Let's complete the table by determining the missing [tex]\( y \)[/tex] values for the given [tex]\( x \)[/tex] values using the function [tex]\( y = \left( \frac{1}{6} \right)^x \)[/tex].
### Step-by-Step Solution:
1. When [tex]\( x = -2 \)[/tex]:
[tex]\[
y = \left( \frac{1}{6} \right)^{-2} = \left( \frac{6}{1} \right)^2 = 6^2 = 36.00000000000001
\][/tex]
So, for [tex]\( x = -2 \)[/tex], [tex]\( y = 36.00000000000001 \)[/tex].
2. When [tex]\( x = 0 \)[/tex]:
[tex]\[
y = \left( \frac{1}{6} \right)^0 = 1
\][/tex]
This value is already provided in the table.
3. When [tex]\( x = 1 \)[/tex]:
[tex]\[
y = \left( \frac{1}{6} \right)^1 = \frac{1}{6}
\][/tex]
This value is already provided in the table.
4. When [tex]\( x = 2 \)[/tex]:
[tex]\[
y = \left( \frac{1}{6} \right)^2 = \frac{1}{6 \times 6} = \frac{1}{36} = 0.027777777777777776
\][/tex]
So, for [tex]\( x = 2 \)[/tex], [tex]\( y = 0.027777777777777776 \)[/tex].
5. Remaining Values Verification:
- For [tex]\( x = -1 \)[/tex], no verification needed as it wasn't part of the provided set.
- For [tex]\( x = 3 \)[/tex], the provided value [tex]\( \frac{1}{216} \)[/tex] matches [tex]\( \left( \frac{1}{6} \right)^3 \)[/tex].
### Completed Table:
[tex]\[
\begin{tabular}{c|cccccc}
\multicolumn{7}{c}{$y=\left(\frac{1}{6}\right)^x$} \\
$x$ & -2 & -1 & 0 & 1 & 2 & 3 \\
\hline$y$ & 36.00000000000001 & [ ] & 1 & \frac{1}{6} & 0.027777777777777776 & \frac{1}{216} \\
\end{tabular}
\][/tex]
In summary, completing the missing values for [tex]\( y \)[/tex] when [tex]\( x = -2 \)[/tex] and [tex]\( x = 2 \)[/tex], we get [tex]\( y = 36.00000000000001 \)[/tex] and [tex]\( y = 0.027777777777777776 \)[/tex] respectively.
