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We're all familiar with the vertex form of a quadratic function,$$a(x-p)^2 +q$$ where $(-p,q)$ represent the coordinates of the maximum or minimum point of the parabola. This is achieved by performing completing the square on the standard form of the quadratic function $ax^2+bx+c$.

My question is: can the same be done for cubic functions? Playing around with graphs have shown me that $$a(x-j)^3 +k$$ graphs a cubic function whose inflection point is $(-j,k)$. However, I have not been able to find a method of converting a standard cubic function $ax^3+bx^2+cx+d$ to its 'vertex form' as stated above that is applicable generally to all cubic functions.

I have tried geometrically 'completing the cube' with a friend and found a way (?) to convert standard cubic expressions to their vertex form, but this method was only applicable to cubics without a linear term. I'm looking for a general method that works for all cubics, I really appreciate the help!

Edit: Okay, so I received a lot of feedback which explained why $a(x-j)^3 +k$ would not work for all cubic equations because it only has one real root, and got pointers in the direction of depressed cubic equations. My question is: is there a general way to show the position of the inflection point using the depressed cubic?

José Carlos Santos
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iciqle
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    Usually one cannot: this is only possible if $3ac=b^2$. – Angina Seng Oct 20 '19 at 06:55
  • Take a look at this Reddit post. In particular, by making the substitution $x = u+v$, you can reduce the cubic into the form $x^3-px=q$. The resulting cubic equation is called a 'depressed cubic'. – Toby Mak Oct 20 '19 at 07:06
  • If $a\ne 0$ and $F(x)=ax^3+bx^2+cx+d $, suppose $F(x)= a(x-j)^3+k$ for all $x$ then $F(x)=0$ would be easy to solve $F(x)=0$. Too easy..... Substituting $x=t-b/3a$ into $a^{-1}F(x)=0$ gives a "depressed cubic" equation of the type $t^3+pt+q=0 .$ See
    https://en.wikipedia.org/wiki/Cubic_equation#Derivation_of_the_roots     – DanielWainfleet Oct 20 '19 at 08:12
  • If substitution $x\to A+\frac{1}{y}$ is acceptable, then for $y$ "complete cubic" is possible. – Dmitry Ezhov Oct 20 '19 at 14:12

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You can't do that in general. Suppose that your cubic had $3$ real roots (example: $x^3-x$). Then you can't do that because $(x-j)^3+k$ has only one real root. So, they cannot be the same cubic.

José Carlos Santos
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Your expression has only one real root but a cubic equation can have at most 3 real roots. To generalise the point of inflection you can use calculus. $ax^3+bx^2+cx+d$ Differentiate it twice and equate it to zero .

You'll get $3ax=-b$

Now you can get co-ordinate of $x=\frac{-b}{3a}$ calculate x and put back that in the cubic equation to get y coordinate

Who am I
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Shifting $x$ by a carefully chosen constant can only delete one term in general, which we usually choose to be the $x^2$ one to obtain a depressed cubic, which if monic after division by $a$ is usually denoted $y^3+py+q,\,y:=x+\frac{b}{3a}$, Therefore, the solution isn't as simple as taking a cube root, However, you can solve it.

J.G.
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