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#.
