Question about Vertical Asymptote: y = x^3/(9-x^2)

frank789

Junior Member
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Sep 16, 2017
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Hi all

was doing a function sketching practice problem and was wondering;

for examples sake lets use:

y = x^3/(9-x^2)

so the domain restrictions on this one would be -3 and 3. Is there ever a case where a domain restriction does not result in a vertical asymptote?
Is it necessary to evaluate the L.H.L. and R.H.L. to verify that the function goes to +- infinity? or is it just implied?

just curious, thanks in advance! :cool:
 
not counting peicewise functions not defined at certain points :lol:
 
y = x^3/(9-x^2)

so the domain restrictions on this one would be -3 and 3. Is there ever a case where a domain restriction does not result in a vertical asymptote?
Is it necessary to evaluate the L.H.L. and R.H.L. to verify that the function goes to +- infinity? or is it just implied?

If by "domain restrictions" you mean points where it is undefined, you can also have "holes" (removable discontinuities), as in (9-x^2)/(9-x^2), to give a very simple example.

It is also possible to have a vertical asymptote that goes the same way on both sides (e.g. both +infinity), as in x^3/(9-x^2)^2. So you do need to consider at least the sign on each side of the asymptote.

I think of a "domain restriction" as an explicit restriction of the domain, as in "f(x) = x^2, x >= 0". But you may be using different terminology.
 
you are right about my poor description but despite that you still managed to answer it haha thanks!
 
Two things:

1a) Make the denominator 0 (zero). These are POTENTIAL vertical asymptotes.
1b) If the numerator is also zero (0) for the same x-value, it might not be an asymptote. It might be a hole. Check it out by reducing common factors.

2) The degree of the of the factor that creates the asymptote dictates the behavior. (Assuming factors are reduced.)
2a) Odd Degree - One side goes up and one side goes down.
2b) Even Degree - both go up or both go down.
2c) Only the sign of one side is needed if you know the degree, but it never hurts to gain more information.
 
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A little fine-tuning:

First, all of this applies specifically to rational functions like yours; other kinds of functions may differ.

Second, it is possible to still have an asymptote when both numerator and denominator are zero; it is best to look at the function after canceling like factors to determine its behavior (keeping in mind that being able to cancel means it is undefined). Ultimately, if the exponent in the denominator was higher, you end up with an asymptote.
 
First, all of this applies specifically to rational functions like yours; other kinds of functions may differ.
We're using the quotient of simple, real-valued polynomials, right? Is there another definition in "Intermediate/Advance Algebra"?

Second, it is possible to still have an asymptote when both numerator and denominator are zero
No example comes to mind. Demonstration?
 
I knew I should have stayed in bed, this morning.

I better check everything else I did all day...

Maybe I didn't get out of bed. Hmmm...
 
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