Hi

y=2x^2-4x+1  |y-intercept Pt(0,1) 

x-intercept when y = 0

2x^2-4x+1 = 0

  = 1 ± .707

Pt(.3,0) and Pt(1.7,0) the x-intercepts (rounded to nearest tenth)

Using the vertex form of a parabola,  where(h,k) is the vertex

y=2x^2-4x+1      |completing square to put into the vertex form

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

y= 2(x-1)^2 -1   Vertex is Pt(1,-1) axis of symmetry is x = 1   



 

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