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

##### 1 Answer

May 1, 2016

Two vertical asymptotes, which are

#### Explanation:

In

vertical asymptotes are obtained by putting denominator equal to zero.

or

Hence, vertical asymptotes are

As the term with highest degree of numerator is

we have one horizontal asymptote given by

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]}