Question 417924
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Hi
y=2x^2-4x+1  |y-intercept Pt(0,1) 
x-intercept when y = 0
2x^2-4x+1 = 0
 {{{x = (4 +- sqrt(8))/(4) = 1 +- sqrt(2)/2}}} = 1 ± .707
Pt(.3,0) and Pt(1.7,0) the x-intercepts (rounded to nearest tenth)
Using the vertex form of a parabola, {{{y=a(x-h)^2 +k}}} 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   
{{{drawing(300,300,   -6, 6, -6, 6,  blue(line(1,6,1,-6))  , grid(1),
circle(1, -1,0.3),
circle(0, 1,0.3),
circle(.3, 0,0.2),
circle(1.7, 0,0.2),
graph( 300, 300, -6, 6, -6, 6,0,2x^2-4x+1))}}}