# How do you find the vertical, horizontal or slant asymptotes for f(x)=(6x) / sqrt( x^2 - 3)?

Mar 27, 2016

Vertical asymptotes$x = \pm \sqrt{3}$.
Horizontal asymptotes: $y = \pm 6$.

#### Explanation:

$y = \frac{6 x}{\sqrt{{x}^{2} - 3}}$.
For real y, $| x | > \sqrt{3}$.
As $x \to \pm \sqrt{3} , y \to \pm \infty$.
So, $x = = - \sqrt{3}$ are asymptotes.

Inversely,
$x = \frac{\sqrt{3} y}{\sqrt{{y}^{2} - 36}}$.
For real x, $| y | > 6$.
As $y \to \pm 6 , x \to \pm \infty$.
So, $y = \pm 6$ are asymptotes.

The graph splits into four branches, in the four quadrants.
The graph is exterior to the region about the origin, bounded by the vertical and horizontal asymptotes, $x = \pm \sqrt{3}$ and $y = \pm 6$.