A) Suppose that f is a linear function satisfying the condition f(kx) = kf(x) for all real numbers k. Prove that the graph of f passes through the origin. B) Let a & b be real numbers, and suppose that inequality ax +b is less than or equal to 0, has no solutions. What can you say about the linear function f(x) = ax +b? Answer in complete sentences and justify what you say. Solution 1. f(kx) = k*f(x) Put k=0, f(0) = 0 Thus, the graph passes through (0,0). 2. ax + b <= 0 Then f(x) = ax + b <= 0 Then, the graph of the function f(x) lies below the x-axis for all real values of x. It never crosses the x-axis. The grapgh touches the x-axis at x = -b/a..