Question

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

Answer 1

Vertical asymptotes at `x=-sqrt5,sqrt5`; 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" "=>" "x^2=5" "=>" "x=+-sqrt5`

Thus, there are two vertical asymptotes: one at `x=-sqrt5` and the other at `x=sqrt5`.

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 `pmsqrt5approxpm2.24`. The horizontal asymptote is at `y=0`. Since we have a horizontal asymptote, the function does not have a slant asymptote.