Question

which graph matches this direct variation equation? y = 2x

Explanation:

Step1: Recall direct variation properties

A direct variation equation \( y = kx \) (here \( k = 2 \)) is a linear equation passing through the origin \((0,0)\) with slope \( k \). The slope \( m = 2 \) means for every \( 1 \) unit increase in \( x \), \( y \) increases by \( 2 \) units.

Step2: Analyze the slope and intercept

• The equation \( y = 2x \) has a \( y \)-intercept of \( 0 \) (so it passes through \((0,0)\)) and a slope of \( 2 \) (rise over run is \( 2/1 \)).
• Let's check points: when \( x = 1 \), \( y = 2(1)=2 \); when \( x = 2 \), \( y = 2(2) = 4 \), etc. The graph should pass through \((0,0)\), \((1,2)\), \((2,4)\) etc., with a steeper slope (since slope \( 2>1 \)).

Step3: Compare with the graphs

• The first graph: Check the slope. If we take two points, say from the grid, but since it's \( y = 2x \), the correct graph should have points like \((0,0)\), \((1,2)\), \((2,4)\). The third graph (assuming the third one has points like when \( x = 1 \), \( y = 2 \); \( x = 2 \), \( y = 4 \)) matches the slope of \( 2 \) and passes through the origin. (Note: Based on the description, the third graph likely has the correct steepness and passes through \((0,0)\) with the correct slope.)

Answer:

The graph (assuming the third one in the display, with points like \((0,0)\), \((1,2)\), \((2,4)\) etc., and a slope of \( 2 \)) that matches \( y = 2x \). (If we consider the options as three graphs, the third graph (right - most) is the correct one as it has the correct slope and passes through the origin with the correct rate of change.)