Question 311192
1. The path of a ball thrown into the air from a height of 3 feet is given by 
y = {{{-1/8}}}x^2 + x + 3, where y is the height of the ball in feet at the
 horizontal distance of x feet from the thrower.
:
a. How high is the ball at its maximum height?
Find the axis of symmetry; x=-b/(2a). In this equation: a=-1/8, b=1
x = {{{(-1)/(2*(-1/8))}}}
x = {{{(-1)/((-2/4)))}}}
x = {{{(-1)/((-1/4))}}}
x = +4
Substitute 4 for x in the original equation to find the max height
y = -{{{1/8}}}(4^2) + 1(4) + 3 
y = -{{{1/8}}}(16) + 1(4) + 3 
y = -2 + 4 + 3
y = 5
:
Looks like this:
{{{ graph( 300, 200, -4, 12, -4, 8, -.125x^2+x+3) }}}
You can see y = 5 (max) when x=4
:
b. estimate the horizontal distance the ball traveled before hitting the ground.
Looks like slightly over 10 ft, doesn't it.