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

1 Answer
Jun 14, 2018

Answer:

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#.