Q. 66 - Precalculus Enhanced with Graphing Utilities

Question

$f\left(x\right)=-4x+1$

(a) Find the average rate of change from $2$ to $5.$

(b) Find the equation of the secant line containing $\left(2,f\left(2\right)\right)$ and $\left(5,f\left(5\right)\right)$.

Step-by-Step Solution

Answer

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

(b) The equation of the secant line is $y=-4x+1.$

Step 1. Given information.

We have:

$f\left(x\right)=-4x+1$

Part a. Step 1. Find the average rate of change from 2 to 5 .

The function is $f\left(x\right)=-4x+1$.

Average rate of change is given by:  

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

Here, $a=2,b=5$:

$$\begin{gathered} \frac{∆y}{∆x}=\frac{f\left(5\right)-f\left(2\right)}{5-2} \\ =\frac{-4\left(5\right)+1-\left(-4\left(2\right)+1\right)}{3} \\ =-4 \end{gathered}$$

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

The equation is $f\left(x\right)=-4x+1$ and the point $\left(2,f\left(2\right)\right)$ and $\left(5,f\left(5\right)\right)$.

Substitute $x=2$ into the equation:

$$\begin{gathered} f\left(2\right)=-4\left(2\right)+1 \\ =-8+1 \\ =-7 \end{gathered}$$

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: 

$$\begin{gathered} y-y_1=m_{\sec }\left(x-x_1\right) \\ y-\left(-7\right)=-4\left(x-2\right) \\ y+7=-4x+8 \\ y=-4x+8-7 \\ y=-4x+1 \end{gathered}$$