1. ## Vertical Asymptotes

f(x) = 4 / (x^2 - 2)^(1/2)

Is zero a vertical Asymptote in this function?

2. ## Re: Vertical Asymptotes

No, it is not! Why would you think it was? (This function is only defined for $x> \sqrt{2}$ or $x< -\sqrt{2}$.)

3. ## Re: Vertical Asymptotes

$x=0$ is not in the domain of this function. Domain is $(-\infty,-\sqrt{2}) \cup (\sqrt{2}, \infty)$

4. ## Re: Vertical Asymptotes

When I plug zero inside the square root, I get a negative value which is undefined. I thought any undefined value in the denominator will be a vertical asymptote for the function.

Then, if it is not a vertical asymptote, what is the correct way to recognize the values of the vertical asymptote?

5. ## Re: Vertical Asymptotes

vertical asymptotes occur when a factor in the denominator equals 0 with no common factor in the numerator

for example ...

$y = \dfrac{(x+1)(x-2)}{(x-3)}$ has a vertical asymptote at $x=3$

$y = \dfrac{(x+1)(x-3)}{(x-3)}$ has a point discontinuity (a hole) at $x=3$

the function you posted

f(x) = 4 / (x^2 - 2)^(1/2)
has vertical asymptotes at $x=\sqrt{2}$ and $x=-\sqrt{2}$

6. ## Re: Vertical Asymptotes

Thanks a lot. I have got the idea.