Question

Find the cubic function that models the data in the table below.
                                                                                                       
x
minus2
minus1
0
1
2
3
4
y
negative 4
0
0
2
12
36
80
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Part 1
yequals
  
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​(Simplify your answer. Do not factor. Use integers or decimals for any numbers in the expression. Round to three decimal places as​ needed.)

Answer 1

To find a cubic function that models the data provided, we can express it in the form:

\[
y = ax^3 + bx^2 + cx + d
\]

Where \(a\), \(b\), \(c\), and \(d\) are the coefficients we need to determine.

Given the data points:

- \((-2, -4)\)
- \((-1, 0)\)
- \((0, 0)\)
- \((1, 2)\)
- \((2, 12)\)
- \((3, 36)\)
- \((4, 80)\)

We can set up a system of equations by substituting the \(x\) and \(y\) values into the cubic function.

Using the first four data points \((-1, 0)\), \((0, 0)\), \((1, 2)\), and \((2, 12)\):

1. For \((-2, -4)\):
\[
-4 = a(-2)^3 + b(-2)^2 + c(-2) + d \quad \Rightarrow \quad -4 = -8a + 4b - 2c + d \quad (1)
\]

2. For \((-1, 0)\):
\[
0 = a(-1)^3 + b(-1)^2 + c(-1) + d \quad \Rightarrow \quad 0 = -a + b - c + d \quad (2)
\]

3. For \((0, 0)\):
\[
0 = a(0)^3 + b(0)^2 + c(0) + d \quad \Rightarrow \quad 0 = d \quad (3)
\]

4. For \((1, 2)\):
\[
2 = a(1)^3 + b(1)^2 + c(1) + d \quad \Rightarrow \quad 2 = a + b + c + d \quad (4)
\]

5. For \((2, 12)\):
\[
12 = a(2)^3 + b(2)^2 + c(2) + d \quad \Rightarrow \quad 12 = 8a + 4b + 2c + d \quad (5)
\]

6. For \((3, 36)\):
\[
36 = a(3)^3 + b(3)^2 + c(3) + d \quad \Rightarrow \quad 36 = 27a + 9b + 3c + d \quad (6)
\]

Substituting \(d = 0\) from equation (3) into the other equations, we get:

1. \(-4 = -8a + 4b - 2c \quad (1.1)\)
2. \(0 = -a + b - c \quad (2)\)
3. \(2 = a + b + c \quad (4.1)\)
4. \(12 = 8a + 4b + 2c \quad (5.1)\)
5. \(36 = 27a + 9b + 3c \quad (6.1)\)

We can solve this system of equations step by step.

From equation (2):
\[
c = -a + b \quad (7)
\]

Substituting (7) into (4.1):
\[
2 = a + b + (-a + b) \quad \Rightarrow \quad 2 = 2b \quad \Rightarrow \quad b = 1
\]

Substituting \(b = 1\) into (7):
\[
c = -a + 1 \quad (8)
\]

Now substituting \(b = 1\) and (8) back into (1.1):
\[
-4 = -8a + 4(1) - 2(-a + 1) \quad \Rightarrow \quad -4 = -8a + 4 + 2a - 2 \quad \Rightarrow \quad -4 = -6a + 2 \quad \Rightarrow \quad -6 = -6a \quad \Rightarrow \quad a = 1
\]

From (8):
\[
c = -1 + 1 = 0
\]

Finally, the coefficients are:
- \(a = 1\)
- \(b = 1\)
- \(c = 0\)
- \(d = 0\)

Thus the cubic function that fits the data is:

\[
y = x^3 + x^2
\]

Summarizing:

\[
\boxed{y = x^3 + x^2}
\]