# How do you find all the asymptotes for function f(x) = (x+4)/(3x^2+5x-2)?

Nov 12, 2016

The vertical asymptotes are $x = \frac{1}{3}$ and $x = - 2$
The vertical asymptote is $y = 0$
There are no slant asymptotes

#### Explanation:

Let's factorise the denominator $3 {x}^{2} + 5 x - 2 = \left(3 x - 1\right) \left(x + 2\right)$

As we cannot divide by $0$, the vertical asymptotes are $x = \frac{1}{3}$ and $x = - 2$

The degree of the numerator is $<$ the degree of the denominator, so we don't have a slant asymptote.

${\lim}_{x \to - \infty} f \left(x\right) = {\lim}_{x \to - \infty} \frac{x}{3 {x}^{2}} = {\lim}_{x \to - \infty} \frac{1}{3 x} = {0}^{-}$

${\lim}_{x \to + \infty} f \left(x\right) = {\lim}_{x \to + \infty} \frac{x}{3 {x}^{2}} = {\lim}_{x \to + \infty} \frac{1}{3 x} = {0}^{+}$

So $y = 0$ is a horizontal asymptote
graph{(x+4)/(3x^2+5x-2) [-11.05, 6.73, -6.2, 2.69]}