Question

How do you find the vertical, horizontal or slant asymptotes for `f(x)= (1)/(x^2-4)`?

Answer 1

The two vertical asymptotes are the vertical lines `x=2` and `x=-2`

The horizontal asymptote is the horizontal line `y=0` for both `x \to \infty` and `x \to -\infty`.

Explanation:

A function has vertical asymptotes where it is not defined. In this case we have a fraction, so the function is not defined where its denominator equals zero.

This means that we must ask that

`x^2-4=0 \iff x^2 = 4 \iff x = \pm sqrt(4)=pm 2`

So, the two vertical asymptotes are the vertical lines `x=2` and `x=-2`

A function has horizontal asymptotes if the limit as `x\to\pm\infty` is finite. In this case,

`lim_{x\to\pm\infty} \frac{1}{x^2-4} = \frac{1}{(\pm\infty)^2-4} = \frac{1}{infty - 4} = \frac{1}{infty}=0`

So, the horizontal asymptote is the horizontal line `y=0` for both `x \to \infty` and `x \to -\infty`.

Note that both passages `(\pm\infty)^2=\infty` and `\infty-4=\infty` are not to be taken as rigorous algebraic passages, but rather as an extimate of the sign of the infinite (first passage), or as a consideration of the fact that the `-4` played no role (second passage).

Since the function has horizontal asymptotes, it can't have slant asymptotes. Actually, if we looked for slany asymptotes we would again find the line `y=0`.