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To solve this problem, we need to determine whether a linear, quadratic, or exponential function best models the data provided in the table. The functions to consider are:

1. **Linear:** \( y = mx + b \)
2. **Quadratic:** \( y = ax^2 + bx + c \)
3. **Exponential:** \( y = a \cdot b^x \)

Let's analyze the dataTo solve this problem, we need to determine whether a linear, quadratic, or exponential function best models the data provided in the table. The functions to consider are:

1. **Linear:** \( y = mx + b \)
2. **Quadratic:** \( y = ax^2 + bx + c \)
3. **Exponential:** \( y = a \cdot b^x \)

Let's analyze the 

\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
2 & -\frac{8}{3} \\
3 & -6 \\
4 & -\frac{32}{3} \\
5 & -\frac{50}{3} \\
6 & -24 \\
\hline
\end{array}
\]

### Step 1: Checking for a Linear Pattern

To check if the data is linear, we need to see if the differences between consecutive \( y \)-values are constant. Let's compute the first differences:

\[
\begin{align*}
\Delta y_1 &= y_3 - y_2 = -6 - \left(-\frac{8}{3}\right) = -6 + \frac{8}{3} = -\frac{10}{3} \\
\Delta y_2 &= y_4 - y_3 = -\frac{32}{3} - (-6) = -\frac{32}{3} + 6 = -\frac{14}{3} \\
\Delta y_3 &= y_5 - y_4 = -\frac{50}{3} - \left(-\frac{32}{3}\right) = -\frac{18}{3} = -6 \\
\Delta y_4 &= y_6 - y_5 = -24 - \left(-\frac{50}{3}\right) = -24 + \frac{50}{3} = -\frac{22}{3} \\
\end{align*}
\]

The first differences are not constant, so the data does not follow a linear pattern.

### Step 2: Checking for a Quadratic Pattern

For a quadratic pattern, the second differences should be constant. Let's calculate the second differences:

\[
\begin{align*}
\Delta^2 y_1 &= \Delta y_2 - \Delta y_1 = -\frac{14}{3} - \left(-\frac{10}{3}\right) = -\frac{4}{3} \\
\Delta^2 y_2 &= \Delta y_3 - \Delta y_2 = -6 - \left(-\frac{14}{3}\right) = -6 + \frac{14}{3} = -\frac{4}{3} \\
\Delta^2 y_3 &= \Delta y_4 - \Delta y_3 = -\frac{22}{3} - (-6) = -\frac{22}{3} + 6 = -\frac{4}{3} \\
\end{align*}
\]

The second differences are constant \((- \frac{4}{3})\), which suggests the data can be modeled by a **quadratic function**.

### Step 3: Finding the Quadratic Function

A quadratic function has the form:

\[
y = ax^2 + bx + c
\]

We have the points \((x_1, y_1) = (2, -\frac{8}{3})\), \((x_2, y_2) = (3, -6)\), and \((x_3, y_3) = (4, -\frac{32}{3})\). Substituting these points into the quadratic equation gives us three equations:

1. \(4a + 2b + c = -\frac{8}{3}\)
2. \(9a + 3b + c = -6\)
3. \(16a + 4b + c = -\frac{32}{3}\)

We can solve this system of equations to find \(a\), \(b\), and \(c\).

Let's solve these equations.The solution to the system of equations gives us the values:

\[
a = -\frac{2}{3}, \quad b = 0, \quad c = 0
\]

Therefore, the quadratic function that models the data is:

\[
y = -\frac{2}{3}x^2
\]

### Conclusion
The data in the table is best modeled by the quadratic function \(y = -\frac{2}{3}x^2\).

---

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### Related Questions:
1. How do you determine if a function is linear, quadratic, or exponential?
2. What is the importance of first and second differences in identifying function types?
3. Can we model the data using an exponential function and compare?
4. How would the model change if additional points were given?
5. How can we verify the accuracy of the quadratic model with the given data?

### Tip:
When analyzing patterns in a data set, start by checking the differences between consecutive values to identify potential linear or quadratic relationships.

Write a linear (y = mx + b), quadratic (y = ax^2), or exponential (y = a(b)^x) function that models the data provided in the table.

Analyze the given data table to find whether a linear, quadratic, or exponential function best models the data. Learn step-by-step how to check for patterns using first and second differences.

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