What are the asymptotes for #(x + 4)/(x^2+x-6)#?

1 Answer
Jul 27, 2015

Answer:

In cases like this, you must often rewrite (factorise)

Explanation:

#=(x+4)/((x-2)(x+3))#

Since the numerator may not be #=0# we conclude:
#x!=2andx!=-3#

So #x=2andx=-3# are the vertical asymptotes.

As #x# gets larger, the #4,-2and3# makes less and less of a difference, so it begins to look like:
#x/(x*x)=1/x# which will get smaller and smaller.

So #y=0# is the horizontal asymptote.
graph{(x+4)/(x^2+x-6) [-10, 10, -5, 5]}