To solve this, let’s analyze the function provided:

The function is: $y = \sqrt{x + 4}$ with the restriction $x \geq -4$.

Step 1: Determine the Domain of $y = \sqrt{x + 4}$

For the square root function to be defined, the expression inside the square root must be non-negative: $x + 4 \geq 0 \Rightarrow x \geq -4$ Thus, the domain of this function is $x \geq -4$.

Step 2: Determine the Range of $y = \sqrt{x + 4}$

For $y = \sqrt{x + 4}$:

• The minimum value of $x$ is $-4$, which makes $y = \sqrt{-4 + 4} = \sqrt{0} = 0$.
• As $x$ increases, $y = \sqrt{x + 4}$ also increases without bound.

Therefore, the range of $y = \sqrt{x + 4}$ is: $y \geq 0$

Step 3: Select the Correct Answer

Looking at the provided answer choices in the dropdown menu, the correct range for this function is: $y \geq 0$

Would you like further details or have any questions about this solution?

Here are some related questions for further practice:

1. What is the inverse of $y = \sqrt{x + 4}$?
2. How does the range of a square root function generally behave?
3. What happens to the domain and range if the function changes to $y = \sqrt{x - 4}$?
4. How would the range change if the function had a restriction $y \leq 4$?
5. Can we define a function with $y = \sqrt{x + 4}$ over all real numbers?

Tip: For functions involving square roots, remember that the expression inside the square root must be non-negative to keep values in the real number domain.