# How do you find the asymptote(s) or hole(s) of #f(x) = (x+3)/(x^2-9)#?

##### 2 Answers

asymptote at

hole at

#### Explanation:

Factor the denominator, it is the difference of squares:

The factor(s) in the denominator that will cancel out, in this case

The factor(s) in the denominator that will not cancel out, in this case

Here is the graph, it shows the asymptote, but not the hole at

graph{(x+3)/(x^2-9) [-6.67, 13.33, -4.44, 5.56]}

VA:

HA:

#### Explanation:

We have the following expression

What I have in blue is a **Difference of Squares** of the form

Now, our new expression is

which simplifies to

Vertical asymptotes are found by setting the denominator equal to zero, because this is the value for which the function is undefined.

Horizontal asymptotes are found by comparing the degrees. Since there is a higher degree in the denominator, we have a horizontal asymptote at

We have a vertical asymptote at

Hope this helps!