# How do you identify all asymptotes or holes for f(x)=1/(2x+6)?

Nov 20, 2017

$f \left(x\right)$ has a vertical asymtote at $x = - 3$
$f \left(x\right)$ has a horizontal asymtote at $y = 0$

#### Explanation:

$f \left(x\right) = \frac{1}{2 x + 6}$

$f \left(x\right)$ is defined for all $x \in \mathbb{R}$ except where $2 x + 6 = 0$

I.e. where $x = - 3$

Consider:

${\lim}_{x \to - {3}^{-}} f \left(x\right) = - \infty$

and

${\lim}_{x \to - {3}^{+}} f \left(x\right) = + \infty$

Hence, $f \left(x\right)$ has a vertical asymtote at $x = - 3$

Now consider:

${\lim}_{x \to + \infty} f \left(x\right) = 0$

and

${\lim}_{x \to - \infty} f \left(x\right) = 0$

Thus, $f \left(x\right)$ has a horizontal asymtote at $y = 0$

These can be seen from the graphic below.

graph{(-y+1/(2x+6))(-0.001y+x+3)=0 [-5.644, 0.514, -1.244, 1.833]}

$f \left(x\right)$ has no other asymtotes or points of discontinuity.