The function given in the image is:

$f(x) = 3\sqrt{-2x} + 1$

I will now provide information for the various required details:

1. Domain: The square root function requires that the expression inside the square root must be non-negative. Hence:

$-2x \geq 0 \implies x \leq 0$

Therefore, the domain of the function is $(-\infty, 0]$.

2. Range: The square root function produces non-negative values. Therefore, $\sqrt{-2x} \geq 0$, and multiplying by 3 gives $3\sqrt{-2x} \geq 0$. So, the function starts at $f(x) = 1$ when $x = 0$, and the values increase as $x$ decreases. Therefore, the range is $[1, \infty)$.

3. X-intercept: To find the x-intercept, we set $f(x) = 0$:

$3\sqrt{-2x} + 1 = 0 \implies 3\sqrt{-2x} = -1 \quad (\text{No solution because a square root cannot be negative}).$

Thus, there are no x-intercepts.

4. Y-intercept: To find the y-intercept, evaluate the function at $x = 0$:

$f(0) = 3\sqrt{-2(0)} + 1 = 3(0) + 1 = 1$

So, the y-intercept is at $(0, 1)$.

5. End behavior: As $x \to -\infty$, $-2x \to \infty$, so $\sqrt{-2x} \to \infty$, and hence $f(x) \to \infty$. As $x \to 0$, $f(x) \to 1$.

6. Increasing/Decreasing intervals: The function is increasing for $x \in (-\infty, 0]$, since the square root function increases as its argument becomes more negative.

Would you like more details on any specific part or clarification on the graph?

Here are 5 related questions for further exploration:

1. How would the function change if the square root term were positive?
2. Can you calculate the derivative of this function to confirm the increasing/decreasing behavior?
3. How does the transformation $+1$ outside the square root affect the graph?
4. What would happen if the coefficient of the square root were different?
5. How would the domain change if we replaced $-2x$ with a different expression?

Tip: For functions involving square roots, always check the domain first by analyzing the expression inside the root!