I have been studying on cubic equations for a while and see that the cubic equation needs to be in the form of $m{x}^{3}+px+q=0$ so that we can find the roots easily. In order to obtain such an equation having no quadratic term for a cubic equation in the form of ${x}^{3}+a{x}^{2}+bx+c=0$, $x$ value needs to be replaced with $t-\frac{a}{3}$.

I realised that the second derivative of any cubic equation in the form of ${x}^{3}+a{x}^{2}+bx+c=0$ is ${y}^{\u2033}=6x+2a$, and when we equalize $y$ to $0$, $x$ is equal to $-\frac{a}{3}$. This had me thinking about any possible relation between the quadratic term and the second derivative.

It is also sort of the same in quadratic functions in the form of ${x}^{2}+ax+b=0$. When we replace $x$ with $-\frac{a}{2}$, which is the first derivative of a quadratic function and the vertex point of it, we obtain the vertex form of the equation which does not have the linear term, ${x}^{1}$.

Is there a relation between the quadratic term and the second derivative of a cubic equation? If the answer is yes, what is it and how is it observed on graphs, in equations?

I realised that the second derivative of any cubic equation in the form of ${x}^{3}+a{x}^{2}+bx+c=0$ is ${y}^{\u2033}=6x+2a$, and when we equalize $y$ to $0$, $x$ is equal to $-\frac{a}{3}$. This had me thinking about any possible relation between the quadratic term and the second derivative.

It is also sort of the same in quadratic functions in the form of ${x}^{2}+ax+b=0$. When we replace $x$ with $-\frac{a}{2}$, which is the first derivative of a quadratic function and the vertex point of it, we obtain the vertex form of the equation which does not have the linear term, ${x}^{1}$.

Is there a relation between the quadratic term and the second derivative of a cubic equation? If the answer is yes, what is it and how is it observed on graphs, in equations?