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 + p x + 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 + b x + c = 0, x value needs to be replaced with t − a 3 .I realised that the second derivative of any cubic equation in the form of x 3 + a x 2 + b x + c = 0 is y ″ = 6 x + 2 a, and when we equalize y to 0, x is equal to − 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 + a x + b = 0. When we replace x with − 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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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    +  p  x  +  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    +  b  x  +  c  =  0,   x value needs to be replaced with   t  −      a    3  .I realised that the second derivative of any cubic equation in the form of       x    3    +  a      x    2    +  b  x  +  c  =  0 is       y    ″    =  6  x  +  2  a, and when we equalize   y to   0,   x is equal to   −      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    +  a  x  +  b  =  0. When we replace   x with   −      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?

mnaonavl · Accepted Answer

If you look at the inflection point of  y  =      x    3    +  a      x    2    +  b  x  +  cwhere the second derivative is   0, you realize that it happens at   x  =  −  a      /    3 while the inflection point of  m      x    3    +  p  x  +  qhappens at   x  =  0Thus the transformation is moving the inflection point to   x  =  0

Bailee Landry · Accepted Answer

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    ″    =  6  x  +  2  a, 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   ∪-shaped if   a  &amp;gt;  0, and   ∩-shaped if   a  &amp;lt;  0. if   a  =  0. then it is neither, because you have the inflection point right there.

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

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