Math 2201
Chapter 6
Final Exam Review
Multiple Choice
____
1. Which relation is quadratic?
y = (x + 5)2
y = (2x2)(x + 1)
y = x2 – x2 + 4x + 2
y = 2x – 6x + 3
a.
b.
c.
d.
____
2. What is the degree of a quadratic function?
a.
b.
c.
d.
____
1
3
2
0
3. What is the y-intercept for y = 3x2 + 2x – 5?
a.
b.
c.
d.
____
–5
5
2
3
4. Which parabola opens upward?
y = 2x – 4x2 – 5
y = 4 – 2x2 –5x
y = 2 + 4x – 5x2
y = –5x + 4x2 + 2
a.
b.
c.
d.
____
5. Which set of data is correct for this graph?
y
3
2
1
–5
–4
–3
–2
–1
–1
1
2
3
4
5
x
–2
–3
–4
–5
–6
–7
Set
A.
B.
C.
D.
a.
b.
c.
d.
____
Axis of Symmetry
x = –2
x = –6
x = –2
x=2
Vertex
(–2, 6)
(–6, –2)
(–2, –6)
(2, 6)
Domain
x∈R
–8 ≤ x ≤ 4
x∈R
–6 ≤ x ≤ 2
Range
y∈R
–8 ≤ y
–6 ≤ y
–6 ≤ y
Set A.
Set C.
Set B.
Set D.
6. Which set of data is correct for this graph?
y
5
4
3
2
1
–5
–4
–3
–2
–1
–1
1
2
3
4
5
x
–2
–3
–4
–5
A.
B.
C.
D.
Axis of Symmetry
x = –2
x = –0.25
x = –0.5
x=3
June 2014
Vertex
(–0.25, –3.125)
(–0.25, 3.125)
(–0.5, 3)
(3, –0.5)
Domain
x∈R
x∈R
–2.5 ≤ x ≤ 1.5
–3 ≤ x ≤ 2
Range
y∈R
y ≤ 3.125
y≤3
y≤5
1
Math 2201
a.
b.
c.
d.
____
7. What are the x- and y-intercepts for the function f(x) =x2 –2x + 3?
x = –3, x = 1, y = 3
no x-intercepts, y = 3
x = 0, x = 3, y = 2
x = –1, x = 3, y = 3
8. What are the x- and y-intercepts for the function f(x) =x2 – 2x – 8?
a.
b.
c.
d.
____
Final Exam Review
Set B.
Set D.
Set C.
Set A.
a.
b.
c.
d.
____
Chapter 6
no x-intercepts, y = –8
x = –2, x = 4, y = –8
x = –2, x = 2, y = –8
x = –4, x = 4, y = –8
9. Which set of ordered pairs satisfy the function f(x) = x2 – 4x + 6?
a.
b.
c.
d.
(–1, 9), (1, 3), (2, 2)
(–3, 27), (0, 6), (5, 11)
(2, 2), (4, 6), (7, 30)
(–2, 18), (–1, 9), (6, 18)
____ 10. The points (–2, 4) and (1, 4) are located on the same parabola. What is the equation for the axis of symmetry
for this parabola?
a.
b.
c.
d.
x = –1.5
x = 0.5
x = –0.5
x = –1
____ 11. The points (–1, 14) and (9, 14) are located on the same parabola. What is the equation for the axis of
symmetry for this parabola?
a.
b.
c.
d.
x=5
x=4
x = –4.5
x=7
____ 12. What is the correct quadratic function for this parabola?
y
3
2
1
–5
–4
–3
–2
–1
–1
1
2
3
4
5
x
–2
–3
–4
–5
–6
–7
a.
b.
c.
d.
f(x) = (x + 2)(x + 3)
f(x) = (x – 2)(x + 3)
f(x) = (x + 2)(x – 3)
f(x) = (x – 2)(x – 3)
____ 13. What is the correct quadratic function for this parabola?
y
5
4
3
2
1
–5
–4
–3
–2
–1
–1
1
2
3
4
5
x
–2
–3
–4
–5
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Math 2201
a.
b.
c.
d.
Chapter 6
Final Exam Review
f(x) = (1 – x)(x + 2)
f(x) = (x – 1)(x + 2)
f(x) = (x + 1)(x + 2)
f(x) = (x – 1)(x – 2)
____ 14. Which set of data is correct for the quadratic relation f(x) = –2(x – 1)(x – 5)?
A.
B.
C.
D.
a.
b.
c.
d.
x-intercepts
(1, 0), (5, 0)
(–1, 0), (–5, 0)
(–1, 0), (–5, 0)
(1, 0), (5, 0)
y-intercept
y = –10
y = 10
y = –10
y = 10
Axis of Symmetry
x=3
x = –3
x=3
x = –3
Vertex
(3, 8)
(–3, 64)
(3, –8)
(3, 64)
Set A.
