# How do you find the asymptotes for f(x)=(6x + 6) / (3x^2 + 1)?

Aug 23, 2016

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 gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

solve: $3 {x}^{2} + 1 = 0 \Rightarrow {x}^{2} = - \frac{1}{3} \text{ which has no real roots}$ graph{(6x+6)/(3x^2+1) [-10, 10, -5, 5]}

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

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