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

Answer:

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]}