Question

determine if the variation is of the form $y = kx$ or $y = \frac{k}{x}$, and find $k$.
how are x and y variables related?
\\(\bigcirc\\) a. $y = \frac{k}{x}$
\\(\bigcirc\\) b. $y = kx$
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$$\begin{tabular}{c|c} $x$ & $y$ \\\\ \\hline 3 & $\\frac{4}{3}$ \\\\ \\hline 4 & 1 \\\\ \\hline 8 & $\\frac{1}{2}$ \\\\ \\hline 16 & $\\frac{1}{4}$ \\\\ \\end{tabular}$$

Explanation:

Step1: Check for direct variation (y = kx)

For direct variation, \( k=\frac{y}{x} \) should be constant. Let's check the first pair \( x = 3,y=\frac{4}{3} \): \( k=\frac{\frac{4}{3}}{3}=\frac{4}{9} \). Second pair \( x = 4,y = 1 \): \( k=\frac{1}{4}=0.25 \). These are not equal, so not direct variation.

Step2: Check for inverse variation (y = k/x)

For inverse variation, \( k = xy \) should be constant. First pair: \( 3\times\frac{4}{3}=4 \). Second pair: \( 4\times1 = 4 \). Third pair (assuming x = 8, y = 1/2): \( 8\times\frac{1}{2}=4 \). Fourth pair: \( 16\times\frac{1}{4}=4 \). So \( k = 4 \), and the relation is \( y=\frac{k}{x} \) (inverse variation).

Answer:

A. \( y=\frac{k}{x} \) (with \( k = 4 \))