Question

A rocket launch from the tower the height of the rocket, y in feet is related to the time after lunch, x in seconds by the giving equation using this equation find the maximum hight reached by the rocket, to the nearest tenth of a foot

y= -16x^2 + 246x +100

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 }}{+246}x\stackrel{\stackrel{c}{\downarrow }}{+100} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right)`

`\left(-\cfrac{ 246}{2(-16)}~~~~ ,~~~~ 100-\cfrac{ (246)^2}{4(-16)}\right)\implies \left(\cfrac{123}{16}~~,~~ 100+\cfrac{15129}{16} \right) \\\\\\ \left( \cfrac{123}{16}~~,~~\cfrac{16729}{16} \right)\implies \stackrel{maximum~height}{\stackrel{\hspace{3em}\downarrow }{\left( 7(11)/(16)~~,~~1045(9)/(16) \right)}}`