1-1 Graphing Quadratic Functions
Graph each function. Then state the domain and range.
1. f(x) = x 2 + 6x + 8
SOLUTION:
For f(x) = x 2 + 6x + 8, a = 1, b = 6, and c = 8. c is the y-intercept, so the y-intercept is 8.
Find the axis of symmetry.
The equation of the axis of symmetry is x = –3, so the x-coordinate of the vertex is –3. Because a > 0, the vertex
is a minimum.
Graph the function.
x
x2 + 6x + 8
(x, f(x))
–5
(–5)2 + 6(–5) + 8
3
–4
(–4)2 + 6(–4) + 8
0
–3
(–3)2 + 6(–3) + 8
–1
–2
(–2)2 + 6(–2) + 8
0
–1
(–1)2 + 6(–1) + 8
3
Analyze the graph.
The parabola extends to positive and negative infinity, so the domain is all real numbers. The range is {y | y ≥ –1}.
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1-1 Graphing Quadratic Functions
ANSWER:
D = {all real numbers}, R = {y | y ≥ –1}
2. f(x) = –x 2 – 2x + 2
SOLUTION:
For f(x) = –x 2 – 2x + 2, a = –1, b = –2, and c = 2. c is the y-intercept, so the y-intercept is 2.
Find the axis of symmetry.
The equation of the axis of symmetry is x = –1, so the x-coordinate of the vertex is –1. Because a < 0, the vertex
is a maximum.
Graph the function.
x
–x2 – 2x + 2
(x, f(x))
–3
–(–3)2 – 2(–3) + 2
–1
–2
–(–2)2 – 2(–2) + 2
2
–1
–(–1)2 – 2(–1) + 2
3
0
–(0)2 – 2(0) + 2
2
1
–(1)2 – 2(1) + 2
–1
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1-1 Graphing Quadratic Functions
Analyze the graph.
The parabola extends to positive and negative infinity, so the domain is all real numbers. The range is {y | y ≤ 3}.
ANSWER:
D = {all real numbers}, R = {y | y ≤ 3}
3. f(x) = 2x 2 – 4x + 3
SOLUTION:
For f(x) = 2x 2 – 4x + 3, a = 2, b = –4, and c = 3. c is the y-intercept, so the y-intercept is 3.
Find the axis of symmetry.
The equation of the axis of symmetry is x = 1, so the x-coordinate of the vertex is 1. Because a > 0, the vertex is a
minimum.
Graph the function.
x
2x2 – 4x + 3
(x, f(x))
–1
2(–1)2 – 4(–1) + 3
9
0
2(0)2 – 4(0) + 3
3
1
2(1)2 – 4(1) + 3
1
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1-1 Graphing Quadratic Functions
2
2(2)2 – 4(2) + 3
3
3
2(3)2 – 4(3) + 3
9
Analyze the graph.
The parabola extends to positive and negative infinity, so the domain is all real numbers. The range is {y | y ≥ 1}.
ANSWER:
D = {all real numbers}, R = {y | y ≥ 1}
4. f(x) = –2x 2
SOLUTION:
For f(x) = –2x 2, a = –2, b = 0, and c = 0. c is the y-intercept, so the y-intercept is 0.
Find the axis of symmetry.
The equation of the axis of symmetry is x = 0, so the x-coordinate of the vertex is 0. Because a < 0, the vertex is a
maximum.
Graph the function.
x
–2x2
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(x, f(x))
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1-1 Graphing Quadratic Functions
–2
–2(–2)2
–8
–1
–2(–1)2
–2
0
–2(0)2
0
1
–2(1)2
–2
2
–2(2)2
–8
Analyze the graph.
The parabola extends to positive and negative infinity, so the domain is all real numbers. The range is {y | y ≤ 0}.
ANSWER:
D = {all real numbers}, R = {y | y ≤ 0}
5. f(x) = x 2 – 4x + 4
SOLUTION:
For f(x) = x 2 – 4x + 4, a = 1, b = –4, and c = 4. c is the y-intercept, so the y-intercept is 4.
Find the axis of symmetry.
The equation of the axis of symmetry is x = 2, so the x-coordinate of the vertex is 2. Because a > 0, the vertex is a
minimum.
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1-1 Graphing Quadratic Functions
Graph the function.
x
x2 – 4x + 4
(x, f(x))
0
(0)2 – 4(0) + 4
4
1
(1)2 – 4(1) + 4
1
2
(2)2 – 4(2) + 4
0
3
(3)2 – 4(3) + 4
1
4
(4)2 – 4(4) + 4
4
Analyze the graph.
The parabola extends to positive and negative infinity, so the domain is all real numbers. The range is {y | y ≥ 0}.
ANSWER:
D = {all real numbers}, R = {y | y ≥ 0}
6. f(x) = x 2 – 6x + 8
SOLUTION:
For f(x) = x 2 – 6x + 8, a = 1, b = –6, and c = 8. c is the y-intercept, so the y-intercept is 8.
