【答案】
分析:(1)直接把B(3,0)、C(0,-3)代入y=x
2+bx+c可得到關(guān)于b、c的方程組,解方程組求得b=-2,c=-3,則二次函數(shù)的表達(dá)式為y=x
2-2x-3;
(2)由于拋物線為軸對(duì)稱圖形,要得到四邊形ABPC為等腰梯形,只有PC∥AB,則點(diǎn)P與點(diǎn)C是拋物線的上的對(duì)稱點(diǎn),可求得拋物線的對(duì)稱軸為直線x=1,于是可得到點(diǎn)C(0,-3)關(guān)于直線x=1對(duì)稱的點(diǎn)P的坐標(biāo)為(2,-3).
(3)作OC的垂直平分線交直線BC下方的拋物線于點(diǎn)P,則PO=PC,根據(jù)翻折的性質(zhì)得OP′=OP,CP′=CP,易得四邊形POP′C為菱形,又E點(diǎn)坐標(biāo)為(0,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512124412334/SYS201311011915121244123025_DA/0.png)
),則點(diǎn)P的縱坐標(biāo)為-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512124412334/SYS201311011915121244123025_DA/1.png)
,再把y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512124412334/SYS201311011915121244123025_DA/2.png)
代入y=x
2-2x-3可求出對(duì)應(yīng)x的值,然后確定滿足條件的P點(diǎn)坐標(biāo).
解答:解:(1)把B(3,0)、C(0,-3)代入y=x
2+bx+c得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512124412334/SYS201311011915121244123025_DA/3.png)
,解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512124412334/SYS201311011915121244123025_DA/4.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512124412334/SYS201311011915121244123025_DA/images5.png)
∴這個(gè)二次函數(shù)的表達(dá)式為y=x
2-2x-3;
(2)∵點(diǎn)P是直線BC下方的拋物線上一動(dòng)點(diǎn),四邊形ABPC為等腰梯形,
∴PC∥AB,
∴點(diǎn)P與點(diǎn)C是拋物線的上的對(duì)稱點(diǎn),
∵拋物線的對(duì)稱軸為直線x=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512124412334/SYS201311011915121244123025_DA/5.png)
=1,
∴點(diǎn)C(0,-3)關(guān)于直線x=1對(duì)稱的點(diǎn)P的坐標(biāo)為(2,-3).
故答案為2,-3;
(3)存在.理由如下:
作OC的垂直平分線交直線BC下方的拋物線于點(diǎn)P,垂足為點(diǎn)E,如圖,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512124412334/SYS201311011915121244123025_DA/images7.png)
則PO=PC,
∵△POC沿CO翻折,得到四邊形POP′C,
∴OP′=OP,CP′=CP,
∴OP′=OP=CP′=CP,
∴四邊形POP′C為菱形,
∵C點(diǎn)坐標(biāo)為(0,-3),
∴E點(diǎn)坐標(biāo)為(0,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512124412334/SYS201311011915121244123025_DA/6.png)
),
∴點(diǎn)P的縱坐標(biāo)為-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512124412334/SYS201311011915121244123025_DA/7.png)
,
把y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512124412334/SYS201311011915121244123025_DA/8.png)
代入y=x
2-2x-3得x
2-2x-3=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512124412334/SYS201311011915121244123025_DA/9.png)
,
解得x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512124412334/SYS201311011915121244123025_DA/10.png)
,
∵點(diǎn)P在直線BC下方的拋物線上,
∴x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512124412334/SYS201311011915121244123025_DA/11.png)
,
∴滿足條件的點(diǎn)P的坐標(biāo)為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512124412334/SYS201311011915121244123025_DA/12.png)
,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512124412334/SYS201311011915121244123025_DA/13.png)
).
點(diǎn)評(píng):本題考查了二次函數(shù)綜合題:二次函數(shù)y=ax
2+bx+c(a、b、c為常數(shù),a≠0)的圖象為拋物線,其頂點(diǎn)式為y=a(x-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512124412334/SYS201311011915121244123025_DA/14.png)
)
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512124412334/SYS201311011915121244123025_DA/15.png)
,拋物線的對(duì)稱軸為x=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512124412334/SYS201311011915121244123025_DA/16.png)
,當(dāng)a>0,y
最小值=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512124412334/SYS201311011915121244123025_DA/17.png)
;當(dāng)a<0,y
最,大值=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191512124412334/SYS201311011915121244123025_DA/18.png)
;拋物線上的點(diǎn)的橫縱坐標(biāo)滿足拋物線的解析式;對(duì)于特殊四邊形的判定與性質(zhì)以及勾股定理要熟練運(yùn)用.