Question

the function $f(x)=x^{2}+2x - 6$ is graphed below. determine the slope of the secant line of $f$ for each of the intervals indicated in the table. find the slope of the secant line of $f$ on the interval $1,x_{2}$ for each value of $x_{2}$ shown in the table. write your answers as decimals.

$x_{2}$	secant slope

|2|
|1.1|
|1.01|
|1.001|

Explanation:

Step1: Recall slope - formula for secant line

The slope of the secant line of a function $y = f(x)$ on the interval $[a,x_2]$ is given by $m=\frac{f(x_2)-f(a)}{x_2 - a}$. Here, $a = 1$ and $f(x)=x^{2}+2x - 6$. So, $f(1)=1^{2}+2\times1 - 6=1 + 2-6=-3$.

Step2: When $x_2 = 2$

First, find $f(2)=2^{2}+2\times2 - 6=4 + 4-6 = 2$. Then, use the slope formula $m=\frac{f(2)-f(1)}{2 - 1}=\frac{2-(-3)}{1}=\frac{2 + 3}{1}=5$.

Step3: When $x_2 = 1.1$

Find $f(1.1)=(1.1)^{2}+2\times1.1 - 6=1.21+2.2 - 6=-2.59$. Then, $m=\frac{f(1.1)-f(1)}{1.1 - 1}=\frac{-2.59-(-3)}{0.1}=\frac{-2.59 + 3}{0.1}=\frac{0.41}{0.1}=4.1$.

Step4: When $x_2 = 1.01$

Find $f(1.01)=(1.01)^{2}+2\times1.01 - 6=1.0201+2.02 - 6=-2.9599$. Then, $m=\frac{f(1.01)-f(1)}{1.01 - 1}=\frac{-2.9599-(-3)}{0.01}=\frac{-2.9599 + 3}{0.01}=\frac{0.0401}{0.01}=4.01$.

Step5: When $x_2 = 1.001$

Find $f(1.001)=(1.001)^{2}+2\times1.001 - 6=1.002001+2.002 - 6=-2.995999$. Then, $m=\frac{f(1.001)-f(1)}{1.001 - 1}=\frac{-2.995999-(-3)}{0.001}=\frac{-2.995999 + 3}{0.001}=\frac{0.004001}{0.001}=4.001$.

Answer:

$x_2$	Secant Slope
1.1	4.1
1.01	4.01
1.001	4.001