I don't understand this proof about graphs of cubic functions

Question

This proof on proofs.wiki shows that: All graphs of cubic functions are transformations of an odd cubic function.

I have no problems with the steps. I just do not understand what is the idea behing this proof.

Answer 1

Forget the linked proof. The idea is the following: All cubic graphs $$\gamma:\quad y=ax^3+bx^2+cx+d\qquad=:f(x)\tag{1}$$ with $a\ne0$ have an inflection point $P:=(p,q)$ whose $x$-coordinate $p=-{b\over3a}$ is found by solving $f''(x)=6ax+2b=0$, and whose $y$-coordinate $q$ is then given by $q:=f(p)$.

We now introduce new coordinates $(\bar x,\bar y)$ whose origin is at $P$. This means that we put $$x:=p+\bar x, \quad y:=q+\bar y$$ in $(1)$. After some calculation it turns out that in the new coordinates $\gamma$ appears as $$\gamma:\quad \bar y=a\>\bar x^3+c'\bar x$$ with a certain new coefficient $c'$. This shows that any cubic graph is point-symmetric with repect to its inflection point.