= −
= −
= − =
= =
Therefore the symmetric points are ( )0,6 and
( )2,6 . Since0 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:( ) ( )
24 1 8 1 6 2 y = − + = . 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
= − +
= − + +
= − + +
= − +
Next, find the y-intercept by substituting 0 for x in
the function:
( ) ( )2
2 0 12 0 33 33 y = − + =
The y-intercept is ( )0,33 . Next find the symmetric
point to ( )0,33 . Substitute 33 for y in the function
and solve for x:
( )
2
2
33 2 12 33
0 2 12
0 2 6
2 0 or 6 0
0 or 6
x x
x x
x x
x x
x x
= − +
= −
= −
= − =
= =
Therefore the symmetric points are ( )0,33 and
( )6,33 . Since0 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:
( ) ( )2
2 3 12 3 33 15 y = − + = . So the vertex is
( )3,15 .
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35. The equation is in vertex form. To sketch the graph
of 2 6 y x= − , which has a vertex of ( )0, 6− ,
translate the graph of 2 y x= down 6 units.
37. First, find the y-intercept by substituting 0 for x in
the function:
( ) ( )2
2.8 0 8.7 0 4 4 y = − + =
The y-intercept is ( )0,4 . Next find the symmetricpoint 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 0
0 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 . Since0 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:
( ) ( )2
2.8 1.56 8.7 1.56 4 2.76 y = − + ≈ − . So the
vertex is ( )1.56, 2.76− .
39. First, find the y-intercept by substituting 0 for x in
the function:
( ) ( )2
3.9 0 6.9 0 3.4 3.4 y = + − = −
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 0
0 or 3.9 6.9
6.90 or 1.773.9
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− − . Since0 ( 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:
( ) ( )2
3.9 1.77 6.9 1.77 3.4 6.45 y = − + − − = − . So the
vertex is ( )1.77, 6.45− − .
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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 7.1
3.6
2.31 7.31 1.97
y x x
y x x
x x
y
y x x
− = −
= + −
+ −=
= + −
Next, find the y-intercept by substituting 0 for x in
the function:
( ) ( )2
2.31 0 7.31 0 1.97 1.97 y = + − = −
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 0
0 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:
( ) ( )2
2.31 1.58 7.31 1.58 1.97 7.75 y = − + − − = − .
So the vertex is ( )1.58, 7.75− − .
43. Since the x-intercepts are symmetric points and
2 64
2
+= , the x-coordinate of the vertex is 4.
45. Since the x-intercepts are symmetric points and
9 4 5
2 2
− += − , the x-coordinate of the vertex is
5
2− .
47. To find the x-intercepts, let 0 y = and solve for x:
( )
20 5 10
0 5 2
5 0 or 2 0
0 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 and0 2
12
+= , the x-
coordinate of the vertex is 1. Substitute 1 for x in
the function and solve for y:
( ) ( )2
5 1 10 1 5 y = − = − . So, the vertex is ( )1, 5− .
49. To find the x-intercepts, let 0 y = and solve for x:
( )
20 4
0 4
0 or 4 0
0 or 4
x x
x x
x x
x x
= −
= −
= − =
= =
The x-intercepts are ( )0,0 and ( )4,0 . The y-
intercept is, therefore, ( )0,0 . Since the x-intercepts
are symmetric points and0 4
22
+= , the x-
coordinate of the vertex is 2. Substitute 2 for x in
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the function and solve for y:
( ) ( )2
5 2 10 2 4 y = − = − . So, the vertex is ( )2, 4− .
51. To find the x-intercepts, let 0 y = and solve for x:
( )( )
20 10 24
0 6 4
6 0 or 4 0
6 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:
( ) ( )2
0 10 0 24 24 y = − + = . The y-intercept is
( )0,24 . Since the x-intercepts are symmetric
points and6 4
52
+= , the x-coordinate of the vertex
is 5. Substitute 5 for x in the function and solve for
y: ( ) ( )2
5 10 5 24 1 y = − + = − . So, the vertex is
( )5, 1− .
