The given function is:

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

To perform a horizontal dilation by a factor of $2$, we need to modify the function's argument (inside the square root) by dividing $x$ by the dilation factor. The general transformation for a horizontal dilation by a factor of $k$ is to replace $x$ with $\frac{x}{k}$.

Step-by-step transformation:

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

2. Apply the horizontal dilation by a factor of 2: $f(x) = \sqrt{\frac{x}{2} - 6}$

Finding the vertex:

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

Now, after the dilation, the function is $f(x) = \sqrt{\frac{x}{2} - 6}$. To find the new vertex, set the expression inside the square root equal to zero:

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

Solving for $x$:

$\frac{x}{2} = 6 \implies x = 12$

So the new vertex is $(12, 0)$.

Final Answer:

After the horizontal dilation by a factor of 2, the vertex of the function is $(12, 0)$.

Would you like more details or have any questions?

Here are 5 related questions to deepen understanding:

1. What happens to the vertex if the dilation factor is less than 1?
2. How would the graph change with a vertical dilation instead?
3. What is the domain of the function $f(x) = \sqrt{\frac{x}{2} - 6}$?
4. How would you graph the original function and the dilated one?
5. How would the function change if the dilation factor were negative?

Tip: Horizontal transformations affect the input $x$, so they can shift or stretch the graph left or right without changing the vertical position of the vertex.