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

1 Answer
Aug 20, 2016

Answer:

#(x^2+1)/(x+1)# has a vertical asymptote #x=-1# and an oblique asymptote #y = x-1#

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 #x->+-oo#, #2/(x+1) -> 0#

So #f(x)# has an oblique asymptote #y=(x-1)#

When #x=-1# the denominator #(x+1)# is zero, but the numerator #(x^2+1)# is non-zero.

Hence #f(x)# has a vertical asymptote at #x=-1#

graph{(y-(x^2+1)/(x+1))(y-x+1) = 0 [-20, 20, -10, 10]}