Certainly! Let's complete the rest of the table step-by-step for the given function rule [tex]\( y = \frac{x}{2} + 5 \)[/tex]:
1. For [tex]\( x = -4 \)[/tex]:
[tex]\[
y = \frac{-4}{2} + 5 = -2 + 5 = 3
\][/tex]
So for [tex]\( x = -4 \)[/tex], [tex]\( y = 3 \)[/tex].
2. For [tex]\( x = -2 \)[/tex]:
[tex]\[
y = \frac{-2}{2} + 5 = -1 + 5 = 4
\][/tex]
So for [tex]\( x = -2 \)[/tex], [tex]\( y = 4 \)[/tex].
3. For [tex]\( x = 0 \)[/tex]:
[tex]\[
y = \frac{0}{2} + 5 = 0 + 5 = 5
\][/tex]
So for [tex]\( x = 0 \)[/tex], [tex]\( y = 5 \)[/tex].
4. For [tex]\( x = 2 \)[/tex]:
[tex]\[
y = \frac{2}{2} + 5 = 1 + 5 = 6
\][/tex]
So for [tex]\( x = 2 \)[/tex], [tex]\( y = 6 \)[/tex].
5. For [tex]\( x = 4 \)[/tex]:
[tex]\[
y = \frac{4}{2} + 5 = 2 + 5 = 7
\][/tex]
So for [tex]\( x = 4 \)[/tex], [tex]\( y = 7 \)[/tex].
Now we fill in the values in the table:
\begin{tabular}{l|lllll}
x & -4 & -2 & 0 & 2 & 4 \\
\hline
y & 3 & 4 & 5 & 6 & 7 \\
\end{tabular}
These are the completed values for the table based on the function rule [tex]\( y = \frac{x}{2} + 5 \)[/tex].
