Download 58. BOATING Miranda has her boat docked on the west

The vertex lies at about (2, 68). Because a is
negative, the graph opens down, and the vertex is a
maximum.
9-1 Graphing Quadratic Functions
58. BOATING Miranda has her boat docked on the
west side of Casper Point. She is boating over to
Casper Marina, which is located directly east of
Step 3 Find the d-intercept.
Use the original equation, and substitute 0 for t.
2
where her boat is docked. The d = −16t + 66t
models the distance she travels north of her starting
point, where d is the number of feet and t the time
traveled in minutes.
The d-intercept is (0, 0).
Step 4 The axis of symmetry divides the parabola
into two equal parts. So if there is a point on one side,
there is a corresponding point on the other side that is
the same distance from the axis of symmetry and has
the same d-value.
t
1
2
3
d
50
68
50
a. Graph this equation.
b. What is the maximum number of feet north that
she traveled?
c. How long did it take her to reach Casper Marina?
Step 5 Connect the points with a smooth curve.
SOLUTION: a. Step 1 Find the equation of the axis of symmetry.
2
d = −16t + 66t, a = –16 and b = 66.
b. The maximum distance Miranda traveled north is
the y-coordinate of the vertex, or about 68 feet.
c. It takes the boat 2 minutes to reach the vertex, and
due to symmetry it will take the boat another 2
minutes to reach Casper Marina. So, it will take
Miranda 2 + 2 or about 4 minutes to reach Casper
Marina.
Step 2 Find the vertex, and determine whether it is a maximum or minimum.
The t-coordinate of the vertex is t = 2. Substitute the
t-coordinate of the vertex into the original equation to
find the value of N.
GRAPHING CALCULATOR Graph each equation. Use the TRACE feature to find the
vertex on the graph. Round to the nearest
thousandth if necessary.
2
59. y = 4x + 10x + 6
SOLUTION: 2
Let = 4x + 10x + 6. Use the WINDOW option to
adjust the viewing window. Use the TRACE option
to move to cursor to the minimum point. The vertex lies at about (2, 68). Because a is
negative, the graph opens down, and the vertex is a
maximum.
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Find the d-intercept.
Use the original equation, and substitute 0 for t.
[-5, 5] scl: 1 by [-5, 5] scl: 1
The vertex is at (–1.25, –0.25).
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c. It takes the boat 2 minutes to reach the vertex, and
due to symmetry it will take the boat another 2
minutes to reach Casper Marina. So, it will take
Miranda 2Quadratic
+ 2 or aboutFunctions
4 minutes to reach Casper
9-1 Graphing
Marina.
GRAPHING CALCULATOR Graph each equation. Use the TRACE feature to find the
vertex on the graph. Round to the nearest
thousandth if necessary.
2
59. y = 4x + 10x + 6
SOLUTION: [-5, 5] scl: 1 by [-2, 18] scl: 2
The vertex is at (0.5, 6).
2
61. y = −5x − 3x − 8
SOLUTION: 2
Let Y1 = −5x − 3x − 8 Use the WINDOW option
to adjust the viewing window. Select the TRACE
option and move the cursor to the maximum point. 2
Let = 4x + 10x + 6. Use the WINDOW option to
adjust the viewing window. Use the TRACE option
to move to cursor to the minimum point. [-5, 5] scl: 1 b [-20, 2] scl: 2
The vertex is at (–0.3, –7.55).
2
62. y = −7x + 12x − 10
[-5, 5] scl: 1 by [-5, 5] scl: 1
The vertex is at (–1.25, –0.25).
2
60. y = 8x − 8x + 8
SOLUTION: SOLUTION: 2
Let Y1 = −7x + 12x − 10 . Use the WINDOW
option to adjust the viewing window. Select the
TRACE option and move the cursor to the
maximum point. 2
Let Y1= 8x − 8x + 8. Use the WINDOW option to
adjust the viewing window. Select the TRACE
option and move the cursor to the minimum point. [-5, 5] scl: 1 by [-20, 2] scl: 2
The vertex is at (0.857, –4.857).
[-5, 5] scl: 1 by [-2, 18] scl: 2
The vertex is at (0.5, 6).
2
61. y = −5x − 3x − 8
SOLUTION: 2
Let Y1 = −5x − 3x − 8 Use the WINDOW option
to adjust the viewing window. Select the TRACE
option and move the cursor to the maximum point. 63. GOLF The average amateur golfer can hit the ball
with an initial upward velocity of 31.3 meters per
second. The height can be modeled by the equation h
2
= −4.9t + 31.3t, where h is the height of the ball, in
feet, after t seconds.
a. Graph this equation.
b. At what height is the ball hit?
c. What is the maximum height of the ball?
d. How long did it take for the ball to hit the ground?
e. State a reasonable range and domain for this
situation.
SOLUTION: a. Step 1 Find the equation of the axis of symmetry.
2
h = −4.9t + 31.3t, a = –4.9 and b = 31.3.
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[-5, 5] scl: 1 b [-20, 2] scl: 2
The vertex is at (–0.3, –7.55).
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e. State a reasonable range and domain for this
situation.
SOLUTION: 9-1 Graphing Quadratic Functions
a. Step 1 Find the equation of the axis of symmetry.
2
h = −4.9t + 31.3t, a = –4.9 and b = 31.3.
Step 2 Find the vertex, and determine whether it is
a maximum or minimum.
