y=ax^2+bx+c
=a(x^2+bx/a)+c
=a[x^2+2*x*(b/2a)+(b/2a)^2-(b/2a)^2]+c
=a[x-(-b/2a)]^2-a*b^2/4a^2+c
=a[x-(-b/2a)]^2-b^2/4a+c
=a[x-(-b/2a)]^2+(4ac-b^2)/4a
因为[x-(-b/2a)]^2>0
所以当a>0时,a[x-(-b/2a)]^2+(4ac-b^2)/4a>=(4ac-b^2)/4a
a<0,a[x-(-b/2a)]^2+(4ac-b^2)/4a<=(4ac-b^2)/4a
为什么二次函数的值域是Y大于等于4A分之4AC-B2我搞不明白,气死我呢,
为什么二次函数的值域是Y大于等于4A分之4AC-B2我搞不明白,气死我呢,
数学人气:436 ℃时间:2019-09-17 08:48:27
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