Set B.
Set D.
Set C.
____ 15. Which set of data is correct for the quadratic relation f(x) = 4(x – 0.5)(x + 1)?
A.
B.
C.
D.
a.
b.
c.
d.
x-intercepts
(–0.5, 0), (1, 0)
(0.5, 0), (–1, 0)
(–0.5, 0), (1, 0)
(0.5, 0), (–1, 0)
y-intercept
y = –2
y = –2
y = 0.5
y = –0.5
Axis of Symmetry
x = 0.25
x = –0.25
x = 0.5
x = –0.5
Vertex
(0.25, –1.25)
(–0.25, –2.25)
(0.5, 0)
(–0.5, –2)
Set A.
Set B.
Set C.
Set D.
____ 16. Which set of data is correct for the quadratic relation f(x) = (x + 1)(x – 2)?
A.
B.
C.
D.
a.
b.
c.
d.
x-intercepts
(1, 0), (–2, 0)
(–1, 0), (2, 0)
(–1, 0), (–2, 0)
(1, 0), (2, 0)
y-intercept
y=2
y = –2
y=2
y=2
Axis of Symmetry
x = –0.5
x = 0.5
x = –1.5
x = 1.5
Vertex
(–0.5, –1.25)
(0.5, –2.25)
(–1.5, 1.75)
(1.5, ––1.25)
Set C.
Set D.
Set A.
Set B.
____ 17. Which relation is the factored form of f(x) = x2 + 2x – 3?
a.
b.
c.
d.
f(x) = (x + 3)(x – 1)
f(x) = x(x + 2) + 3
f(x) = (x – 2)2
f(x) = (x – 3)(x + 1)
____ 18. Which relation is the factored form of f(x) = –2x2 – 5x – 2?
a.
b.
c.
d.
f(x) = –2(x + 2)(x + 0.5)
f(x) = 2(x – 2)(x – 0.5)
f(x) = 2(x – 2)(x + 0.5)
f(x) = –2(x – 1)(x – 5)
____ 19. Which relation is the factored form of f(x) = x2 + 2x – 8?
a.
b.
c.
d.
f(x) = (x + 2)(x – 4)
f(x) = 2(x + 2)(x – 2)
f(x) = (x – 1)(x + 8)
f(x) = (x – 2)(x + 4)
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Math 2201
Chapter 6
Final Exam Review
____ 20. Which relation is the factored form of f(x) = x2 – 4x + 4?
a.
b.
c.
d.
f(x) = (x – 2)2
f(x) = (x + 4)(x – 1)
f(x) = 4(x – 1)2
f(x) = (x – 2)(x + 2)
____ 21. Which set of data is correct for the quadratic relation f(x) = –2(x – 12)2 + 15?
Direction parabola opens
Vertex
Axis of Symmetry
downward
(15, –12)
x = 15
A.
downward
(12, 15)
x = 12
B.
upward
(–12, 15)
x = –12
C.
upward
(15, 12)
x = 15
D.
a.
b.
c.
d.
Set D.
Set A.
Set C.
Set B.
____ 22. Which set of data is correct for the quadratic relation f(x) = 5(x – 27)2 – 9?
Direction parabola opens
Vertex
Axis of Symmetry
upward
(27, –9)
x = 27
A.
downward
(–27, 9)
x = –27
B.
upward
(–27, –9)
x = –9
C.
downward
(27, 9)
x=9
D.
a.
b.
c.
d.
Set D.
Set B.
Set A.
Set C.
____ 23. Which function has a maximum value?
a.
b.
c.
d.
f(x) = –14(x + 4)2 – 7
f(x) = 4(x + 7)2 + 9
f(x) = (x – 9)2 –14
f(x) = 7(x + 14)2 – 4
____ 24. Which function has a maximum value?
a.
b.
c.
d.
f(x) = –3(x – 12)2 + 5
f(x) = 2(x – 15)2 – 3
f(x) = (x – 13)2 + 12
f(x) = 1.2(x + 3)2 + 1.5
____ 25. Which function has a minimum value?
a.
b.
c.
d.
f(x) = –3.2(x – 4.2)2 + 1.6
f(x) = –3(x – 7.5)2 – 2.6
f(x) = –2(x + 4)2 + 6
f(x) = 0.5(x – 2.2)2 + 6.1
____ 26. Which function has a minimum value?
a.
b.
c.
d.
f(x) = –(x – 15)2 + 5
f(x) = (x – 5)2 + 15
f(x) = –(x + 1)2 – 5
f(x) = –(x – 5)2 + 10
____ 27. How many zeros does f(x) = a(x – 2)2 + 5 have if a > 0?
a.
b.
c.
d.
