求二次函数顶点坐标的公式!
人气:419 ℃ 时间:2020-01-26 14:25:24
解答
一般式:y=ax^2+bx+c(a,b,c为常数,a≠0)
顶点式:y=a(x-h)^2+k
[抛物线的顶点P(h,k)]
对于二次函数y=ax^2+bx+c
其顶点坐标为 (-b/2a,(4ac-b^2)/4a)
交点式:y=a(x-x₁)(x-x ₂) [仅限于与x轴有交点A(x₁ ,0)和 B(x₂,0)的抛物线]
其中x1,2= -b±√b^2-4ac
注:在3种形式的互相转化中,有如下关系:
______
h=-b/2a= (x₁+x₂)/2 k=(4ac-b^2)/4a 与x轴交点:x₁,x₂=(-b±√b^2-4ac)/2a
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