SOLUTION: find the asymptotes: f (x)=2 --- x squared -9 i hope that equation made sense it should be right on top of eachother but i c

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Question 47800: find the asymptotes: f (x)=2
---
x squared -9
i hope that equation made sense
it should be right on top of eachother but i cant seem to do that.
thanks for your help
Found 2 solutions by longjonsilver, venugopalramana:
Answer by longjonsilver(2297) About Me  (Show Source):
You can put this solution on YOUR website!
+f%28x%29+=+2%2F%28x%5E2-9%29+ can be written as
+f%28x%29+=+2%2F%28%28x%2B3%29%28x-3%29%29+

Now, in ANY fraction, if the denominator is zero you are screwed!. So we want to avoid that at all costs. If x was either 3 or -3, then one of the brackets would be zero..and we would be screwed.

So (vertical) asymptotes are x=3 and x=-3.

Now, to find any horizontal asymptotes, we need to re-write the function in terms of y...

+f%28x%29+=+2%2F%28x%5E2-9%29+
+y+=+2%2F%28x%5E2-9%29+
+y%28x%5E2-9%29+=+2+
+x%5E2y-9y+=+2+
+x%5E2y+=+2%2B9y+
+x%5E2+=+%28%282%2B9y%29%2Fy%29+
+x+=+sqrt%28%282%2B9y%29%2Fy%29+ taking just the positive version (taking square roots gives you 2 versions: a positive and a negative version.)

and so we get in function notation:
+f%28y%29+=+sqrt%28%282%2B9y%29%2Fy%29+ which is just the same as writing +f%28x%29+=+sqrt%28%282%2B9x%29%2Fx%29+.

And so again we ask the question: at what value(s) of y does the denominator become zero? answer is y=0.

So vertical asymptote is at y=0

These are your 3 asymptotes in this question.

jon

Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
find the asymptotes: f (x)=2
---
x squared -9
i hope that equation made sense
it should be right on top of eachother but i cant seem to do that.
thanks for your help
y=2/(x^2-9)=2/(x+3)(x-3)
hence y tends to infinity as x+3 tends to zero or x tends to -3
and also as x-3 tends to zero...that is x tends to 3.
hence the asymptotes are
x=-3 and x=3