2
0
1
It is impossible to determine.
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Math 2201
Chapter 6
Final Exam Review
____ 28. How many zeros does f(x) = a(x – 5)2 have if a < 0?
a.
b.
c.
d.
It is impossible to determine.
1
0
2
____ 29. How many zeros does f(x) = (x – c)2 + d have if d > 0?
a.
b.
c.
d.
1
0
It is impossible to determine.
2
____ 30. Which quadratic function represents this parabola?
y
5
4
3
2
1
–5
–4
–3
–2
–1
–1
1
2
3
4
5
x
–2
–3
–4
–5
a.
b.
c.
d.
f(x) = –(x + 2)2 + 1
f(x) = –(x – 2)2 + 1
f(x) = –(x + 2)2 – 1
f(x) = (x – 2)2 + 1
____ 31. Which quadratic function represents this parabola?
y
5
4
3
2
1
–5
–4
–3
–2
–1
–1
1
2
3
4
5
x
–2
–3
–4
–5
a.
b.
c.
d.
f(x) = 4(x + 1.5)2 – 2
f(x) = –4(x + 1.5)2 + 2
f(x) = 4(x + 1.5)2 + 2
f(x) = 4(x – 1.5)2 – 2
____ 32. Which quadratic function does not have the range y ≥ 0?
a.
b.
c.
d.
y = (x + 14)2
y = x2 – 14x + 49
y = x2 + 14x + 49
y = x2 + 14x – 49
____ 33. Which quadratic function does not have the domain x ∈ R and the range y ≤ 4?
a.
b.
c.
d.
y = –0.25(x – 2)(x – 11)
y = –(x – 5)(x – 9)
y = –2(x – 7)2 + 4
y = –x2 + 14x – 45
____ 34. Which quadratic function does not have the domain x ∈ R and the range y ≥ –6?
a.
b.
c.
d.
y = 6(x – 4)(x – 2)
y = 0.5(x – 3)2 – 6
y = 0.5x2 – 2x – 4
y = x2 – 2x – 6
____ 35. Which quadratic function does not have a = –1 and vertex (5, 1)?
a.
b.
c.
d.
y = (4 – x)(x – 6)
y = –(x + 4)(x – 6)
y = –x2 + 10x – 24
y = –(x – 5)2 + 1
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Math 2201
Chapter 6
Final Exam Review
____ 36. Which quadratic function does not have a = 2 and vertex (–3, –8)?
a.
b.
c.
d.
y = (2x + 2)(x + 5)
y = 2x2 + 8x + 12
y = 2(x + 3)2 – 8
y = 2(x + 5)(x + 1)
____ 37. Which quadratic function does not have a line of symmetry at x = 0?
y = x2 + 10
y = 0.5(x + 2)2
y = 3x2 – 12
y = 2(x – 1)(x + 1)
a.
b.
c.
d.
____ 38. Which quadratic function does not have vertex (–5, 4) and does not pass through (–1, 12)?
y = –x2 – 10x – 21
y = –(x + 7)(x + 3)
y = (5 – x)2 + 4
y = –(x + 5)2 + 4
a.
b.
c.
d.
____ 39. Which equation represents the quadratic function y = 0.5(x + 4)(x – 3) in standard form?
y = 0.5x2 + 0.5x – 6
y = 0.5x2 – 3.5x + 6
y = 0.5x2 – 0.5x – 6
y = 0.5x2 + 3.5x + 6
a.
b.
c.
d.
____ 40. Which equation represents the quadratic function y = –2(x + 1)(x – 5) in standard form?
y = –2x2 + 4x + 8
y = –2x2 + 12x – 10
y = –2x2 + 8x – 12
y = –2x2 + 8x + 10
a.
b.
c.
d.
____ 41. Which equation represents the quadratic function y = 2(x – 0.5)2 – 9 in standard form?
y = 2x2 + 2x + 8.5
y = 2x2 – 0.5x – 8.5
y = 2x2 – 2x – 8.5
y = 2x2 – 2x – 9.5
a.
b.
c.
d.
____ 42. Which equation represents the quadratic function y = –(x + 7)2 – 8 in standard form?
y = –x2 – 14x – 49
y = –x2 – 14x – 41
y = –x2 – 14x – 57
y = –(x2 – 14x + 41)
a.
b.
c.
d.
