# Vertical and Horizontal Asymptotes

1. Sep 14, 2009

### Loppyfoot

1. The problem statement, all variables and given/known data
What are the vertical and horizontal asymptotes for:
2x2/ SQRT(x4-81)?

2. Relevant equations

3. The attempt at a solution
For one VA, I got x=3. Are there any more? I am troubled with this square root on the bottom.

2. Sep 14, 2009

### Office_Shredder

Staff Emeritus
Practically, what the square root on the bottom does is two things:
1) It limits where the function can be defined. If x4>81, then the function is not defined. It also has the potential to make asymptotic behavior different... an example is something like

$$\frac{x-1}{ \sqrt{(x^4-1)(x-1)}}$$

At first glance it appears to have an asymptote at x=1 and x=-1, but the x-1 at the top cancels with the x-1 in the denominator (you can see this by factoring the denominator) in the limit as x approaches 1, so no asymptotic behavior occurs.

What's required for a zero in the denominator to cause a vertical asymptote is for there to be no canceling zeroes, or not enough canceling zeroes at least, in the numerator. Having a square root can effect this as it effectively "halves" the power of the zero in the denominator.

You definitely have at least one more vertical asymptote. For what x is x4-81=0? You should be able to fully solve this (either by factoring or by noticing you can set x2=y and solving y2-81=0 first)

3. Sep 15, 2009

### njama

You got x4-81=x4-34=(x2-32)(x2+32)

Can you determine the other values for x for which the denominator is 0 ?

Last edited: Sep 15, 2009
4. Sep 15, 2009

### HallsofIvy

Staff Emeritus
For very large x (but negative since x cannot be larger than $\sqrt[4]{81}$) we can ignore the "81" in comparison with the "x4". Then $2x^2/\sqrt{x^4- 81}$ is approximately $2x^2/\sqrt[4]{x^4}$.

5. Sep 15, 2009

### Mentallic

Oh? So x must be less than $\sqrt[4]{81}$? I didn't know we were searching for complex solutions as well :tongue2: Please consider revising this sentence.

$x^4$ is positive for all real x, not just negative x.

Therefore, the domain is only defined where $$\sqrt{x^4-81}> 0$$

If you know how to solve this, it leaves x<-3, x>3

Just to fix the denominator up slightly: for large $\pm x$ the function tends towards $$y=\frac{2x^2}{\sqrt{x^4}}$$ so you can simplify this and this will give you the horizontal asymptote