How do you find the vertex of #y=3x^2-6x-2#?

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Answer 1 · 2016-12-24T08:17:33

The vertex is #(1,-5)#

We need
#(a+b)^2=a^2+2ab+b^2#

The vertex form of a parabola, vertex #(h,k)# is

#y-k=a(x-h)^2#

Let's complete the squares and write in the equation in the vertex form

#y=3x^2-6x-2#

#y=3(x^2-2x)-2#

#y=3(x^2-2x+1)-2-3#

#y=3(x-1)^2-5#

#y+5=3(x-1)^2#

The vertex is #(1,-5)#

graph{3x^2-6x-2 [-16.02, 16.02, -8.01, 8.01]}