For f(x)=x−6−−−−√
, what is the vertex of f(x)
 after being horizontally dilated by a factor of 2
?

Write your answer as an ordered pair. Do not include spaces.

The function given is \( f(x) = \sqrt{x - 6} \). To horizontally dilate the function by a factor of 2, we replace \( x \) with \( \frac{x}{2} \). So the new function becomes:

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

Next, we determine the vertex. The vertex of the original function \( f(x) = \sqrt{x - 6} \) is at \( (6, 0) \), since the square root function \( \sqrt{x - h} \) has a vertex at \( (h, 0) \).

For the transformed function \( f(x) = \sqrt{\frac{x}{2} - 6} \), set the expression inside the square root equal to 0 to find the x-coordinate of the new vertex:

\[
\frac{x}{2} - 6 = 0
\]

Solve for \( x \):

\[
\frac{x}{2} = 6
\]

\[
x = 12
\]

Thus, the vertex of the new function is \( (12, 0) \).

The final answer is:
\[
(12,0)
\]

Would you like more details or have any questions?

Here are 5 related questions to explore further:
1. What happens to the graph when the function is vertically stretched instead?
2. How do you find the domain of a square root function?
3. How does a horizontal translation affect the vertex of a function?
4. What is the effect of a negative dilation factor on the function?
5. How does combining horizontal dilation and translation affect the graph?

**Tip:** When dealing with horizontal dilations, always divide \( x \) by the dilation factor in the function.

For f(x)=√(x−6), what is the vertex of f(x) after being horizontally dilated by a factor of 2? Write your answer as an ordered pair. Do not include spaces.

Learn how to find the vertex of the square root function f(x) = √(x−6) after being horizontally dilated by a factor of 2. This solution covers key concepts such as horizontal dilation and transformations of functions, suitable for students in Grades 9-12.

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