Question

Complete parts​ (a)-(c).
x
1
2
3
4
5
6

y
negative 2
negative 1
0
4
8
16
Question content area bottom
Part 1
​a) Find a cubic function that models the data in the table. Report the model with three decimal places.

y=

Answer 1

To find a cubic function that models the data from the given points, we are looking for a function of the form:

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

We have the following data points from the table:

| \(x\) | \(y\) |
|-------|-------|
| 1 | -2 |
| 2 | -1 |
| 3 | 0 |
| 4 | 4 |
| 5 | 8 |
| 6 | 16 |

We can set up a system of equations using the values of \(y\) for \(x = 1, 2, 3, 4\):

1. For \(x = 1\):
\[
a(1)^3 + b(1)^2 + c(1) + d = -2 \quad (1)
\]
2. For \(x = 2\):
\[
a(2)^3 + b(2)^2 + c(2) + d = -1 \quad (2)
\]
3. For \(x = 3\):
\[
a(3)^3 + b(3)^2 + c(3) + d = 0 \quad (3)
\]
4. For \(x = 4\):
\[
a(4)^3 + b(4)^2 + c(4) + d = 4 \quad (4)
\]

This gives us a system of four equations:

1. \(a + b + c + d = -2\)
2. \(8a + 4b + 2c + d = -1\)
3. \(27a + 9b + 3c + d = 0\)
4. \(64a + 16b + 4c + d = 4\)

Now, we can solve this system of equations to find the coefficients \(a\), \(b\), \(c\), and \(d\).

### Step 1: Eliminate \(d\)

Subtracting equation (1) from the others to eliminate \(d\):

- From (2) - (1):
\[
7a + 3b + c = 1 \quad (5)
\]

- From (3) - (1):
\[
26a + 8b + 2c = 2 \quad (6)
\]

- From (4) - (1):
\[
63a + 15b + 3c = 6 \quad (7)
\]

### Step 2: Further Solve

Now we will solve equations (5), (6), and (7):

From (6), we can express \(c\):
\[
c = 1 - 7a - 3b \quad (8)
\]

Substituting (8) into (6):
\[
26a + 8b + 2(1 - 7a - 3b) = 2
\]
\[
26a + 8b + 2 - 14a - 6b = 2
\]
\[
12a + 2b = 0
\]
\[
6a + b = 0 \quad (9)
\]

From (9):
\[
b = -6a \quad (10)
\]

Substituting (10) into (8):
\[
c = 1 - 7a - 3(-6a) = 1 - 7a + 18a = 1 + 11a \quad (11)
\]

Substituting (10) and (11) into (1):
\[
a - 6a + (1 + 11a) + d = -2
\]
\[
a - 6a + 1 + 11a + d = -2
\]
\[
6a + d + 1 = -2
\]
\[
d = -2 - 1 - 6a = -3 - 6a \quad (12)
\]

### Step 3: Substitute into (7)

Now use (10), (11), and (12) in equation (7):
\[
63a + 15(-6a) + 3(1 + 11a) = 6
\]
\[
63a - 90a + 3 + 33a = 6
\]
\[
6a + 3 = 6
\]
\[
6a = 3 \implies a = \frac{1}{2}
\]

### Step 4: Find \(b\), \(c\), and \(d\)

Now we can find \(b\), \(c\), and \(d\):
- From (10):
\[
b = -6a = -6 \cdot \frac{1}{2} = -3
\]
- From (11):
\[
c = 1 + 11a = 1 + 11 \cdot \frac{1}{2} = 1 + 5.5 = 6.5
\]
- From (12):
\[
d = -3 - 6\left(\frac{1}{2}\right) = -3 - 3 = -6
\]

### Final Function

Thus, our coefficients are:
- \(a = \frac{1}{2}\)
- \(b = -3\)
- \(c = 6.5\)
- \(d = -6\)

Putting them together, we find the cubic function:

\[
y = \frac{1}{2}x^3 - 3x^2 + 6.5x - 6
\]

To report the coefficients with three decimal places:

\[
y = 0.500x^3 - 3.000x^2 + 6.500x - 6.000
\]

Thus, the final cubic function is:

\[
\boxed{y = 0.500x^3 - 3.000x^2 + 6.500x - 6.000}
\]