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

Explanation:

Step1: Recall average - rate - of - change formula

The average rate of change of a function $y = g(t)$ over the interval $[a,b]$ is $\frac{g(b)-g(a)}{b - a}$, where $a=-4$ 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\ &=-4^2+5\ &=-16+5\ &=-11 \end{align*} ]

Step4: Calculate the average rate of change

[ \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:

$1$