College Algebra, Chapter 3, 3.4, Section 3.4, Problem 22

A linear function $g(x) = -4x + 2$.

a.) Find the average rate of change of the function between $x = a$ and $x = a + h$.


$\begin{equation} \begin{aligned} \text{average rate of change } =& \frac{g(b) - g(a)}{b - a} && \text{Model} \\ \\ \text{average rate of change } =& \frac{g(a + h) - g(a)}{a + h - a} && \text{Substitute } b = a + h \text{ and } a = a \\ \\ \text{average rate of change } =& \frac{-4(a + h) + 2 - [-4 (a) + 2]}{h} && \text{Simplify} \\ \\ \text{average rate of change } =& \frac{-4a - 4h + 2 + 4a - 2}{h} && \text{Combine like terms} \\ \\ \text{average rate of change } =& \frac{-4 \cancel{h}}{\cancel{h}} && \text{Cancel out like terms} \\ \\ \text{average rate of change } =& -4 && \text{Answer} \end{aligned} \end{equation}$


b.) Show that the average rate of change is the same as the slope of the line.

Using slope of the line,

$\displaystyle m = \frac{y_2 - y_1}{x_2 - x_1}$, where $y = g(x)$


$\begin{equation} \begin{aligned} & \text{Solving for } y_2 && \text{Solving for } y_1 \\ \\ & y_2 = -4x_2 + 2 && y_1 = -4x_1 + 2 \\ \\ & y_2 = -4 (a + h) + 2 && y_1 = -4(a) + 2 \\ \\ & y_2 = -4a - 4h + 2 && y_1 = -4a + 2 \end{aligned} \end{equation}$



$\begin{equation} \begin{aligned} m =& \frac{y_2 - y_1}{x_2 - x_1} && \text{Model} \\ \\ m =& \frac{-4a - 4h + 2 - (-4a + 2)}{a + h - a} && \text{Substitute } x_1 = a, x_2 = a + h, y_1 = -4a + 2 \text{ and } y_2 = -4a - 4h + 2 \\ \\ m =& \frac{-4a - 4h + 2 + 4a - 2}{h} && \text{Combine like terms} \\ \\ m =& \frac{-4 \cancel{h}}{\cancel{h}} && \text{Cancel out like term} \\ \\ m =& -4 && \text{Answer} \end{aligned} \end{equation}$