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

##### 1 Answer

#### Answer:

Vertical asymptotes at

#### Explanation:

**Finding vertical asymptotes:**

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

#x^2-5=0" "=>" "x^2=5" "=>" "x=+-sqrt5#

Thus, there are two vertical asymptotes: one at

**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

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 *is* at