The function (2x^3+11x^2+5x-1)/(x^2+6x+5)2x3+11x2+5x−1x2+6x+5 has two vertical asymptotes x=-5x=−5 and x=-1x=−1 and an oblique asymptote y=2x-1y=2x−1.
First of all, we'll have a look on how to find asymptotes in general:
Let ff be a function of xx (f(x)=text(something with )x)(f(x)=something with x).
1. Vercital asymptotes
We look for vertical asymptotes in points cc that are on the "ends" of domain of ff.
The line x=cx=c is a vertical asymptote when
lim_(x -> c^+) f(x)=+-infty or lim_(x -> c^-) f(x)=+-infty
2. Horizontal asymptotes
We look for horizontal asymptotes in +-infty but only when its sensible, meaning the domain of f "streches" towards +-infty.
The line y=d is a horizontal asymptote when
d=lim_(x -> +infty) f(x) or d=lim_(x -> -infty) f(x) is a constant (is not +-infty).
3. Oblique, or slant, asymptotes
We look for oblique asymptotes in +infty or -infty, one at a time.
The line y=ax+b is an oblique asymptote when both
a=lim_(x -> +-infty) f(x)/x and b=lim_(x -> +-infty) [f(x)-ax] are constants.
In our example f(x)=(2x^3+11x^2+5x-1)/(x^2+6x+5).
The domain is D_f=RR setminus {-5,-1}.
1.
lim_(x -> -5^(+-)) (2x^3+11x^2+5x-1)/(x^2+6x+5)=+-infty
so we have a vertical asymptote x=-5
and
lim_(x -> -1^(+-)) (2x^3+11x^2+5x-1)/(x^2+6x+5)=+-infty
so we have a vertical asymptote x=-1
2.
lim_(x -> +-infty) (2x^3+11x^2+5x-1)/(x^2+6x+5)=+-infty
no horizontal asymptotes
3.
a=lim_(x -> +-infty) (2x^3+11x^2+5x-1)/(x(x^2+6x+5))=2
and
b=lim_(x -> +-infty) [(2x^3+11x^2+5x-1)/(x^2+6x+5)-2x]=-1
so we have an oblique asymptote y=2x-1.
