The function you are given is:

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

A horizontal dilation by a factor of 2 means that the input $x$ is stretched, affecting the function horizontally. To apply this, we replace $x$ with $\frac{x}{2}$, resulting in the transformed function:

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

Next, let's find the vertex of the original and the transformed functions.

Original Function:

The original function $f(x) = \sqrt{x - 6}$ has a vertex when the expression inside the square root is zero. This happens at:

$x - 6 = 0 \quad \Rightarrow \quad x = 6$

Thus, the vertex of the original function is $(6, 0)$.

Transformed Function:

For the transformed function $f(x) = \sqrt{\frac{x}{2} - 6}$, the vertex occurs when the expression inside the square root is zero:

$\frac{x}{2} - 6 = 0 \quad \Rightarrow \quad \frac{x}{2} = 6 \quad \Rightarrow \quad x = 12$

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

Final Answer:

The vertex of the function after being horizontally dilated by a factor of 2 is $(12, 0)$.

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Do you need more details or have any questions? Here are 5 related questions you might find helpful:

1. How does horizontal dilation affect the graph of a square root function in general?
2. What happens to the vertex if the function is dilated horizontally by a factor of 1/2 instead?
3. How can you find the domain of the transformed function?
4. How do vertical dilations differ from horizontal ones in terms of their effect on the graph?
5. Can we apply both horizontal and vertical transformations to this function simultaneously?

Tip: Horizontal dilations modify the input, causing the graph to stretch or compress without affecting the output values directly.