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

1 Answer
May 4, 2016

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.