Re: calculate vertex of a parabola

*To*: mathgroup at smc.vnet.net*Subject*: [mg112675] Re: calculate vertex of a parabola*From*: "Sjoerd C. de Vries" <sjoerd.c.devries at gmail.com>*Date*: Sun, 26 Sep 2010 02:43:37 -0400 (EDT)*References*: <i7k4ch$lgl$1@smc.vnet.net>

I'm not sure that I understand you correctly. The vertex, or minimum or maximum value of a parabola can be found by setting the derivative of its equation equal to zero and solving for x: In[8]:= D[a*x^2 + b*x + c, x] Out[8]= b + 2 a x In[9]:= Solve[b + 2 a x == 0, x] Out[9]= {{x -> -(b/(2 a))}} So the x-coordinate of the vertex is -b/(2a) and the y-coordinate is given by: In[10]:= a*x^2 + b*x + c /. x -> -(b/(2 a)) Out[10]= -(b^2/(4 a)) + c So the vertex doesn't depend on x as you seem to assume. Your problem is not really a Mathematica problem, it's more a problem of mathematics (or actually your grasp of it). Cheers -- Sjoerd On Sep 25, 8:21 am, Momo K <momok1... at googlemail.com> wrote: > Hello, > > please excuse me for my language but I am German. > > What I wanna do is to calculate the vertex of a parabola or its mathematic > equoation of its vertex. > E. g. if I have a equation of the form "a*x^2 + b*x + c = f(x)", I want an > output like the following: > "a*(x-g)+h = f(x)" > > I already tried with the command factor, but I didn'T found any possibility > to transform the equoation from above into the last one. > > Thank you > Momo > > PS: I am really surprised; I thought, that this would be the first or second > hit in documentation of Mathematica, but even with Google, I didn't found > anything.