# How do you find vertical, horizontal and oblique asymptotes for h(x) = (2x^2-5x-12)/(3x^2-11x-4 )?

Aug 13, 2018

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

#### Explanation:

$\text{factorise numerator/denominator and simplify}$

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

The denominator of h(x) cannot be zero as this would make h(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.

$\text{solve "3x+1=0rArrx=-1/3" is the asymptote}$

$\text{Horizontal asymptotes occur as }$

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

$\text{divide terms on numerator/denominator by } x$

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

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

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

Oblique asymptotes occur when the degree of the numerator is greater than the degree of the denominator. This is not the case here hence there is no oblique asymptote.
graph{(2x+3)/(3x+1) [-10, 10, -5, 5]}