Question

for the function $f(x)=x^{2}-1$, find the slope of the secant line between $x = - 3$ and $x = 4$.

Explanation:

Step1: Find function values at given points

First, find $f(-3)$ and $f(4)$.
For $x=-3$, $f(-3)=(-3)^2 - 1=9 - 1 = 8$.
For $x = 4$, $f(4)=4^2-1=16 - 1=15$.

Step2: Use slope formula for secant line

The slope $m$ of the secant line between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $m=\frac{y_2 - y_1}{x_2 - x_1}$. Here, $x_1=-3,y_1 = f(-3)=8,x_2 = 4,y_2=f(4)=15$.
So $m=\frac{f(4)-f(-3)}{4-(-3)}=\frac{15 - 8}{4 + 3}=\frac{7}{7}=1$.

Answer:

$1$