Q. 64 - Precalculus Enhanced with Graphing Utilities

Question

Find the average rate of change of $h\left(x\right)=x^2-2x+3$

(a) From $-1$ to $1$

(b) From $0$ to $2$

(c) From $2$ to $5$

Step-by-Step Solution

Answer

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

(b) The average rate of change is $0.$

(c) The average rate of change is $5.$

Step 1. Given information.

We have:

$h\left(x\right)=x^2-2x+3$

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

The function is $h\left(x\right)=x^2-2x+3$.

Average rate of change is given by:  

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

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

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

Part b. Step 1. Find the average rate of change from 0 to 2 .

The function is $h\left(x\right)=x^2-2x+3$.

Average rate of change is given by:  

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

Here, $a=0,b=2$:

$$\begin{gathered} \frac{∆y}{∆x}=\frac{f\left(2\right)-f\left(0\right)}{2-0} \\ =\frac{{\left(2\right)}^2-2\left(2\right)+1-\left({\left(0\right)}^2-2\left(0\right)+1\right)}{2} \\ =0 \end{gathered}$$

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

The function is $h\left(x\right)=x^2-2x+3$.

Average rate of change is given by:  

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

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

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