To construct a cubic function $f(x)$ with roots at $x = -3$, $x = -1$, and $x = 2$, and passing through the point $(-2, 4)$, follow these steps:

1. Form the general cubic function: Since the roots are $-3$, $-1$, and $2$, the function can be written as: $f(x) = a(x + 3)(x + 1)(x - 2)$ Here, $a$ is a constant to be determined using the additional point.

2. Substitute the point $(-2, 4)$ into the function: Substitute $x = -2$ and $f(x) = 4$: $4 = a(-2 + 3)(-2 + 1)(-2 - 2)$

3. Simplify to solve for $a$: $4 = a(1)(-1)(-4)$ $4 = 4a$ $a = 1$

4. Write the final cubic function: Substituting $a = 1$ into the general equation gives: $f(x) = (x + 3)(x + 1)(x - 2)$

Final Answer:

$f(x) = (x + 3)(x + 1)(x - 2)$

You can expand this if needed: $f(x) = x^3 + 2x^2 - 5x - 6$

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Would you like more details on this process or further clarification?
Here are some related questions you might explore:

1. How can the cubic function be graphed to verify the points?
2. How do you factorize a polynomial if roots are not integers?
3. What is the role of the constant $a$ in polynomial scaling?
4. How can we check if a cubic function passes through a given point?
5. What changes in the function if one root is a double root?

Tip: Verify your cubic function by substituting all given points to ensure accuracy.