Find the axis of symmetry.
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1-1 Graphing Quadratic Functions
The equation of the axis of symmetry is x = 3, so the x-coordinate of the vertex is 3. Because a > 0, the vertex is a
minimum.
Graph the function.
x
x2 – 6x + 8
(x, f(x))
1
(1)2 – 6(1) + 8
3
2
(2)2 – 6(2) + 8
0
3
(3)2 – 6(3) + 8
–1
4
(4)2 – 6(4) + 8
0
5
(5)2 – 6(5) + 8
3
Analyze the graph.
The parabola extends to positive and negative infinity, so the domain is all real numbers. The range is {y | y ≥ –1}.
ANSWER:
D = {all real numbers}, R = {y | y ≥ –1}
7. Compare the graph of f(x) to a quadratic function g(x) with a y-intercept of 1 and a vertex at (1, 3). Which
function has a greater maximum? Explain.
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1-1 Graphing Quadratic Functions
SOLUTION:
From the graph, f(x) appears to have a maximum of –1. Graph g(x) using the given information. The vertex is at
(1, 3), so the axis of symmetry is x = 1.
The y-intercept is 1, so (0, 1) is on the graph. Reflect (0, 1) in the axis of symmetry. So, (2, 1) is also on the graph.
Connect the points with a smooth curve. g(x) has the greater maximum. Sample answer: Its vertex is a maximum
point at (1, 3), which is 4 units above the vertex of f(x) which is (1, –1).
ANSWER:
g(x); Sample answer: Its vertex is a maximum point at (1, 3), which is 4 units above the vertex of f(x) which is (1,
–1).
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1-1 Graphing Quadratic Functions
8. Compare the graph of f(x) to a quadratic function g(x) with a y-intercept of 0.5 and a vertex at (–1, –5). Which
function has a lesser minimum? Explain.
SOLUTION:
From the graph, f(x) appears to have a minimum of –1. Graph g(x) using the given information. The vertex is at (–
1, –5), so the axis of symmetry is x = –1.
The y-intercept is 0.5, so (0, 0.5) is on the graph. Reflect (0, 0.5) in the axis of symmetry. So, (–2, –0.5) is also on
the graph.
Connect the points with a smooth curve. g(x) has the lesser minimum. Sample answer: Its vertex is a minimum
point at (–1, –5), which is 4 units below the vertex of f(x) which is (–1, –1).
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1-1 Graphing Quadratic Functions
ANSWER:
g(x); Sample answer: Its vertex is a minimum point at (–1, –5), which is 4 units below the vertex of f(x) which is (–
1, –1).
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1-1 Graphing Quadratic Functions
9. Compare f(x) = x 2 – 10x + 5 to the quadratic function g(x) shown in the table. Which function has the lesser
minimum? Explain.
x
–10
–5
0
5
10
g(x)
170
70
10
–10
10
SOLUTION:
Find the vertex of f(x) by first finding the axis of symmetry, then substituting to find the y-value of the vertex.
The equation of the axis of symmetry is x = 5, so the x-coordinate of the vertex is 5. Because a > 0, the vertex is a
minimum.
The vertex of f(x) is a minimum point at (5, –20).
The points in the table are symmetric around point (5, –10), so the vertex of g(x) is at (5, –10). The points to the
left and right of the vertex are above the vertex, so the vertex is a minimum point.
f(x) has the lesser minimum. Its vertex is a minimum point at (5, –20), which is 10 units below the vertex of g(x)
which is (5, –10).
ANSWER:
f(x); Sample answer: Its vertex is a minimum point at (5, –20), which is 10 units below the vertex of g(x) which is
(5, –10).
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1-1 Graphing Quadratic Functions
10. Compare f(x) = –x 2 + 6x – 15 to the quadratic function g(x) shown in the table. Which function has the greater
maximum? Explain.
x
–6
–3
0
3
6
g(x)
–26
–11
–2
1
–2
SOLUTION:
Find the vertex of f(x) by first finding the axis of symmetry, then substituting to find the y-value of the vertex.
The equation of the axis of symmetry is x = 3, so the x-coordinate of the vertex is 3. Because a < 0, the vertex is a
maximum.
The vertex of f(x) is a maximum point at (3, –6).
The points in the table are symmetric around point (3, 1), so the vertex of g(x) is at (3, 1). The points to the left
and right of the vertex are below the vertex, so the vertex is a maximum point.
g(x) has the greater maximum. Its vertex is a maximum point at (3, 1), which is 7 units above the vertex of f(x)
which is (3, –6).
ANSWER:
g(x); Sample answer: Its vertex is a maximum point at (3, 1), which is 7 units above the vertex of f(x) which is (3,
–6).
