The given function is:

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

To horizontally dilate the function by a factor of 2, we replace $x$ with $\frac{x}{2}$. This gives us the new function:

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

Now, let's find the vertex of the transformed function.

Step 1: Identify the domain of the transformed function

The expression inside the square root must be non-negative:

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

Solving for $x$:

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

So, the function is defined for $x \geq 12$.

Step 2: Find the vertex

The vertex of the original function $f(x) = \sqrt{x - 6}$ occurs when $x = 6$, because that makes the expression inside the square root zero.

For the horizontally dilated function $f(x) = \sqrt{\frac{x}{2} - 6}$, the vertex occurs when:

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

Vertex of the transformed function:

The vertex of the function after the horizontal dilation is at $x = 12$. Since at $x = 12$, $f(x) = \sqrt{0} = 0$, the vertex is:

$(12, 0)$

Would you like more details or have any further questions?

Here are 5 related questions to deepen your understanding:

1. How does horizontal dilation affect the graph of any function?
2. How would the vertex change if the dilation factor was $\frac{1}{2}$ instead of 2?
3. How do vertical shifts affect the position of the vertex?
4. What would happen to the function if it were vertically stretched instead?
5. How does the domain of a function change with horizontal dilation?

Tip: Horizontal transformations, such as dilations and translations, affect the input $x$ and can be counterintuitive since they involve reciprocals (e.g., $x \to \frac{x}{2}$).