____ 43. Which quadratic function defines this parabola in vertex form?
y
5
4
3
2
1
–5
–4
–3
–2
–1
–1
1
2
3
4
5
x
–2
–3
–4
–5
a.
b.
c.
d.
y = –(x + 2)2 + 5
y = (–2x + 0.5)2 + 2
y = –2(x – 1)2 + 3
y = –2(x + 1)2 + 4
Short Answer
44. If a parabola with equation y = ax2 + bx + c opens downward, will a be positive or negative?
45. If a parabola with equation y = ax2 + bx + c has a y-intercept above the x-axis, will c be positive or negative?
June 2014
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Math 2201
Chapter 6
Final Exam Review
46. If a parabola with equation y = ax2 + bx + c has a line of symmetry to the left of the y-axis, will b be positive
or negative?
47. If a parabola with equation y = ax2 + bx + c opens upward, will a be positive or negative?
48. If a parabola with equation y = ax2 + bx + c has a y-intercept below the x-axis, will c be positive or negative?
49. Make a table of values, then sketch the graph of the relation y = x2 + 2x + 11.
50. Fill in the table for the relation y = x2 + 2x + 11.
y-intercept
x-intercept(s)
Axis of symmetry
Vertex
Domain
Range
51. Fill in the table for the relation y = –0.5x2 + 0.5x + 3.
y-intercept
x-intercept(s)
Axis of symmetry
Vertex
Domain
Range
52. Fill in the table for the relation y = x2 – 1.5x + 2.
Maximum or minimum
Axis of symmetry
Vertex
53. Determine the domain and range for the relation y = x2 – 6x – 4.
54. Determine the domain and range for the relation y = –x2 – 3x.
55. Fill in the table for the relation y = –2x2 + 2x – 1.
Maximum or minimum
Axis of symmetry
Vertex
56. A quadratic function has an equation that can be written in the form f(x) = a(x – r)(x – s). The graph of the
function has x-intercepts at (–4, 0) and (2, 0) and passes through the point (–2, –8). Write the equation of the
function.
57. A quadratic function has an equation that can be written in the form f(x) = a(x – r)(x – s). The graph of the
function has x-intercepts at (3, 0) and (6, 0) and passes through the point (7, –4). Write the equation of the
function.
58. Fill in the table for the quadratic function f(x) = –2(x + 5)(x + 2).
y-intercept
Zeros
Axis of symmetry
Vertex
59. Sketch the graph of the relation y = –2(x + 5)(x + 2), then state the domain and range.
60. Fill in the table for the quadratic function f(x) = 3(x – 0.5)(x – 1).
y-intercept
Zeros
Axis of symmetry
Vertex
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Math 2201
Chapter 6
Final Exam Review
61. Use the graph to determine the equation of the parabola.
y
9
8
7
6
5
4
3
2
1
–1
–1
1
2
3
4
5
6
7
8
9
x
62. Use the graph to determine the equation of the parabola.
1
–2
y
–1
1
2
x
–1
–2
–3
63. Determine the equation that defines a quadratic function with x-intercepts located at (–9, 0) and (–2, 0) and a
y-intercept of (0, 18). Provide a sketch to support your work.
64. Determine the equation that defines a quadratic function with x-intercepts located at (4, 0) and (12, 0) and a
y-intercept of (0, –24). Provide a sketch to support your work.
65. Sketch the graph of f(x) = –(x – 4)2 + 2, then state the domain and range of the function.
66. Sketch the graph of f(x) = 2(x – 1.5)2 + 3, then state the domain and range of the function.
67. Determine the value of a given that (7, –6) satisfies the quadratic function
f(x) = a(x – 5)2 + 6.
68. Determine the value of a given that (1, 20) satisfies the quadratic function
f(x) = a(x + 3)2 + 12.
69. Use the graph to determine the equation of the parabola.
y
25
20
15
10
5
–6
–4
–2
–5
2
4
6
8
10
12
14
x
–10
–15
–20
–25
70. How many zeros does f(x) = 0.5(x – 3)2 + 18 have? Test your prediction by sketching the graph.
71. How many zeros does f(x) = –0.5(x + 3)2 + 18 have? Test your prediction by sketching the graph.
72. How many zeros does f(x) = –0.5(x – 3)2 – 18 have? Test your prediction by sketching the graph.
73. Given a = –3 and vertex (–2, 0.5), determine the quadratic function in vertex form:
y = a(x – h)2 + k.
74. Given a = 8 and vertex (0.25, –4), determine the quadratic function in vertex form:
y = a(x – h)2 + k.
75. Given point (2, 2) and vertex (5, –2.5), determine the quadratic function in vertex form:
y = a(x – h)2 + k.
76. Given point (–11, –11) and vertex (–9, 1), determine the quadratic function in vertex form:
y = a(x – h)2 + k.
