二次函数y=(x-1)2+2的最小值是( )
A. -2
B. 2
C. -1
D. 1
人气:176 ℃ 时间:2019-08-19 07:03:35
解答
∵(x-1)2≥0,∴y=(x-1)2+2≥2,
∴最小值为2.
故选B.
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