# How do you find vertical, horizontal and oblique asymptotes for R(x) = (6x^2 + x + 12)/(3x^2 - 5x - 2)?

Aug 2, 2016

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

#### Explanation:

The denominator of R(x) cannot be zero as this is 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.

solve: $3 {x}^{2} - 5 x - 2 = 0 \Rightarrow \left(3 x + 1\right) \left(x - 2\right) = 0$

$\Rightarrow x = - \frac{1}{3} \text{ and " x=2" are the asymptotes}$

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 power of x, that is ${x}^{2}$

$\frac{\frac{6 {x}^{2}}{x} ^ 2 + \frac{x}{x} ^ 2 + \frac{12}{x} ^ 2}{\frac{3 {x}^{2}}{x} ^ 2 - \frac{5 x}{x} ^ 2 - \frac{2}{x} ^ 2} = \frac{6 + \frac{1}{x} + \frac{12}{x} ^ 2}{3 - \frac{5}{x} - \frac{2}{x} ^ 2}$

as $x \to \pm \infty , R \left(x\right) \to \frac{6 + 0 + 0}{3 - 0 - 0}$

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

Oblique 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 oblique asymptotes.
graph{(6x^2+x+12)/(3x^2-5x-2) [-10, 10, -5, 5]}