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. How far does the ball travel horizontally before hitting the ground? (Round to the nearest hundredth.) 

Answer 1

0 = -0.6x² + 2.7x + 6

Can you solve for x and answer?

Answer 2

18.15 = max

18.15 feet

f(x) = -.6x^2 +2.7x + 6 = horizontal distance (odds are you really meant vertical distance)

= -.6(x^2 - 2.7x/.6 + (2.7/.12)^2) + 6 + .6(2.7/.12)^2

= -.6(x - 2.7/.12)^2 + 6 + (2.7)^2/(.24)

= -.6(x - 22.5)^2 + 36.375

vertex = (22.5, 36.375) with 36.375 feet = maximum