What is the:

Question

- vertical asymptote
-horizontal asymptote
-non permissible value

of y=(x^2+x-12)/(x+4) ?

So this simplifies to a linear function y=x-3. BUT the oblique asymptote is also y=x-3. So does this graph not exist? I am really confused.

Answer 1

in the simplification process you would have to state it this way ....
y=(x^2+x-12)/(x+4)
= (x-3)(x+4)/(x+4)
= x - 3 , x ≠ -4

So in effect
y=(x^2+x-12)/(x+4)
is the same as
y = x - 3 for every value of x except -4

If we sub x = -4 into y = x-3, we get y = -7
If we sub x = -4 into y=(x^2+x-12)/(x+4) , we get 0/0

So the function y=(x^2+x-12)/(x+4)
is the same as y = x-3, with a hole at (-4,-7)
with no vertical or horizontal asymptotes

Verification:
http://www.wolframalpha.com/input/?i=y%3D%28x%5E2%2Bx-12%29%2F%28x%2B4%29

Answer 2

Why do we have to sub it back into the original function to get 0/0? If we got 0/x then it means it's a vertical asymptote right?

and what about the oblique asymptote? We don't worry about it then?