IDNLearn.com: Your trusted source for finding accurate answers. Discover reliable answers to your questions with our extensive database of expert knowledge.

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]