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1-1 Graphing Quadratic Functions
11. FISHING A county park sells annual permits to its fishing lake. Last year, the county sold 480 fishing permits for
$80 each. This year, the park is considering a price increase. They estimate that for each $4 increase, they will sell
16 fewer annual fishing permits.
a. Define variables and write a function to represent the situation.
b. How much would the park have to charge for each permit in order to maximize its revenue from fishing permits?
c. Find the domain and range in the context of this situation.
SOLUTION:
a. Let x = the number of price increases, and let y = revenue. P(x) is equal to the price of each permit(80 + 4x)
times the total number of passes sold (480 – 16x). So, the equation is P(x) = (80 + 4x)(480 – 16x).
b. Simplify the equation, find the axis of symmetry, and substitute to find the amount the park should charge.
Find the axis of symmetry. Because a is negative, the vertex is a maximum.
The park will make the most money with 5 price increases, so they should charge 80 + 4(5) or $100 for each
permit.
c. Because the number of price increases cannot be negative, the domain must be greater than zero. The number of
permits sold cannot be negative, 480 ÷ 16 = 30. So the domain is {x | 0 ≤ x ≤ 30}. The range is {y | 0 ≤ y ≤
40,000} because the amount of money generated cannot be negative, and the maximum amount of money
generated is P(5) = −64(5)2 + 640(5) + 38,400 or 40,000.
ANSWER:
a. For x = the number of price increases and y = revenue, y = (80 + 4x)(480 – 16x).
b. $100 per permit
c. D = {x | 0 ≤ x ≤ 30}, R = {y | 0 ≤ y ≤ 40,000}
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1-1 Graphing Quadratic Functions
12. SALES Last month, a retailer sold 120 jar candles at $30 per candle. This month the retailer is considering
putting the candles on sale. They estimate that for each $2 decrease in price, they will sell 10 additional candles.
a. How much should the retailer charge for each candle in order to maximize its profit?
b. Find the domain and the range in the context of this situation.
SOLUTION:
a. Let x = the number of price decreases, and let y = revenue. P(x) is equal to the price of each candle (30 – 2x)
times the total number of candles sold (120 + 10x). So, the equation is P(x) = (30 – 2x)(120 + 10x).
Simplify the equation, find the axis of symmetry, and substitute to find the amount the park should charge.
Find the axis of symmetry. Because a is negative, the vertex is a maximum.
The retailer will make the most money with 1.5 price decreases, so they should charge 30 – 2(1.5) or $27 for each
candle.
b. The domain is {x | 0 ≤ x ≤ 15} because the number of price decreases and the number of candles sold cannot
be negative. The range is {y | 0 ≤ y ≤ 3645} because the amount of money generated cannot be negative, and the
maximum amount of money generated is P(1.5) = −20(1.5)2 + 60(1.5) + 3600 or 3645.
ANSWER:
a. $27
b. D = {x | 0 ≤ x ≤ 15}, R = {y | 0 ≤ y ≤ 3645}
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1-1 Graphing Quadratic Functions
Determine the average rate of change of f(x) over the specified interval.
13. f(x) = x 2 – 10x + 5; interval [–4, 4]
SOLUTION:
The average rate of change is equal to
.
First find f(4) and f(−4).
Then substitute to find the average rate of change.
The average rate of change of the function over the interval [–4, 4] is –10.
ANSWER:
–10
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1-1 Graphing Quadratic Functions
14. f(x) = 2x 2 + 4x – 6; interval [–3, 3]
SOLUTION:
The average rate of change is equal to
.
First find f(3) and f(−3).
Then substitute to find the average rate of change.
The average rate of change of the function over the interval [–3, 3] is 4.
ANSWER:
4
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1-1 Graphing Quadratic Functions
15. f(x) = 3x 2 – 3x + 1; interval [–5, 5]
SOLUTION:
The average rate of change is equal to
.
First find f(5) and f(−5).
Then substitute to find the average rate of change.
The average rate of change of the function over the interval [–5, 5] is –3.
ANSWER:
–3
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1-1 Graphing Quadratic Functions
16. f(x) = 4x 2 + x + 3; interval [–2, 2]
SOLUTION:
The average rate of change is equal to
.
First find f(2) and f(−2).
Then substitute to find the average rate of change.
The average rate of change of the function over the interval [–2, 2] is 1.
ANSWER:
1
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1-1 Graphing Quadratic Functions
17. f(x) = 2x 2 – 11; interval [–3, 3]
SOLUTION:
The average rate of change is equal to
.
First find f(3) and f(−3).
Then substitute to find the average rate of change.
The average rate of change of the function over the interval [–3, 3] is 0.
ANSWER:
0
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1-1 Graphing Quadratic Functions
18. f(x) = –2x 2 + 8x + 7; interval [–4, 4]
SOLUTION:
The average rate of change is equal to
.
First find f(4) and f(−4).
Then substitute to find the average rate of change.
The average rate of change of the function over the interval [–4, 4] is 8.
