How do you find the line of symmetry for the parabola whose equation is #y=2x^2-4x+1#?

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Answer 1 · 2018-05-16T13:49:43

Axis (or line) of symmetry is #x-1=0# or #x=1#

When the equation of parabola in vertex form i.e.

#y=a(x-h)^2+k#, #(h,k)# is the vertex and #x-h=0# is the axis (or line) of symmetry.

Now we can write #y=2x^2-4x+1# as

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

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

Hence, vertex is #(1,-1)# and axis (or line) of symmetry is #x-1=0# or #x=1#

graph{(y-2x^2+4x-1)(x-1)((x-1)^2+(y+1)^2-0.01)=0 [-3.625, 6.375, -1.78, 3.22]}