To transform the function $f(x) = \sqrt{x}$ according to the given operations, we'll proceed step by step:

1. Shifting Six Units to the Left:

• When a function is shifted to the left by $c$ units, you replace $x$ with $x + c$ in the function.
• So, shifting $f(x) = \sqrt{x}$ six units to the left results in: $f(x) = \sqrt{x + 6}$

2. Reflecting in the x-axis:

• Reflecting a function across the x-axis involves multiplying the function by $-1$.
• Applying this reflection to $f(x) = \sqrt{x + 6}$ gives: $f(x) = -\sqrt{x + 6}$

3. Reflecting in the y-axis:

• Reflecting a function across the y-axis involves replacing $x$ with $-x$ in the function.
• Applying this reflection to $f(x) = -\sqrt{x + 6}$ gives: $f(x) = -\sqrt{-x + 6}$
• This can also be written as: $f(x) = -\sqrt{6 - x}$

Final Answer:

The function $f(x) = \sqrt{x}$, after being shifted six units to the left and reflected in both the x-axis and y-axis, becomes: $f(x) = -\sqrt{6 - x}$

Would you like a detailed explanation of any of these steps or more examples?

Here are 8 related questions to expand on this topic:

1. What happens to the domain and range of $\sqrt{x}$ after these transformations?
2. How does reflecting a function in the y-axis differ from reflecting in the x-axis?
3. What is the graph of the original function $\sqrt{x}$ and how do these transformations affect it?
4. What other transformations can be applied to functions like $\sqrt{x}$?
5. Can you provide the transformation for shifting a function 4 units up and reflecting it across the x-axis?
6. How would you find the inverse of the transformed function $-\sqrt{6 - x}$?
7. How does the vertex of a parabolic function shift under similar transformations?
8. What are the key properties of square root functions and how are they altered by transformations?

Tip: Understanding how basic transformations (shifts, reflections, stretches) affect the graph of a function will help you visualize complex functions more easily.