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Let's solve these problems step-by-step. The questions are asking to find all stationary points, state their nature, and sketch the graph of each function. 

### 1. Problem Set 1

#### **a. \( y = 9x^2 - x^3 \)**

1. **Find the derivative:**
   \[
   y' = \frac{d}{dx}(9x^2 - x^3) = 18x - 3x^2.
   \]
2. **Find stationary points by setting \( y' = 0 \):**
   \[
   18x - 3x^2 = 0 \implies 3x(6 - x) = 0 \implies x = 0 \text{ or } x = 6.
   \]
3. **Determine the nature of stationary points:**
   - **Second derivative test:**
     \[
     y'' = \frac{d}{dx}(18x - 3x^2) = 18 - 6x.
     \]
   - Evaluate at \( x = 0 \) and \( x = 6 \):
     - \( y''(0) = 18 > 0 \) (local minimum).
     - \( y''(6) = 18 - 36 = -18 < 0 \) (local maximum).

   - **Stationary points:**
     - \( x = 0 \) is a local minimum.
     - \( x = 6 \) is a local maximum.

#### **b. \( y = x^3 - 3x^2 - 9x \)**

1. **Find the derivative:**
   \[
   y' = \frac{d}{dx}(x^3 - 3x^2 - 9x) = 3x^2 - 6x - 9.
   \]
2. **Find stationary points by setting \( y' = 0 \):**
   \[
   3x^2 - 6x - 9 = 0 \implies x^2 - 2x - 3 = 0 \implies (x - 3)(x + 1) = 0.
   \]
   - So, \( x = 3 \) and \( x = -1 \).
3. **Determine the nature of stationary points:**
   - **Second derivative test:**
     \[
     y'' = \frac{d}{dx}(3x^2 - 6x - 9) = 6x - 6.
     \]
   - Evaluate at \( x = 3 \) and \( x = -1 \):
     - \( y''(3) = 6(3) - 6 = 12 > 0 \) (local minimum).
     - \( y''(-1) = 6(-1) - 6 = -12 < 0 \) (local maximum).

   - **Stationary points:**
     - \( x = 3 \) is a local minimum.
     - \( x = -1 \) is a local maximum.

#### **c. \( y = x^4 - 4x^3 \)**

1. **Find the derivative:**
   \[
   y' = \frac{d}{dx}(x^4 - 4x^3) = 4x^3 - 12x^2.
   \]
2. **Find stationary points by setting \( y' = 0 \):**
   \[
   4x^2(x - 3) = 0 \implies x = 0 \text{ or } x = 3.
   \]
3. **Determine the nature of stationary points:**
   - **Second derivative test:**
     \[
     y'' = \frac{d}{dx}(4x^3 - 12x^2) = 12x^2 - 24x.
     \]
   - Evaluate at \( x = 0 \) and \( x = 3 \):
     - \( y''(0) = -24 < 0 \) (local maximum).
     - \( y''(3) = 36 - 24 = 12 > 0 \) (local minimum).

   - **Stationary points:**
     - \( x = 0 \) is a local maximum.
     - \( x = 3 \) is a local minimum.

### 2. Problem Set 2

Let's solve these step-by-step next.

Would you like me to proceed with problem set 2, or provide more details about any of these solutions? 

Here are some follow-up questions to expand on this:

1. How can we graph these functions to visualize the stationary points better?
2. What is the importance of the second derivative test in determining the nature of stationary points?
3. How can we apply these concepts to real-world optimization problems?
4. How does the behavior of polynomial functions change near their stationary points?
5. What are some examples of functions with only one stationary point?

**Tip:** Always check both the first and second derivatives to ensure the nature of each stationary point is correctly identified.

For each of the following, find all stationary points and state their nature. Sketch the graph of each function.

Learn how to find stationary points and determine their nature using the first and second derivatives. This detailed solution involves polynomial functions, differentiation, and the second derivative test. Suitable for advanced high school students.

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