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The problem given involves the first derivative \( y' = x(x - 12)^2 \) of a continuous function \( y = f(x) \), and we are asked to:

1. Find the second derivative \( y'' \).
2. Use graphing procedures to sketch the general shape of the graph of \( f(x) \).

### Step 1: Find the second derivative \( y'' \)

The first derivative is:

\[
y' = x(x - 12)^2
\]

To find \( y'' \), we apply the product rule. The product rule for derivatives is \( \frac{d}{dx}[u \cdot v] = u'v + uv' \), where \( u = x \) and \( v = (x - 12)^2 \).

First, compute the derivatives of \( u \) and \( v \):
- \( u = x \Rightarrow u' = 1 \)
- \( v = (x - 12)^2 \Rightarrow v' = 2(x - 12) \) (by the power rule and chain rule)

Now, applying the product rule:

\[
y'' = \frac{d}{dx} \left[ x(x - 12)^2 \right] = 1 \cdot (x - 12)^2 + x \cdot 2(x - 12)
\]

Simplify each term:

\[
y'' = (x - 12)^2 + 2x(x - 12)
\]

Now expand both terms:

\[
y'' = (x - 12)^2 + 2x(x - 12) = (x^2 - 24x + 144) + 2x(x - 12)
\]
\[
y'' = x^2 - 24x + 144 + 2x^2 - 24x
\]

Combine like terms:

\[
y'' = (x^2 + 2x^2) - (24x + 24x) + 144
\]
\[
y'' = 3x^2 - 48x + 144
\]

So the second derivative is:

\[
y'' = 3x^2 - 48x + 144
\]

### Step 2: Analyze the graph using the derivatives

To sketch the general shape of the graph of \( f(x) \), let's analyze the behavior of the function based on the first and second derivatives.

1. **Critical points**: Critical points occur where \( y' = 0 \). Set \( y' = x(x - 12)^2 = 0 \).
   - \( x = 0 \) or \( x - 12 = 0 \Rightarrow x = 12 \).

   Therefore, the critical points are \( x = 0 \) and \( x = 12 \).

2. **Concavity and inflection points**: To determine the concavity of the graph, we analyze \( y'' = 3x^2 - 48x + 144 \).
   - Find the inflection points where \( y'' = 0 \). Set \( 3x^2 - 48x + 144 = 0 \), and divide by 3:
     \[
     x^2 - 16x + 48 = 0
     \]

     Solve using the quadratic formula:
     \[
     x = \frac{-(-16) \pm \sqrt{(-16)^2 - 4(1)(48)}}{2(1)}
     \]
     \[
     x = \frac{16 \pm \sqrt{256 - 192}}{2} = \frac{16 \pm \sqrt{64}}{2} = \frac{16 \pm 8}{2}
     \]

     So \( x = 12 \) or \( x = 4 \). These are the potential inflection points.

3. **Behavior at critical points**:
   - At \( x = 0 \): \( y' = 0 \) and the graph touches the x-axis but does not cross it since \( (x - 12)^2 \) is squared, making the slope zero and positive on both sides of \( x = 12 \).
   - At \( x = 12 \): The slope changes, suggesting a critical behavior (possibly a local minimum or maximum).

### Conclusion:
The general shape of \( f(x) \):
- The graph has critical points at \( x = 0 \) and \( x = 12 \).
- The graph is concave down between \( x = 0 \) and \( x = 12 \) (since \( y'' \) is positive), and it changes concavity at the inflection points, which occur at \( x = 4 \) and \( x = 12 \).
- The graph touches but does not cross the x-axis at \( x = 0 \), and there is a local minimum at \( x = 12 \).

Do you have any questions, or would you like further clarification on any step?

---

Here are five related questions for practice:
1. What is the graph's behavior as \( x \to \infty \) based on \( y' \)?
2. How would you confirm if the point \( x = 12 \) is a local minimum or maximum using the second derivative test?
3. What would the behavior of the function be at negative values of \( x \)?
4. Can you describe the relationship between the first and second derivatives in determining the function's shape?
5. What would happen if the coefficient of \( x \) in \( y' \) was different?

**Tip**: Always check both the first and second derivatives to fully understand the function's behavior!

The first derivative of a continuous function y = f(x) is y' = x(x - 12)^2. Find y'' and then use the graphing procedure to sketch the general shape of the graph of f.

Learn how to find the second derivative of the function y = f(x) with y' = x(x - 12)^2 using calculus techniques such as the product rule and quadratic formula. Also, sketch the graph by analyzing the critical points and inflection points. Suitable for students in advanced high school or introductory college-level calculus.

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