求y=2x^2-4x+1的单调区间？？？？

y=2x^2-4x+1
=2(x^2-2x+1-1)+1
=2[(x-1)^2-1]+1
=2(x-1)^2-1
对称轴是x=1,函数开口向上,则单调递增区间是:[1,+无穷),单调递减区间是(-无穷,1]

y=2x^2-4x+1
=2x^2-4x+2-1
=2(x^2-2x+1)-1
=2(x-1)^2-1
所以x<1,单调递减
x>1,单调递增

y=2x^2-4x+1
=2(x^2-2x)+1
=2(x^2-2x+1-1)+1
=2[(x-1)^2-1)+1
=2(x-1)^2-2*1+1
=y=2（x-1）^2-1
对称轴是x=1,函数开口向上,则单调递增区间是:[1,+无穷),单调递减区间是(-无穷,1

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