The given function is $f(x) = \sqrt{x - 6}$.

When the function undergoes a horizontal dilation by a factor of 2, the transformation affects the $x$-coordinate of the function. For a horizontal dilation by a factor of $2$, the transformation is applied as $x \to \frac{x}{2}$. So, the transformed function becomes:

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

The vertex of the original function $\sqrt{x - 6}$ is at $(6, 0)$, since $f(6) = \sqrt{6 - 6} = 0$.

To find the new vertex, we solve for $x$ in the equation:

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

Solving for $x$:

$\frac{x}{2} = 6 \implies x = 12$

Thus, the new vertex is $(12, 0)$.

The answer is $(12,0)$.

Do you need any further clarifications or details?

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Here are 5 related questions you might explore:

1. How does a vertical dilation affect the vertex of a square root function?
2. What happens to the domain of $f(x)$ when it undergoes a horizontal dilation?
3. How would the vertex change if the function were reflected horizontally?
4. Can you find the axis of symmetry for the transformed function?
5. How would the graph change if a vertical shift was also applied?

Tip: For horizontal transformations, always remember that they affect the $x$-coordinate in the opposite way to vertical transformations.