Calculus I: Mean Value Theorem proof to prove slope of a graph and it's points

Here's the info: Suppose f(x) is a differentiable function on the interval [a,b]. Let y=mx+b be the line passing through the points (a, f(a)) and (b, f(b)). Suppose that f(x) intersects y exactly 4 times.
Use the Mean Value Theorem to prove that there exists at least three values of c such that the instantaneous rate of change at c equals the average rate of change on [a,b]
(Phew, that's a mouth full)
So I tried just showing that the slope at any point is equal to the diffinition of average slope, but my teacher didn't go for it. Late he hinted that we need to use three different points and then use the MVT three different times but I'm confused as to what points I'm supposed to use exactly.
Please help!

Here's a small beef with your statement of the question. Did it really say "intersects y exactly 4 times" or did it in fact say "intersects y=mx+b exactly four times" ? I hope it said the latter (and you should've copied it, if it did). What is meant, I think, is that the curve intersects the line four times. Thus, the intersection points are not only (a,f(a)) and (b,f(b)) but also, call 'em whatever you want, but how about (p,f(p)) and (q,f(q)) for the other two intersection points. We don't know whether p and q are less than a or greater than b or between a and b...but it really doesn't matter. Pick one case, and do it in detail; then "wave your hands" about how the other possible cases would all be similar. Let's say p < a < q < b. Since all four of the points lie on y=mx+b, the Mean Value Thm is going to guarantee you one point in each of the three intervals (p to a, a to q, q to b) where the instantaneous slope is equal to m.