The height y (in feet) of a ball thrown by a child is < br /> < br /> where x is the horizontal distance in feet from the point at which the ball is thrown< br /> < br /> a).how high is the ball when it leaves the child's hand< br /> < br /> b).what is the maximum high of the ball< br /> < br /> c).how far from the child does the ball strike the ground< br />

I will assume you mistyped this question. For y = -1/16x^2 + 4x + 3, the answers to this question are
a) 3 feet
b) 67 feet
c) 64.741 feet

For a) we note that at x = 0, that is the instant where the ball leaves the hand. y(0) = 0.

For b), we find the vertex of y = -1/16x^2 + 4x + 3
                               y = -1/16x^2 + 4x + 3
                               y = -1/16(x^2 - 64x) + 3
                               y = -1/16(x^2 - 64x + 1024 - 1024) + 3
                               y = -1/16((x-32)^2 - 1024) + 3
                               y = -1/16(x-32)^2 + 64 + 3
                               y = -1/16(x-32)^2 + 67
The vertex is at (32,67) so 67 is the maximum height.

For c), we find the x-intercepts with the quadratic formula on
y = -1/16x^2 + 4x + 3=0:
                     x = [ -b ± √b^2 - 4ac ] / (2a)
                     x = [ -4 ± √4^2 - 4(-1/16)(3) ] / (2(-1/16))   
                     x = -0.741, 64.741
Only the positive solution, so 64.741 feet