ANSWER:
8
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1-1 Graphing Quadratic Functions
Determine the average rate of change of f(x) over the specified interval.
19. interval [–3, 3]
x
–3
–2
–1
0
1
2
3
f(x)
0
–3
–4
–3
0
5
12
SOLUTION:
The average rate of change is equal to
.
First find f(3) and f(−3) from the table.
f(3) = 12
f(–3) = 0
Then substitute to find the average rate of change.
The average rate of change of the function over the interval [–3, 3] is 2.
ANSWER:
2
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1-1 Graphing Quadratic Functions
20. interval [–4, 4]
x
–4
–2
0
2
4
f(x)
–27
–3
5
–3
–27
SOLUTION:
The average rate of change is equal to
.
First find f(4) and f(−4) from the table.
f(4) = –27
f(–4) = –27
Then substitute to find the average rate of change.
The average rate of change of the function over the interval [–4, 4] is 0.
ANSWER:
0
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1-1 Graphing Quadratic Functions
21. interval [–2, 2]
x
–2
–1
0
1
2
f(x)
–3
–3
–1
3
9
SOLUTION:
The average rate of change is equal to
.
First find f(2) and f(−2) from the table.
f(2) = 9
f(–2) = –3
Then substitute to find the average rate of change.
The average rate of change of the function over the interval [–2, 2] is 3.
ANSWER:
3
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1-1 Graphing Quadratic Functions
22. interval [–5, 5]
x
–5
–3
–1
0
1
3
5
f(x)
–39
–15
1
6
9
9
1
SOLUTION:
The average rate of change is equal to
.
First find f(5) and f(−5) from the table.
f(5) = 1
f(–5) = –39
Then substitute to find the average rate of change.
The average rate of change of the function over the interval [–5, 5] is 4.
ANSWER:
4
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1-1 Graphing Quadratic Functions
23. interval [–3, 3]
x
–3
–2
–1
0
1
2
3
f(x)
27
12
3
0
3
12
27
SOLUTION:
The average rate of change is equal to
.
First find f(3) and f(−3) from the table.
f(3) = 27
f(–3) = 27
Then substitute to find the average rate of change.
The average rate of change of the function over the interval [–3, 3] is 0.
ANSWER:
0
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1-1 Graphing Quadratic Functions
24. interval [–2, 2]
x
–2
–1
0
1
2
f(x)
12
5
0
–3
–4
SOLUTION:
The average rate of change is equal to
.
First find f(2) and f(−2) from the table.
f(2) = –4
f(–2) = 12
Then substitute to find the average rate of change.
The average rate of change of the function over the interval [–2, 2] is –4.
ANSWER:
–4
25. FOOD The graph shows the number of people in the U.S. who consumed between 8 and 11 bags of potato chips
in a year since 2011.
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1-1 Graphing Quadratic Functions
a. Use the graph to estimate the average rate of change of consumption from 2011 to 2016. Then check your
results algebraically.
b. Interpret your results in the context of the situation.
SOLUTION:
a. From the graph, the change in the y-values is approximately 12.5, and the change in the x-values is 5. So, the
rate of change is approximately
The average rate of change is equal to
or 2.5.
.
b. Sample answer: From 2011–2016, the number of people in the U.S. who consumed between 8 and 11 bags
increased by an average of 2.476 million people per year.
ANSWER:
a.
or 2.5;
b. Sample answer: From 2011–2016, the number of people in the U.S. who consumed between 8 and 11 bags
increased by an average of 2.476 million people per year.
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1-1 Graphing Quadratic Functions
26. EARNINGS The graph shows the amount of money Sheila earned each year since 2008.
a. Use the graph to estimate the average rate of change of Sheila’s earnings from 2008 to 2018. Then check your
results algebraically.
b. Interpret your results in the context of the situation.
SOLUTION:
a. From the graph, the change in the y-values is approximately 5, and the change in the x-values is 10. So, the rate
of change is approximately
since the units are measured in ten thousands.
The average rate of change is equal to
.
or $5100 since the units are measured in ten thousands.
b. Sample answer: From 2008–2018, the amount of money Sheila earned increased by an average of $5100 per
year.
ANSWER:
a.
;
b. Sample answer: From 2008–2018, the amount of money Sheila earned increased by an average of $5100 per
year.
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1-1 Graphing Quadratic Functions
Complete parts a-c for each quadratic function.
a. Find the y-intercept, the equation of the axis of symmetry, and the x-coordinate of the vertex.
b. Make a table of values that includes the vertex.
c. Use this information to graph the function.
27. f(x) = 2x 2 – 6x – 9
SOLUTION:
a. For f(x) = 2x 2 – 6x – 9, a = 2, b = –6, and c = –9. c is the y-intercept, so the y-intercept is –9.
Find the axis of symmetry.
