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The height y (in feet) of a ball thrown by a child is

y=−1/12x^2+6x+3

where x is the horizontal distance in feet from the point at which the ball is thrown.


(a) How high is the ball when it leaves the child's hand? (Hint: Find y when x=0)

Your answer is y=_______


(b) What is the maximum height of the ball? _______


(c) How far from the child does the ball strike the ground? ______

Respuesta :

hasanahmad368

Answer:

Step-by-step explanation:

a)

y=−1/12x^2+6x+3

y=−1/12(0)^2+6(0)+3

y = 3

b)

y=−1/12x^2+6x+3

y = -1/12 (x^2-72x) + 3

y = =-1/12 (x^2-72x+1296-1296) +3

y = -1/12(x^2 -72x +1296) + 108 + 3

y = -1/12 (x - 36)^2 +111

maximum height of the ball is 111 feets

c)

y = -1/12 (x - 36)^2 +111

0 = -1/12 (x - 36)^2 +111

-111 = -1/12 (x - 36)^2

1332 = (x - 36)^2

36.497 = x - 36

x = 72.497

How far from the child does the ball strike the ground = 72.497 feets

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