Certainly! Let's complete the table step by step according to the given function rule [tex]\( y = 4 - \frac{3}{4} x \)[/tex].
1. Determine [tex]\( y \)[/tex] when [tex]\( x = 4 \)[/tex]:
[tex]\[
y = 4 - \frac{3}{4} \times 4
\][/tex]
Substituting [tex]\( x = 4 \)[/tex]:
[tex]\[
y = 4 - \frac{3 \times 4}{4} = 4 - 3 = 1
\][/tex]
Thus, when [tex]\( x = 4 \)[/tex], [tex]\( y = 1 \)[/tex].
2. Determine [tex]\( y \)[/tex] when [tex]\( x = 8 \)[/tex]:
[tex]\[
y = 4 - \frac{3}{4} \times 8
\][/tex]
Substituting [tex]\( x = 8 \)[/tex]:
[tex]\[
y = 4 - \frac{3 \times 8}{4} = 4 - 6 = -2
\][/tex]
Thus, when [tex]\( x = 8 \)[/tex], [tex]\( y = -2 \)[/tex].
3. Determine [tex]\( y \)[/tex] when [tex]\( x = 12 \)[/tex]:
[tex]\[
y = 4 - \frac{3}{4} \times 12
\][/tex]
Substituting [tex]\( x = 12 \)[/tex]:
[tex]\[
y = 4 - \frac{3 \times 12}{4} = 4 - 9 = -5
\][/tex]
Thus, when [tex]\( x = 12 \)[/tex], [tex]\( y = -5 \)[/tex].
Now, let's fill in the table with these values:
[tex]\[
\begin{tabular}{c|ccccc}
$x$ & -4 & 0 & 4 & 8 & 12 \\
\hline
$y$ & 7 & 4 & 1 & -2 & -5 \\
\end{tabular}
\][/tex]
So the completed table is:
[tex]\[
\begin{tabular}{c|ccccc}
$x$ & -4 & 0 & 4 & 8 & 12 \\
\hline
$y$ & 7 & 4 & 1 & -2 & -5 \\
\end{tabular}
\][/tex]
