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What is the constant of variation for the following table? If your answer is in the form of a fraction, write your answer as [tex]$\frac{a}{b}$[/tex] (where [tex]$a$[/tex] is the numerator and [tex][tex]$b$[/tex][/tex] is the denominator).

\begin{tabular}{|c|c|}
\hline
[tex]$x$[/tex] & [tex]$f(x)$[/tex] \\
\hline
-8 & -2 \\
\hline
-4 & -1 \\
\hline
0 & 0 \\
\hline
4 & 1 \\
\hline
8 & 2 \\
\hline
\end{tabular}


Sagot :

To determine the constant of variation for the given table, we need to use the fact that for a linear relationship [tex]\( f(x) = kx \)[/tex], where [tex]\( k \)[/tex] is the constant of variation.

Let's start by examining two pairs of points from the table to find [tex]\( k \)[/tex]:

1. The points [tex]\((-8, -2)\)[/tex]
2. The points [tex]\((-4, -1)\)[/tex]

Using the general form [tex]\( f(x) = kx \)[/tex] (or [tex]\( y = kx \)[/tex]), we can find [tex]\( k \)[/tex] by dividing [tex]\( f(x) \)[/tex] by [tex]\( x \)[/tex]:

For the point [tex]\((-8, -2)\)[/tex]:
[tex]\[ k = \frac{f(x)}{x} = \frac{-2}{-8} = \frac{1}{4} \][/tex]

For the point [tex]\((-4, -1)\)[/tex]:
[tex]\[ k = \frac{f(x)}{x} = \frac{-1}{-4} = \frac{1}{4} \][/tex]

Both calculations give us the same constant of variation [tex]\( k \)[/tex].

Therefore, the constant of variation for the given table is
[tex]\[ \boxed{\frac{1}{4}} \][/tex]
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