Question

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 = -16x² + 240x + 96

Explanation:

Step1: Identify the coefficients

The quadratic - function is in the form $y = ax^{2}+bx + c$, where $a=-16$, $b = 240$, and $c = 96$.

Step2: Find the x - value of the vertex

The x - value of the vertex of a quadratic function $y = ax^{2}+bx + c$ is given by $x=-\frac{b}{2a}$. Substitute $a=-16$ and $b = 240$ into the formula:
$x=-\frac{240}{2\times(-16)}=\frac{240}{32}=7.5$.

Step3: Find the y - value of the vertex

Substitute $x = 7.5$ into the equation $y=-16x^{2}+240x + 96$.
$y=-16\times(7.5)^{2}+240\times7.5 + 96$.
First, calculate $-16\times(7.5)^{2}=-16\times56.25=-900$.
Then, calculate $240\times7.5 = 1800$.
So, $y=-900 + 1800+96=996$.

Answer:

$996.0$