We are given the function $f(x) = \frac{x^2 + 2x}{\sqrt{-2x - 6}}$, and the task is to determine the domain (daerah asal) of the function.

Steps to solve:

1. Denominator condition: The square root, $\sqrt{-2x - 6}$, must have a non-negative argument. Hence, we need: $-2x - 6 \geq 0$ Simplifying this inequality: $-2x \geq 6$ $x \leq -3$ So, $x$ must be less than or equal to $-3$.

2. Numerator condition: The numerator, $x^2 + 2x$, is a quadratic expression and doesn't impose any restrictions on $x$. It is defined for all real numbers.

Conclusion:

The domain of the function is determined by the condition on the square root, which requires $x \leq -3$.

Thus, the correct answer is: $\boxed{B: \{ x \in \mathbb{R}, x \leq -3 \}}$

Would you like more details on this, or do you have any questions?

Here are 5 related questions to deepen your understanding:

1. How would the domain change if the square root were removed from the denominator?
2. What happens to the domain if the denominator becomes $\sqrt{2x + 6}$?
3. Can you graph the function and its domain?
4. What are the critical points of the function, and how do they affect its behavior?
5. How do you determine the range of a function like this?

Tip: Always remember that the square root function requires the argument inside to be non-negative, which heavily influences the domain of the function.