Question

on of the secant line passing through (-6, f(-6)) and (2, f(2)). write your answer in the form y = mx + b.

Explanation:

Step1: Calculate the slope $m$.

The slope formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $m=\frac{y_2 - y_1}{x_2 - x_1}$. Here $x_1=-6,y_1 = f(-6),x_2 = 2,y_2=f(2)$. So $m=\frac{f(2)-f(-6)}{2-(-6)}=\frac{f(2)-f(-6)}{8}$.

Step2: Use the point - slope form $y - y_1=m(x - x_1)$ to find the equation of the line.

Let's use the point $(2,f(2))$. The point - slope form is $y - f(2)=\frac{f(2)-f(-6)}{8}(x - 2)$.
Expand the right - hand side: $y - f(2)=\frac{f(2)-f(-6)}{8}x-\frac{2(f(2)-f(-6))}{8}$.
Then $y=\frac{f(2)-f(-6)}{8}x-\frac{2(f(2)-f(-6))}{8}+f(2)$.
Simplify the constant term:
\[

$$\begin{align*} -\frac{2(f(2)-f(-6))}{8}+f(2)&=-\frac{f(2)}{4}+\frac{f(-6)}{4}+f(2)\\ &=\frac{-f(2)+f(-6) + 4f(2)}{4}\\ &=\frac{3f(2)+f(-6)}{4} \end{align*}$$

\]
So the equation of the line is $y=\frac{f(2)-f(-6)}{8}x+\frac{3f(2)+f(-6)}{4}$.

Answer:

$y=\frac{f(2)-f(-6)}{8}x+\frac{3f(2)+f(-6)}{4}$