# How do you find the asymptotes for h(x) = (2x^2-5x-12)/(3x^2-11x-4)?

Apr 6, 2016

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

#### Explanation:

First step is to factorise h(x).

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

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

Vertical asymptotes occur when the denominator of a rational function tends to zero. To find the equation let the denominator equal zero.

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

Horizontal asymptotes occur as ${\lim}_{x \to \pm \infty} f \left(x\right) \to 0$

divide all terms on numerator/denominator by x

$\frac{\frac{2 x}{x} + \frac{3}{x}}{\frac{3 x}{x} + \frac{1}{x}} = \frac{2 + \frac{3}{x}}{3 + \frac{1}{x}}$

as $x \to \pm \infty , \frac{3}{x} \text{ and } \frac{1}{x} \to 0$

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

here is the graph of h(x).
graph{(2x+3)/(3x+1) [-10, 10, -5, 5]}