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

Question

1822 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-01T05:55:28

We have a vertical asymptote at #x=-1# and

an oblique asymptote #y=x+1#

As in #f(x)=(x^2+2x)/(x+1)# we have the denominator #x+1#,

when #x->-1#, #(x+1)->0# and #f(x)->oo#

As such we have a vertical asymptote at #x=-1#

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

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

Hence when #x->oo#, #f(x)->x+1#

Hence, we have an oblique asymptote #y=x+1#

There is no other asymptote.
graph{(x^2+2x)/(x+1) [-10, 10, -5, 5]}