53. To find the x-intercepts, let 0 y = and solve for x:
( )( )
20 8 7
0 7 1
7 0 or 1 0
7 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:
( ) ( )2
0 8 0 7 7 y = − + = . The y-intercept is ( )0,7 .
Since the x-intercepts are symmetric points and
7 14
2
+= , the x-coordinate of the vertex is 4.
Substitute 4 for x in the function and solve for y:
( ) ( )2
4 8 4 7 9 y = − + = − . So, the vertex is ( )4, 9− .
55. To find the x-intercepts, let 0 y = and solve for x:
( )( )
20 9
0 3 3
3 0 or 3 0
3 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 9 y = − = − . The y-intercept is ( )0, 9− .
Since the x-intercepts are symmetric points and
( )3 30
2
+ −
= , the x-coordinate of the vertex is 0.
So the vertex is ( )0, 9− .
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57. To find the x-intercepts, let 0 y = and solve for x:
( )( )
20 2 11 21
0 2 3 7
2 3 0 or 7 0
2 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:
( ) ( )2
2 0 11 0 21 21 y = − − = − . The y-intercept is
( )0, 21− . Since the x-intercepts are symmetric
points and( )7 1.5
2.75
2
+ −
= , the x-coordinate of
the vertex is 2.75. Substitute 2.75 for x in the
function and solve for y:
( ) ( )2
2 2.75 11 2.75 21 36.125 y = − − = − . So, the
vertex is ( )2.75, 36.125− .
59. a. Begin by finding the h-coordinate of the h-
intercept:
( ) ( ) ( )2
0 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 form2
( )h t at bt c= + + and a = -16 < 0, the vertexis 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 0
0 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:
( ) ( ) ( )2
4.37 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 form2( )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,
( ) ( ) ( )
2
0 0.09 0 1.65 0 9.72 9.72 f = − + =
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 :
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( )
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 0
0 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:
( ) ( ) ( )2
9.17 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,
( ) ( ) ( )2
0 0.00248 0 0.29 0 31.48 31.48 f = − + = .
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 0
0 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:
( ) ( ) ( )2
58.47 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,
( ) ( ) ( )2
0 5.07 0 102.93 0 460.40 460.40 f = − + − = − .
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 0
0 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 theaverage 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:
( ) ( ) ( )2
10.15 5.07 10.15 102.93 10.15 460.4
62.02
f = − + −
≈
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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 12 f = + − = − .
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 0
0 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 is0 ( 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 0
2 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 is2 ( 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.
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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 2 y x= − , translated up 4 units
and to the right 4 units, so ( )2
4 4 y x= − − + . (The
vertex is (4, 4)).
7. 23 ( 2) ( 3 ) ( 3 )2 3 6 x 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 10 x 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 21 1 1 1 1 1
5 8 5 8 5 8 x x x
− + = −
21 1
25 64 x= −
14. 2 2 4 2 2( 3)( 7) 7 3 21 x x x x x− − = − − +
4 210 21 x x= − +
15.
2 3 2 2
( 2)( 5 4) 5 4 2 10 8 x x x x x x x x+ − + = − + + − + 3 23 6 8 x 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= − + :
2
3 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) 3 f 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 1
4 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+ − = + −
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27. 2 2
6 33 36 3(2 11 12) x x x x − + = − +
3(2 3)( 4) x x = − −
28.
