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

\begin{tabular}{|c|c|}
\hline
[tex]$x$[/tex] & [tex]$y$[/tex] \\
\hline
-2 & [tex]$\frac{1}{9}$[/tex] \\
\hline
-1 & [tex]$\frac{1}{3}$[/tex] \\
\hline
0 & 1 \\
\hline
1 & 3 \\
\hline
2 & 9 \\
\hline
\end{tabular}

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.

Sagot :

GinnyAnsweridn
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."
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