Question

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+147x+118

Answer 1

Answer:  Max height = 455.6 feet

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Explanation:

The general equation

y = ax^2 + bx + c

has the vertex (h,k) such that

h = -b/(2a)

In this case, a = -16 and b = 147. This means,

h = -b/(2a)

h = -147/(2*(-16))

h = 4.59375

The x coordinate of the vertex is x = 4.59375

Plug this into the original equation to find the y coordinate of the vertex.

y = -16x^2+147x+118

y = -16(4.59375)^2+147(4.59375)+118

y = 455.640625

The vertex is located at (h,k) = (4.59375, 455.640625)

The max height of the rocket occurs at the vertex point. Therefore, the max height is y = 455.640625 feet which rounds to y = 455.6 feet