Converting a x 3 + b x 2 + c x + d to a ( x − j ) 3 + kWe're all familiar with the vertex form of a quadratic function, a ( x − p ) 2 + qwhere (-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 a x 2 + b x + c.My question is: can the same be done for cubic functions? Playing around with graphs have shown me that a ( x − j ) 3 + kgraphs 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 a x 3 + b x 2 + c x + 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?

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Converting   a      x    3    +  b      x    2    +  c  x  +  d to   a  (  x  −  j      )    3    +  kWe&#039;re all familiar with the vertex form of a quadratic function,  a  (  x  −  p      )    2    +  qwhere (-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   a      x    2    +  b  x  +  c.My question is: can the same be done for cubic functions? Playing around with graphs have shown me that  a  (  x  −  j      )    3    +  kgraphs 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   a      x    3    +  b      x    2    +  c  x  +  d to its &#039;vertex form&#039; as stated above that is applicable generally to all cubic functions.I have tried geometrically &#039;completing the cube&#039; 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&#039;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?

Phiplyrhypelw0 · Accepted Answer

Step 1You can&#039;t do that in general. Suppose that your cubic had 3 real roots (example:       x    3    −  x).Step 2Then you can&#039;t do that because   (  x  −  j      )    3    +  k has only one real root. So, they cannot be the same cubic.

Anton Huynh · Accepted Answer

Step 1Your 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.   a      x    3    +  b      x    2    +  c  x  +  d Differentiate it twice and equate it to zero .Step 2You&#039;ll get   3  a  x  =  −  bNow you can get co-ordinate of   x  =            −      b              3      a       calculate x and put back that in the cubic equation to get y coordinate

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

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