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

To apply a horizontal dilation by a factor of 2, we replace $x$ with $\frac{x}{2}$. Thus, the new function becomes:

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

Now, let's find the vertex. For the original function, $f(x) = \sqrt{x - 6}$, the vertex occurs when $x = 6$ (since the square root function starts at $x = 6$).

After the horizontal dilation, we need to solve for when the expression inside the square root equals zero:

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

Solving for $x$:

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

Thus, the vertex of the horizontally dilated function is $(12, 0)$.

The answer is: $(12,0)$.

Would you like further details or have any other questions?

Here are 5 related questions:

1. How does horizontal dilation affect the shape of a function?
2. What is the effect of vertical dilation on the vertex of a function?
3. How do you find the domain of a square root function?
4. What changes occur in the graph of a function when both horizontal and vertical dilations are applied?
5. How would you horizontally translate the function $f(x) = \sqrt{x - 6}$?

Tip: Remember, a horizontal dilation stretches or compresses the graph depending on the factor applied.