77. Determine the quadratic function that contains the factors (x + 5) and (x – 9) and the point (5, –20). Express
your answer in vertex form.
78. a) Sketch the graph of y = (x + 1)(x – 7).
b) State the maximum or minimum value of the function.
c) Express the function in standard form.
79. a) Sketch the graph of y = 3(x + 0.5)2 + 2.
b) State the maximum or minimum value of the function.
c) Express the function in standard form.
June 2014
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Math 2201
Chapter 6
Final Exam Review
80. Determine the quadratic function that defines the parabola in vertex form.
y
16
14
12
10
8
6
4
2
–4
2
–2
–2
4
6
8
10
12
14
16
x
–4
81. Determine the quadratic function that defines the parabola in vertex form.
y
10
8
6
4
2
–4
–2
–2
2
4
6
8
10
12
14
16
x
–4
–6
–8
–10
82. A juice bar sells 480 1-L jugs of fruit juice per day at a price of $4.50. The previous year sales have revealed
that for every $0.50 increase in price, the store will sell 20 fewer 1-L jugs. What price will maximize the juice
bar's revenue? Include any quadratic functions you use in your answer.
83. A juice bar sells 200 fresh fruit smoothies per day at a price of $6.00. The previous year sales have revealed
that for every $0.50 decrease in price, the store will sell 50 additional smoothies. What price will maximize
the juice bar's revenue? Include any quadratic functions you use in your answer.
Problem
84. The height of a soccer ball above the ground, y, in metres, is modelled by the function
y = –4.9x2 + 5x + 1, where x is the time in seconds after the ball is kicked.
a) Use technology to determine the maximum height the ball will reach. Round your answer to the nearest
tenth of a metre.
b) State any restrictions on the domain and range of the function.
c) For how long is the ball in the air?
85. The height of a golf ball above the ground, y, in metres, is modelled by the function
y = –4.9x2 + 10x, where x is the time in seconds after the ball is hit.
a) Use technology to determine the maximum height the ball will reach. Round your answer to the nearest
tenth of a metre.
b) State any restrictions on the domain and range of the function.
c) For how long is the ball in the air?
86. The height, in metres, of a fireworks rocket is modelled by the function
h(t) = –4.9t2 + 20t + 4, where t is the time in seconds after the rocket is fired.
a) Determine the domain and range of the function to the nearest tenth.
b) Use technology to graph the function.
87. The height, in metres, of a fireworks rocket is modelled by the function
h(t) = –4.9t2 + 25t + 10, where t is the time in seconds after the rocket is fired.
a) Determine the domain and range of the function to the nearest tenth.
b) Use technology to graph the function.
88. A skier’s jump can be modelled by the function y = –4.9x2 + 3.2x + 2.5, where y is the skier's height above the
ground, in metres, and x is the time, in seconds, that the skier is in the air.
a) Use technology to graph the function.
b) Determine the coordinates of the vertex.
c) Determine the skier’s maximum height and state the range of this function.
89. A skier’s jump can be modelled by the function y = –4.9x2 + 3x + 4, where y is the skier's height above the
ground, in metres, and x is the time, in seconds, that the skier is in the air.
a) Use technology to graph the function.
b) Determine the coordinates of the vertex.
c) Determine the skier’s maximum height and state the range of this function.
90. A fire hose placed on the ground sprays water in a path modelled by the quadratic function f(x) = –0.18x2 +
2x, where x is the horizontal distance and f(x) the height of the sprayed water, both in metres. How high and
how far did the hose spray the water?
91. A fire hose placed on the ground sprays water in a path modelled by the quadratic function f(x) = –0.16x2 +
2x, where x is the horizontal distance and f(x) the height of the sprayed water, both in metres. How high and
how far did the hose spray the water?
June 2014
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Math 2201
Chapter 6
Final Exam Review
92. Rachel took a series of photos at 0.25 s intervals of a friend throwing a horseshoe. The table of values shows
the height of the horseshoe at each time. What was the maximum height of the horseshoe?
Time (s) Height
(ft)
0.00
1.5
0.25
4.5
0.50
6.0
0.75
6.0
1.00
4.5
1.25
1.5
93. Omar took a series of photos at 0.20 s intervals of a friend practicing shot put. The table of values shows the
height of the weighted ball at each time. What was the maximum height of the ball?
Time (s)
Height (m)
0.00
1.8
0.20
5.4
0.40
7.2
0.60
7.2
0.80
5.4
1.00
1.8
94. Rosa is building three enclosed gardens as shown. She bought 200 m of fencing and wants to maximize the
total area for the gardens. She wrote the function A(x) = –x2 + 100x to represent the total area of the gardens,
A(x), in square metres, if each garden is x metres wide.
a) Determine the maximum total area of the three gardens.
b) State the domain and range of the variables in her equation.
c) What are the dimensions of one garden?
