# How do you find the vertical, horizontal or slant asymptotes for  r(x)= ((2x^2+14x-36)/(x^2+x-12))?

Jul 13, 2016

vertical asymptotes x = -4 , x = 3
horizontal asymptote y = 2

#### Explanation:

The denominator of r(x) cannot equal zero as this is undefined. Setting the denominator equal 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.

solve: ${x}^{2} + x - 12 = 0 \Rightarrow \left(x + 4\right) \left(x - 3\right) = 0$

$\Rightarrow x = - 4 , x = 3 \text{ are the asymptotes}$

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by the highest exponent of x , that is ${x}^{2}$

$\frac{\frac{2 {x}^{2}}{x} ^ 2 + \frac{14 x}{x} ^ 2 - \frac{36}{x} ^ 2}{{x}^{2} / {x}^{2} + \frac{x}{x} ^ 2 - \frac{12}{x} ^ 2} = \frac{2 + \frac{14}{x} - \frac{36}{x} ^ 2}{1 + \frac{1}{x} - \frac{12}{x} ^ 2}$

as $x \to \pm \infty , r \left(x\right) \to \frac{2 + 0 - 0}{1 + 0 - 0}$

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

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (both of degree 2) Hence there are no slant asymptotes.
graph{(2x^2+14x-36)/(x^2+x-12) [-20, 20, -10, 10]}