The equation of the axis of symmetry is x = 1.5, so the x-coordinate of the vertex is 1.5.
b. Complete the table.
x
2x2 – 6x – 9
(x, f(x))
0
2(0)2 – 6(0) – 9
–9
1
2(1)2 – 6(1) – 9
–13
1.5
2(1.5)2 – 6(1.5) – 9
–13.5
2
2(2)2 – 6(2) – 9
–13
3
2(3)2 – 6(3) – 9
–9
c. Graph the function.
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1-1 Graphing Quadratic Functions
ANSWER:
a. y-int = −9; axis of symmetry: x = 1.5; x-coordinate of vertex = 1.5
b.
x
f(x)
0
–9
1
–13
1.5
–13.5
2
–13
3
–9
c.
28. f(x) = –3x 2 – 9x + 2
SOLUTION:
a. For f(x) = –3x 2 – 9x + 2, a = –3, b = –9, and c = 2. c is the y-intercept, so the y-intercept is 2.
Find the axis of symmetry.
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1-1 Graphing Quadratic Functions
The equation of the axis of symmetry is x = –1.5, so the x-coordinate of the vertex is –1.5.
b. Complete the table.
x
–3x2 – 9x + 2
(x, f(x))
–3
–3(–3)2 – 9(–3) + 2
2
–2
–3(–2)2 – 9(–2) + 2
8
–1.5
–3(–1.5)2 – 9(–1.5) + 2
8.75
–1
–3(–1)2 – 9(–1) + 2
8
0
–3(0)2 – 9(0) + 2
2
c. Graph the function.
ANSWER:
a. y-inter = 2; axis of symmetry: x = –1.5; x-coordinate of vertex = –1.5
b.
x
f(x)
–3
2
–2
8
–1.5
8.75
–1
8
0
2
c.
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1-1 Graphing Quadratic Functions
29. f(x) = –4x 2 + 5x
SOLUTION:
a. For f(x) = –4x 2 + 5x, a = –4, b = 5, and c = 0. c is the y-intercept, so the y-intercept is 0.
Find the axis of symmetry.
The equation of the axis of symmetry is x =
, so the x-coordinate of the vertex is
.
b. Complete the table.
x
–4x2 + 5x
(x, f(x))
–6
1
1.5625
1
–4(1)2 + 5(1)
1
2
–4(2)2 + 5(2)
–6
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1-1 Graphing Quadratic Functions
c. Graph the function.
ANSWER:
a. y-int = 0; axis of symmetry: x =
; x-coordinate of vertex =
b.
x
f(x)
–6
1
1.5625
1
1
2
–6
c.
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30. f(x) = 2x 2 + 11x
SOLUTION:
a. For f(x) = 2x 2 + 11x, a = 2, b = 11, and c = 0. c is the y-intercept, so the y-intercept is 0.
Find the axis of symmetry.
The equation of the axis of symmetry is x = –2.75, so the x-coordinate of the vertex is –2.75.
b. Complete the table.
x
2x2 + 11x
(x, f(x))
–4
2(–4)2 + 11(–4)
–12
–3
2(–3)2 + 11(–3)
–15
–2.75
2(–2.75)2 + 11(–2.75)
–15.125
–2.5
2(–2.5)2 + 11(–2.5)
–15
–1.5
2(–1.5)2 + 11(–1.5)
–9
c. Graph the function.
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ANSWER:
a. y-int = 0; axis of symmetry: x = –2.75; x-coordinate of vertex = –2.75
b.
x
f(x)
–4
–12
–3
–15
–2.75
–15.125
–2.5
–15
–1.5
–12
c.
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31. f(x) = 0.25x 2 + 3x + 4
SOLUTION:
a. For f(x) = 0.25x 2 + 3x + 4, a = 0.25, b = 3, and c = 4. c is the y-intercept, so the y-intercept is 4.
Find the axis of symmetry.
The equation of the axis of symmetry is x = 6, so the x-coordinate of the vertex is 6.
b. Complete the table.
x
0.25x2 + 3x + 4
(x, f(x))
–10
0.25(0)2 + 3(0) + 4
–1
–8
0.25(0)2 + 3(0) + 4
–4
–6
0.25(0)2 + 3(0) + 4
–5
–4
0.25(0)2 + 3(0) + 4
–4
–2
0.25(0)2 + 3(0) + 4
–1
c. Graph the function.
ANSWER:
a. y-int = 4; axis of symmetry: x = –6; x-coordinate of vertex = –6
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b.
x
f(x)
–10
–1
–8
–4
–6
–5
–4
–4
–2
–1
c.
32. f(x) = –0.75x 2 + 4x+ 6
SOLUTION:
a. For f(x) = –0.75x 2 + 4x + 6, a = –0.75, b = 4, and c = 6. c is the y-intercept, so the y-intercept is 6.
Find the axis of symmetry.
The equation of the axis of symmetry is x =
, so the x-coordinate of the vertex is
.
b. Complete the table.
