答:y=-x²+3x+4=-(x-4)(x+1)交点A(4,0),B(-1,0),C(0,4)直线AC为y=-x+4与AC垂直的直线斜率k=11)当CP⊥AC时:CP直线为y=x+4,与抛物线联立:y=x+4=-x²+3x+4,x²-2x=0,x=2点P(2,6)2)当AP⊥AC时:AP直线为y=x-4,与抛物线联立:y=x-4=-x²+3x+4,x²-2x-8=0,x=-2点P为(-2,-6)综上所述,点P为(2,6)或者(-2,-6)
(-2,-6)或(2,6)