How do you find all the asymptotes for #(2x^3+11x^2+5x-1)/(x^2+6x+5 )#?

Question

Elias Ackerman · Accepted Answer

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#.

Sophie Adams · Answer

undefined To find the asymptotes of the given rational function, you first need to determine whether there are any vertical, horizontal, or slant asymptotes.

1. Vertical asymptotes occur where the denominator equals zero but the numerator does not. So, set the denominator equal to zero and solve for x. Any values of x that make the denominator zero but not the numerator will result in vertical asymptotes.

2. Horizontal asymptotes occur when the degree of the numerator is less than or equal to the degree of the denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degrees are equal, the horizontal asymptote is the ratio of the leading coefficients.

3. Slant asymptotes occur when the degree of the numerator is exactly one greater than the degree of the denominator. To find the slant asymptote, perform polynomial long division or use another appropriate method to divide the numerator by the denominator. The quotient obtained will be the equation of the slant asymptote.

Once you have determined the type of asymptotes, you can find their equations and graph the function accordingly.