The given function is:

$f(x) = \sqrt{x - 6}$

To horizontally dilate this function by a factor of 2, we replace $x$ with $\frac{x}{2}$. The new function becomes:

$f(x) = \sqrt{\frac{x}{2} - 6}$

The vertex of the original function $f(x) = \sqrt{x - 6}$ occurs where the expression inside the square root equals zero, i.e., $x - 6 = 0$, so $x = 6$. Therefore, the vertex of the original function is $(6, 0)$.

For the horizontally dilated function $f(x) = \sqrt{\frac{x}{2} - 6}$, we set the expression inside the square root to zero:

$\frac{x}{2} - 6 = 0$

Solving for $x$:

$\frac{x}{2} = 6 \quad \Rightarrow \quad x = 12$

At this point, the value of $f(x)$ is $0$, so the vertex is at $(12, 0)$.

Thus, the vertex of the horizontally dilated function is $\boxed{(12, 0)}$.

Would you like more details or have any questions?

Here are 5 related questions to deepen understanding:

1. How do vertical dilations affect the vertex of a function?
2. What is the domain of the horizontally dilated function $f(x) = \sqrt{\frac{x}{2} - 6}$?
3. How would the vertex change if we applied a vertical dilation instead of a horizontal one?
4. Can you explain why the horizontal dilation caused the vertex to shift from $(6, 0)$ to $(12, 0)$?
5. What happens to the graph of $f(x) = \sqrt{x - 6}$ if it is reflected horizontally?

Tip: Remember, horizontal transformations (like dilations and translations) affect the $x$-coordinate of the vertex, while vertical transformations affect the $y$-coordinate.