# How do you find vertical, horizontal and oblique asymptotes for (x+2)/sqrt(6x^2+5x+4)?

Jul 28, 2017

The horizontal asymptote is $y = \frac{1}{\sqrt{6}}$
There are no oblique or vertical asymptote

#### Explanation:

Let $f \left(x\right) = \frac{x + 2}{\sqrt{6 {x}^{2} + 5 x + 4}}$

$\forall x \in \mathbb{R}$, $\left(6 {x}^{2} + 5 x + 4\right) > 0$, $\implies$, no vertical asymptote

Let rewrite the function,

$f \left(x\right) = \frac{\cancel{x} \left(1 + \frac{2}{x}\right)}{\cancel{x} \sqrt{6 + \frac{5}{x} + \frac{4}{x} ^ 2}}$

Therefore,

${\lim}_{x \to \pm \infty} f \left(x\right) = {\lim}_{x \to \pm \infty} \frac{\left(1 + \frac{2}{x}\right)}{\sqrt{6 + \frac{5}{x} + \frac{4}{x} ^ 2}} = \frac{1}{\sqrt{6}}$

The horizontal asymptote is $y = \frac{1}{\sqrt{6}}$

graph{(y-(x+2)/sqrt(6x^2+5x+4))(y-1/sqrt6)=0 [-5.546, 5.555, -2.773, 2.774]}