How do you find the asymptotes of #f(x)= (x^2+1)/(x+1)#?

1 Answer
Oct 6, 2015

Answer:

#y=x-1# is an oblique asymptote.

Explanation:

#(x^2+1)/(x+1) = x-1+2/(x+1)# #" "#(by division)

So, as #x# increases or decreases without bound, the difference between #f(x)# and the #y# value of #y=x-1# approaches #0#.

So the graph of #f(x)# approaches the line #y=x-1#.