问题
填空题
已知二次函数y=(x-1)2+(x-3)2,当x=______时,函数达到最小值.
答案
因为原式可化为y=2x2-8x+10=2(x-2)2+2,所以当x=2时,函数达到最小值.
已知二次函数y=(x-1)2+(x-3)2,当x=______时,函数达到最小值.
因为原式可化为y=2x2-8x+10=2(x-2)2+2,所以当x=2时,函数达到最小值.
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