To determine the domain of the function $y = \sqrt{x} + 4$, we need to identify the values of $x$ for which the function is defined.

The function involves a square root, which is only defined for non-negative values of $x$. This is because the square root of a negative number is not a real number. Therefore, $x$ must be greater than or equal to 0. There is no upper limit to the values that $x$ can take.

So, the domain of the function $y = \sqrt{x} + 4$ is: $0 \leq x < \infty$

Thus, the correct answer is: $\boxed{0 \leq x < \infty}$

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

Here are some related questions for further exploration:

1. What is the range of the function $y = \sqrt{x} + 4$?
2. How would the domain change if the function were $y = \sqrt{x-4} + 4$?
3. What is the domain of the function $y = \sqrt{4 - x}$?
4. How does the addition of a constant outside the square root affect the domain of the function?
5. What is the domain of the function $y = \sqrt{x^2 - 4}$?

Tip: When dealing with square root functions, always consider the values of $x$ that make the expression inside the square root non-negative.