3 2
5 45 5 ( 9) x x x x
− = − 2 2
5 ( 3 )
5 ( 3)( 3)
x x
x x x
= −= + −
29. 3 2 2
3 27 42 3 ( 9 14) x x x x x x − + = − +
3 ( 2)( 7) x x x = − −
30. 3 2 2
4 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 0 x x − − =
( 6)( 4) 0
6 0 or 4 0
6 or 4
x x
x x
x x
− + =− = + == = −
32. 2
25 0 x − =
2 25 0
( 5)( 5) 0
5 0 or 5 0
5 or 5
x
x x
x x
x x
− =− + =
− = + == = −
33. 2 1
081
x − =
2
2 10
9
1 10
9 9
1 10 or 0
9 9
1 1 or
9 9
x
x x
x x
x x
− =
− + =
− = + =
= = −
34. 2
3 7 2 0 x x − + =
(3 1)( 2) 0
3 1 0 or 2 0
3 1 or 21
or 23
x x
x x
x x
x x
− − =− = − =
= == =
35. 2
10 100 x x =
210 100 0
10 ( 10) 0
10 0 or 10 0
0 or 10
x x
x x
x x
x x
− =− =
= − == =
36. 2 2
3 10 5 5 x x x x − + = − + +
22 9 5 0
(2 1)( 5) 0
2 1 0 or 5 0
2 1 or 5
1 or 5
2
x x
x x
x x
x x
x x
+ − =− + =
− = + == = −
= = −
37. 2 3
3 30 6 x x x + =
3 2
2
6 3 30 0
3 (2 10) 0
3 (2 5)( 2) 0
3 0 or 2 5 0 or 2 0
0 or 2 5 or 2
50 or or 2
2
x x x
x x x
x x x
x x x
x x x
x x x
− − =
− − =− + =
= − = + == = = −
= = = −
38. 3 2
4 12 3 x 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 0
3 or 2 or 2
x x x
x x x
x x
x x
x x x
x x x
x x x
+ − + =
+ − + =
+ − =+ − =+ − + =
+ = − = + == − = = −
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39. 23 ( 7) 13 2 7 x x x− + = −
2 2
2
3 21 13 2 7
21 20 0
( 20)( 1) 0
20 0 or 1 0
20 or 1
x x x
x x
x x
x x
x x
− + = −
− + =
− − =− = − == =
40. 2 220 24 4 9 x x x− = −
216 24 9 0
(4 3)(4 3) 0
4 3 0
4 3
3
4
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 5
0 (3 5)(2 1)
3 5 0 or 2 1 0
3 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 are5
,03
and1
,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 ) 7
0 (8 7)(8 7)
8 7 0 or 8 7 0
8 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 are7
,08
and7
,08
−
.
43. 2(0) 5(0) (0) 4 4 f = − − = −
44.
21 1 1
5 45 5 5
f = − −
1 15 4
25 51 1
45 5
4
= − −
= − −
= −
45. ( ) ( ) ( )2
2 5 2 2 4 f 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 4 f x x x= − −
25 4 0
(5 4)( 1) 0
5 4 0 or 1 0
5 4 or 1
4 or 1
5
x x
x x
x x
x x
x x
− − =+ − =
+ = − == − =
= − =
47. ( ) 25 4 f x x x= − −
2
2
5 4 2
5 6 0
(5 6)( 1) 0
5 6 0 or 1 0
5 6 or 1
6 or 1
5
x x
x x
x x
x x
x x
x x
− − =
− − =− + =
− = + == = −
= = −
48. ( ) 25 4 f x x x= − −
2
2
5 4 4
5 0
(5 1) 0
0 or 5 1 0
0 or 5 1
10 or
5
x x
x x
x x
x x
x x
x x
− − = −
− =− =
= − == =
= =
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SSM: Intermediate Algebra Chapter 6 Review Exercises
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49. Since2 9 7
2 2
− += the x-coordinate of the vertex
must be7
2.
50. First, find the y-intercept by substituting 0 for x in
the function:
( )22(0) 8 0 5 5 y = − + + =
The y-intercept is ( )0,5 . Next find the symmetric
point to ( )0,5 . Substitute 5 for y in the function
and solve for x:
( )
2
2
5 2 8 5
0 2 8
0 2 4
2 0 or 4 0
0 or 4
x x
x x
x x
x x
x x
= − + +
= − +
= − −
− = − =
= =
Therefore the symmetric points are ( )0,5 and
( )4,5 . Since0 4
22
+= , the x-coordinate of the
vertex is 2. To find the y-coordinate of the vertex,
substitute 2 for x and solve for y:
( )
22(2) 8 2 5 13 y = − + + = . So the vertex is
( )2,13 .
