To determine whether the given table represents a linear or nonlinear function, we need to analyze the given values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex], and calculate the rates of change (slopes) between consecutive points.
The given table is:
[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
-2 & \frac{1}{9} \\
\hline
-1 & \frac{1}{3} \\
\hline
0 & 1 \\
\hline
1 & 3 \\
\hline
2 & 9 \\
\hline
\end{array}
\][/tex]
First, let's find the changes in [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
### Change in [tex]\( x \)[/tex] values ([tex]\(\Delta x\)[/tex]):
Calculating the differences between consecutive [tex]\( x \)[/tex]-values:
[tex]\[
\Delta x = x_{i+1} - x_i
\][/tex]
[tex]\[
\begin{array}{|c|c|c|}
\hline
\text{Interval} & \text{Points} & \Delta x \\
\hline
1 & (-2, -1) & -1 - (-2) = 1 \\
\hline
2 & (-1, 0) & 0 - (-1) = 1 \\
\hline
3 & (0, 1) & 1 - 0 = 1 \\
\hline
4 & (1, 2) & 2 - 1 = 1 \\
\hline
\end{array}
\][/tex]
So, [tex]\(\Delta x = [1, 1, 1, 1]\)[/tex].
### Change in [tex]\( y \)[/tex] values ([tex]\(\Delta y\)[/tex]):
Calculating the differences between consecutive [tex]\( y \)[/tex]-values:
[tex]\[
\Delta y = y_{i+1} - y_i
\][/tex]
[tex]\[
\begin{array}{|c|c|c|}
\hline
\text{Interval} & \text{Points} & \Delta y \\
\hline
1 & \left(\frac{1}{9}, \frac{1}{3}\right) & \frac{1}{3} - \frac{1}{9} = \frac{3}{9} - \frac{1}{9} = \frac{2}{9} \approx 0.222 \\
\hline
2 & \left(\frac{1}{3}, 1\right) & 1 - \frac{1}{3} = \frac{3}{3} - \frac{1}{3} = \frac{2}{3} \approx 0.667 \\
\hline
3 & (1, 3) & 3 - 1 = 2 \\
\hline
4 & (3, 9) & 9 - 3 = 6 \\
\hline
\end{array}
\][/tex]
So, [tex]\(\Delta y = [0.222, 0.667, 2, 6]\)[/tex].
### Rates of change (slopes):
Calculating the rate of change (slope) for each interval:
[tex]\[
\text{Rate of change} = \frac{\Delta y}{\Delta x}
\][/tex]
[tex]\[
\begin{array}{|c|c|c|}
\hline
\text{Interval} & \Delta y & \Delta x & \text{Rate of Change} \\
\hline
1 & 0.222 & 1 & \frac{0.222}{1} = 0.222 \\
\hline
2 & 0.667 & 1 & \frac{0.667}{1} = 0.667 \\
\hline
3 & 2 & 1 & \frac{2}{1} = 2 \\
\hline
4 & 6 & 1 & \frac{6}{1} = 6 \\
\hline
\end{array}
\][/tex]
So, the rates of change are [tex]\(0.222, 0.667, 2.0, 6.0\)[/tex].
### Determining constancy of the rate of change:
The rates of change are [tex]\(0.222, 0.667, 2.0, 6.0\)[/tex], which are not constant.
Thus, the table represents a nonlinear function because the rate of change is not constant.
The correct statement is:
"The table represents a nonlinear function because the rate of change is not constant."
