How do you find the axis of symmetry, and the maximum or minimum value of the function #y = 2x^2 - 4x + 1#?

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Answer 1 · 2017-01-20T08:03:44

The axis is #x=1#
The minimum value is #y=-1#

Let`s complete the squares

#y=2x^2-4x+1#

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

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

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

This is the vertex form of the equation

The axis of symmetry is #x=1#

The vertex is #(1,-1)#

The minimum value is #y=-1#

graph{(y-(2x^2-4x+1))(y-100x+100)((x-1)^2+(y+1)^2-0.002)=0 [-5.546, 5.55, -2.773, 2.774]}