f(x) = 4 / (x^2 - 2)^(1/2)
Is zero a vertical Asymptote in this function?
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?
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
has vertical asymptotes at $x=\sqrt{2}$ and $x=-\sqrt{2}$f(x) = 4 / (x^2 - 2)^(1/2)