the height y (in feet) of a ball thrown by a child is [tex]y = \frac{1}{16} {x}^{2} + 4x + 3[/tex]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 handb).what is the maximum high of the ballc).how far from the child does the ball strike the ground

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