Q. 70 - Precalculus Enhanced with Graphing Utilities

Question

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

(a) Find the average rate of change from $0$ to $3.$

(b) Find the equation of the secant line containing $\left(0,h\left(0\right)\right)$ and $\left(3,h\left(3\right)\right)$.

Step-by-Step Solution

Answer

(a) The average rate of change is $-5.$

(b) The equation of the secant line is $y=-5x.$

Step 1. Given information.

We have:

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

Part a. Step 1. Find the average rate of change from 0 to 3 .

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

Average rate of change is given by:      

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

Here, $a=0,b=3:$

$$\begin{gathered} \frac{∆y}{∆x}=\frac{f\left(3\right)-f\left(0\right)}{3-0} \\ \frac{∆y}{∆x}=\frac{-2{\left(3\right)}^2+3-\left(-2{\left(0\right)}^2+0\right)}{3} \\ =-5 \end{gathered}$$

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

The function is $h\left(x\right)=-2x^2+x$ and the points $\left(0,h\left(0\right)\right)$ and $\left(3,h\left(2\right)\right)$.

Substitute $x=0:$

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

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

$m=-5$

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

$$\begin{gathered} y-y_1=m_{\sec }\left(x-x_1\right) \\ y-0=-5\left(x-0\right) \\ y=-5x \end{gathered}$$