The height y (in feet) of a ball thrown by a child is y=−112x^2+6x+3

Question

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

where  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?  _______ feet

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

 ______feet

(c) How far from the child does the ball strike the ground? Round your answers to the nearest 0.01.

_______ feet

PLEASE ANSWER A,B AND C!! THANK YOU!!!

Answer 1

(a) x=0 , so 3 ft

(b) 0 = dy/dx = -224x + 6

x = -6/-224 = 3/112

plugging this into the function, you get 3.08

(c) quadratic formula says 0.193

Answer 2

a. for x=0 y=3ft

b. to find the maximum height the ball goes take the derivative of the function and set to 0 which gives

0=-224x + 6 x= 6/224 = 3/112 substitute this into the equation to find the height y= -112(3/112)² +6(3/112)

+3. This gives y= -9/112 +18/112 +3 =9(-1/112 + 1/56) +3 =9/112 +3 = 3.08ft. So this is the maximum height

that the ball goes.

c. The ball goes a distance x from the child. To find x let y=0 so 0=-112x² +6x +3 or 0=112x² -6x -3 use the

quadratic formula to find x. where x= (6 ±√(36 +1344))/224 =(6±37.15)/224 = 43.15/224 = .19ft