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

1 Answer
Dec 8, 2016

Answer:

The vertical asymptote is #x=-1#
The oblique asymptote is #y=x#
No horizontal asymptote

Explanation:

Let #f(x)=(x^2-2x)/(x+1)#

The domain of #f(x)# is #D_f(x)=RR-{-1} #

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

So, the vertical asymptote is #x=-1#

The degree of the numerator is #># to the degree of the denominator, so we expect an oblique asymptote.

Let's do a long division

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

#color(white)(aaaa)##x^2+x##color(white)(aaaaa)##∣##x#

#color(white)(aaaaa)##0-3x#

So,

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

Therefore

The oblique asymptote is #y=x#

To calculate the limits as #x->+-oo#, we take the terms of highest degree in the numerator and the denominator

#lim_(x->+-oo)f(x)=lim_(x->+-oo)x^2/x=lim_(x->+-oo)x=+-oo#

graph{(y-(x^2-x)/(x+1))(y-x)=0 [-36.53, 36.57, -18.26, 18.27]}