# How do you find the asymptotes of a rational function?

Sep 15, 2014

To Find Vertical Asymptotes:

In order to find the vertical asymptotes of a rational function, you need to have the function in factored form. You also will need to find the zeros of the function. For example, the factored function $y = \frac{x + 2}{\left(x + 3\right) \left(x - 4\right)}$ has zeros at x = - 2, x = - 3 and x = 4.

*If the numerator and denominator have no common zeros, then the graph has a vertical asymptote at each zero of the denominator. In the example above $y = \frac{x + 2}{\left(x + 3\right) \left(x - 4\right)}$, the numerator and denominator do not have common zeros so the graph has vertical asymptotes at x = - 3 and x = 4.

*If the numerator and denominator have a common zero, then there is a hole in the graph or a vertical asymptote at that common zero.
Examples:
1. $y = \frac{\left(x + 2\right) \left(x - 4\right)}{x + 2}$ is the same graph as y = x - 4, except it has a hole at x = - 2.

2.$y = \frac{\left(x + 2\right) \left(x - 4\right)}{\left(x + 2\right) \left(x + 2\right) \left(x - 4\right)}$ is the same as the graph of $y = \frac{1}{x + 2} ,$ except it has a hole at x = 4. The vertical asymptote is x = - 2.

To Find Horizontal Asymptotes:

• The graph has a horizontal asymptote at y = 0 if the degree of the denominator is greater than the degree of the numerator. Example: In $y = \frac{x + 1}{{x}^{2} - x - 12}$ (also $y = \frac{x + 1}{\left(x + 3\right) \left(x - 4\right)}$ ) the numerator has a degree of 1, denominator has a degree of 2. Since the degree of the denominator is greater, the horizontal asymptote is at $y = 0$.

• If the degree of the numerator and the denominator are equal, then the graph has a horizontal asymptote at $y = \frac{a}{b}$, where a is the coefficient of the term of highest degree in the numerator and b is the coefficient of the term of highest degree in the denominator. Example: In $y = \frac{3 x + 3}{x - 2}$ the degree of both numerator and denominator are both 1, a = 3 and b = 1 and therefore the horizontal asymptote is $y = \frac{3}{1}$ which is $y = 3$

• If the degree of the numerator is greater than the degree of the denominator, then the graph has no horizontal asymptote.