95. Explain how you would graph the function y = 4(x + 1)2 – 4 without using a table of values or a graphing
calculator.
96. A parabola has the vertex (–7, –2).
a) Write an equation to describe all parabolas with this vertex.
b) A parabola with the given vertex passes through the point (–9, 10). Determine the equation for this
parabola.
c) State the domain and range of the function.
97. A parabolic arch has zeros located at (2, 0) and (32, 0). The parabola has a maximum height of 112.5 ft.
a) Define the equation of the parabola in vertex form. Explain your reasoning.
b) State the domain and range of the function describing the arch.
98. A parabolic arch has zeros located at (–10, 0) and (22, 0). The parabola has a maximum height of 512 cm.
a) Define the equation of the parabola in vertex form. Explain your reasoning.
b) State the domain and range of the function describing the arch.
99. Gordon and Hanna are standing 10 ft apart, playing badminton. They use a video camera to determine that the
path of the birdie on one volley is defined by the function
h(x) = –0.04(x – 5)2 + 7, where x is the horizontal distance, measured in feet, from Gordon toward Hanna.
a) Determine the axis of symmetry of the parabola.
b) What was the highest point of the birdie's path?
c) How high was the birdie from the ground when it was 2.5 ft from Gordon?
d) What is the range for this function? Justify your answer.
100. A subway tunnel forms a parabolic arch. The arch is 8 m wide at the base and 8 m high in the centre. A 0.4 m
wide emergency walkway is built along the side of the tunnel. Can a 6 ft tall person walk upright along the
walkway? (1 ft = 0.3048 m
June 2014
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Math 2201
Chapter 6
Final Exam Review
MULTIPLE CHOICE
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
31.
32.
33.
34.
35.
36.
37.
38.
39.
40.
41.
42.
43.
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
A
C
A
D
B
A
B
B
B
C
C
B
A
A
B
D
A
A
D
B
D
C
A
A
D
B
B
B
B
A
A
D
A
D
B
B
B
C
A
D
C
C
D
SHORT ANSWER
44. ANS:
negative
45.
ANS:
positive
46.
ANS:
negative
47.
ANS:
positive
48.
ANS:
negative
49.
ANS:
x
y
–3
–2
–1
0
1
14
11
10
11
14
y
20
18
16
14
12
10
8
6
4
2
–5
50.
51.
–4
–3
–2
–1
1
2
3
4
5
x
ANS:
y-intercept
x-intercept(s)
Axis of symmetry
Vertex
Domain
Range
(0, 11)
none
x = –1
(–1, 10)
x∈R
y ≥ 10
ANS:
y-intercept
x-intercept(s)
Axis of symmetry
Vertex
Domain
Range
(0, 3)
(–2, 0), (3, 0)
x = 0.5
(0.5, 3.125)
x∈R
y ≤ 3.125
June 2014
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Math 2201
52.
Chapter 6
ANS:
Maximum or minimum
Axis of symmetry
Vertex
minimum
x = 0.75
(0.75, 1.4375)
53.
ANS:
Domain: x ∈ R; Range: y ≥ –13
54.
ANS:
Domain: x ∈ R; Range: y ≤ 2.25
55.
ANS:
Maximum or minimum
Axis of symmetry
Vertex
56.
ANS:
f(x) = (x + 4)(x – 2)
57.
ANS:
f(x) = –(x – 3)(x – 6)
58.
ANS:
y-intercept
Zeros
Axis of symmetry
Vertex
59.
Final Exam Review
maximum
x = –0.5
(0.5, –0.5)
y = –20
(–5, 0), (–2, 0)
x = –3.5
(–3.5, 4.5)
ANS:
y
5
4
3
2
1
–7
–6
–5
–4
–3
–2
–1
–1
1
2
x
3
–2
–3
–4
–5
x ∈ R, y ≤ 4.5
60.
ANS:
y-intercept
y = 1.5
(0.5, 0), (1, 0)
x = 0.75
(0.75, –0.1875)
Zeros
Axis of symmetry
Vertex
61.
ANS:
y = 0.5(x – 4)(x – 5)
62.
ANS:
y = –3(x – 1)2
63.
ANS:
y = (x + 9)(x + 2)
y
8
6
4
2
–16 –14 –12 –10 –8
–6
–4
–2
–2
2
6 x
4
–4
–6
–8
–10
–12
64.
ANS:
y = –0.5(x – 4)(x – 12)
y
8
6
4
2
–6
–2
–2
–4
2
4
6
8
10
12
x
14
–4
–6
–8
–10
–12
65.
