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}
$