The function #(2x^3+11x^2+5x-1)/(x^2+6x+5)# has two vertical asymptotes #x=-5# and #x=-1# and an oblique asymptote #y=2x-1#.

First of all, we'll have a look on how to find asymptotes in general:

Let #f# be a function of #x# #(f(x)=text(something with )x)#.

**1. Vercital asymptotes**

We look for vertical asymptotes in points #c# that are on the "ends" of domain of #f#.

The line #x=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#.