二次函数y=x²+(2m+1)x+m²-1的最小值是0,求m的值
人气:421 ℃ 时间:2019-12-03 10:29:55
解答
a>0开口向上
化为顶点式
y=(x+(2m+1)/2)^2+m^2-1-((2m+1)/2)^2
最小值为0
所以m^2-1-((2m+1)/2)^2=0
求得m=-5/4
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