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Identify the data set that could be quadratic.

\begin{tabular}{|c|c|c|c|c|c|}
\hline
[tex]$x$[/tex] & 1 & 4 & 7 & 10 & 13 \\
\hline
[tex]$y$[/tex] & 2 & 8 & 17 & 29 & 44 \\
\hline
\end{tabular}

\begin{tabular}{|c|c|c|c|c|c|}
\hline
[tex]$x$[/tex] & -1 & -2 & -3 & -4 & -5 \\
\hline
[tex]$y$[/tex] & -1 & -8 & -27 & -64 & -125 \\
\hline
\end{tabular}

\begin{tabular}{|c|c|c|c|c|c|}
\hline
[tex]$x$[/tex] & 3 & 6 & 9 & 12 & 15 \\
\hline
[tex]$y$[/tex] & 9 & 18 & 27 & 36 & 45 \\
\hline
\end{tabular}

\begin{tabular}{|c|c|c|c|c|c|}
\hline
[tex]$x$[/tex] & -1 & -2 & -3 & -4 & -5 \\
\hline
[tex]$y$[/tex] & 4 & 8 & 12 & 16 & 20 \\
\hline
\end{tabular}


Sagot :

To determine which data set is quadratic, we will examine the first and second differences of the [tex]\( y \)[/tex] values in each data set. A quadratic relationship will have constant second differences. Let's go through the step-by-step process for each data set.

### Data Set 1
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline x & 1 & 4 & 7 & 10 & 13 \\ \hline y & 2 & 8 & 17 & 29 & 44 \\ \hline \end{array} \][/tex]

1. First Differences:
[tex]\[ 8 - 2 = 6, \quad 17 - 8 = 9, \quad 29 - 17 = 12, \quad 44 - 29 = 15 \][/tex]
So, the first differences are: [tex]\([6, 9, 12, 15]\)[/tex]

2. Second Differences:
[tex]\[ 9 - 6 = 3, \quad 12 - 9 = 3, \quad 15 - 12 = 3 \][/tex]
So, the second differences are: [tex]\([3, 3, 3]\)[/tex]

Since the second differences are constant (all equal to 3), this data set can be quadratic.

### Data Set 2
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline x & -1 & -2 & -3 & -4 & -5 \\ \hline y & -1 & -8 & -27 & -64 & -125 \\ \hline \end{array} \][/tex]

1. First Differences:
[tex]\[ -8 - (-1) = -7, \quad -27 - (-8) = -19, \quad -64 - (-27) = -37, \quad -125 - (-64) = -61 \][/tex]
So, the first differences are: [tex]\([-7, -19, -37, -61]\)[/tex]

2. Second Differences:
[tex]\[ -19 - (-7) = -12, \quad -37 - (-19) = -18, \quad -61 - (-37) = -24 \][/tex]
So, the second differences are: [tex]\([-12, -18, -24]\)[/tex]

Since the second differences are not constant, this data set is not quadratic.

### Data Set 3
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline x & 3 & 6 & 9 & 12 & 15 \\ \hline y & 9 & 18 & 27 & 36 & 45 \\ \hline \end{array} \][/tex]

1. First Differences:
[tex]\[ 18 - 9 = 9, \quad 27 - 18 = 9, \quad 36 - 27 = 9, \quad 45 - 36 = 9 \][/tex]
So, the first differences are: [tex]\([9, 9, 9, 9]\)[/tex]

Since the first differences are constant, this data set represents a linear relationship, not a quadratic one. Therefore, we do not need to check the second differences.

### Data Set 4
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline x & -1 & -2 & -3 & -4 & -5 \\ \hline y & 4 & 8 & 12 & 16 & 20 \\ \hline \end{array} \][/tex]

1. First Differences:
[tex]\[ 8 - 4 = 4, \quad 12 - 8 = 4, \quad 16 - 12 = 4, \quad 20 - 16 = 4 \][/tex]
So, the first differences are: [tex]\([4, 4, 4, 4]\)[/tex]

Similarly, since the first differences are constant, this data set is linear, not quadratic.

### Conclusion
Based on the analysis, only Data Set 1 has constant second differences, indicating that it represents a quadratic relationship. Thus, Data Set 1 is the quadratic one.