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 the maximum height reached by the rocket, to the nearest tenth of a foot.
y=-16x^2+228x+71

Respuesta :

Answer:

883.3

Step-by-step explanation:

thats all.

An equation is formed of two equal expressions. The maximum height reached by a rocket, to the nearest tenth of a foot is 883.3 feet.

What is an equation?

An equation is formed when two equal expressions are equated together with the help of an equal sign '='.

To find the maximum height through which the rocket will reach, we need to differentiate the given function, therefore, we can write,

y=-16x²+228x+71

dy/dx = -16(2x)+228

Substitute the value of dy/dx as 0, to get the value of x,

0 = -32x + 228

228 = 32x

x =7.125

Substitute the value of x in the equation to get the maximum height,

y=-16x²+228x+71

y=-16(7.125²)+228(7.125)+71

y=883.3feet

Hence, the maximum height reached by the rocket, to the nearest tenth of a foot is 883.3 feet.

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