# I have been studying on cubic equations for a while and see that the cubic equation needs to be in t

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 ${y}^{″}=6x+2a$, 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?
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mnaonavl
If you look at the inflection point of
$y={x}^{3}+a{x}^{2}+bx+c$
where the second derivative is $0$, you realize that it happens at $x=-a/3$ while the inflection point of
$m{x}^{3}+px+q$
happens at $x=0$
Thus the transformation is moving the inflection point to $x=0$
###### Not exactly what you’re looking for?
Bailee Landry
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 $y$, 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.