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

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Answer 1 · 2016-06-05T19:30:28

the secant line has slope $0$

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

$f(0)=0^2-4(0)=0" "=>" point at (0,0)"$

$f(4)=4^2-4(4)=0" "=>" point at (4,0)"$

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

$m=(0-0)/(4-0)=0/4=0$

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 $f(x)$ at $x=a$ and $x=b$ is

$m=(f(b)-f(a))/(b-a)$

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