The problem asks for the domain of the function $h(x) = \sqrt{x + 6}$.

To find the domain, we need to determine the values of $x$ for which the expression under the square root is non-negative, since the square root of a negative number is not defined in the real number system.

Step 1: Set the expression inside the square root greater than or equal to zero:

$x + 6 \geq 0$

Step 2: Solve the inequality:

$x \geq -6$

Conclusion:

The domain of the function is all values of $x$ greater than or equal to $-6$. In interval notation, the domain is: $[-6, \infty)$ Or, you could write it as: $x \geq -6$

Would you like any further clarification on this? Let me know if you'd like to explore related topics.

Here are 5 additional questions for further practice:

1. What would the domain be if the function were $h(x) = \sqrt{x - 3}$?
2. How does the domain change if the square root expression is inside a fraction, like $h(x) = \frac{1}{\sqrt{x+5}}$?
3. How do you find the domain of a function involving cube roots, such as $h(x) = \sqrt[3]{x + 2}$?
4. What happens if the function is $h(x) = \sqrt{x^2 - 4}$? How do you find its domain?
5. Can the domain ever be negative for a square root function? Why or why not?

Tip: Always check for restrictions that come from square roots, fractions, or logarithmic functions when determining domains!