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.