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?
Best Answer
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.