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

May 1, 2016

Two vertical asymptotes, which are $x = - \frac{1}{3}$ and $x = 2$ and one horizontal asymptote given by $y = 2$

#### Explanation:

In $G \left(x\right) = \frac{6 {x}^{2} + x + 12}{3 {x}^{2} - 5 x - 2}$,

vertical asymptotes are obtained by putting denominator equal to zero.

or $3 {x}^{2} - 5 x - 2 = 0$ or$3 {x}^{2} - 6 x + x - 2 = 0$ or

$3 x \left(x - 2\right) + 1 \left(x - 2\right) = 0$ or $\left(3 x + 1\right) \left(x - 2\right) = 0$.

Hence, vertical asymptotes are $x = - \frac{1}{3}$ and $x = 2$

As the term with highest degree of numerator is $6 {x}^{2}$ and that of denominator $3 {x}^{2}$ are equal,

we have one horizontal asymptote given by $y = \frac{6 {x}^{2}}{3 {x}^{2}} = 2$

There no oblique asymptote (for which highest degree of numerator has to exceed that of denominator by one).

graph{(6x^2+x+12)/(3x^2-5x-2) [-20, 20, -10, 10]}