# How do you find the vertical, horizontal or slant asymptotes for f(x) = (2x^2-5x-12)/(3x^2-11x-4 )?

Oct 4, 2016

vertical asymptote at $x = - \frac{1}{3}$
horizontal asymptote at $y = \frac{2}{3}$

#### Explanation:

The first step is to factorise and simplify f(x).

$f \left(x\right) = \frac{\left(2 x + 3\right) \cancel{\left(x - 4\right)}}{\left(3 x + 1\right) \cancel{\left(x - 4\right)}} = \frac{2 x + 3}{3 x + 1}$

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.

solve : $3 x + 1 = 0 \Rightarrow x = - \frac{1}{3} \text{ is the asymptote}$

Horizontal asymptotes occur as

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

divide numerator/denominator by x

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

as $x \to \pm \infty , f \left(x\right) \to \frac{2 + 0}{3 + 0}$

$\Rightarrow y = \frac{2}{3} \text{ is the asymptote}$

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