Question

what is the average rate of change of $y = -\frac{1}{2}(x - 3)(x + 6)$ over the interval $0 < x < 9$? the average rate of change is $square$.

Explanation:

Step1: Expand the function

\[

$$\begin{align*} y&=-\frac{1}{2}(x - 3)(x + 6)\\ &=-\frac{1}{2}(x^{2}+6x-3x - 18)\\ &=-\frac{1}{2}(x^{2}+3x - 18)\\ &=-\frac{1}{2}x^{2}-\frac{3}{2}x + 9 \end{align*}$$

\]

Step2: Find \(y(0)\) and \(y(9)\)

When \(x = 0\), \(y(0)=-\frac{1}{2}(0 - 3)(0 + 6)=9\)
When \(x = 9\),
\[

$$\begin{align*} y(9)&=-\frac{1}{2}(9 - 3)(9 + 6)\\ &=-\frac{1}{2}\times6\times15\\ &=-45 \end{align*}$$

\]

Step3: Calculate the average rate of change

The formula for the average rate of change of a function \(y = f(x)\) over the interval \([a,b]\) is \(\frac{f(b)-f(a)}{b - a}\). Here \(a = 0\), \(b = 9\), \(f(0)=9\) and \(f(9)=-45\).
\[

$$\begin{align*} \text{Average rate of change}&=\frac{y(9)-y(0)}{9 - 0}\\ &=\frac{-45 - 9}{9}\\ &=\frac{-54}{9}\\ &=- 6 \end{align*}$$

\]

Answer:

\(-6\)