# How do you find the vertical, horizontal and slant asymptotes of: f(x) = (6x + 6) / (3x^2 + 1)?

Apr 15, 2017

The horizontal asymptote is $y = 0$
No vertical asymptote
No slant asymptote

#### Explanation:

The domain of $f \left(x\right)$ is ${D}_{f} \left(x\right) = \mathbb{R}$

$\forall x \in \mathbb{R} , \left(3 {x}^{2} + 1\right) > 0$

So, there is no vertical asymptote.

As the degree of the numerator is $<$ than the degree of the denominator, there is no slant asymptote.

${\lim}_{x \to + \infty} f \left(x\right) = {\lim}_{x \to + \infty} \frac{6 x}{3 {x}^{2}} = {\lim}_{x \to + \infty} \frac{2}{x} = {0}^{+}$

${\lim}_{x \to - \infty} f \left(x\right) = {\lim}_{x \to - \infty} \frac{6 x}{3 {x}^{2}} = {\lim}_{x \to - \infty} \frac{2}{x} = {0}^{-}$

The horizontal asymptote is $y = 0$

graph{(6x+6)/(3x^2+1) [-18.02, 18.03, -9.01, 9.01]}

Apr 15, 2017

$\text{horizontal asymptote at } y = 0$

#### Explanation:

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

$\text{solve } 3 {x}^{2} + 1 = 0 \Rightarrow {x}^{2} = - \frac{1}{3}$

This has no real roots hence there are no vertical asymptotes.

Horizontal asymptotes occur as

${\lim}_{x \to \pm \infty} , f \left(x\right) \to c \text{ (a constant)}$

divide terms on numerator/denominator by the highest power of x, that is ${x}^{2}$

$f \left(x\right) = \frac{\frac{6 x}{x} ^ 2 + \frac{6}{x} ^ 2}{\frac{3 {x}^{2}}{x} ^ 2 + \frac{1}{x} ^ 2} = \frac{\frac{6}{x} + \frac{6}{x} ^ 2}{3 + \frac{1}{x} ^ 2}$

as $x \to \pm \infty , f \left(x\right) \to \frac{0 + 0}{3 + 0}$

$\Rightarrow y = 0 \text{ is the asymptote}$

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here hence there are no slant asymptotes.
graph{(6x+6)/(3x^2+1) [-10, 10, -5, 5]}