待定系数法求二次函数解析式

1yax2bxc2ya(xh)2k3xxya(x1)(x2)12x1A10B45C03214233x1030231.1913352.=2=6,=2=10,=3=24,3.121104.41035.=32016.=244127.5020348.1849.x6612810y=x-3xAyBABx=1111=ax2+bx+c2=mx+nA-25B14yy=2x+31242.53.53.0515(1)(2)1.80.2513ABCB90AB12cmBC24cmPAABB2cm/sQBBCCcm/sPQABPBQStt1416m40m()(15AOBAOB90AOBOA3,11B2AOB3B1AB1B