【答案】
分析:(1)分別利用待定系數(shù)法求兩函數(shù)的解析式:把A(3,0)B(0,-3)分別代入y=x
2+mx+n與y=kx+b,得到關(guān)于m、n的兩個(gè)方程組,解方程組即可;
(2)設(shè)點(diǎn)P的坐標(biāo)是(t,t-3),則M(t,t
2-2t-3),用P點(diǎn)的縱坐標(biāo)減去M的縱坐標(biāo)得到PM的長(zhǎng),即PM=(t-3)-(t
2-2t-3)=-t
2+3t,然后根據(jù)二次函數(shù)的最值得到
當(dāng)t=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164116916244483/SYS201310221641169162444025_DA/0.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164116916244483/SYS201310221641169162444025_DA/1.png)
時(shí),PM最長(zhǎng)為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164116916244483/SYS201310221641169162444025_DA/2.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164116916244483/SYS201310221641169162444025_DA/3.png)
,再利用三角形的面積公式利用S
△ABM=S
△BPM+S
△APM計(jì)算即可;
(3)由PM∥OB,根據(jù)平行四邊形的判定得到當(dāng)PM=OB時(shí),點(diǎn)P、M、B、O為頂點(diǎn)的四邊形為平行四邊形,然后討論:當(dāng)P在第四象限:PM=OB=3,PM最長(zhǎng)時(shí)只有
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164116916244483/SYS201310221641169162444025_DA/4.png)
,所以不可能;當(dāng)P在第一象限:PM=OB=3,(t
2-2t-3)-(t-3)=3;當(dāng)P在第三象限:PM=OB=3,t
2-3t=3,分別解一元二次方程即可得到滿(mǎn)足條件的t的值.
解答:解:(1)把A(3,0)B(0,-3)代入y=x
2+mx+n,得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164116916244483/SYS201310221641169162444025_DA/images5.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164116916244483/SYS201310221641169162444025_DA/5.png)
解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164116916244483/SYS201310221641169162444025_DA/6.png)
,
所以?huà)佄锞€(xiàn)的解析式是y=x
2-2x-3.
設(shè)直線(xiàn)AB的解析式是y=kx+b,
把A(3,0)B(0,-3)代入y=kx+b,得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164116916244483/SYS201310221641169162444025_DA/7.png)
,
解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164116916244483/SYS201310221641169162444025_DA/8.png)
,
所以直線(xiàn)AB的解析式是y=x-3;
(2)設(shè)點(diǎn)P的坐標(biāo)是(t,t-3),則M(t,t
2-2t-3),
因?yàn)閜在第四象限,
所以PM=(t-3)-(t
2-2t-3)=-t
2+3t,
當(dāng)t=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164116916244483/SYS201310221641169162444025_DA/9.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164116916244483/SYS201310221641169162444025_DA/10.png)
時(shí),二次函數(shù)的最大值,即PM最長(zhǎng)值為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164116916244483/SYS201310221641169162444025_DA/11.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164116916244483/SYS201310221641169162444025_DA/12.png)
,
則S
△ABM=S
△BPM+S
△APM=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164116916244483/SYS201310221641169162444025_DA/13.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164116916244483/SYS201310221641169162444025_DA/14.png)
.
(3)存在,理由如下:
∵PM∥OB,
∴當(dāng)PM=OB時(shí),點(diǎn)P、M、B、O為頂點(diǎn)的四邊形為平行四邊形,
①當(dāng)P在第四象限:PM=OB=3,PM最長(zhǎng)時(shí)只有
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164116916244483/SYS201310221641169162444025_DA/15.png)
,所以不可能有PM=3.
②當(dāng)P在第一象限:PM=OB=3,(t
2-2t-3)-(t-3)=3,解得t
1=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164116916244483/SYS201310221641169162444025_DA/16.png)
,t
2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164116916244483/SYS201310221641169162444025_DA/17.png)
(舍去),所以P點(diǎn)的橫坐標(biāo)是
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164116916244483/SYS201310221641169162444025_DA/18.png)
;
③當(dāng)P在第三象限:PM=OB=3,t
2-3t=3,解得t
1=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164116916244483/SYS201310221641169162444025_DA/19.png)
(舍去),t
2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164116916244483/SYS201310221641169162444025_DA/20.png)
,所以P點(diǎn)的橫坐標(biāo)是
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164116916244483/SYS201310221641169162444025_DA/21.png)
.
所以P點(diǎn)的橫坐標(biāo)是
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164116916244483/SYS201310221641169162444025_DA/22.png)
或
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164116916244483/SYS201310221641169162444025_DA/23.png)
.
點(diǎn)評(píng):本題考查了二次函數(shù)的綜合題:先利用待定系數(shù)法求函數(shù)的解析式,然后根據(jù)解析式表示點(diǎn)的坐標(biāo),再利用坐標(biāo)表示線(xiàn)段的長(zhǎng),利用二次函數(shù)的性質(zhì)求線(xiàn)段的最大值.同時(shí)考查了平行四邊形的判定定理以及一元二次方程的解法.