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calculate vertex of a parabola
*To*: mathgroup at smc.vnet.net
*Subject*: [mg112721] calculate vertex of a parabola
*From*: Momo K <momok1994 at googlemail.com>
*Date*: Tue, 28 Sep 2010 06:04:55 -0400 (EDT)
Hello,
after many misunderstandings, I present the "pen and paper-method":
a * x^2 + b*x + c
=a (*x^2 + (b*x)/2*) + c
Now I set (from the underlined from above):
z = x ^ 2 + (b * x)/2
= x ^ 2 + (b * x)/2 + (b/2)^2 - (b/2)^2 ;add term to have form of
a binomial theorem (a+b)^2 = a^2 + 2*a*b + b^2 = (a+b)^2
= (x+(b/2))^2 - (b/2)^2
Now I continue with both terms:
a ((x+(b/2)^2)^2 - (b/2)^2) + c
=a(x+(b/2))^2 - a(b/2)^2 + c
So, my vertex is at X: -(b/2)^2
Y: -a(b/2)^2 + c
I hope this is correct and I didn't miscalculated...
My question now is if there is a comfortable way to calculate form of the
vertex of the parabola.
Thanks
PS: Yes, I've never used derivats because I'm 16 and in 11th form/grade.
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