question a rocket is launched from a tower. t...
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 + 235x + 67
Answer
# Explanation:
## Step1: Identify the coefficients
The quadratic - function is \(y = ax^{2}+bx + c\), where \(a=-16\), \(b = 235\), \(c = 67\).
## Step2: Find the x - coordinate of the vertex
The x - coordinate of the vertex of a quadratic function \(y = ax^{2}+bx + c\) is given by \(x=-\frac{b}{2a}\).
Substitute \(a=-16\) and \(b = 235\) into the formula: \(x=-\frac{235}{2\times(-16)}=\frac{235}{32}=7.34375\).
## Step3: Find the y - coordinate of the vertex
Substitute \(x = 7.34375\) into the equation \(y=-16x^{2}+235x + 67\).
\[
\begin{align*}
y&=-16\times(7.34375)^{2}+235\times7.34375 + 67\\
&=-16\times53.93046875+1725.78125+67\\
&=-862.8875+1725.78125+67\\
&=929.89375
\end{align*}
\]
Rounding to the nearest tenth, \(y\approx929.9\).
# Answer:
929.9