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

Jun 14, 2018

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

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

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

${x}^{2} - 4 = 0 \setminus \iff {x}^{2} = 4 \setminus \iff x = \setminus \pm \sqrt{4} = \pm 2$

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

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

${\lim}_{x \setminus \to \setminus \pm \setminus \infty} \setminus \frac{1}{{x}^{2} - 4} = \setminus \frac{1}{{\left(\setminus \pm \setminus \infty\right)}^{2} - 4} = \setminus \frac{1}{\infty - 4} = \setminus \frac{1}{\infty} = 0$

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

Note that both passages ${\left(\setminus \pm \setminus \infty\right)}^{2} = \setminus \infty$ and $\setminus \infty - 4 = \setminus \infty$ 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 $y = 0$.