# How do you find the asymptotes for f(x)=(x+2)/sqrt(6x^2+5x+4) ?

Sep 25, 2017

Deleted - see the other posting.

#### Explanation:

incorrect original.

Sep 25, 2017

We have two horizontal asymptotes given by $y = \pm \frac{1}{\sqrt{6}}$

#### Explanation:

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

= $\frac{1 + \frac{2}{x}}{\sqrt{6 + \frac{5}{x} + \frac{4}{x} ^ 2}}$

Hence when $x \to \infty$, $f \left(x\right) \to \frac{1}{\sqrt{6}} = \frac{\sqrt{6}}{6} \cong 0.41$

However, when $x \to - \infty$, while numerator is negative, denominator is positive and hence $f \left(x\right) \to - \frac{1}{\sqrt{6}}$

Hence we have two horizontal asymptotes given by $y = \pm \frac{1}{\sqrt{6}}$

graph{(x+2)/sqrt(6x^2+5x+4) [-5.647, 4.353, -0.84, 4.16]}