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This is exercise $1$ on page $45$ of Tristan Needham's Visual Complex Analysis:

1. The roots of a general cubic equation in $X$ may be viewed (in the $XY$- plane) as the intersection of thhe $X$- 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 $X = -\dfrac A3$

$(ii)$ Deduce (geometrically) that the substitution $X = (x-\dfrac A3)$ will reduce the above equation to the form $Y = x^3 + bx + c$.

$(i)$ is easy enough by using the second derivative. (is there another way though?)

$(ii)$ I know that making said substitution will center the graph so that the inflection point will lie on the $Y$ axis. However, I'm not sure what to do beyond that. By staring at the graph below, I realized just now that it would be reasonable visually to argue $Y'(x) = Y'(-x)$, so the derivative has to be a parabola of the form $x^2 + \gamma$, which means that $Y$ is a depressed cubic. However, I'm not sure if this is a geometric enough approach.

I also tried to view of $0 = X^3 + AX^2 + BX + C$ as $x^3$ intersecting a parabola, and of $0 = x^3 + bx+c$ as $x^3$ intersecting a line; but I'm not sure how helpful this is because then the substitution $X = x - \dfrac A3$ loses its easy visual meaning.

Cubic:

Depressed Cubic:

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    Possibly useful is my answer to the Mathematics Educators StackExchange question [The sum - product problem](http://matheducators.stackexchange.com/questions/8020/the-sum-product-problem/8027) (see especially the last part of my answer). – 2017-02-24
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    The geometric intuition is that the point of inflection $(x_0, y_0)$ is a center of symmetry for the entire curve, so writing the equation as $y+y_0=f(x+x_0)$ the RHS will be an odd function in $x+x_0\,$, which means that it will have only odd powers of $x+x_0$ i.e. no square term. Of course, this alone does *not* prove that the point of inflection is in fact a center of symmetry. – 2017-02-24
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    @dxiv Thanks, that seems like a better version of my argument $\forall x f'(x) = f'(-x)$. – 2017-02-24
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    @dxiv What do you mean though that this does not prove the point of inflection is not a center of symmetry? What is your definition of a center of symmetry? – 2017-02-24
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    My previous comment stated the implication "*if the cubic has a center of symmetry, then the respective change of variables gives a depressed cubic*". You can use it the other way around, and if you accept (or prove algebraically) that the given substitution gives a depressed cubic, then it follows that the inflection point is the center of symmetry. As for the definition, it's just the usual one - if point $(x,y)$ is on the curve then so is its symmetric $(2x_0-x,2y_0-y)$ with respect to $(x_0,y_0)$.. – 2017-02-24
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    Re (i). By def'n, $ p$ is an inflection point of $f$ iff $f''(p)=0$. So there is no "other way". – 2017-02-24
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    @user254665 Well inflection points have to do with concavity, which was probably studied before calculus; that's why I thought there might be a geometric way. – 2017-02-24
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    You can take the radius $R(r)$ of the circle thru the 3 points $(p-r, f(p-r)), (p,f(p)), (p+r,f(p+r) .$ If $R(r)\to S$ as $r\to 0$, we say that the radius of curvature of the graph of $f$ at the point $(p,f(p))$ is $S.$ After a fair bit of calculation you can infer that if $f''$ is continuous then $R(r)\to \infty$ as $r\to 0$ just when $f''(p)=0.$ This is not exactly a better way to do it! – 2017-02-24
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    @user254665 Youre right! :) – 2017-02-24
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    @user254665: Regarding inflection points, there are several different definitions that can be found in textbooks. See the following for more details: Andrew M. Bruckner, [*Some nonequivalent definitions of inflection points*](http://www.jstor.org/stable/2310782); George M. Ewing, [*On the definition of inflection point*](http://www.jstor.org/stable/2302439); Temple H. Fay, [*Convex functions*](http://dl.acm.org/citation.cfm?id=305524); A. W. Walker, [*What is a point of inflection?*](http://www.jstor.org/stable/2306661) – 2017-02-24

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