How do you find vertical, horizontal and oblique asymptotes for #(x^2+1)/(x+1)#?
1 Answer
Aug 20, 2016
Explanation:
#f(x) = (x^2+1)/(x+1)#
#=(x^2-1+2)/(x+1)#
#=((x-1)(x+1)+2)/(x+1)#
#=x-1+2/(x+1)#
As
So
When
Hence
graph{(y-(x^2+1)/(x+1))(y-x+1) = 0 [-20, 20, -10, 10]}