While working on part (c) of Can you find… curvy cubics edition, which asked: “Can you find a cubic curve that has a local minimum when \(x=-1\)?”, two students had the following conversation:
A: I can find a cubic with a stationary point at \(x=-1\) by having two of the \(x\)-axis intersection points at \(x=-2\) and \(x=0\), because the stationary point is half-way between the two intersection points. I need a third intersection point, too, not between these, so I may as well choose \(x=4\), so my cubic is \(y=-x(x+2)(x-4)\) (with a minus sign so the cubic is the right way up).
B: I’m not sure that’s right; is the stationary point really half-way between the \(x\)-axis intersection points?
Can you resolve their debate?
If student A is right in this case, does this approach always work for finding a cubic with a given stationary (turning) point, or only sometimes?
If student B is right in this case, does student A’s approach ever work?
