如图,已知抛物线与x轴交于A(-1,0),B(3,0)两点,与y轴交于点C（0,3） （1）求抛物线的解析式

(1) y = ax^2 + bx + c代入A,B,C的坐标:A:a -b + c = 0B:9a + 3b + c = 0C:c = 3a = -1,b = 2,c = 3y = -x^2 + 2x + 3(2) y = -x^2 + 2x + 3 = -(x - 1)^2 +4D(1,4)抛物线对称轴为x = 1P是C的以x = 1为对称轴的对称点时,|DC| = |DP|,△PDC是等腰三角形P的横坐标为2,P(2,3)(3) BC斜率k1 = (3-0)/(0-3) = -1CD斜率k2 = (4-3)/(1-0) = 1k1*k2 = -1BC与CD垂直,DM与CB平行,DM斜率-1,方程为 y - 4 = -(x -1) (点斜式)y = -x + 5 = -x^2 + 2x + 3x^2 -3x + 2 = 0(x-1)(x-2) = 0x = 1 (顶点D,舍去)x = 2与抛物线交点M(2,3)