Question

如图,抛物线Y=-1/4x^2+x+3与x轴交与点A、B,与Y轴交于点C,顶点为D,对称轴l与直线

Answer 1

(1)解析：∵y=-1/4x^2+x+3∴C(0,3),A(-2,0),B(6,0)∴BC 方程为y=-1/2(x-6)=-1/2x+3(2)解析：由抛物线可求出D(2,4)D到直线BC的距离d=|2+2*4-6|/√5=4√5/5∴圆半径r的取值范围为r>=4√5/5(3)解析：∵r=5设P(x,y) d=|x+2y-6|/√5=|3x-1/2x^2|/√5当0