51. The equation is in vertex form. To sketch the graph
of
2 29 9 y x x= − = − + , which has a vertex of
( )0,9 , translate the graph of 2 y x= − up 9 units.
52. To find the x-intercepts, let 0 y = and solve for x:
( )
20 3 6
0 3 2
3 0 or 2 0
0 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 and0 2
12
+= , the x-
coordinate of the vertex is 1. Substitute 1 for x in
the function and solve for y: ( ) ( )2
3 1 6 1 3 y = − = − .
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
+ + = − +
= − +
Next, find the y-intercept by substituting 0 for x in
the function:
( ) ( )2
4.1 0 11.7 0 2.1 2.1 y = − + =
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:
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Chapter 6 Review Exercises SSM : Intermediate Algebra
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 0
0 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 . Since0 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:
( ) ( )2
4.1 1.43 11.7 1.43 2.1 6.25 y = − + ≈ − . So the
vertex is ( )1.43, 6.25− .
54. Answers may vary. Example: 2( 5) y x= −
55. Answers may vary. Example:
2( 5)( 6) 8 12 y x x x x= − − = − +
56. a. Since the function is in the form2( )h t at bt c= + + and a = -16 < 0, the vertex
is the maximum point. Find the h(t )-coordinate
of the vertex.
( ) ( ) ( )2
0 16 0 100 0 3 3h = − + + =
So, the h-intercept is (0, 3). Next, find thesymmetric 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 0
0 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 6.253.125
2
+= , 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:
( ) ( ) ( )2
3.125 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 minimumpoint of the parabola where the median age is at a
minimum. Find the vertex. Begin by finding the f -
intercept: let t = 0,
( ) ( ) ( )2
0 0.0027 0 0.31 0 29.29 29.29 f = − + = .
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SSM: Intermediate Algebra Chapter 6 Test
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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 0
0 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) 7 y 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 49 x 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) 11 f 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 0 x x− − =
( 5)( 2) 0
5 0 or 2 0
2 or 2
x x
x x
x x
− + =
− = + =
= = −
12. (2 7)( 3) 10 x x− − =
2
2
2
2 6 7 21 10
2 13 21 10
2 13 11 0
(2 11)( 1) 0
2 11 0 or 1 0
2 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 0 x − =
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Chapter 6 Test SSM : Intermediate Algebra
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( )
( )( )
22(5 ) 4 0
5 4 5 4 0
5 4 0 or 5 4 0
5 4 or 5 44 4
or5 5
x
x x
x x
x x
x x
− =
− + =
− = + =
= = −
= = −
14. 3 22 3 18 27 x 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) 0
2 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) 7 y 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) 7
3 9 7
4 9
4
9
a
a
a
a
a
= − +
= − +
= +
− =
= −
So the equation of the parabola is
24 ( 4) 79
y 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
( ) 0 f x =
20 5 24
0 ( 8)( 3)
8 0 or 3 0
8 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
is8 ( 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 30.25 y = − − = −
So, the vertex is (2.5, -30.25).
17. ( ) ( ) ( )2
0 3 0 10 0 8 8 f = + − = −
18. ( ) ( ) ( )2
3 3 3 10 3 8 11 f − = − + − − = −
19. 20 3 10 8 x x= + −
0 (3 2)( 4)
3 2 0 or 4 0
3 2 or 4
2 or 4
3
x x
x x
x x
x x
= − +
− = + =
= = −
= = −
20. 21 3 10 8 x x= + −
2
2
5 3 10 8
0 3 10 13
0 (3 13)( 1)
3 13 0 or 1 0
3 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) 0.75 f = .
22. Since (-3, 0) lies on the curve, ( 3) 0 f − = .
23. This is not true for any value of x.
24. Since (1, 1) lies on the curve, x = 1 when ( ) 1 f x = .
25. Since (-3, -3) and (5, -3) lie on the curve, x = 3 and
x = 5 when ( ) 3 f x = −
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