Question

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

(a) Find the average rate of change from role="math" localid="1646225596666" $-1$to role="math" localid="1646225601567" $2.$

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

Short answer

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

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

Step-by-step solution

Step 1. Given information.

We have:

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

Part a. Step 1. Find the average rate of change from -1to 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, $a=-1,b=2$:

role="math" localid="1646231606728" $$\begin{gathered} \frac{∆y}{∆x}=\frac{g\left(2\right)-g(-1)}{2-\left(-1\right)} \\ =\frac{{\left(2\right)}^2+1-\left({\left(-1\right)}^2+1\right)}{2+1} \\ =1 \end{gathered}$$

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:

localid="1646231928057" $$\begin{gathered} y-y_1=m_{\sec }\left(x-x_1\right) \\ y-2=1\left(x-\left(-1\right)\right) \\ y+7=x+1 \\ y=x+1-7 \\ y=x-6 \end{gathered}$$