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

1 Answer
Dec 15, 2016

Answer:

The vertical asymptote is #x=3#
No horizontal asymptote.
The oblique asymptote is #y=x+2#

Explanation:

As you cannot divide by #0#, #x!=3#

Therefore,

The vertical asymptote is #x=3#

As the degree of the numerator is #># than the degree of the denominator, we expect an oblique asymptote.

Let's do a long division

#color(white)(aaaa)##x^2-x+1##color(white)(aaaa)##∣##x-3#

#color(white)(aaaa)##x^2-3x##color(white)(aaaaaaa)##∣##x+2#

#color(white)(aaaaa)##0+2x+1#

#color(white)(aaaaaaa)##+2x-6#

#color(white)(aaaaaaa)##+0+7#

Therefore,

#(x^2-x+1)/(x-3)=x+2+7/(x-3)#

The oblique asymptote is #y=x+2#

There is no horizontal asymptote. When #x->+-oo#,

#f(x)->+-oo#

graph{(y-(x^2-x+1)/(x-3))(y-x-2)=0 [-41.1, 41.1, -20.56, 20.56]}