How do you find the slope of the secant lines of f(x)=(x^2)-4x at [0,4]?

1 Answer
Jun 5, 2016

the secant line has slope $0$

Explanation:

Find the points at $x = 0$ and $x = 4$.

$f \left(0\right) = {0}^{2} - 4 \left(0\right) = 0 \text{ "=>" point at (0,0)}$

$f \left(4\right) = {4}^{2} - 4 \left(4\right) = 0 \text{ "=>" point at (4,0)}$

The slope of the line connecting the points $\left(0 , 0\right)$ and $\left(4 , 0\right)$ is

$m = \frac{0 - 0}{4 - 0} = \frac{0}{4} = 0$

The secant line is a horizontal line with slope $0$:

graph{(y-x^2+4x)(y-0)=0 [-9.375, 10.625, -5.12, 4.88]}

We can generalize the previous process to say that the slope of the secant line of $f \left(x\right)$ at $x = a$ and $x = b$ is

$m = \frac{f \left(b\right) - f \left(a\right)}{b - a}$