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0.75x2 + 4x + 6
x
(x, f(x))
10
11.25
3
–0.75(3)2 + 4(3) + 6
11.25
4
–0.75(4)2 + 4(4) + 6
10
c. Graph the function.
ANSWER:
a. y-int = 6; axis of symmetry: x =
; x-coordinate of vertex =
b.
x
f(x)
10
11.25
3
11.25
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4
10
c.
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Determine whether each function has a maximum or a minimum value. Then find and use the x-coordinate
of the vertex to determine the maximum or minimum.
33. f(x) = −9x 2 – 12x +19
SOLUTION:
For f(x) = −9x 2 – 12x + 19, a = −9, b = −12, and c = 19.
Find the axis of symmetry.
The equation of the axis of symmetry is x =
, so the x-coordinate of the vertex is
. Because a < 0, the
vertex is a maximum.
The maximum is
.
ANSWER:
max;
, 23
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34. f(x) = 7x 2 – 21x + 8
SOLUTION:
For f(x) = 7x 2 – 21x + 8, a = 7, b = −21, and c = 8.
Find the axis of symmetry.
The equation of the axis of symmetry is x = 1.5, so the x-coordinate of the vertex is 1.5. Because a > 0, the vertex
is a minimum.
The minimum is f(x) = 7x 2 – 21x + 8 = 7(1.5)2 – 21(1.5)+ 8 = −7.75.
ANSWER:
min; 1.5; −7.75
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35. f(x) = −5x 2 + 14x – 6
SOLUTION:
For f(x) = −5x 2 + 14x – 6, a = −5, b = 14, and c = −6.
Find the axis of symmetry.
The equation of the axis of symmetry is x = 1.4, so the x-coordinate of the vertex is 1.4. Because a < 0, the vertex
is a maximum.
The maximum is f(x) = −5x 2 + 14x – 6 = −5(1.4)2 + 14(1.4) – 6 = 3.8.
ANSWER:
max; 1.4; 3.8
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36. f(x) = 2x 2 – 13x – 9
SOLUTION:
For f(x) = 2x 2 – 13x – 9, a = 2, b = −13, and c = −9.
Find the axis of symmetry.
The equation of the axis of symmetry is x = 3.25, so the x-coordinate of the vertex is 3.25. Because a > 0, the
vertex is a minimum.
The minimum is f(x) = 2x 2 – 13x – 9 = 2(3.25)2 – 13(3.25) – 9 = −30.125.
ANSWER:
min; 3.25; −30.125
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37. f(x) = 9x – 1 – 18x 2
SOLUTION:
For f(x) = 9x – 1 – 18x 2, a = −18, b = 9, and c = −1.
Find the axis of symmetry.
The equation of the axis of symmetry is x = 0.25, so the x-coordinate of the vertex is 0.25. Because a < 0, the
vertex is a maximum.
The maximum is f(x) = 9x – 1 – 18x 2 = 9(0.25) – 1 – 18(0.25)2 = 0.125.
ANSWER:
max; 0.25; 0.125
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38. f(x) = −16 – 18x – 12x 2
SOLUTION:
For f(x) = −16 – 18x – 12x 2, a = −12, b = −18, and c = −16.
Find the axis of symmetry.
The equation of the axis of symmetry is x = −0.75, so the x-coordinate of the vertex is −0.75. Because a < 0, the
vertex is a maximum.
The maximum is f(x) = −16 – 18x – 12x 2 = −16 – 18(−0.75) – 12(−0.75)2 = −9.25.
ANSWER:
max; −0.75; −9.25
39. HEALTH CLUBS Last year, the Sports Time Athletic Club charged $20 per month to participate in an aerobics
class. Seventy people attended the classes. The club wants to increase the class price. They expect to lose one
customer for each $1 increase in the price.
a. Define variables and write an equation for a function that represents the situation. Make a table and graph the
function.
b. Find the vertex of the function and interpret it in the context of the situation. Does it seem reasonable? Explain.
SOLUTION:
a. Let x = the number of price increases, and let y = revenue. P(x) is equal to the price of participating in an
aerobics (20 + x) times the total number of passes sold (70 – x). So, the equation is P(x) = (20 + x)(70 – x).
x
y
0
1400
5
1625
10
1800
15
1925
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1-1 Graphing Quadratic Functions
20
2000
25
2025
30
2000
35
1925
40
1800
45
1625
50
1400
b. The vertex of the function is (25, 2025). The club should increase the price by $25 to make a maximum profit of
$2025. Sample answer: Increasing the price by $25 is more than twice the current price, so this may be
unreasonable. The club may want to revisit their assumptions.
