Sure! Let's analyze each table to determine the type of relation between [tex]\( x \)[/tex] and [tex]\( y \)[/tex].
### First Table Analysis
| [tex]\( x \)[/tex] | [tex]\( y \)[/tex] |
|---------|---------|
| -2 | -2 |
| -1 | 2 |
| 0 | 6 |
| 1 | 10 |
| 2 | 14 |
1. Examining Linearity:
- Calculate the differences between consecutive [tex]\( y \)[/tex]-values:
[tex]\[
\begin{align*}
y(-1) - y(-2) &= 2 - (-2) = 4, \\
y(0) - y(-1) &= 6 - 2 = 4, \\
y(1) - y(0) &= 10 - 6 = 4, \\
y(2) - y(1) &= 14 - 10 = 4.
\end{align*}
\][/tex]
- Since the differences (Δy) are constant (all equal to 4), this suggests a linear relationship.
2. Conclusion:
- The relation between [tex]\( x \)[/tex] and [tex]\( y \)[/tex] in the first table is Linear.
### Second Table Analysis
| [tex]\( x \)[/tex] | [tex]\( y \)[/tex] |
|---------|---------|
| -2 | -2 |
| -1 | 1 |
| 0 | 6 |
| 1 | 13 |
| 2 | 22 |
1. Examining Linearity:
- Calculate the differences between consecutive [tex]\( y \)[/tex]-values:
[tex]\[
\begin{align*}
y(-1) - y(-2) &= 1 - (-2) = 3, \\
y(0) - y(-1) &= 6 - 1 = 5, \\
y(1) - y(0) &= 13 - 6 = 7, \\
y(2) - y(1) &= 22 - 13 = 9.
\end{align*}
\][/tex]
- The differences (Δy) are not constant (they are 3, 5, 7, and 9).
2. Examining Exponentiality:
- Calculate the ratios between consecutive [tex]\( y \)[/tex]-values:
[tex]\[
\begin{align*}
\text{Ratio}(-1, -2) &= \frac{1}{-2} = -0.5, \\
\text{Ratio}(0, -1) &= \frac{6}{1} = 6, \\
\text{Ratio}(1, 0) &= \frac{13}{6} \approx 2.17, \\
\text{Ratio}(2, 1) &= \frac{22}{13} \approx 1.69.
\end{align*}
\][/tex]
- The ratios are not constant either (they are -0.5, 6, approximately 2.17, and approximately 1.69).
3. Conclusion:
- Since the differences and the ratios are not constant, the relation does not fit standard linear or exponential models. Therefore, the relation between [tex]\( x \)[/tex] and [tex]\( y \)[/tex] in the second table is Other.
### Summary
- The type of relation for the first table is Linear.
- The type of relation for the second table is Other.
