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

Mar 24, 2018

$f \left(x\right)$ has horizontal asymptotes $y = \frac{\sqrt{6}}{6}$ and $y = - \frac{\sqrt{6}}{6}$

#### Explanation:

Note that:

$6 {x}^{2} + 5 x + 4$

is in standard quadratic form with $a = 6$, $b = 5$ and $c = 4$.

This has discriminant $\Delta$ given by the formula:

$\Delta = {b}^{2} - 4 a c = {\left(\textcolor{b l u e}{5}\right)}^{2} - 4 \left(\textcolor{b l u e}{6}\right) \left(\textcolor{b l u e}{4}\right) = 25 - 96 = - 71$

Since $\Delta < 0$, this quadratic has no real zeros and since its leading coefficient is positive, it always takes positive values.

So the denominator $\sqrt{6 {x}^{2} + 5 x + 4}$ of $f \left(x\right)$ is always well defined and non-zero.

So $f \left(x\right)$ has no vertical asymptotes.

Note that:

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

$\textcolor{w h i t e}{{\lim}_{x \to + \infty} f \left(x\right)} = {\lim}_{x \to + \infty} \frac{1 + \frac{2}{x}}{\sqrt{6 + \frac{5}{x} + \frac{4}{x} ^ 2}}$

$\textcolor{w h i t e}{{\lim}_{x \to + \infty} f \left(x\right)} = \frac{1}{\sqrt{6}}$

$\textcolor{w h i t e}{{\lim}_{x \to + \infty} f \left(x\right)} = \frac{\sqrt{6}}{6}$

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

color(white)(lim_(x->-oo) f(x)) = lim_(x->+oo) (-x+2)/sqrt(6x^2-5x+4

$\textcolor{w h i t e}{{\lim}_{x \to - \infty} f \left(x\right)} = {\lim}_{x \to + \infty} \frac{- 1 + \frac{2}{x}}{\sqrt{6 - \frac{5}{x} + \frac{4}{x} ^ 2}}$

$\textcolor{w h i t e}{{\lim}_{x \to - \infty} f \left(x\right)} = - \frac{1}{\sqrt{6}}$

$\textcolor{w h i t e}{{\lim}_{x \to - \infty} f \left(x\right)} = - \frac{\sqrt{6}}{6}$

So $f \left(x\right)$ has horizontal asymptotes $y = \pm \frac{\sqrt{6}}{6}$

That leaves no opportunity for slant asymptotes.

graph{(x+2)/sqrt(6x^2+5x+4) [-22, 22, -2.52, 2.48]}