### Solution:
To find a cubic function that models the given data, we need to determine the cubic polynomial [tex]\(y = ax^3 + bx^2 + cx + d\)[/tex] that fits the data points [tex]\((x, y)\)[/tex].
Here, the given data points are:
[tex]\[ \begin{aligned}
x: & \quad -2, -1, 0, 1, 2, 3, 4, \\
y: & \quad -20, -4, 0, -2, -4, 0, 16.
\end{aligned} \][/tex]
We will define the function in the form:
[tex]\[ y = ax^3 + bx^2 + cx + d. \][/tex]
Using a method of polynomial fitting (e.g., least squares fit), we can find the coefficients [tex]\(a\)[/tex], [tex]\(b\)[/tex], [tex]\(c\)[/tex], and [tex]\(d\)[/tex]. The best-fit cubic polynomial through these points is:
[tex]\[ y = 1.000x^3 - 3.000x^2 + 0x + 0 \][/tex]
This gives us the function:
[tex]\[ y = x^3 - 3x^2 \][/tex]
### Evaluation of the Function:
Let's check the values by plugging in the given [tex]\(x\)[/tex]-values:
1. For [tex]\(x = -2\)[/tex]:
[tex]\[ y = (-2)^3 - 3(-2)^2 = -8 - 12 = -20 \][/tex]
2. For [tex]\(x = -1\)[/tex]:
[tex]\[ y = (-1)^3 - 3(-1)^2 = -1 - 3 = -4 \][/tex]
3. For [tex]\(x = 0\)[/tex]:
[tex]\[ y = (0)^3 - 3(0)^2 = 0 \][/tex]
4. For [tex]\(x = 1\)[/tex]:
[tex]\[ y = (1)^3 - 3(1)^2 = 1 - 3 = -2 \][/tex]
5. For [tex]\(x = 2\)[/tex]:
[tex]\[ y = (2)^3 - 3(2)^2 = 8 - 12 = -4 \][/tex]
6. For [tex]\(x = 3\)[/tex]:
[tex]\[ y = (3)^3 - 3(3)^2 = 27 - 27 = 0 \][/tex]
7. For [tex]\(x = 4\)[/tex]:
[tex]\[ y = (4)^3 - 3(4)^2 = 64 - 48 = 16 \][/tex]
The function [tex]\( y = x^3 - 3x^2 \)[/tex] matches all the given [tex]\( (x, y) \)[/tex] pairs perfectly.
### Final Simplified Answer:
So, the cubic function that models the provided data is:
[tex]\[ y = x^3 - 3x^2 \][/tex]
No further simplification is needed, as this equation is already in its most simplified form.
