Q. 67 - Precalculus Enhanced with Graphing Utilities

Question

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

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

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

Step-by-Step Solution

Answer

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

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

Step 1. Given information.

We have:

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

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

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

Average rate of change is given by:   

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

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

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

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

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

Substitute $x=-2$:

$$\begin{gathered} f(-2)={(-2)}^2-2=4-2 \\ =2 \end{gathered}$$

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:  

$$\begin{gathered} y-y_1=m_{\sec }\left(x-x_1\right) \\ y-2=-1\left(x-(-2)\right) \\ y+2=-1\left(x+2\right) \\ y=-x-2-2 \\ y=-x-4 \end{gathered}$$