# How do you find the Vertical, Horizontal, and Oblique Asymptote given x/(x^2+x-6)?

Jun 19, 2018

$\text{vertical asymptotes at "x--3" and } x = 2$
$\text{horizontal asymptote at } y = 0$

#### Explanation:

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

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

$\text{solve } {x}^{2} + x - 6 = 0 \Rightarrow \left(x + 3\right) \left(x - 2\right) = 0$

$x = - 3 \text{ and "x=2" are the asymptotes}$

$\text{horizontal asymptotes occur as}$

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

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

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

$\text{as } x \to \pm \infty , f \left(x\right) \to \frac{0}{1 + 0 - 0}$

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

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 are no oblique asymptotes.
graph{x/(x^2+x-6) [-10, 10, -5, 5]}