Find the cubic function that models the data in the table below.

Question

                                                                                                       
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:

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