How do you graph $f(x)=(x^2-2x)/(x^2-4)$ using holes, vertical and horizontal asymptotes, x and y intercepts?

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Answer 1 · 2017-06-18T17:22:48

We can factor $f(x)$ .

$f(x) = (x(x - 2))/((x + 2)(x - 2))$

$f(x) = x/(x+ 2)$

This means that there will be a hole at $x = 2$ . There will be a vertical asymptote at $x = -2$ .

The horizontal asymptote will occur at the ratio between the highest power in the numerator and in the denominator (only if the powers are equal).

Hence, there will be a horizontal asymptote at $y = 1/1 = 1$ .

As for intercepts, the graph will pass through the origin, and the origin will serve as the x and y intercept.

Here is the graph:

graph{(x^2 - 2x)/(x^2 - 4) [-10, 10, -5, 5]}

Hopefully this helps!

How do you graph f(x)=(x^2-2x)/(x^2-4)f(x)=(x^2-2x)/(x^2-4) using holes, vertical and horizontal asymptotes, x and y intercepts?