The given function is:

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

To find the vertex of $f(x)$ after a horizontal dilation by a factor of 2, we follow these steps:

Step 1: Horizontal dilation transformation

A horizontal dilation by a factor of $k = 2$ affects the input (i.e., $x$) of the function. To apply this transformation, replace $x$ with $x/2$ in the original function:

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

This is the new equation of the function after the dilation.

Step 2: Identifying the vertex

The vertex of the function occurs when the expression inside the square root is 0, since that’s the smallest value the square root can take.

Set the inside of the square root to zero:

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

Solve for $x$:

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

Step 3: Calculating the function value at the vertex

Now, substitute $x = 12$ into the new function:

$f(12) = \sqrt{\frac{12}{2} - 6} = \sqrt{6 - 6} = \sqrt{0} = 0$

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

Final Answer:

The vertex of the function after the horizontal dilation by a factor of 2 is:

$(12, 0)$

Would you like further details or have any questions?

Here are five related questions for further exploration:

1. How do horizontal translations affect the vertex of a square root function?
2. What is the effect of a vertical dilation on $f(x) = \sqrt{x - 6}$?
3. How would you reflect the function horizontally and what would the new vertex be?
4. How do you find the domain and range of a square root function after transformations?
5. How would the graph of $f(x) = \sqrt{x - 6}$ change if it is shifted upwards by 3 units?

Tip: Horizontal dilations involve changing the input of the function, while vertical dilations involve scaling the output.