Certainly! Let's analyze each table step-by-step and match it with its appropriate equation.
Table 1:
[tex]\[
\begin{array}{|r|r|}
\hline
\text{Input} & \text{Output} \\
\hline
-2 & 2 \\
-1 & 1 \\
0 & 0 \\
1 & 1 \\
2 & 2 \\
3 & 3 \\
\hline
\end{array}
\][/tex]
Analysis:
If we look at the data in this table, we notice that the output is the same as the input.
Equation for Table 1:
[tex]\[ y = x \][/tex]
Table 2:
[tex]\[
\begin{array}{|r|r|}
\hline
\text{Input} & \text{Output} \\
\hline
-2 & 4 \\
-1 & 1 \\
0 & 0 \\
1 & 1 \\
2 & 4 \\
3 & 9 \\
\hline
\end{array}
\][/tex]
Analysis:
In this table, the output is the square of the input value.
Equation for Table 2:
[tex]\[ y = x^2 \][/tex]
Table 3:
[tex]\[
\begin{array}{|r|r|}
\hline
\text{Input} & \text{Output} \\
\hline
-2 & -0.5 \\
-1 & -1 \\
0 & - \\
1 & 1 \\
2 & 0.5 \\
3 & 0.33 \\
\hline
\end{array}
\][/tex]
Analysis:
The output in this table is the reciprocal of the input. Note: [tex]\(0.33\)[/tex] is approximately [tex]\(\frac{1}{3}\)[/tex].
Equation for Table 3:
[tex]\[ y = \frac{1}{x} \][/tex]
Table 4:
[tex]\[
\begin{array}{|r|r|}
\hline
\text{Input} & \text{Output} \\
\hline
-2 & -8 \\
-1 & -1 \\
0 & 0 \\
1 & 1 \\
2 & 8 \\
3 & 27 \\
\hline
\end{array}
\][/tex]
Analysis:
Here, the output is the cube of the input value.
Equation for Table 4:
[tex]\[ y = x^3 \][/tex]
Table 5:
[tex]\[
\begin{array}{|r|r|}
\hline
\text{Input} & \text{Output} \\
\hline
-2 & -2 \\
-1 & -1 \\
0 & 0 \\
1 & -1 \\
2 & -2 \\
3 & -3 \\
\hline
\end{array}
\][/tex]
Analysis:
In this table, the output is the negative of the input value.
Equation for Table 5:
[tex]\[ y = -x \][/tex]
Summary of Matching:
1. Table 1: [tex]\( y = x \)[/tex]
2. Table 2: [tex]\( y = x^2 \)[/tex]
3. Table 3: [tex]\( y = \frac{1}{x} \)[/tex]
4. Table 4: [tex]\( y = x^3 \)[/tex]
5. Table 5: [tex]\( y = -x \)[/tex]
This concludes the detailed step-by-step matching of each table with its respective equation.
