Q. 69 - Precalculus Enhanced with Graphing Utilities

Question

$h\left(x\right)=x^2-2x$

(a) Find the average rate of change from $2$ to $4.$

(b) Find the equation of the secant line containing $\left(2,h\left(2\right)\right)$ and $\left(4,h\left(4\right)\right)$.

Step-by-Step Solution

Answer

(a) The average rate of change is $4.$

(b) The equation of the secant line is $y=4x-8.$

Step 1. Given information.

We have:

$h\left(x\right)=x^2-2x$

Part a. Step 1. Find the average rate of change from 2 to 4 .

The function is $h\left(x\right)=x^2-2x$.

Average rate of change is given by:     

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

Here, $a=2,b=4$:

$$\begin{gathered} \frac{∆y}{∆x}=\frac{f\left(4\right)-f\left(2\right)}{4-2} \\ \frac{∆y}{∆x}=\frac{{\left(4\right)}^2-2\left(4\right)-\left({\left(2\right)}^2-2\left(2\right)\right)}{2} \\ =4 \end{gathered}$$

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

The function is $h\left(x\right)=x^2-2x$ and the points $\left(2,h\left(2\right)\right)$ and $\left(4,h\left(4\right)\right)$.

Substitute $x=2:$

$$\begin{gathered} h\left(2\right)={\left(2\right)}^2-2\left(2\right) \\ =4-4 \\ =0 \end{gathered}$$

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

$m=4$

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

$$\begin{gathered} y-y_1=m_{\sec }\left(x-x_1\right) \\ y-0=4\left(x-2\right) \\ y=4x-8 \end{gathered}$$