Q. 68 - Precalculus Enhanced with Graphing Utilities

Question

$g\left(x\right)=x^2+1$

(a) Find the average rate of change from $-1$ to $2.$

(b) Find the equation of the secant line containing $\left(-1,g(-1)\right)$ and $\left(2,g\left(2\right)\right)$.

Step-by-Step Solution

Answer

(a) The average rate of change is $1.$

(b) The equation of the secant line is $y=x-6.$

Step 1. Given information.

We have:

$g\left(x\right)=x^2+1$

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

The function is $g\left(x\right)=x^2+1$.

Average rate of change is given by:    

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

Here, $a=-1,b=2$:

$$\begin{gathered} \frac{∆y}{∆x}=\frac{g\left(2\right)-g(-1)}{2-\left(-1\right)} \\ =\frac{{\left(2\right)}^2+1-\left({\left(-1\right)}^2+1\right)}{2+1} \\ =1 \end{gathered}$$

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

The function is $f\left(x\right)=x^2+1$ and the points $\left(-1,g\left(-1\right)\right)$ and $\left(2,g\left(2\right)\right)$.

Substitute $x=-1$:

$$\begin{gathered} g(-1)={\left(-1\right)}^2+1 \\ =1+1 \\ =2 \end{gathered}$$

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

$m=1$

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

$$\begin{gathered} y-y_1=m_{\sec }\left(x-x_1\right) \\ y-2=1\left(x-\left(-1\right)\right) \\ y+7=x+1 \\ y=x+1-7 \\ y=x-6 \end{gathered}$$