This is probably too late now to be of any use to you - sorry!
You seem to have worked out the equation to the parabola y = (-192/17)x^2 + (2112/17)x + 5922/17 and to do that you must have made the gradients of the cubic and the parabola equal to each other AND have used the points (-3, -126) and (14, 126).
Therefore you have found a parabola that satisfies the conditions that it smoothly join at A and passes through points A and D.
I think you have been given a trick (or poorly worded) question. From Question e (which talks about completing the track circuit from D back to A) , it appears as though the required parabola may also have been intended to complete the track from D to A. If that is the case, then they would want a positive parabola which would have a turning point at x = 5.5. However, that would have a negative gradient at x=-3 which would produce the "elbow" since the cubic has a positive gradient at that point.
You could argue that the parabola y = (-192/17)x^2 + (2112/17)x + 5922/17 fits the specification except it would be one helluva mess with half a dozen runners trying to turn at the resulting cusp at point A.