# How do you find the slope of the secant lines of f(x) = -3x + 2  through the points: (-4,(f(-4)) and (1,f(1))?

Mar 1, 2016

$- 3$

#### Explanation:

Find the points' $y$-values by evaluating $f \left(- 4\right)$ and $f \left(1\right)$:

$f \left(- 4\right) = - 3 \left(- 4\right) + 2 = 12 + 2 = 14$

$f \left(1\right) = - 3 \left(1\right) + 2 = - 3 + 2 = - 1$

The two points on the secant line are $\left(- 4 , 14\right)$ and $\left(1 , - 1\right)$.

The slope $m$ can then be found using the slope equation:

$m = \frac{\Delta y}{\Delta x} = \frac{14 - \left(- 1\right)}{- 4 - 1} = \frac{15}{- 5} = - 3$

This should make sense, since $f \left(x\right)$ is a line. The secant line, which passes through two points on $f \left(x\right)$, will have to be the exact same as $f \left(x\right)$--there's no other way a line can intercept two points.

Since the secant line is identical to $f \left(x\right)$, we can tell that they will have the same slope, and the slope of $f \left(x\right) = - 3 x + 2$ is $- 3$.