Let's complete the table based on the function [tex]\( y = -\frac{2}{3} x + 7 \)[/tex] and the given domain [tex]\(\{-12, -6, 3, 15\}\)[/tex].
1. For [tex]\( x = -6 \)[/tex]:
[tex]\[
y = -\frac{2}{3}(-6) + 7 = 4 + 7 = 11
\][/tex]
So, when [tex]\( x = -6 \)[/tex], [tex]\( y = 11 \)[/tex].
2. When [tex]\( y = 5 \)[/tex]:
[tex]\[
5 = -\frac{2}{3} x + 7
\][/tex]
[tex]\[
\Rightarrow -\frac{2}{3} x = 5 - 7
\][/tex]
[tex]\[
\Rightarrow -\frac{2}{3} x = -2
\][/tex]
[tex]\[
\Rightarrow x = 3
\][/tex]
So, when [tex]\( y = 5 \)[/tex], [tex]\( x = 3 \)[/tex].
3. For [tex]\( x = 15 \)[/tex]:
[tex]\[
y = -\frac{2}{3}(15) + 7 = -10 + 7 = -3
\][/tex]
So, when [tex]\( x = 15 \)[/tex], [tex]\( y = -3 \)[/tex].
4. When [tex]\( y = 15 \)[/tex]:
[tex]\[
15 = -\frac{2}{3} x + 7
\][/tex]
[tex]\[
\Rightarrow -\frac{2}{3} x = 15 - 7
\][/tex]
[tex]\[
\Rightarrow -\frac{2}{3} x = 8
\][/tex]
[tex]\[
\Rightarrow x = -12
\][/tex]
So, when [tex]\( y = 15 \)[/tex], [tex]\( x = -12 \)[/tex].
The completed table is:
\begin{tabular}{|c|c|}
\hline[tex]$x$[/tex] & [tex]$y$[/tex] \\
\hline-6 & 11 \\
\hline 3 & 5 \\
\hline 15 & -3 \\
\hline-12 & 15 \\
\hline
\end{tabular}
