How do you find vertical, horizontal and oblique asymptotes for (6x + 6) / (3x^2 + 1)?

Jul 20, 2018

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

Explanation:

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

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

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

$\text{This has no real solutions hence there are no vertical}$
$\text{asymptotes}$

$\text{Horizontal asymptotes occur as}$

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

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

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

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

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

$\text{Oblique asymptotes occur when the degree of the}$
$\text{numerator is greater than the degree of the denominator. }$
$\text{This is not the case here hence no oblique asymptotes}$

graph{(6x+6)/(3x^2+1) [-10, 10, -5, 5]}