To complete the equation that describes how [tex]\( x \)[/tex] and [tex]\( y \)[/tex] are related given the table, we will determine the slope ([tex]\( dy/dx \)[/tex]) between consecutive values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex].
The given table is:
[tex]\[
\begin{array}{|c|c|c|}
\hline
x & y & dy \\
\hline
-5 & 17 & \\
\hline
-4 & 14 & \quad=[?] x \\
\hline
-3 & 11 & \\
\hline
-2 & 8 & \\
\hline
-1 & 5 & \\
\hline
0 & 2 & \\
\hline
\end{array}
\][/tex]
First, we can look at the change in [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex] for each interval:
[tex]\[
\begin{aligned}
dy_1 &= \frac{y(-4) - y(-5)}{-4 - (-5)} = \frac{14 - 17}{-4 + 5} = \frac{-3}{1} = -3, \\
dy_2 &= \frac{y(-3) - y(-4)}{-3 - (-4)} = \frac{11 - 14}{-3 + 4} = \frac{-3}{1} = -3, \\
dy_3 &= \frac{y(-2) - y(-3)}{-2 - (-3)} = \frac{8 - 11}{-2 + 3} = \frac{-3}{1} = -3, \\
dy_4 &= \frac{y(-1) - y(-2)}{-1 - (-2)} = \frac{5 - 8}{-1 + 2} = \frac{-3}{1} = -3, \\
dy_5 &= \frac{y(0) - y(-1)}{0 - (-1)} = \frac{2 - 5}{0 + 1} = \frac{-3}{1} = -3.
\end{aligned}
\][/tex]
As can be seen from the calculations, the rate of change [tex]\( dy/dx \)[/tex] in each interval is consistently [tex]\(-3\)[/tex].
Therefore, for the row where [tex]\( x = -4 \)[/tex]:
[tex]\[
dy = -3 \times \Delta x
\][/tex]
Given [tex]\(\Delta x = 1\)[/tex] (because the difference in [tex]\( x \)[/tex] values is [tex]\(-4 - (-5) = 1\)[/tex]), we can write:
[tex]\[
dy = -3 \times (-4 - (-5)) = -3 \times 1 = -3
\][/tex]
So the relation for [tex]\( dy \)[/tex] when [tex]\( x = -4 \)[/tex] is:
[tex]\[
dy = -3 \times 1 = -3
\][/tex]
Thus, for the row where [tex]\( x = -4 \)[/tex], the completed equation is:
[tex]\[
\begin{array}{|c|c|c|}
\hline
x & y & dy \\
\hline
-5 & 17 & \\
\hline
-4 & 14 & = -3 \times 1 \\
\hline
-3 & 11 & \\
\hline
-2 & 8 & \\
\hline
-1 & 5 & \\
\hline
0 & 2 & \\
\hline
\end{array}
\][/tex]
