How do you find the equation of the secant line through the points where X has the given values: $f(x)=(x^2)-4x$ ; x=0, x=4?

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Answer 1 · 2015-09-20T23:43:36

See the explanation.

$y = f(x) = x^2-4x$

At $x=0$ , we get $y = f(0) = (0)^2-4(0) = 0$

At $x=4$ , we get $y = f(4) = (4)^2-4(4) = 0$

This gives us two points on the secant line:

$(0,0)$ and $(4,0)$

The slope of the line through those points is $(0-0)/(4-0) = 0/4=0$

The equation of the line through $(0,0)$ with slope $0$ is $y=0$ . (It is the horizontal axis).

How do you find the equation of the secant line through the points where X has the given values: f(x)=(x^2)-4xf(x)=(x^2)-4x; x=0, x=4?