y=a(x-x₁)(x-x₂)的顶点坐标(x₁+x₂/2 ,)
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/21 07:51:38
![y=a(x-x₁)(x-x₂)的顶点坐标(x₁+x₂/2 ,)](/uploads/image/z/7099814-38-4.jpg?t=y%3Da%28x-x%26%238321%3B%29%28x-x%26%238322%3B%29%E7%9A%84%E9%A1%B6%E7%82%B9%E5%9D%90%E6%A0%87%EF%BC%88x%26%238321%3B%2Bx%26%238322%3B%2F2+%2C%EF%BC%89)
y=a(x-x₁)(x-x₂)的顶点坐标(x₁+x₂/2 ,)
y=a(x-x₁)(x-x₂)的顶点坐标(x₁+x₂/2 ,)
y=a(x-x₁)(x-x₂)的顶点坐标(x₁+x₂/2 ,)
将x=(x₁+x₂)/2代入y=a(x-x₁)(x-x₂)得:
y=a[(x₁+x₂)/2 - x₁]*[(x₁+x₂)/2 -x₂]
=a[[(x₂ -x₁)/2]*[(x₁-x₂)/2]
=-a(x₁-x₂)²/4
所以y=a(x-x₁)(x-x₂)的顶点坐标为( (x₁+x₂)/2 ,-a(x₁-x₂)²/4)