Question

$f\left(x\right)=5x-2$

(a) Find the average rate of change from $1$to $3.$

(b) Find the equation of the secant line containing$\left(1,f\left(1\right)\right)$and$\left(3,f\left(3\right)\right)$.

Short answer

(a) The average rate of change is $5.$

(b) The equation of the secant line isrole="math" localid="1646224686021" $y=5x-8$.

Step-by-step solution

Step 1.  Given information.

We have:

$f\left(x\right)=5x-2$

Part a. Step 1. Find the average rate of change from 1 to 3.

The equation is $f\left(x\right)=5x-2$.

Average rate of change is given by:

localid="1646229963744" $\frac{∆y}{∆x}=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

Here, $a=1,b=3$

$$\begin{gathered} \frac{∆y}{∆x}=\frac{f\left(3\right)-f\left(1\right)}{3-1} \\ =\frac{5\left(3\right)-2-\left(5\left(1\right)-2\right)}{2} \\ =5 \end{gathered}$$

Part b. Step 1.  Find the equation of the secant line.

The equation is $f\left(x\right)=5x-2$and the point $\left(2,f\left(2\right)\right)$, $\left(3,f\left(3\right)\right)$.

Substitute $x=2$in the equation:

$$\begin{gathered} f\left(2\right)=5\left(2\right)-2 \\ =10-2 \\ =8 \end{gathered}$$

The slope of the secant line is equal to the average rate of change of a function, it follows:

$m_{\sec }=5$

Therefore, use the point-slope form to equation of the secant line:

localid="1646230369239" $$\begin{gathered} y-y_1=m_{\sec }\left(x-x_1\right) \\ y-2=5\left(x-2\right) \\ y=5x-10+2 \\ y=5x-8 \end{gathered}$$