Converting $ax^3 + bx^2 +cx + d$ to $a(x-j)^3 +k$

Question

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?

Answer 1

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.

Answer 2

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

Answer 3

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.