A rocket is launched from a tower. The height of the rocket, y in feet, is related to the time after launch, x in seconds, by the given equation. Using this equation, find out the time at which the rocket will reach its max, to the nearest 100th of a second. y=-16x^2+228x+116

You have a nice equation y(x) which tell us the height of the rocket.
y(x) =-16x²+187x+113
We get the height of the tower for x=0
y(0) = 0 -16*0 + 187*0 + 113 = 113
y(x) =-16x²+187x+113 derive to get max
y´(x) = -32x + 187
y' = 0, when -32x + 187 = 0 => x = 187/32= 5.84 is this max? Derive again
y'' = -32 < 0 => max , ok.
y(5.84) = -16*5.84² + 187*5.84 + 113 = 1111.64 ≈ 1111.6 ( from the tower)
Relatively the ground: 1111.6 + 113 = 1224.6 ft