# How to find the asymptotes of f(x) = (x+6)/(2x+1) ?

Jan 21, 2016

This function has vertical asymptote $x = - \frac{1}{2}$ and horizontal asymptote $y = \frac{1}{2}$

#### Explanation:

To check if a rational function has a vertical asymtote(s) you have to look for zeroes of the denominator.

In this case there is one zero ${x}_{0} = - \frac{1}{2}$. So $x = - \frac{1}{2}$ is a vertical asymptote.

To look for the horizontal asymptotes you have to calculate

${\lim}_{x \to - \infty} f \left(x\right)$ and ${\lim}_{x \to + \infty} f \left(x\right)$. If the limits are finite and

equal to $l$, then the line $y = l$ is the asymptote.

In this example we have:

${\lim}_{x \to - \infty} \frac{x + 6}{2 x - 1} = {\lim}_{x \to - \infty} \frac{1 + \frac{6}{x}}{2 - \frac{1}{x}} = \frac{1}{2}$

${\lim}_{x \to + \infty} \frac{x + 6}{2 x - 1} = {\lim}_{x \to + \infty} \frac{1 + \frac{6}{x}}{2 - \frac{1}{x}} = \frac{1}{2}$

The limits are finite and equal, so $y = \frac{1}{2}$ is a horizontal asymptote.