To determine the domain of the function $h(x) = \sqrt{3x + 9}$, we need to ensure the expression under the square root is non-negative because the square root of a negative number is not defined in the set of real numbers.

Step-by-Step Solution:

1. The condition for the square root to be defined is: $3x + 9 \geq 0$

2. Solve for $x$: $3x \geq -9$ $x \geq -3$

Domain:

The domain of $h(x)$ is all values of $x$ such that $x \geq -3$.

Correct Answer:

$\boxed{x \geq -3}$

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Here are some follow-up questions for practice:

1. What is the domain of $h(x) = \sqrt{2x - 4}$?
2. Solve for $x$ in $5x + 10 \geq 0$.
3. What is the domain of $h(x) = \sqrt{x^2 - 4}$?
4. How does the square root function affect the range of a function?
5. What would happen to the domain if $h(x) = \sqrt{3x + 9}$ were inside a fraction?

Tip:

Always remember: for square root functions, the expression inside the square root must be greater than or equal to zero for the domain in real numbers.