ANS:
y
4
3
2
1
–3
–2
–1
–1
1
2
3
4
5
6
7
x
1
2
3
4
5
x
–2
–3
–4
–5
y ≤ 4, x ∈ R
66.
ANS:
y
8
7
6
5
4
3
2
1
–5
–4
–3
–2
–1
–1
–2
y ≥ 1.5, x ∈ R
67.
ANS:
a = –3
68.
ANS:
a = 0.5
June 2014
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Math 2201
Chapter 6
69.
ANS:
y = 5(x – 10)2 – 20
70.
ANS:
Vertex is at (3, 18) and a > 0 so graph opens upward. No zeroes; the graph is above the x-axis.
71.
ANS:
Vertex is at (–3, 18) and a < 0 so graph opens downward. Two zeroes: (–9, 0) and (3, 0).
72.
ANS:
Vertex is at (3, –18) and a < 0 so graph opens downward. No zeroes; the graph is below the x-axis.
73.
ANS:
y = –3(x + 2)2 + 0.5
74.
ANS:
y = 8(x – 0.25)2 – 4
75.
ANS:
y = 0.5(x – 5)2 – 2.5
76.
ANS:
y = –3(x + 9)2 + 1
77.
ANS:
y = 0.5(x + 2)2 – 24.5
78.
ANS:
a)
Final Exam Review
y
4
2
–10 –8
–6
–4
2
–2
–2
4
6
8
x
10
–4
–6
–8
–10
–12
–14
–16
b) (3, –16)
c) y = x2 – 6x – 7
79.
ANS:
a)
y
4
3
2
1
–2
–1
1
2
x
b) (–0.5, 2)
c) y = 3x2 +3x + 2.75
80.
ANS:
y = –(x – 9)2 + 15
81.
ANS:
y = –0.25(x – 8)2 + 6
82.
ANS:
$8.25; Since the maximum value of R(x) = (4.50 + 0.50x)(480 – 20x) is (7.5, 2722.5).
83.
ANS:
$4.00; Since the maximum value of R(x) = (6 – 0.50x)(200 + 50x) is (4, 1600).
PROBLEM
84.
ANS:
a)
b) 0 ≤ x ≤ 2, 0 ≤ y ≤ 2.3
c) 1.2 s
85.
ANS:
a)
b) 0 ≤ x ≤ 2, 0 ≤ y ≤ 5.1
c) 2.0 s
86.
ANS:
a) 0 ≤ x ≤ 4.27, 0 ≤ y ≤ 24.4
b)
87.
ANS:
a) 0 ≤ t ≤ 5.5, 0 ≤ h(t) ≤ 41.89
b)
June 2014
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Math 2201
88.
Chapter 6
Final Exam Review
ANS:
a)
b) (0.33, 3.0)
c) The skier’s maximum height is 3.0 m. 0 ≤ y ≤ 3.0
89.
ANS:
a)
b) (0.31, 4.5)
c) The skier’s maximum height is 4.5 m. 0 ≤ y ≤ 4.5
90.
ANS:
The coefficient of x2 is negative, so the parabola has a maximum. Make a table of values.
x
y
0.00
0.00
1.00
1.82
2.00
3.28
3.00
4.38
4.00
5.12
5.00
5.50
5.10
5.52
6.00
5.52
7.00
5.18
8.00
4.48
9.00
3.42
10.00
2.00
11.00
0.22
12.00
–1.92
The maximum value is between (5.10, 5.52) and (6, 5.52). Use these points to determine the equation of the axis of symmetry.
x = (5.10 + 6) ÷ 2
x = 5.55
Substitute into the function.
f(x) = –0.18x2 + 2x
f(x) = –0.18(5.55)2 + 2(5.55)
f(x) = 5.56
The maximum height of the water is 5.56 m.
The maximum horizontal distance the water will spray is (5.55)(2) = 11.1 m.
91.
92.
ANS:
The coefficient of x2 is negative, so the parabola has a maximum. Make a table of values.
x
y
0
0
1
1.84
2
3.36
3
4.56
4
5.44
5
6.00
6
6.24
6.5
6.24
7
6.16
8
5.76
9
5.04
10
4.00
11
2.64
12
0.96
13
–1.04
The maximum value is between (6, 6.24) and (6.5, 6.24). Use these points to determine the equation of the axis of symmetry.
x = (6 + 6.5) ÷ 2
x = 6.25
Substitute into the function.
f(x) = –0.16x2 + 2x
f(x) = –0.16(6.25)2 + 2(6.25)
f(x) = 6.25
The maximum height of the water is 6.25 m.
The maximum horizontal distance the water will spray is (6.25)(2) = 12.5 m.
