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

Question

Chloe Alexander · Accepted Answer

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

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#.

Ariana Alvarez · Answer

undefined To find the vertical asymptote(s) of a rational function, determine the values of x for which the denominator equals zero. These values will give the vertical asymptote(s) of the function.

For the given function f(x) = (3x^2 + 2x - 3) / (x - 1), the vertical asymptote occurs when the denominator, x - 1, equals zero. Therefore, set x - 1 = 0 and solve for x:

x - 1 = 0
x = 1

So, the vertical asymptote is x = 1.

To find the horizontal asymptote of a rational function, compare the degrees of the numerator and denominator. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is y = 0. If the degrees are equal, divide the leading coefficients to find the horizontal asymptote. If the degree of the numerator is greater, there is no horizontal asymptote.

In this case, the degree of the numerator is 2, and the degree of the denominator is 1. Therefore, since the degree of the denominator is greater, the horizontal asymptote is y = 0.

To find any oblique asymptotes, perform polynomial long division. However, since the degree of the numerator is greater than the degree of the denominator by 1, there is no oblique asymptote for this function.