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

1 Answer
Mar 29, 2016

Answer:

First, simplify the expression by canceling:
#f(x) = 1/((x+1)(x-3))#
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: my screenshot 4