I then ask them to show that the point of inflexion of is at the midpoint of the turning points. This involves using the equations for the roots of a quadratic: and but the algebra isn’t particularly nice (details are here).

It dawned on me that there’s a better way which is not only simpler to prove but gives a more powerful result: a cubic is (rotationally) symmetric about its point of inflexion.

To see this, take the cubic and translate it so that the point of inflexion is at the origin. This clearly has no effect on the symmetry so it is sufficient to prove the result in this case.

If has its point of inflexion at the origin then clearly and, as , it follows that so . Rotate this by about the origin, for example by applying the usual matrix , and you get which is clearly the same cubic. Hence it has rotational symmetry about the origin, and in particular, the point of inflexion at the origin is the midpoint of the stationary points.

the following appears:

Putting gives an angle , confirming the condition for the lines to be perpendicular

This is of course complete nonsense. As I’ve said before doesn’t exist and is only defined on ie for
The textbook was written by the examiners (which is one reason why we use it); this worries me even more.
I suppose this is better than one well-known textbook back in the eighties which solved the equation by putting then ‘showing’ or . This seems to show that all linear equations are quadratics in disguise; or cubics, quartics – who knows where this nonsense leads
See also Division by zero shock!