已知一交函數(shù)y=-2x+6的圖像與x軸交于點(diǎn)A,與y軸交于點(diǎn)C,二次函數(shù)y=ax²+bx+c（a≠0）的圖像過(guò)A,C
（1）當(dāng)S△ABC=4S△BOC時(shí),求拋物線y=ax²+bx+c的解析式和此函數(shù)頂點(diǎn)坐標(biāo).
（2）以O(shè)A的長(zhǎng)為直徑作圓M,試判定圓M與直線AC的位置關(guān)系,并說(shuō)明理由.
我找到了第一二問(wèn)的解法如下：
(1) A(3,0),C(0,6)
代入y=ax²+bx+c c = 6 b = -3a -2
y = ax² -(3a+2)x + 6
設(shè)B(p,0),3(A的橫坐標(biāo)),p為ax² -(3a+2)x + 6 = 0的解
y = ax² -(3a+2)x + 6 = a(x-p)(x-3) = ax² -(p+3)ax + 3ap
3a+2 = (p+3)a,ap = 2 S△ABC = 4S△BOC
△ABC 和△BOC高均|OC|,故|AB| = 4|OB |AB| = 3 - p
|OB| = 0-p = -p p = -1 a = 2/p = -2 ,b= -3a -2 = 4
得：拋物線的解析式:y = -2x² + 4x + 6
(2)以O(shè)A的長(zhǎng)為直徑作圓M,M(3/2,0),圓半徑 r = 3/2
M與y=-2x+6 (2x + y - 6 = 0)的距離為d = |(3/2)*2 + 0 - 6|/√(2²+1²) = 3/√5 = 3√5/5 < 3/2
圓M與直線AC相交.