Question

which graph matches this direct variation equation?
y = -4x

Explanation:

Step1: Recall Direct Variation Graph Properties

A direct variation equation is of the form \( y = kx \), where \( k \) is the constant of variation. The graph of \( y = kx \) is a straight line passing through the origin \((0,0)\) (since when \( x = 0 \), \( y = 0 \)). The slope of the line is \( k \). For \( y=-4x \), the slope \( k = - 4 \), which is negative. A negative slope means the line falls from left to right (as \( x \) increases, \( y \) decreases).

Step2: Analyze Each Graph

- First graph: The line has a positive slope (rises from left to right), so it can't be \( y = - 4x \) (which has a negative slope).
- Second graph: The line passes through the origin (good for direct variation) and has a negative slope (falls from left to right). Let's check the slope magnitude. The slope \( k=\frac{\Delta y}{\Delta x}\). If we take two points, say from the graph, when \( x = 0 \), \( y = 0 \); when \( x = 4 \), \( y=-2\) (approximate). Then slope \( k=\frac{-2 - 0}{4 - 0}=-\frac{1}{2}\), which is not \(-4\). Wait, maybe my approximation is wrong. Wait, the equation is \( y=-4x \), so the slope is \(-4\), which is a steep negative slope.
- Third graph: Wait, no, wait the third graph—wait, no, let's re - check. Wait, the second graph: maybe I misread. Wait, the equation \( y = - 4x \) has a slope of \(-4\), which is a very steep negative slope (since the absolute value of 4 is large). So the line should be very steep, falling from left to right, passing through the origin.

Wait, maybe I made a mistake in the second graph. Let's re - express:

The direct variation \( y=-4x \) has:
- Passes through \((0,0)\) (origin).
- Slope \( m=-4\), so for every 1 unit increase in \( x \), \( y \) decreases by 4 units. So the line is steep and decreasing.

Looking at the graphs:

First graph: positive slope (increasing), so eliminate.

Third graph: Wait, does it pass through the origin? No, wait the third graph seems to have a y - intercept (crosses y - axis at non - zero). Wait, no, direct variation \( y = kx \) has no y - intercept (y - intercept is 0). So the line must pass through the origin.

Wait, maybe the second graph: let's check the points. If \( x = 1 \), \( y=-4(1)=-4 \); \( x = 0 \), \( y = 0 \); \( x=-1 \), \( y = 4 \). So the line should pass through \((0,0)\), \((1, - 4)\), \((-1,4)\). So it's a steep line, passing through origin, negative slope.

The second graph: let's see, if we take \( x = 0 \), \( y = 0 \); when \( x = 1 \), \( y=-4\) (so the point \((1, - 4)\) should be on it). So the second graph—wait, maybe the second graph is the one with the line passing through the origin, negative slope, and steep? Wait, maybe the first graph is positive slope (so no), third graph has a y - intercept (so no, since direct variation has y - intercept 0). Wait, maybe I mis - identified the graphs.

Wait, the correct graph should be the one that:

- Passes through the origin (since \( y = kx \) has \( b = 0 \) in slope - intercept form \( y=mx + b \)).
- Has a negative slope (since \( k=-4<0 \)).
- Has a steep slope (since \( |k| = 4 \) is large).

So among the three graphs, the second graph (the middle one) passes through the origin, has a negative slope. Wait, but the slope of the second graph—let's calculate it properly. Suppose in the middle graph, when \( x = 0 \), \( y = 0 \); when \( x = 2 \), \( y=-8 \)? No, maybe the grids are different. Wait, maybe the middle graph is the one with the line passing through the origin and having a negative slope, and the steepness: since \( k=-4 \), the line is steep. So the middle graph (second graph) is the one that matches \( y=-4x \) because:

- It passes through the origin (required for direct variation \( y = kx \)).
- It has a negative slope (matches \( k=-4<0 \)).
- The steepness is consistent with a slope of \(-4\) (since a slope of \(-4\) means for \( x = 1 \), \( y=-4\), so the line is steeply decreasing).

Answer:

The middle graph (the second graph among the three) matches the equation \( y=-4x \).