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Which statement best explains whether the following table represents a linear or nonlinear function?

[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
-2 & \frac{1}{y} \\
\hline
-1 & \frac{1}{3} \\
\hline
0 & 1 \\
\hline
1 & 3 \\
\hline
2 & 0 \\
\hline
\end{array}
\][/tex]

A. The table represents a linear function because the rate of change is constant.

B. The table represents a linear function because the rate of change is not constant.

C. The table represents a nonlinear function because the rate of change is constant.

D. The table represents a nonlinear function because the rate of change is not constant.



Answer :

Let's determine whether the given table represents a linear or nonlinear function. To do this, we need to analyze the rate of change (slope) between each pair of consecutive points.

Here is the table of points:

[tex]\[ \begin{tabular}{|c|c|} \hline $x$ & $y$ \\ \hline -2 & \frac{1}{-2} \\ \hline -1 & \frac{1}{3} \\ \hline 0 & 1 \\ \hline 1 & 3 \\ \hline 2 & 0 \\ \hline \end{tabular} \][/tex]

### Step-by-Step Calculation of the Rate of Change

1. Between [tex]$(-2, -\frac{1}{2})$[/tex] and [tex]$(-1, \frac{1}{3})$[/tex]:

[tex]\[ \Delta x = -1 - (-2) = 1, \quad \Delta y = \frac{1}{3} - \left(-\frac{1}{2}\right) = \frac{1}{3} + \frac{1}{2} = \frac{2}{6} + \frac{3}{6} = \frac{5}{6} \][/tex]

[tex]\[ \text{Rate of change} = \frac{\Delta y}{\Delta x} = \frac{\frac{5}{6}}{1} = \frac{5}{6} \approx 0.833 \][/tex]

2. Between [tex]$(-1, \frac{1}{3})$[/tex] and [tex]$(0, 1)$[/tex]:

[tex]\[ \Delta x = 0 - (-1) = 1, \quad \Delta y = 1 - \frac{1}{3} = \frac{3}{3} - \frac{1}{3} = \frac{2}{3} \][/tex]

[tex]\[ \text{Rate of change} = \frac{\Delta y}{\Delta x} = \frac{\frac{2}{3}}{1} = \frac{2}{3} \approx 0.667 \][/tex]

3. Between [tex]$(0, 1)$[/tex] and [tex]$(1, 3)$[/tex]:

[tex]\[ \Delta x = 1 - 0 = 1, \quad \Delta y = 3 - 1 = 2 \][/tex]

[tex]\[ \text{Rate of change} = \frac{\Delta y}{\Delta x} = \frac{2}{1} = 2 \][/tex]

4. Between [tex]$(1, 3)$[/tex] and [tex]$(2, 0)$[/tex]:

[tex]\[ \Delta x = 2 - 1 = 1, \quad \Delta y = 0 - 3 = -3 \][/tex]

[tex]\[ \text{Rate of change} = \frac{\Delta y}{\Delta x} = \frac{-3}{1} = -3 \][/tex]

### Analysis of Results

The calculated rates of change between each pair of points are:

1. Between [tex]$x = -2$[/tex] and [tex]$x = -1$[/tex]: Approximately 0.833
2. Between [tex]$x = -1$[/tex] and [tex]$x = 0$[/tex]: Approximately 0.667
3. Between [tex]$x = 0$[/tex] and [tex]$x = 1$[/tex]: Exactly 2
4. Between [tex]$x = 1$[/tex] and [tex]$x = 2$[/tex]: Exactly -3

Since the rate of change is not consistent (it varies between the pairs of points), the table does not represent a linear function because a linear function must have a constant rate of change (slope).

### Conclusion

Based on our analysis:

The table represents a nonlinear function because the rate of change is not constant.

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