Question 1198223
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Given equation: y = 2x^2-4x+1


It is of the form y = ax^2+bx+c
a = 2
b = -4
c = 1


Use the first two values to find that
h = -b/(2a)
h = -(-4)/(2*2)
h = 1
This is the x coordinate of the vertex (h,k)


Plug this into the given equation to find the y coordinate of the vertex.
y = 2x^2-4x+1
y = 2(1)^2-4(1)+1
y = -1
Therefore k = -1 is the y coordinate of the vertex (h,k)


Vertex = (h,k) = (1,-1)
This represents the lowest point on the parabola.
We know the parabola opens upward because a > 0.


Summary:
a = 2
h = 1
k = -1
Vertex form y = a(x-h)^2+k then leads to <font color=red>y = 2(x-1)^2-1</font>


Graph:
{{{drawing(500,500,-7,7,-7,7,
graph(500,500,-7,7,-7,7,0,2x^2-4x+1),
circle(1,-1,0.1),
circle(1,-1,0.12),
circle(1,-1,0.15),
locate(1,-1,"(1,-1)")
)}}}
I recommend using free graphing tools like Desmos and GeoGebra. 
Or you can use a TI83 or TI84 calculator (or similar).
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