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

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Answer 1 · 2016-05-01T07:42:01

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

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]}

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