Question

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

Answer 1

Two vertical asymptotes, which are `x=-1/3` and `x=2` and one horizontal asymptote given by `y=2`

Explanation:

In `G(x)=(6x^2+x+12)/(3x^2-5x-2)`,

vertical asymptotes are obtained by putting denominator equal to zero.

or `3x^2-5x-2=0` or`3x^2-6x+x-2=0` or

`3x(x-2)+1(x-2)=0` or `(3x+1)(x-2)=0`.

Hence, vertical asymptotes are `x=-1/3` and `x=2`

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

we have one horizontal asymptote given by `y=(6x^2)/(3x^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]}