g(t)=-(t - 1)^{2}+5
what is the average rate of change of g over the interval -4 ≤ t ≤ 5?

Question

Accepted Answer

# Explanation:
## Step1: Recall the formula for average rate of change
The average rate of change of a function $y = g(t)$ over the interval $[a,b]$ is given by $\frac{g(b)-g(a)}{b - a}$. Here, $a=-4$ and $b = 5$.
## Step2: Calculate $g(-4)$
Substitute $t=-4$ into $g(t)=-(t - 1)^{2}+5$.
$$
\begin{align*}
g(-4)&=-(-4 - 1)^{2}+5\
&=-( - 5)^{2}+5\
&=-25 + 5\
&=-20
\end{align*}
$$
## Step3: Calculate $g(5)$
Substitute $t = 5$ into $g(t)=-(t - 1)^{2}+5$.
$$
\begin{align*}
g(5)&=-(5 - 1)^{2}+5\
&=-16+5\
&=-11
\end{align*}
$$
## Step4: Calculate the average rate of change
Using the formula $\frac{g(b)-g(a)}{b - a}$, with $a=-4$, $b = 5$, $g(-4)=-20$ and $g(5)=-11$
$$
\begin{align*}
\frac{g(5)-g(-4)}{5-(-4)}&=\frac{-11-(-20)}{5 + 4}\
&=\frac{-11 + 20}{9}\
&=\frac{9}{9}\
&=1
\end{align*}
$$

# Answer:
$1$

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QUESTION IMAGE

Question

Explanation:

Step1: Recall the formula for average rate of change

Step2: Calculate \(g(-4)\)

Step3: Calculate \(g(5)\)

Step4: Calculate the average rate of change

Answer: