Pre-cal with Algebra

Question

A ball is thrown upward and outward from a height of 6 feet. The height of the ball, 𝑓(𝑥), in feet, can be modeled by 𝑓(𝑥) = −0.6𝑥^2 + 2.7𝑥 + 6 where 𝑥 is the ball’s horizontal distance, in feet, from where it was thrown. What is the maximum height of the ball and how far from where it was thrown does this occur? (Round your values to the nearest hundredth.) 

Answer 1

Asked and answered.

Answer 2

f = -.6t^2 +2.7t +6

f'= -1.2t +2.7=0

solve for t, then plug back into f(t)

f(2.7/1.2) = -.6(2.7/1.2)^2 + 2.7(2.7/1.2) + 6 = -2.7^2/.6 +2(2.7)^2/.6 + 6 = (2.7)^2/.6 + 6 =

18.15

but

36.375 = 36 3/8 feet is more likely the answer you want

problem with the problem. you may have miscopied or had a typo or a mis print, switching horizontal with vertical distance