1. The problem statement, all variables and given/known data Find a cubic function ax^3 + bx^2 + cx + d whose graph has horizontal tangents at the points (-2, 6) and (2. 0). 2. Relevant equations No idea. 3. The attempt at a solution I realize that the trick in here somewhere is to work "backwards". I completed a problem earlier that asked me to find the points at which a given function had horizontal tangents. I tried to go back to that and construct a path to the solution for this problem by going in reverse but did not get anywhere. So far I've taken the first and second derivatives of the given cubic function to arrive at y' = 3ax^2 + 2bx + c y'' = 6ax + 2b Since the points are given, I figure perhaps I need to set something equal to a point-slope equation.. something like this, basically: y - 6 = (x - (-2)) for the first point, for example. I'm confused as to how exactly to proceed and whether how I thought about this so far is the right way of thinking about it. Some pointers would be appreciated. Thanks!