(1)令x2+4x+3=0,即(x+1)(x+3)=0,解得x=-1,x=-3.故A(-3,0),B(-1,0);因为y=x2+4x+3=x2+4x+4-1=(x+2)2-1,故顶点坐标为P(-2,-1).(2)如图当-3<x<0时,-1<y<3.