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

Question

Chloe Alexander · Accepted 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.

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:

Madelyn Anderson · Answer

undefined To find the vertical asymptotes, set the denominator equal to zero and solve for $x$. In this case, $x+1=0$, $x+3=0$, and $x-3=0$. So, the vertical asymptotes are $x=-1$ and $x=3$.

To find the horizontal asymptote, compare the degrees of the numerator and denominator. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is $y=0$. If the degrees are equal, divide the leading coefficients. If the degree of the numerator is greater, there is no horizontal asymptote. In this case, since the degree of the numerator is less than the degree of the denominator, there is a horizontal asymptote at $y=0$.

To find oblique asymptotes, divide the numerator by the denominator using polynomial long division. If the result is a polynomial, there is no oblique asymptote. If the result has a quotient with a non-zero slope, the oblique asymptote exists, which is represented by the quotient polynomial. In this case, $f(x) = \frac{x+3}{(x+1)(x+3)(x-3)}$. After polynomial long division, if we find a non-zero quotient, that would represent the oblique asymptote.