Given the function $f(x) = \sqrt{x - 6}$, we need to determine the vertex after applying a horizontal dilation by a factor of 2.

Step-by-Step Process:

1. Original Function: The function is $f(x) = \sqrt{x - 6}$. This is a square root function, which typically has the form $f(x) = \sqrt{x - h}$, where $h$ determines the horizontal shift. For the given function, $h = 6$, so the vertex of the original function is $(6, 0)$.

2. Horizontal Dilation: A horizontal dilation by a factor of 2 means that the input to the function is multiplied by $\frac{1}{2}$. Therefore, the dilated function becomes: $f_{\text{dilated}}(x) = \sqrt{\frac{x}{2} - 6}$ or equivalently: $f_{\text{dilated}}(x) = \sqrt{\frac{1}{2}(x - 12)}$ Thus, the vertex of the new function occurs when the expression inside the square root is zero, i.e.: $\frac{x}{2} - 6 = 0 \quad \Rightarrow \quad x = 12$

3. New Vertex: After the horizontal dilation, the vertex shifts to $(12, 0)$.

Final Answer:

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

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

Here are some related questions to deepen understanding:

1. How would the function change with a vertical dilation instead of horizontal?
2. What happens to the domain of $f(x) = \sqrt{x - 6}$ after the dilation?
3. How would you find the range of the dilated function?
4. What effect would a reflection across the y-axis have on the vertex?
5. How does shifting the function vertically affect the vertex?

Tip: Horizontal dilations affect the x-values, scaling them by the reciprocal of the dilation factor.