【答案】
分析:(1)由題意可知拋物線的頂點(diǎn)就是A點(diǎn),因此可將A的坐標(biāo)代入拋物線的解析式中,并根據(jù)對稱軸x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191457923055386/SYS201311011914579230553025_DA/0.png)
=1,聯(lián)立方程組即可求出a,c的值,進(jìn)而可得出拋物線的解析式.
(2)四邊形OPEF是個(gè)直角梯形,可先求出AD,AB所在直線的解析式,根據(jù)AD所在直線的解析式設(shè)出P的坐標(biāo),又由于PE∥x軸,P、E兩點(diǎn)的縱坐標(biāo)相同,然后根據(jù)AB所在直線的解析式得出E點(diǎn)的坐標(biāo),進(jìn)而可求出F點(diǎn)的坐標(biāo).根據(jù)求出的P、E、F三點(diǎn)坐標(biāo),可得出梯形的上下底OF、EP的長以及直角梯形的高EF的長(即E點(diǎn)縱坐標(biāo)的絕對值),根據(jù)梯形的面積公式即可得出關(guān)于梯形的面積與P點(diǎn)坐標(biāo)的函數(shù)解析式,然后將S=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191457923055386/SYS201311011914579230553025_DA/1.png)
代入函數(shù)中即可求出P點(diǎn)的坐標(biāo).
解答:![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191457923055386/SYS201311011914579230553025_DA/images2.png)
解:(1)由題意,知點(diǎn)A(1,-4)是拋物線的頂點(diǎn),
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191457923055386/SYS201311011914579230553025_DA/2.png)
∴a=1,c=-3,
∴拋物線的函數(shù)關(guān)系式為y=x
2-2x-3.
(2)由(1)知,點(diǎn)C的坐標(biāo)是(0,-3).
設(shè)直線AC的函數(shù)關(guān)系式為y=kx+b,
則
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191457923055386/SYS201311011914579230553025_DA/3.png)
∴b=-3,k=-1,
∴y=-x-3.
由y=x
2-2x-3=0,得x
1=-1,x
2=3,
∴點(diǎn)B的坐標(biāo)是(3,0).
設(shè)直線AB的函數(shù)關(guān)系式是y=mx+n,
則
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191457923055386/SYS201311011914579230553025_DA/4.png)
解得m=2,n=-6.
∴直線AB的函數(shù)關(guān)系式是y=2x-6.
設(shè)P點(diǎn)坐標(biāo)為(x
P,y
P),則y
P=-x
P-3.
∵PE∥x軸,
∴E點(diǎn)的縱坐標(biāo)也是-x
P-3.
設(shè)E點(diǎn)坐標(biāo)為(x
E,y
E),
∵點(diǎn)E在直線AB上,
∴-x
P-3=2x
E-6,
∴x
E=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191457923055386/SYS201311011914579230553025_DA/5.png)
.
∵EF⊥x軸,
∴F點(diǎn)的坐標(biāo)為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191457923055386/SYS201311011914579230553025_DA/6.png)
,0),
∴PE=x
E-x
P=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191457923055386/SYS201311011914579230553025_DA/7.png)
,OF=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191457923055386/SYS201311011914579230553025_DA/8.png)
,EF=-(-x
P-3)=x
P+3,
∴S
四邊形OPEF=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191457923055386/SYS201311011914579230553025_DA/9.png)
(PE+OF)•EF=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191457923055386/SYS201311011914579230553025_DA/10.png)
(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191457923055386/SYS201311011914579230553025_DA/11.png)
+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191457923055386/SYS201311011914579230553025_DA/12.png)
)•(x
P+3)=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191457923055386/SYS201311011914579230553025_DA/13.png)
,
2x
P2+3x
P-2=0,
∴x
P=-2,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191457923055386/SYS201311011914579230553025_DA/14.png)
,
當(dāng)y=0時(shí),x=-3,
而-3<-2<1,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191457923055386/SYS201311011914579230553025_DA/15.png)
,
∴P點(diǎn)坐標(biāo)為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191457923055386/SYS201311011914579230553025_DA/16.png)
和(-2,-1)
點(diǎn)評:本題著重考查了待定系數(shù)法求二次函數(shù)解析式及二次函數(shù)的綜合應(yīng)用.