Is there a relationship between the second derivative and the quadratic term of a cubic equation?

Question

I have been studying on cubic equations for a while and see that the cubic equation needs to be in the form of $mx^3+px+q=0$ 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 $x^3+ax^2+bx+c=0$, $x$ value needs to be replaced with $t-\frac a{3}$.

I realised that the second derivative of any cubic equation in the form of $x^3+ax^2+bx+c=0$ is $y''=6x+2a$, 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?

Answer 1

Setting the second derivative equal to zero is going to find you the point of inflection, which for a cubic function will be halfway between the two turning points (if there are any). So any reduced cubic will have the property that $(0,c)$ is a point of rotational symmetry of the graph.

Answer 2

There is not one relation, but two.

You noticed the first one: $y''$ is equal to zero exactly once, and this happens at the centre of symmetry of the curve, which is also its only inflection point. If you look at the abscissa ($x$-value) of this point, it will tell you what $a$ is (by the formula that you obtained).

The second relation is that in $y''=6x+2a$, you can also equal not $y''$, but $x$, to $0$. This tells you that at the point where the curve intersects the $y$-axis, the curve is $\cup$-shaped if $a>0$, and $\cap$-shaped if $a<0$. if $a=0$, then it is neither, because you have the inflection point right there.

Answer 3

If you look at the inflection point of $$ y=x^3+ax^2+bx+c$$ where the second derivative isw $0$, you realize that it happens at $x=-a/3$ while the inflection point of $$mx^3 +px+q$$ happens at $x=0$

Thus the transformation is moving the inflection point to $x=0$