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

Oct 25, 2017

See below.

#### Explanation:

Vertical asymptotes occur at the values of $x$ for which the function is undefined. In this case if $x = 2$

$\frac{\sqrt{{x}^{2} + 2}}{3 x - 6} = \frac{\sqrt{{x}^{2} + 2}}{3 \left(2\right) - 6} = \frac{\sqrt{{x}^{2} + 2}}{0}$ ( undefined division by 0).

So the line $x = 2$ is a vertical asymptote:

We now need to see what happens as $x$ increases without bound in the positive direction and the negative direction.

As $x \to \infty$ both numerator and denominator are positive and approach infinity, so the limit is 0.

${\lim}_{x \to \infty} \frac{\sqrt{{x}^{2} + 2}}{3 x - 6} = 0$

As $x \to - \infty$ the numerator is positive and the denominator is negative and both approach infinity, so the limit is 0.

${\lim}_{x \to - \infty} \frac{\sqrt{{x}^{2} + 2}}{3 x - 6} = 0$

This shows that the x axis is a horizontal asymptote. So the line:

$y = 0$ is an asymptote.

Graph:

graph{(sqrt(x^2+2))/(3x-6) [-14.23, 14.25, -7.12, 7.11]}