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Re: calculate vertex of a parabola

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> 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.

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