Answer :
To determine which type of function best models the given data, we need to analyze the relationship between the \(x\) and \(y\) values in the table. The table provided is:
[tex]\[ \begin{tabular}{|c|c|} \hline[tex]$x$[/tex] & [tex]$y$[/tex] \\
\hline 2 & 4 \\
\hline 4 & 2 \\
\hline 6 & 1 \frac{1}{3} \\
\hline 8 & 1 \\
\hline 10 & \frac{4}{5} \\
\hline
\end{tabular}
\][/tex]
### Step-by-Step Analysis:
1. Linear Function:
For a linear relationship, the rate of change between \(x\) and \(y\) should be constant. If \(y = mx + b\), then the difference \( y_2 - y_1 \) should be the same for equal differences \( x_2 - x_1 \). Examining the data roughly, the \(y\) values don't change linearly with the \(x\) values, so this is unlikely.
2. Quadratic Function:
For a quadratic relationship, the function follows the form \( y = ax^2 + bx + c \). Again, the changes in \(y\) values here are not fitting a quadratic pattern when correlated to the given \(x\) values as each increase in \(x\) does not correspond to a polynomial fit visually estimated from the given values.
3. Logarithmic Function:
A logarithmic function follows the form \( y = a \log(x) + b \). This sort of function grows quickly for small \(x\) and slowly for large \(x\), which does not align with our provided data points particularly well.
4. Inverse Variation:
For inverse variation, \(xy = k\) for a constant \(k\). We can check if each product \( x \cdot y \) yields a constant value:
[tex]\[ 2 \cdot 4 = 8 \\ 4 \cdot 2 = 8 \\ 6 \cdot 1\frac{1}{3} = 6 \cdot \frac{4}{3} = 8 \\ 8 \cdot 1 = 8 \\ 10 \cdot \frac{4}{5} = 8 \][/tex]
Each product of \( x \cdot y \) is indeed a constant value, 8. This confirms the data satisfies the condition for an inverse variation.
### Conclusion:
Given the data in the table and the analysis, the best type of function that models the data is an inverse variation function.
[tex]\[ \begin{tabular}{|c|c|} \hline[tex]$x$[/tex] & [tex]$y$[/tex] \\
\hline 2 & 4 \\
\hline 4 & 2 \\
\hline 6 & 1 \frac{1}{3} \\
\hline 8 & 1 \\
\hline 10 & \frac{4}{5} \\
\hline
\end{tabular}
\][/tex]
### Step-by-Step Analysis:
1. Linear Function:
For a linear relationship, the rate of change between \(x\) and \(y\) should be constant. If \(y = mx + b\), then the difference \( y_2 - y_1 \) should be the same for equal differences \( x_2 - x_1 \). Examining the data roughly, the \(y\) values don't change linearly with the \(x\) values, so this is unlikely.
2. Quadratic Function:
For a quadratic relationship, the function follows the form \( y = ax^2 + bx + c \). Again, the changes in \(y\) values here are not fitting a quadratic pattern when correlated to the given \(x\) values as each increase in \(x\) does not correspond to a polynomial fit visually estimated from the given values.
3. Logarithmic Function:
A logarithmic function follows the form \( y = a \log(x) + b \). This sort of function grows quickly for small \(x\) and slowly for large \(x\), which does not align with our provided data points particularly well.
4. Inverse Variation:
For inverse variation, \(xy = k\) for a constant \(k\). We can check if each product \( x \cdot y \) yields a constant value:
[tex]\[ 2 \cdot 4 = 8 \\ 4 \cdot 2 = 8 \\ 6 \cdot 1\frac{1}{3} = 6 \cdot \frac{4}{3} = 8 \\ 8 \cdot 1 = 8 \\ 10 \cdot \frac{4}{5} = 8 \][/tex]
Each product of \( x \cdot y \) is indeed a constant value, 8. This confirms the data satisfies the condition for an inverse variation.
### Conclusion:
Given the data in the table and the analysis, the best type of function that models the data is an inverse variation function.
