Inflection Point Calculation: Reduction of Cubic with Second Derivative Method

[Hill]

Homework Statement

The roots of a general cubic equation in X may be viewed (in the XY-plane) as the intersections of the X-axis with the graph of a cubic of the form, Y = X^3 + AX^2 + BX + C.
(i) Show that the point of inflection of the graph occurs at X = −A/3.
(ii) Deduce (geometrically) that the substitution X = x − A/3 will reduce the above equation to the form Y = x^3 + bx + c.
(iii) Verify this by calculation.

Relevant Equations

Y = X^3 + AX^2 + BX + C
Y = x^3 + bx + c

(i) I take the second derivative of Y: Y'' = 6X + 2A. Y'' = 0 when X = -A/3. Moreover, as Y'' is linear it changes sign at this X. Thus, it is the point of inflection.

(iii) After the substitution, the term x^2 appears twice: one, from X^3 as -3(x^2)(A/3), and another from AX^2 as Ax^2. They cancel. Thus, there is no x^2 term.

(ii) Here I am not sure. The only "geometrical" reasoning I can think of is as follows. The substitution, X = x - A/3 translates the graph of Y in such a way that the inflection point is now at x=0. If the new graph is Y = x^3 + ax^2 + bx +c, its inflection point is at x = -a/3. Thus, a = 0 and Y = x^3 + bx + c.
Is there any other "geometrical deduction"?

 

[anuttarasammyak]

If the new graph is Y = x^3 + ax^2 + bx +c, its inflection point is at x = -a/3. Thus, a = 0 and Y = x^3 + bx + c.

$$Y'=3x^2+2ax+b$$
$$Y^"=6x+2a$$
Its reflection point is at x=0 there Y"(0)=0.so a=0
The graph is convex upward/downward bordered by inflection point x=0. x^3 and x follow this, but x^2 violates it.