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

When a function undergoes a horizontal dilation by a factor of $k$, the transformation is represented by replacing $x$ with $\frac{x}{k}$. So, a horizontal dilation by a factor of 2 modifies the function as follows:

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

Now, to find the vertex of this function:

1. The original vertex of $f(x) = \sqrt{x - 6}$ is at $(6, 0)$, because the square root function $\sqrt{x - 6}$ starts at $x = 6$ when the expression under the square root equals 0.

2. Applying the horizontal dilation by a factor of 2, the new vertex is at $(2 \times 6, 0) = (12, 0)$.

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

Answer:

$(12,0)$

Would you like more details, or do you have any questions?

Here are five related questions to consider:

1. How does vertical dilation affect the vertex of a function?
2. What happens to the graph of $f(x) = \sqrt{x - 6}$ if it's shifted vertically?
3. How do you find the domain of a square root function after horizontal dilation?
4. How do horizontal and vertical dilations combine when applied simultaneously to a function?
5. What is the geometric meaning of dilating a function horizontally?

Tip: Horizontal dilations affect the $x$-coordinates by scaling them, while vertical dilations scale the $y$-coordinates.