How do you find vertical, horizontal and oblique asymptotes for #f(x)=(3x^2 + 2x - 3 )/( x - 1)#?

1 Answer
May 4, 2016

Answer:

This function #f# has:
a vertical asymptote #x=1#,
no horizontal asymptotes,
and an oblique asymptote #y=3x+5#.

Explanation:

I'm going to use these formulas:

1. Function #f# has a vertical asymptote in #c# (equation: #x=c#) iff #c !in D_f# (#c# is not in the domain of the function) and #lim_(x -> c^pm)f(x)=pm infty#.

2. Function #f# has a horizontal asymptote in #pm infty# (equation: #y=c#) iff #lim_(x -> pm infty) f(x)=c< infty# (this limit must be a finite number).

3. Function #f# has an oblique asymptote in #pm infty# (equation: #y=ax+b#) iff both of these limits are finite:
#a=lim_(x -> pm infty) f(x)/x#
#b=lim_(x -> pm infty) (f(x)-ax)#

Important note: A horizontal asymptote is a special case of an qblique one (where #a=0#).

For #f(x)=(3x^2+2x-3)/(x-1)# we have:
#D_f=mathbb(R)\setminus {1}#

1.
#lim_(x -> 1^+) f(x)=[(2)/(0^+)]=+infty#
#lim_(x -> 1^-) f(x)=[(2)/(0^-)]=-infty#
which gives us a vertical asymptote #x=1#.

2.
#lim_(x -> +infty) f(x)=lim_(x -> +infty) (3x+2-3/x)/(1-1/x)=+infty#
#lim_(x -> -infty) f(x)=lim_(x -> -infty) (3x+2-3/x)/(1-1/x)=-infty#
which gives us no horizontal asymptotes.

3.
#lim_(x -> pm infty) f(x)/x=lim_(x -> pm infty) (3x^2+2x-3)/(x^2-x)=#
#=lim_(x -> pm infty) (3+2/x-3/x^2)/(1-1/x)=3=a#
#lim_(x -> pm infty) (f(x)-ax)=lim_(x -> pm infty) ((3x^2+2x-3)/(x-1)-3x)=#
#=lim_(x -> pm infty) (3x^2+2x-3-3x^2+3x)/(x-1)=#
#=lim_(x -> pm infty) (5x-3)/(x-1)=lim_(x -> pm infty) (5-3/x)/(1-1/x)=5=b#
which gives us an oblique asymptote #y=3x+5#.