【答案】
分析:(1)根據(jù)-x
2+2x+3=0,解得x
1=3、x
2=-1,即點(diǎn)A(-1,0),B(3,0),根據(jù)拋物線(xiàn)y=-x
2+2x+3交y軸于點(diǎn)C,可知當(dāng)x=0時(shí),y=3,所以C(0,3)
(2)拋物線(xiàn)y=-x
2+2x+3的點(diǎn)頂為M,根據(jù)頂點(diǎn)公式可知M(1,4),過(guò)點(diǎn)M作ME⊥AB于E,則ME=4,OE=1,BE=2,OC=3,所以S
△BCM=S
四邊形COBM-S
△BOC=3
(3)分情況討論,共有4個(gè)點(diǎn).
(1)以AC為腰:
①當(dāng)以點(diǎn)A為圓心,AC長(zhǎng)為半徑畫(huà)弧交x軸于點(diǎn)P
1,p
2(p
1在p
2的右側(cè))
可知P
1(
,0)P
2(-
,0),交y軸于一點(diǎn)p
5;②以點(diǎn)C為圓心,AC為半徑畫(huà)弧交x軸于點(diǎn)P
3,點(diǎn)P
3與點(diǎn)A關(guān)于y軸對(duì)稱(chēng),則點(diǎn)P
3坐標(biāo)為(1,0),交y軸于兩點(diǎn)p
6,p
7,
(2)以AC為底邊:作AC的垂直平分線(xiàn)交x軸于點(diǎn)p
4垂足為F,利用△AOC∽△AFP
4可求AP
4=5,OP
4=5-1=4,所以P
4(4,0).
解答:解:(1)∵拋物線(xiàn)y=-x
2+2x+3交x軸于A,B兩點(diǎn)
∴-x
2+2x+3=0,
解得x
1=3,x
2=-1
∴點(diǎn)A(-1,0),B(3,0)
又∵拋物線(xiàn)y=-x
2+2x+3交y軸于點(diǎn)C,
∴點(diǎn)C(0,3)
(2)∵拋物線(xiàn)y=-x
2+2x+3的頂點(diǎn)為M
∴x=
=1
y=
∴M(1,4)
過(guò)點(diǎn)M作ME⊥AB于E,則ME=4,OE=1,
∴BE=OB-OE=3-1=2,OC=3
∴S
△BCM=S
△△BOC=3.
(3)存在點(diǎn)P
1)以AC為腰:
①當(dāng)以點(diǎn)A為圓心,AC長(zhǎng)為半徑畫(huà)弧交x軸于點(diǎn)P
1,p
2(p
1在p
2的右側(cè))
AC=
=
,
∴P
1O=
,P
2O=
∴P
1(
,0)P
2(-
,0)
交y軸于p
5與C點(diǎn)關(guān)于x軸對(duì)稱(chēng),坐標(biāo)為:(0,-3)
②以點(diǎn)C為圓心,AC為半徑畫(huà)弧交x軸于點(diǎn)P
3∴點(diǎn)P
3與點(diǎn)A關(guān)于y軸對(duì)稱(chēng),則點(diǎn)P
3坐標(biāo)為(1,0),
交y軸于點(diǎn)p
6,p
7兩點(diǎn),p
6(0,3-
),p
7(0,3+
)
2)以AC為底邊:作AC的垂直平分線(xiàn)交x軸于點(diǎn)p
4垂足為F,則AF=
∵∠AFP
4=∠AOC=90°
∠CAO=∠P
4AF
∴△AOC∽△AFP
4∴
∴
=
∴AP
4=5,
∴OP
4=5-1=4,
∴P
4(4,0)
∴點(diǎn)P的坐標(biāo)為:P
1(
,0)P
2(-
,0)P
3(1,0),P
4(4,0),p
5(0,-3),p
6(0,
-3),p
7(0,3+
).
點(diǎn)評(píng):本題是二次函數(shù)的綜合題型,其中涉及到的知識(shí)點(diǎn)有拋物線(xiàn)的頂點(diǎn)公式和三角形的面積求法.在求有關(guān)動(dòng)點(diǎn)問(wèn)題時(shí)要注意分析題意分情況討論結(jié)果.