Re: calculate vertex of a parabola

*To*: mathgroup at smc.vnet.net*Subject*: [mg112697] Re: calculate vertex of a parabola*From*: Helen Read <readhpr at gmail.com>*Date*: Mon, 27 Sep 2010 05:47:36 -0400 (EDT)*References*: <i7k4ch$lgl$1@smc.vnet.net> <i7mq3c$out$1@smc.vnet.net>

On 9/26/2010 2:43 AM, Sjoerd C. de Vries wrote: > On Sep 25, 8:21 am, Momo K<momok1... at googlemail.com> wrote: >> >> 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'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] > >> 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). A more elementary (non-calculus) method of finding the vertex is to complete the square, which is what the OP asked about. I think the OP meant to write a*(x-g)^2+h = f(x) and accidentally left off the square. Here's one way to do it. If we want to write y=a x^2 + b x + c in the form y=a (x - u)^2 + v Try expanding the second form, than match coefficients. Expand[a (x - u)^2 + v] This gives: a u^2 + v - 2 a u x + a x^2 Now match coefficients and solve for u and v. Solve[{-2 a u == b, a u^2 + v == c}, {u, v}] -- Helen Read University of Vermont