(1)设抛物线解析式为y=a(x+1)(x-3),
∵抛物线过点(0,3),
∴-3=a(0+1)(0-3),
∴a=1,
∴抛物线解析式为y=(x+1)(x-3)=x
2-2x-3,
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/c2fdfc039245d6889f350715a7c27d1ed21b244f?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
∵y=x
2-2x-3=(x-1)
2-4,
∴M(1,-4).
(2)如图1,连接BC、BM、CM,作MD⊥x轴于D,
∵S
△BCM=S
梯形OCMD+S
△BMD-S
△BOC=
×(3+4)×1+
×2×4-
×3×3
=
+
-
=3
S
△ABC=
?AB?OC=
×4×3=6,
∴S
△BCM:S
△ABC=3:6=1:2.