Answer :
To determine the type of function representing the given table of values, we need to analyze the relationship between the [tex]\( x \)[/tex] and [tex]\( y \)[/tex] values:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -2 & 8 \\ \hline -1 & 6 \\ \hline 0 & 4 \\ \hline 1 & 2 \\ \hline 2 & 0 \\ \hline \end{array} \][/tex]
### Step-by-Step Solution:
1. Observe the Differences in [tex]\( y \)[/tex]-Values:
First, calculate the differences between consecutive [tex]\( y \)[/tex]-values to check if they form a constant ratio.
[tex]\[ \Delta y = y_{i+1} - y_i \][/tex]
Let's calculate these:
[tex]\[ \begin{aligned} &\Delta y_1 = y_2 - y_1 = 6 - 8 = -2 \\ &\Delta y_2 = y_3 - y_2 = 4 - 6 = -2 \\ &\Delta y_3 = y_4 - y_3 = 2 - 4 = -2 \\ &\Delta y_4 = y_5 - y_4 = 0 - 2 = -2 \\ \end{aligned} \][/tex]
The differences in [tex]\( y \)[/tex]-values are [tex]\( -2 \)[/tex], [tex]\( -2 \)[/tex], [tex]\( -2 \)[/tex], and [tex]\( -2 \)[/tex].
2. Check the Differences in [tex]\( x \)[/tex]-Values:
Now, check the differences between consecutive [tex]\( x \)[/tex]-values.
[tex]\[ \Delta x = x_{i+1} - x_i \][/tex]
Since [tex]\( x \)[/tex]-values are evenly spaced [tex]\( (\Delta x = 1) \)[/tex]:
[tex]\[ \begin{aligned} &\Delta x_1 = x_2 - x_1 = -1 - (-2) = 1 \\ &\Delta x_2 = x_3 - x_2 = 0 - (-1) = 1 \\ &\Delta x_3 = x_4 - x_3 = 1 - 0 = 1 \\ &\Delta x_4 = x_5 - x_4 = 2 - 1 = 1 \\ \end{aligned} \][/tex]
The differences in [tex]\( x \)[/tex]-values are [tex]\( 1 \)[/tex], [tex]\( 1 \)[/tex], [tex]\( 1 \)[/tex], and [tex]\( 1 \)[/tex].
3. Calculate the Ratio of Differences:
The slope [tex]\( m \)[/tex] of a linear function [tex]\( y = mx + b \)[/tex] can be determined by the constant ratio of [tex]\( \frac{\Delta y}{\Delta x} \)[/tex].
For this data:
[tex]\[ \frac{\Delta y_1}{\Delta x_1} = \frac{-2}{1} = -2 \][/tex]
[tex]\[ \frac{\Delta y_2}{\Delta x_2} = \frac{-2}{1} = -2 \][/tex]
[tex]\[ \frac{\Delta y_3}{\Delta x_3} = \frac{-2}{1} = -2 \][/tex]
[tex]\[ \frac{\Delta y_4}{\Delta x_4} = \frac{-2}{1} = -2 \][/tex]
The ratio of differences (or slope [tex]\( m \)[/tex]) is constant and equals [tex]\( -2 \)[/tex] for each pair. This means the data forms a straight line, indicating a linear relationship.
### Conclusion:
Since the differences in [tex]\( y \)[/tex]-values divided by differences in [tex]\( x \)[/tex]-values are constant, the function that matches the given data is Linear.
Thus, the answer is:
[tex]\[ \boxed{1} \][/tex]
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -2 & 8 \\ \hline -1 & 6 \\ \hline 0 & 4 \\ \hline 1 & 2 \\ \hline 2 & 0 \\ \hline \end{array} \][/tex]
### Step-by-Step Solution:
1. Observe the Differences in [tex]\( y \)[/tex]-Values:
First, calculate the differences between consecutive [tex]\( y \)[/tex]-values to check if they form a constant ratio.
[tex]\[ \Delta y = y_{i+1} - y_i \][/tex]
Let's calculate these:
[tex]\[ \begin{aligned} &\Delta y_1 = y_2 - y_1 = 6 - 8 = -2 \\ &\Delta y_2 = y_3 - y_2 = 4 - 6 = -2 \\ &\Delta y_3 = y_4 - y_3 = 2 - 4 = -2 \\ &\Delta y_4 = y_5 - y_4 = 0 - 2 = -2 \\ \end{aligned} \][/tex]
The differences in [tex]\( y \)[/tex]-values are [tex]\( -2 \)[/tex], [tex]\( -2 \)[/tex], [tex]\( -2 \)[/tex], and [tex]\( -2 \)[/tex].
2. Check the Differences in [tex]\( x \)[/tex]-Values:
Now, check the differences between consecutive [tex]\( x \)[/tex]-values.
[tex]\[ \Delta x = x_{i+1} - x_i \][/tex]
Since [tex]\( x \)[/tex]-values are evenly spaced [tex]\( (\Delta x = 1) \)[/tex]:
[tex]\[ \begin{aligned} &\Delta x_1 = x_2 - x_1 = -1 - (-2) = 1 \\ &\Delta x_2 = x_3 - x_2 = 0 - (-1) = 1 \\ &\Delta x_3 = x_4 - x_3 = 1 - 0 = 1 \\ &\Delta x_4 = x_5 - x_4 = 2 - 1 = 1 \\ \end{aligned} \][/tex]
The differences in [tex]\( x \)[/tex]-values are [tex]\( 1 \)[/tex], [tex]\( 1 \)[/tex], [tex]\( 1 \)[/tex], and [tex]\( 1 \)[/tex].
3. Calculate the Ratio of Differences:
The slope [tex]\( m \)[/tex] of a linear function [tex]\( y = mx + b \)[/tex] can be determined by the constant ratio of [tex]\( \frac{\Delta y}{\Delta x} \)[/tex].
For this data:
[tex]\[ \frac{\Delta y_1}{\Delta x_1} = \frac{-2}{1} = -2 \][/tex]
[tex]\[ \frac{\Delta y_2}{\Delta x_2} = \frac{-2}{1} = -2 \][/tex]
[tex]\[ \frac{\Delta y_3}{\Delta x_3} = \frac{-2}{1} = -2 \][/tex]
[tex]\[ \frac{\Delta y_4}{\Delta x_4} = \frac{-2}{1} = -2 \][/tex]
The ratio of differences (or slope [tex]\( m \)[/tex]) is constant and equals [tex]\( -2 \)[/tex] for each pair. This means the data forms a straight line, indicating a linear relationship.
### Conclusion:
Since the differences in [tex]\( y \)[/tex]-values divided by differences in [tex]\( x \)[/tex]-values are constant, the function that matches the given data is Linear.
Thus, the answer is:
[tex]\[ \boxed{1} \][/tex]
