The height y (in feet) of a ball thrown by a child is y=−116x2+4x+3

Question

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

y=−116x2+4x+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?     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

Answer 1

Notice that is a quadratic facing down so the highest point will be the vertex, the initial height will be the y-intercept, and the distance from the child will be the positive x-intercept.

(a) Find y-intercept (let x = 0 and solve for y)

y = -116(0)²+4(0)+3 = 3 ft

(b) Find the y-value of the vertex since the question is asking for the max height

x-value of vertex: x = -b/2a = -4/(2*-116) = 1/58

y-value of vertex (plug x-value back into the equation): y = -116(1/58)²+4(1/58)+3 = 3.03 ft

(c) Find the positive x-intercept (y = 0)

0 = -116x²+4x+3

*can't be factored so use the quadratic formula

x = [-4 ± √(4²-4(-116)(3))]/(2*-116) = -0.14 or 0.18 ft

x = 0.18 ft