一道函数题抛物线y=ax²+bx+c(a<0)交x轴于点A(-1,0)、B(3,0),交y轴于点C,顶点为D,

1个回答

  • (1)点A(-1,0)、B(3,0)代入抛物线得

    a - b+c=0···············(1)

    9a+3b+c=0···············(2)

    结合(1)(2)得 b=-2a c=-3a·····(3)

    因为抛物线y=a(x+b/2a)^2-(b^2-4ac)/4a ··(4)

    结合(3)(4) 得顶点 D(1,-4a)

    (2)因为BD是圆的直径 根据勾股定理有

    [-4a-(-3a)]^2 + 1^2 +(-3a)^2 + 3^2=(3-1)^2 + (-4a)^2

    => a^2 = 1

    => a =+-1

    又因为a