ANS:
Graph the data, determine the axis of symmetry (x = 0.625), and use it to determine the vertex. At 0.625 s, the horseshoe reached its
June 2014
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Math 2201
Chapter 6
Final Exam Review
maximum height of about 6.2 ft.
93.
ANS:
94.
ANS:
a)
Graph the data, determine the axis of symmetry (x = 0.5), and use it to determine the vertex. At 0.5 s, the ball reached its maximum height of
about 7.4 ft.
A(x)
x
0
0
10
900
20
1600
30
2100
40
2400
50
2500
60
2400
70
2100
80
1600
90
900
100
0
The maximum area in 2500 m2.
b) x ≤ 100, 0 ≤ A(x) ≤ 2500
c) x = 50 m, so dimensions are 50 m by
m or 16.7 m.
95.
ANS:
Use different forms of the equation to determine points on the graph.
From y = 4(x + 1)2 – 4, the vertex is (–1, –4) and a > 0, so the graph opens upward. The equation for the line of symmetry is x = –1.
Expand the equation and then factor it to determine the y-intercept and x-intercepts.
y = 4(x + 1)2 – 4
y = 4(x2 + 2x + 1) – 4
y = 4x2 + 8x + 4 – 4
y = 4x2 + 8x
The y-intercept is (0, 0).
y = 4x2 + 8x
y = 4x(x + 2)
The x-intercepts are (0, 0) and (–2, 0).
To draw the graph, plot the vertex and intercepts, and draw a curve through the points.
96.
ANS:
a) y = a(x + 7)2 – 2, where a ∈R.
b) Substitute (–9, 10). into the equation and solve for a.
y = a(x + 7)2 – 2
10 = a(–9 + 7)2 – 2
12 = a(4)
3 =a
The equation is y = 3(x + 7)2 – 2.
c) y ≥ –7, x ∈R
97.
ANS:
a) Use the zeros to write the factored form of the equation.
y = a(x – 2)(x – 32)
Use the x-values of the zeros to determine the axis of symmetry.
x=
x = 17
The maximum height is 112.5 ft and the axis of symmetry is x = 17, so the vertex is
(17, 112.5). Substitute the vertex into the equation and solve for a.
y = a(x – 2)(x – 32)
112.5 = a(17 – 2)(17 – 32)
112.5 = a(–225)
–0.5 = a
The factored equation is y = –0.5(x – 2)(x – 32).
Use a = –0.5 and the vertex to write the vertex form: y = –0.5(x – 17)2 + 112.5.
b) 0 ≤ y ≤ 112.5, 2 ≤ x ≤ 32
98.
ANS:
a) Use the zeros to write the factored form of the equation.
y = a(x + 10)(x – 22)
Use the x-values of the zeros to determine the axis of symmetry.
x=
99.
x=6
The maximum height is 512 cm and the axis of symmetry is x = 60, so the vertex is (6, 512). Substitute the vertex into the equation and solve for a.
y = a(x + 10)(x – 22)
512 = a(6 + 10)(60 – 22)
512 = a(–256)
–2 = a
The factored equation is y = –2(x + 10)(x – 22).
Use a = –2 and the vertex to write the vertex form: y = –2(x – 6)2 + 512.
b) 0 ≤ y ≤ 512, –10 ≤ x ≤ 22
ANS:
June 2014
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Math 2201
Chapter 6
Final Exam Review
a) The vertex is (5, 7), so the axis of symmetry is x = 5.
b) The vertex is (5, 7), so the highest point of the birdie's path is 7 ft.
c) Substitute 2.5 for x in the equation and solve for h(2.5).
h(2.5) = –0.04(2.5 – 5)2 + 7
h(2.5) = –0.04(6.25) + 7
h(2.5) = 6.75
The birdie is 6.75 ft from the ground.
d) 0 ≤ x ≤ 10; Gordon and Hanna are standing 10 ft apart.
100.
ANS:
Set the left base of the arch at (0, 0). The arch is 8 m wide, so the other zero is at (8, 0). Use the zeros to write the factored equation.
y = a(x – 0)(x – 8) or y = ax(x – 8)
The x-coordinate of the vertex is half the base length and the y-coordinate is the arch height:
or (4, 8). Use the vertex to determine the value of
a.
y = ax(x – 8)
8 = a(4)(4 – 8)
8 = a(–16)
a = –0.5
So the equation is y = –0.5x(x – 8).
Substitute x = 0.4 into the equation.
y = –0.5x(x – 8)
y = –0.5(0.4)(0.4 – 8)
y = 1.52
The headroom of the walkway is about 1.5 m.
Convert 6 ft to metres: (6)(0.3048) = 1.8288.
The person could not walk upright along the walkway.
June 2014
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