the function $f(x)=x^{2}-x - 40$ is graphed below. determine the slope of the secant line of $f$ for each of…

Explanation:

Step1: Recall secant - slope formula

The slope of the secant line of the function $y = f(x)$ on the interval $[x_1,x_2]$ is given by $m=\frac{f(x_2)-f(x_1)}{x_2 - x_1}$. Here, $x_1 = 1$ and $f(x)=x^{2}-x - 40$. So $f(1)=1^{2}-1 - 40=-40$.

Step2: When $x_2 = 2$

First, find $f(2)=2^{2}-2 - 40=4 - 2-40=-38$. Then, use the slope formula $m=\frac{f(2)-f(1)}{2 - 1}=\frac{-38-(-40)}{1}=\frac{-38 + 40}{1}=2$.

Step3: When $x_2 = 1.1$

Find $f(1.1)=(1.1)^{2}-1.1 - 40=1.21-1.1 - 40=-39.89$. Then, $m=\frac{f(1.1)-f(1)}{1.1 - 1}=\frac{-39.89-(-40)}{0.1}=\frac{-39.89 + 40}{0.1}=\frac{0.11}{0.1}=1.1$.

Step4: When $x_2 = 1.01$

Find $f(1.01)=(1.01)^{2}-1.01 - 40=1.0201-1.01 - 40=-39.9899$. Then, $m=\frac{f(1.01)-f(1)}{1.01 - 1}=\frac{-39.9899-(-40)}{0.01}=\frac{-39.9899 + 40}{0.01}=\frac{0.0101}{0.01}=1.01$.

Step5: When $x_2 = 1.001$

Find $f(1.001)=(1.001)^{2}-1.001 - 40=1.002001-1.001 - 40=-39.998999$. Then, $m=\frac{f(1.001)-f(1)}{1.001 - 1}=\frac{-39.998999-(-40)}{0.001}=\frac{-39.998999 + 40}{0.001}=\frac{0.001001}{0.001}=1.001$.

Answer:

$x_2$	Secant Slope
2	2
1.1	1.1
1.01	1.01
1.001	1.001