# How do you find the asymptotes for f(x)=1/(x^2+4)?

Mar 23, 2016

This function only has a horizontal asymptote: $y = 0$

#### Explanation:

Notice that $f \left(x\right) \ne 0$ for all $x \in \mathbb{R}$, since if $f \left(x\right) = 0$ then $1 = 0 \left({x}^{2} + 4\right) = 0$, which is false.

As $x \to \pm \infty$ we find $\frac{1}{{x}^{2} + 4} \to 0$

So we have a horizontal asymptote $y = 0$

On the other hand ${x}^{2} + 4 \ge 4 > 0$ for all $x \in \mathbb{R}$, so the denominator is non-zero for all Real values of $x$.

So there is no vertical asymptote.

graph{1/(x^2+4) [-2.5, 2.5, -1.25, 1.25]}