ANSWER:
a. Let x = the number of price increases, and let y = revenue. P(x) is equal to the price of participating in an
aerobics (20 + x) times the total number of passes sold (70 – x). So, the equation is P(x) = (20 + x)(70 – x).
x
y
0
1400
5
1625
10
1800
15
1925
20
2000
25
2025
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1-1 Graphing Quadratic Functions
30
2000
35
1925
40
1800
45
1625
50
1400
b. (25, 2025); The club should increase the price by $25 to make a maximum profit of $2025. Sample answer:
Increasing the price by $25 is more than twice the current price, so this may be unreasonable. The club may want
to revisit their assumptions.
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40. TICKETS The manager of a community symphony estimates that the symphony will earn −40P2 + 1100P dollars
per concert if they charge P dollars for tickets. What ticket price should the symphony charge in order to maximize
its profits?
SOLUTION:
Find the axis of symmetry.
The equation of the axis of symmetry is x = 13.75, so the x-coordinate of the vertex is 13.75. Because a < 0, the
vertex is a maximum. Since P represents the price of tickets, the symphony should charge $13.75 for tickets to
maximize profits.
ANSWER:
$13.75
41. REASONING On Friday nights, the local cinema typically sells 200 tickets at $6.00 each. The manager
estimates that for each $0.50 increase in the ticket price, 10 fewer people will come to the cinema.
a. Define the variables x and y. Then write and graph a function to represent the expected revenue from ticket
sales, and determine the domain of the function for the situation.
b. What price should the manager set for a ticket in order to maximize the revenue? Justify your reasoning.
c. Explain why the graph decreases from x = 4 to x = 20, and interpret the meaning of the x-intercept of the graph.
SOLUTION:
a. Let x = number of price increases, and let y = the revenue. R(x) is equal to the price of each ticket (6 + 0.50x)
times the total number of tickets sold (200 – 10x). So, the equation R(x) = (6 + 0.50x)(200 – 10x) = 1200 +
100x – 60x – 5x 2 = –5x 2 + 40x + 1200. Because the price changes and the revenue will not be negative, the
domain is {x | 0 ≤ x ≤ 20}.
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Because the price changes and the revenue will not be negative, the domain is {x | 0 ≤ x ≤ 20}.
b. The graph shows that the maximum revenue value occurs at x = 4, which corresponds to a ticket price of $6.00
+ 4($0.50) or $8.00. The maximum revenue is R(4) = $1280.00.
c. As price increases continue, the demand for tickets will decrease. The x-intercept indicates a price that is too
high; so fewer people will attend the cinema.
ANSWER:
a. If x = number of price increases, the revenue is R(x) = –5x 2 + 40x + 1200. Because the price changes and the
revenue will not be negative, the domain is {x | 0 ≤ x ≤ 20}.
b. The graph shows that the maximum revenue value occurs at x = 4, which corresponds to a ticket price of $6.00
+ 4($0.50) or $8.00. The maximum revenue is R(4) = $1280.00.
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c. As price increases continue, the demand for tickets will decrease. The x-intercept indicates a price that is too
high; so fewer people will attend the cinema.
42. USE A MODEL From 4 feet above a swimming pool, Tomas throws a ball upward with a velocity of 32 feet per
second. The height h(t) of the ball t seconds after Tomas throws it is given by h(t) = –16t 2 + 32t + 4. For t ≥ 0,
find the maximum height reached by the ball and the time that this height is reached.
SOLUTION:
For h(t) = –16t 2 + 32t + 4, a = –16, b = 32, and c = 4.
Find the axis of symmetry.
The equation of the axis of symmetry is x = 1, so the x-coordinate of the vertex is 1. Because a < 0, the vertex is a
maximum.
The maximum is
.
ANSWER:
20 ft;1 second
43. TRAJECTORIES At a special ceremonial reenactment, a cannonball is launched from a cannon on the wall of
Fort Chambly, Quebec. If the path of the cannonball is traced on a graph so that the cannon is situated on the yaxis, the equation that describes the path is
, where x is the horizontal distance from the
cliff and y is the vertical distance above the ground in feet. How high above the ground is the cannon?
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SOLUTION:
For
,a=
,b=
, and c = 20.
Find the axis of symmetry.
The equation of the axis of symmetry is x = 400, so the x-coordinate of the vertex is 400. Because a < 0, the
vertex is a maximum.
The maximum is
.
ANSWER:
20 ft
44. CONSTRUCT ARGUMENTS Which function has a greater maximum: f(x) = –2x 2 + 6x – 7 or the function
shown in the graph? Explain your reasoning using a graph.
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SOLUTION:
From the graph, g(x) appears to have a maximum of –2.
Graph f(x) using the given information. For f(x) = – 2x 2 + 6x – 7, a = –2, b = 6, and c = –7.
Find the axis of symmetry.
The equation of the axis of symmetry is x = 1.5, so the x-coordinate of the vertex is 1.5. Because a < 0, the vertex
is a maximum.