The t-coordinate of the vertex is t = 3.2. Substitute
the t-coordinate of the vertex into the original
equation to find the value of h.
The vertex lies at about (3.2, 50). Because a is
negative, the graph opens down, and the vertex is a
maximum.
Step 3 Find the h-intercept.
Use the original equation, and substitute 0 for t.
The h-intercept is (0, 0).
Step 4 The axis of symmetry divides the parabola
into two equal parts. So if there is a point on one side,
there is a corresponding point on the other side that is
the same distance from the axis of symmetry and has
the same h-value.
t
0
3.2
6.4
h
0
50
0
Step 5 Connect the points with a smooth curve.
b. The ball is hit when the time is zero, or at the teSolutions
ManualSince
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Cognero is (0, 0), the ball is hit
intercept.
the by
t-intercept
at 0 meters.
c. The maximum height of the ball is at the vertex.
b. The ball is hit when the time is zero, or at the tintercept. Since the t-intercept is (0, 0), the ball is hit
at 0 meters.
c. The maximum height of the ball is at the vertex.
The vertex is (3.2, 50), so the maximum height of the
ball is 50 meters.
d. It takes the ball about 3.2 seconds to reach the
vertex, and another 3.2 seconds to come down.
Therefore, it takes about 3.2 + 3.2 or about 6.4
seconds to reach the ground.
e . Since the time is zero when the ball is hit and 6.4
when it reaches the ground, the domain is D = {t|0 ≤ t ≤ 6.4}. The ball starts at 0 meters and reaches a maximum height of about 50 meters, so the R = {h|0
≤ h ≤ 50.0}
64. FUNDRAISING The marching band is selling
poinsettias to buy new uniforms. Last year the band
charged $5 each, and they sold 150. They want to
increase the price this year, and they expect to lose
10 sales for each $1 increase. The sales revenue R,
in dollars, generated by selling the poinsettias is
predicted by the function R = (5 + p )(150 - 10p ),
where p is the number of $1 price increases.
a. Write the function in standard form.
b. Find the maximum value of the function.
c. At what price should the poinsettias be sold to
generate the most sales revenue? Explain your
reasoning.
SOLUTION: a.
b. The maximum value of the function occurs at the
2
vertex. In the function R = −10p + 100p + 750, a =
–10, b = 100, and c = 750.
The x-coordinate of the vertex is
.
The p -coordinate of the vertex is p = 5. Substitute
the p -coordinate of the vertex into the original
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equation to find the value of R.
e . Since the time is zero when the ball is hit and 6.4
when it reaches the ground, the domain is D = {t|0 ≤ t ≤ 6.4}. The ball starts at 0 meters and reaches a maximum Quadratic
height of about
50 meters, so the R = {h|0
9-1 Graphing
Functions
≤ h ≤ 50.0}
64. FUNDRAISING The marching band is selling
poinsettias to buy new uniforms. Last year the band
charged $5 each, and they sold 150. They want to
increase the price this year, and they expect to lose
10 sales for each $1 increase. The sales revenue R,
in dollars, generated by selling the poinsettias is
predicted by the function R = (5 + p )(150 - 10p ),
where p is the number of $1 price increases.
a. Write the function in standard form.
b. Find the maximum value of the function.
c. At what price should the poinsettias be sold to
generate the most sales revenue? Explain your
reasoning.
original price was $5, the new price should be 5 + 5
or $10.
65. FOOTBALL A football is kicked up from ground
level at an initial upward velocity of 90 feet per
2
second. The equation h = −16t + 90t gives the
height h of the football after t seconds.
a. What is the height of the ball after one second?
b. When is the ball 126 feet high?
c. When is the height of the ball 0 feet? What do
these points represent in the context of the situation?
SOLUTION: 2
a. Substitute t = 1 into the equation h = −16t + 90t.
The height of the football after one second is 74 feet.
b. Substitute h = 126 into the equation h = −16t2 +
90t and then factor.
SOLUTION: a.
b. The maximum value of the function occurs at the
2
vertex. In the function R = −10p + 100p + 750, a =
–10, b = 100, and c = 750.
The x-coordinate of the vertex is
.
This equation is now in the form ab = 0.
The ball will reach a height of 126 feet on the way up
The p -coordinate of the vertex is p = 5. Substitute
the p -coordinate of the vertex into the original
equation to find the value of R.
The maximum value of the function is 1000.
c. The maximum revenue of $1000, is generated
when p = 5, so by five $1 increases. Since the
original price was $5, the new price should be 5 + 5
or $10.
65. FOOTBALL A football is kicked up from ground
level at an initial upward velocity of 90 feet per
2
second. The equation h = −16t + 90t gives the
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height
h of-the
football
after t seconds.
a. What is the height of the ball after one second?
b. When is the ball 126 feet high?
at t =
or 2.625 seconds and on the way down at t
= 3 seconds.
2
c. Substitute h = 0 into the equation h = −16t + 90t
and then factor.
Solve for t.
The height of the ball is 0 feet when t = 0 or t =
5.625. The height of the ball is 0 feet before the ball
is kicked, or when t = 0 seconds, and it is 0 again
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when the ball hits the ground after the kick, or when
t
= 5.625 seconds.
9-1 Graphing Quadratic Functions
The height of the ball is 0 feet when t = 0 or t =
5.625. The height of the ball is 0 feet before the ball
is kicked, or when t = 0 seconds, and it is 0 again
when the ball hits the ground after the kick, or when t
= 5.625 seconds.
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