How do you find the slant asymptote of #(x^2) / (x+1)#?

1 Answer
Jan 4, 2016

Answer:

Re-express as the sum of a linear polynomial and a rational expression that tends to zero as #x->+-oo# to find slant asymptote:

#y = x-1#

Explanation:

#x^2/(x+1)#

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

#=((x-1)(x+1)+1)/(x+1)#

#=x-1+1/(x+1)#

Then #1/(x+1) -> 0# as #x->+-oo#

So the slant asymptote (a.k.a. oblique asymptote) is #y = x-1#