The maximum is f(x) = –2x 2 + 6x – 7 = –2(1.5)2 + 6(1.5) – 7 = –2.5. The vertex is at (1.5, –2.5). The yintercept is –7, so (0, –7) is on the graph. Reflect (0, –7) in the axis of symmetry. So, (3, –7) is also on the
graph. Connect the points with a smooth curve.
g(x) has the greater maximum. Sample answer: Its vertex is a maximum point at (–6, –2), which is 0.5 units above
the vertex of f(x) which is (1.5, –2.5).
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ANSWER:
Graph f(x) on the same coordinate plane and compare; g(x) has the greater maximum value.
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45. FIND THE ERROR Lucas thinks that the functions f(x) and g(x) have the same maximum. Madison thinks that
g(x) has a greater maximum. Is either of them correct? Explain your reasoning.
g(x) is a quadratic function
with x-intercepts of 4 and 2
and a y-intercept of –8.
SOLUTION:
From the graph it appears that f(x) has a maximum of –2.
Since g(x) is a quadratic function with x-intercepts of 4 and 2, the axis of symmetry will be the midpoint of the xintercepts, 3.
Use the fact that the parabola faces down and the x-intercepts to create an equation.
Substitute 3 for x to get the maximum.
So, Madison is correct. Sample answer: f(x) has a maximum of –2. g(x) has a maximum of 1.
ANSWER:
Madison; sample answer: f(x) has a maximum of –2. g(x) has a maximum of 1.
46. PERSEVERE The table at the right represents some points on the graph of a quadratic function.
a. Find the values of a, b, c, and d.
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b. What is the x-coordinate of the vertex of the function?
c. Does the function have a maximum or a minimum?
x
–20
c
–5
–1
d–1
5
7
15
14 – c
f(x)
–377
–13
–2
22
a
a – 24
–b
–202
–377
SOLUTION:
a. Using the four points given in the table and the quadratic regression function on the graphing calculator we get
the quadratic function y = –x 2 + 23. This function is used to find the missing values a, b, c, d. For each missing
value, substitute what is known from the table into the quadratic function and solve.
In this case, we would use c = –6 to make sense in the table of values.
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In this case, we use d = 2 because the result of using d = 0 is already in the table.
b. The x-coordinate of the vertex of the function is 0 because it is the midpoint of the x-values of the two
symmetric points in the table, (–1, 22) and (1, 22).
c. The function has a maximum because the y-values increase as x increase to 0, then the y-values decrease as the
x-values increase to infinity.
ANSWER:
a. a = 22; b = 26; c = –6; d = 2
b. 0
c. maximum
47. WRITE Describe how you determine whether a function is quadratic and if it has a maximum or minimum value.
SOLUTION:
A function is quadratic if it has no other terms than a quadratic term, linear term, and constant term. The function
has a maximum if the coefficient of the quadratic term is negative and has a minimum if the coefficient of the
quadratic term is positive.
ANSWER:
Sample answer: A function is quadratic if it has no other terms than a quadratic term, linear term, and constant
term. The function has a maximum if the coefficient of the quadratic term is negative and has a minimum if the
coefficient of the quadratic term is positive.
48. CREATE Give an example of a quadratic function with each characteristic.
a. maximum of 8
b. minimum of –4
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c. vertex of (–2, 6)
SOLUTION:
a. The y-intercept will be the maximum when the axis of symmetry is at x = 0 and the parabola faces down. This
happens when b is 0 and the leading coefficient is negative. So, the simplest function with a maximum of 8 is f(x) =
–x 2 + 8.
a = –1, b = 0, and c = 8.
Axis of symmetry:
The equation of the axis of symmetry is x = 0, so the x-coordinate of the vertex is 0. Because a < 0, the vertex is a
maximum.
The maximum is f(x) = –x 2 + 8 = –(0)2 + 8 = 8.
b. The y-intercept will be the minimum when the axis of symmetry is at x = 0 and the parabola faces up. This
happens when b is 0 and the leading coefficient is positive. So, the simplest function with a minimum of –4 is f(x)
= x2 – 4
a = 1, b = 0, and c = –4.
Axis of symmetry:
The equation of the axis of symmetry is x = 0, so the x-coordinate of the vertex is 0. Because a > 0, the vertex is a
minimum.
The minimum is f(x) = x 2 – 4 = (0)2 – 4 = –4.
c. Sample answer: f(x) = x 2 + 4x + 10
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This sample answer was arrived at using the formula for axis of symmetry. This is solved below, but imagine we
only knew we wanted the axis to be –2. We would have to use number sense to reason out values for a and b that
would arrive at that value. In this case a = 1 and b = 4.
Axis of symmetry:
Then using the values for a and b, the standard form of a quadratic, and the value of the vertex, solve to find c.
ANSWER:
a. Sample answer: f(x) = –x 2 + 8
b. Sample answer: f(x) = x 2 – 4
c. Sample answer: f(x) = x 2 + 4x + 10
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