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+254x+79

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Answer 1

1018.4 feet

Explanation:

The maximum height reached by the rocket can be found by using the formula for the vertex of a parabola.

The vertex of the parabola y = ax^2 + bx + c is given by (-b/2a, c - b^2/4a).

In this case, the equation is y = -16x^2 + 254x + 79.

The maximum height is reached at x = -b/2a = -254/(2*-16) = 7.9375 seconds.

Plugging this value into the equation gives y = -16(7.9375)^2 + 254(7.9375) + 79 = 1018.4 feet.

Therefore, the maximum height reached by the rocket is 1018.4 feet to the nearest tenth of a foot.

Hope this helps!