The height y (in feet) of a ball thrown by a child isy=−1/12x^2+6x+3where 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? ______

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) +3y = -1/12(x^2 -72x +1296) + 108 + 3y = -1/12 (x - 36)^2 +111maximum height of the ball is 111 feetsc) y = -1/12 (x - 36)^2 +1110 = -1/12 (x - 36)^2 +111-111 = -1/12 (x - 36)^21332 = (x - 36)^236.497 = x - 36 x = 72.497How far from the child does the ball strike the ground = 72.497 feets