# How do you find the vertical, horizontal or slant asymptotes for (2x+1)/(x^2-5)?

Mar 25, 2016

Vertical asymptotes at $x = - \sqrt{5} , \sqrt{5}$; horizontal asymptote at $y = 0$

#### Explanation:

Finding vertical asymptotes:

In a rational function such as this one, vertical asymptotes will occur when the denominator of the function equals $0$.

${x}^{2} - 5 = 0 \text{ "=>" "x^2=5" "=>" } x = \pm \sqrt{5}$

Thus, there are two vertical asymptotes: one at $x = - \sqrt{5}$ and the other at $x = \sqrt{5}$.

Finding horizontal asymptotes:

Again, for a rational function, when the degree of the denominator is larger than the degree of the numerator, the function's horizontal asymptote is at $y = 0$.

We can graph the function to see if we're on the right track:

graph{(2x+1)/(x^2-5) [-18.09, 17.95, -8.5, 9.52]}

The vertical asymptotes seem to be around or approaching $\pm \sqrt{5} \approx \pm 2.24$. The horizontal asymptote is at $y = 0$. Since we have a horizontal asymptote, the function does not have a slant asymptote.