Pearson Algebra 2 Common Core, 2011
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Pearson Algebra 2 Common Core, 2011 View details
5. Quadratic Equations
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Exercise 59 Page 231

Practice makes perfect
a The function below models the height h of the tennis ball in meters from the point of the toss after t seconds.
h(t)=-5t^2+7tTo find the maximum height reached by the ball, we need to find the maximum point on the graph of y=h(t). Since the leading coefficient of h(t) is negative (-5), its graph is a parabola that opens downward. This means the maximum point is the vertex. Let's find its t-coordinate. t = -b/2a ⇒ t = -7/2(-5) = 0.7 To find the h-coordinate, we substitute t=0.7 into h(t).
h(t)=-5t^2+7t
h( 0.7)=-5( 0.7)^2+7( 0.7)
h(0.7) = -2.45 + 4.9
h(0.7) = 2.45
Therefore the vertex is ( 0.7, 2.45), which means the ball took 0.7 seconds to reach the maximum height of 2.45 meters from the point of the toss.
b This time, we need to find the time at which the ball was 0.6 meters from the point of the toss. To do that, we must solve the equation h(t)= 0.6.
h(t) = -5t^2 + 7t = 0.6Let's group all the terms on the left-hand side and use the Quadratic Formula to solve this equation for t.
-5t^2 + 7t = 0.6
-5t^2 + 7t - 0.6 = 0
Solve for t
t_(1,2)=- 7±sqrt(7^2-4( -5)( -0.6))/2( -5)
t_(1,2)=-7±sqrt(49 - 12)/-10
t_(1,2)=-7±sqrt(37)/-10
t_(1,2)=-7 ± 6.08/-10
t_(1,2)=7 ± 6.08/10
lct_1 = .0.92 /10. & (I) t_2 = .13.08 /10. & (II)

(I), (II):Calculate quotient

lt_1 = 0.1 t_2 = 1.3
We got two different times at which the ball was 0.6 meters from the point of the toss. From Part A, we know the ball reached the maximum height at 0.7 seconds. This means that at t_1=0.1 seconds the ball was going up, and at t_2=1.3 seconds the ball was going down.

In conclusion, the ball remained in the air for approximately 1.3 seconds between the toss and the serve.