A function exhibits direct variation if it takes the form \(y = kx\), where \(k\) is a non-zero constant. For a graph to represent a function with direct variation, it must pass through the origin (0, 0) because direct variation implies that when \(x = 0\), \(y\) must also equal \(0\).
Let's analyze the options:
-
A line passing through (negative 4, 0) and (0, negative 2): This does not pass through the origin (0, 0).
-
A line passing through (negative 5, 4) and (0, 3): This also does not pass through the origin (0, 0).
-
A line passing through (negative 4, negative 6) and (0, 3): This does not pass through the origin (0, 0).
-
A line passing through (negative 1, negative 4), (0, 0), and (1, 4): This does pass through the origin (0, 0).
Therefore, the graph that represents a function with direct variation is the one that passes through the points (negative 1, negative 4), (0, 0), and (1, 4).
1 answer