The vertical asymptotes occur where the denominator of the function equals zero.

For the given function:

$f(x) = \frac{2x}{x^2 - 9}$

We find the vertical asymptotes by solving the equation:

$x^2 - 9 = 0$

This can be factored as:

$(x - 3)(x + 3) = 0$

So, the solutions are:

$x = 3 \quad \text{and} \quad x = -3$

Thus, the vertical asymptotes are at $x = 3$ and $x = -3$.

Final answer:

Vertical asymptotes at $x = 3, -3$.

Would you like a more detailed explanation, or do you have any questions?

Here are five related questions to deepen your understanding:

1. How do you find horizontal asymptotes of a rational function?
2. What is the domain of the function $f(x) = \frac{2x}{x^2 - 9}$?
3. What happens to the function $f(x)$ as $x$ approaches the asymptotes?
4. How do you determine if a rational function has any holes (removable discontinuities)?
5. Can a rational function have more than two vertical asymptotes?

Tip: Always check if the denominator can be factored, as this simplifies finding vertical asymptotes.