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The problem you provided is in German and involves analyzing a family of functions. Let's break it down step by step:

### Problem Summary (Translated):
You are given the family of functions:

\[
f_a(x) = -\frac{1}{8}(a^2x^3 - 12ax^2 + 36x)
\]
where \(a \in \mathbb{R}^+\) and \(x \in \mathbb{R}\).

- **Task 1**: Investigate the function \( G_a \) for symmetry with respect to the y-axis or the origin.
- **Task 2**: Determine the zeros of \( f_a \).
- **Task 3**: Determine the location and nature of the extremum points of \( G_a \).
- **Task 4**: Sketch the graph of \( G_2 \) (the graph for \( a = 2 \)).

### Step-by-Step Solution:

1. **Symmetry**:
   To check for symmetry, we determine if the function is even, odd, or neither by substituting \( x \) with \( -x \) into \( f_a(x) \).

   - For **y-axis symmetry**, the function must be even, meaning \( f_a(-x) = f_a(x) \).
   - For **origin symmetry**, the function must be odd, meaning \( f_a(-x) = -f_a(x) \).

   Substituting \( -x \) into \( f_a(x) \):

   \[
   f_a(-x) = -\frac{1}{8}(a^2(-x)^3 - 12a(-x)^2 + 36(-x)) = -\frac{1}{8}(-a^2x^3 - 12ax^2 - 36x)
   \]

   This expression does not match \( f_a(x) \) nor \( -f_a(x) \), so the function is **neither even nor odd**, indicating no symmetry.

2. **Zeros (Nullstellen)**:
   To find the zeros, solve \( f_a(x) = 0 \):

   \[
   -\frac{1}{8}(a^2x^3 - 12ax^2 + 36x) = 0
   \]

   Simplifying:

   \[
   a^2x^3 - 12ax^2 + 36x = 0
   \]

   Factor the equation:

   \[
   x(a^2x^2 - 12ax + 36) = 0
   \]

   One solution is \( x = 0 \). For the other solutions, solve the quadratic:

   \[
   a^2x^2 - 12ax + 36 = 0
   \]

   Using the quadratic formula:

   \[
   x = \frac{12a \pm \sqrt{(12a)^2 - 4(a^2)(36)}}{2a^2}
   \]

   Simplifying the discriminant:

   \[
   x = \frac{12a \pm \sqrt{144a^2 - 144a^2}}{2a^2} = \frac{12a}{2a^2} = \frac{6}{a}
   \]

   Thus, the zeros are \( x = 0 \) and \( x = \frac{6}{a} \).

3. **Extrema (Extrempunkte)**:
   To find the extrema, take the derivative \( f_a'(x) \) and set it equal to zero.

   The derivative is:

   \[
   f_a'(x) = -\frac{1}{8}(3a^2x^2 - 24ax + 36)
   \]

   Setting \( f_a'(x) = 0 \):

   \[
   3a^2x^2 - 24ax + 36 = 0
   \]

   Divide by 3:

   \[
   a^2x^2 - 8ax + 12 = 0
   \]

   Use the quadratic formula to solve for \( x \):

   \[
   x = \frac{8a \pm \sqrt{(8a)^2 - 4a^2(12)}}{2a^2} = \frac{8a \pm \sqrt{64a^2 - 48a^2}}{2a^2} = \frac{8a \pm \sqrt{16a^2}}{2a^2} = \frac{8a \pm 4a}{2a^2}
   \]

   Thus, the solutions are \( x_1 = \frac{12}{a} \) and \( x_2 = \frac{4}{a} \).

   To determine the nature of these points, compute the second derivative or analyze the sign of \( f_a'(x) \) around these points.

4. **Sketch of \( G_2 \)**:
   Substitute \( a = 2 \) into the function:

   \[
   f_2(x) = -\frac{1}{8}(4x^3 - 24x^2 + 36x)
   \]

   Simplify:

   \[
   f_2(x) = -\frac{1}{2}(x^3 - 6x^2 + 9x)
   \]

   You can now plot the curve by calculating key points, the zeros, and the extrema.

Would you like further details on any of these steps or help with the sketch? Let me know!

### Follow-up Questions:
1. How do we interpret symmetry tests mathematically?
2. Can you explain the quadratic formula and its application to find the zeros again?
3. How does the second derivative test help in classifying extrema?
4. What does it mean for a function to be even or odd?
5. How do we plot cubic functions effectively?

### Tip:
Always remember to check for extraneous solutions when solving higher-degree polynomials!

Gegeben ist die Funktionenschar f_a(x): x ↦ -1/8(a^2x^3 - 12ax^2 + 36x) mit a ∈ R+ und x ∈ R. Ihre Graphen werden mit Ga bezeichnet. Untersuchen Sie Ga auf Symmetrie zur y-Achse bzw. zum Ursprung. Bestimmen Sie die Nullstellen von fa sowie Lage und Art der Extrempunkte von Ga. Skizzieren Sie anschließend G2.

Solve the polynomial function f_a(x) = -1/8(a^2x^3 - 12ax^2 + 36x) with detailed steps. Learn how to check for symmetry, find zeros, and determine extrema. Sketch the graph for a = 2. Suitable for Grade 12 students.

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