# How do you find vertical, horizontal and oblique asymptotes for  f (x)= (x+3)/( (x+1) (x+3) (x-3))?

Mar 29, 2016

First, simplify the expression by canceling:
$f \left(x\right) = \frac{1}{\left(x + 1\right) \left(x - 3\right)}$
This will create a "hole" at x = -3 since x + 3 is the removable factor.

#### Explanation:

Now, set the remaining factors of the denominator equal to 0 to find vertical asymptotes:
x + 1 =0 and x - 3 = 0.
x = -1 and x - 3 are the vertical asymptotes!

A horizontal asymptote will be found at y = 0 since very large values of x (both positive and negative) will produce very small y-values that approach 0.

There will be no oblique asymptotes because the degree of the numerator is smaller than the degree of the denominator, thus giving you a horizontal asymptote.

Here is the graph: