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The problem asks to find the equation for a cubic polynomial that could generate the given graph, passing through the point \((0, 12)\). The provided form of the equation is:

\[
y = 3(x + 1)^2(x - 2)
\]

### Steps to verify and adjust the equation:
1. The general form of the cubic polynomial is \(y = a(x + 1)^2(x - 2)\), where \(a\) is a constant that we need to determine.
2. The graph passes through the point \((0, 12)\), so we can substitute \(x = 0\) and \(y = 12\) into the equation to find \(a\).

### Substitution:
\[
12 = a(0 + 1)^2(0 - 2)
\]
\[
12 = a(1^2)(-2)
\]
\[
12 = -2a
\]
\[
a = \frac{12}{-2} = -6
\]

### Final Equation:
Thus, the equation for the cubic polynomial is:

\[
y = -6(x + 1)^2(x - 2)
\]

This equation should match the graph and pass through the point \((0, 12)\).

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Would you like further details or have any questions?

Here are some related questions:
1. What is the general form of a cubic polynomial?
2. How do you find the value of a constant in a polynomial equation given a point on the curve?
3. Why does the point \((0, 12)\) help determine the constant in this problem?
4. How do transformations affect the shape of cubic polynomials?
5. Can cubic polynomials have more than two real roots?

**Tip:** Always substitute known points into the polynomial to solve for unknown constants when the form of the equation is partially given.

Find the equation for a cubic polynomial that could generate the graph given that it passes through the point (0, 12).

Learn how to solve for the equation of a cubic polynomial that passes through the point (0, 12). This problem involves key algebraic concepts like cubic polynomials and finding constants in polynomial equations.

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