The gradient of the function is constant s the independent variable (x) varies The graph passes through the origin. That is to say when x = 0, y = 0. Clearly A and D pass through the origin, and the gradient is constant because they are linear functions, so they are direct variations. The graph of 1/x does not have a constant gradient, so any stretch of this graph (to y = k/x for some constant k) will similarly not be direct variation. Indeed there is a special name for this function, inverse proportion/variation. It appears both B and C are inverse proportion, however if I interpret B as y = (2/5)x instead, it is actually linear. I believe the answer is C. Hope I helped!