Asked by Beth
                What is the:
- 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.
            
        - 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.
Answers
                    Answered by
            Reiny
            
    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
    
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
                    Answered by
            Beth
            
    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?
    
and what about the oblique asymptote? We don't worry about it then?
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