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+201x+132

Answer 1

Check the picture below.

`\textit{vertex of a vertical parabola, using coefficients} \\\\ y=\stackrel{\stackrel{a}{\downarrow }}{-16}x^2\stackrel{\stackrel{b}{\downarrow }}{+201}x\stackrel{\stackrel{c}{\downarrow }}{+132} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right)`

`\left(-\cfrac{ 201}{2(-16)}~~~~ ,~~~~ 132-\cfrac{ (201)^2}{4(-16)}\right) \implies \left( - \cfrac{ 201 }{ -32 }~~,~~132 - \cfrac{ 40401 }{ -64 } \right) \\\\\\ \left( \cfrac{201}{32}~~,~~\cfrac{48849}{64} \right)\implies (6.3~~,~~\stackrel{feet}{\text{\LARGE 763.3}})`