【答案】
分析:(1)由平行四邊形的性質(zhì)和點(diǎn)A、B的坐標(biāo)便可求出C點(diǎn)坐標(biāo),將C點(diǎn)坐標(biāo)代入正比例函數(shù)即可求得直線l的解析式;
(2)根據(jù)題意,得OP=t,AQ=2t,根據(jù)t的取值范圍不同分三種情況分別進(jìn)行討論,得到三種S關(guān)于t的函數(shù),解題時(shí)注意t的取值范圍;
(3)分別根據(jù)三種函數(shù)解析式求出當(dāng)t為何值時(shí),S最大,然后比較三個(gè)最大值,可知當(dāng)t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190754070841345/SYS201311011907540708413027_DA/0.png)
時(shí),S有最大值,最大值為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190754070841345/SYS201311011907540708413027_DA/1.png)
;
(4)根據(jù)題意并細(xì)心觀察圖象,分兩種情況討論可知:當(dāng)t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190754070841345/SYS201311011907540708413027_DA/2.png)
時(shí),△QMN為等腰三角形.
解答:解:(1)由題意知:點(diǎn)A的坐標(biāo)為(8,0),點(diǎn)B的坐標(biāo)為(11.4),
且OA=BC,故C點(diǎn)坐標(biāo)為C(3,4),
設(shè)直線l的解析式為y=kx,
將C點(diǎn)坐標(biāo)代入y=kx,
解得k=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190754070841345/SYS201311011907540708413027_DA/3.png)
,
∴直線l的解析式為y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190754070841345/SYS201311011907540708413027_DA/4.png)
x;
故答案為:(3,4),y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190754070841345/SYS201311011907540708413027_DA/5.png)
x;
(2)根據(jù)題意,得OP=t,AQ=2t.分三種情況討論:
①當(dāng)0<t≤
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190754070841345/SYS201311011907540708413027_DA/6.png)
時(shí),如圖1,M點(diǎn)的坐標(biāo)是(t,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190754070841345/SYS201311011907540708413027_DA/7.png)
t).
過(guò)點(diǎn)C作CD⊥x軸于D,過(guò)點(diǎn)Q作QE⊥x軸于E,可得△AEQ∽△ODC,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190754070841345/SYS201311011907540708413027_DA/8.png)
,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190754070841345/SYS201311011907540708413027_DA/9.png)
,
∴AE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190754070841345/SYS201311011907540708413027_DA/10.png)
,EQ=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190754070841345/SYS201311011907540708413027_DA/11.png)
t,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190754070841345/SYS201311011907540708413027_DA/images12.png)
∴Q點(diǎn)的坐標(biāo)是(8+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190754070841345/SYS201311011907540708413027_DA/12.png)
t,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190754070841345/SYS201311011907540708413027_DA/13.png)
t),
∴PE=8+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190754070841345/SYS201311011907540708413027_DA/14.png)
t,
∴S=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190754070841345/SYS201311011907540708413027_DA/15.png)
t,
②當(dāng)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190754070841345/SYS201311011907540708413027_DA/16.png)
<t≤3時(shí),如圖2,過(guò)點(diǎn)Q作QF⊥x軸于F,
∵BQ=2t-5,
∴OF=11-(2t-5)=16-2t,
∴Q點(diǎn)的坐標(biāo)是(16-2t,4),
∴PF=16-2t-t=16-3t,
∴S=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190754070841345/SYS201311011907540708413027_DA/17.png)
t,
③當(dāng)點(diǎn)Q與點(diǎn)M相遇時(shí),16-2t=t,解得t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190754070841345/SYS201311011907540708413027_DA/18.png)
.
當(dāng)3<t<
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190754070841345/SYS201311011907540708413027_DA/19.png)
時(shí),如圖3,MQ=16-2t-t=16-3t,MP=4.
S=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190754070841345/SYS201311011907540708413027_DA/20.png)
•4•(16-3t)=-6t+32,
所以S=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190754070841345/SYS201311011907540708413027_DA/21.png)
;
(3)①當(dāng)0<t≤
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190754070841345/SYS201311011907540708413027_DA/22.png)
時(shí),S=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190754070841345/SYS201311011907540708413027_DA/23.png)
,
∵a=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190754070841345/SYS201311011907540708413027_DA/24.png)
>0,拋物線開(kāi)口向上,t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190754070841345/SYS201311011907540708413027_DA/25.png)
時(shí),最大值為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190754070841345/SYS201311011907540708413027_DA/26.png)
;
②當(dāng)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190754070841345/SYS201311011907540708413027_DA/27.png)
<t≤3時(shí),S=-2t
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190754070841345/SYS201311011907540708413027_DA/28.png)
.
∵a=-2<0,拋物線開(kāi)口向下.
∴當(dāng)t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190754070841345/SYS201311011907540708413027_DA/29.png)
時(shí),S有最大值,最大值為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190754070841345/SYS201311011907540708413027_DA/30.png)
.
③當(dāng)3<t<
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190754070841345/SYS201311011907540708413027_DA/31.png)
時(shí),S=-6t+32,
∵k=-6<0.
∴S隨t的增大而減小.
又∵當(dāng)t=3時(shí),S=14.當(dāng)t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190754070841345/SYS201311011907540708413027_DA/32.png)
時(shí),S=0.
∴0<S<14.
綜上所述,當(dāng)t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190754070841345/SYS201311011907540708413027_DA/33.png)
時(shí),S有最大值,最大值為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190754070841345/SYS201311011907540708413027_DA/34.png)
.
(4)當(dāng)M點(diǎn)在線段CB上運(yùn)動(dòng)時(shí),點(diǎn)Q一定在線段CB上,
①點(diǎn)Q在點(diǎn)M右側(cè),QM=xQ-xM=16-2t-t=16-3t,NM=NP-MP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190754070841345/SYS201311011907540708413027_DA/35.png)
t-4
則有16-3t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190754070841345/SYS201311011907540708413027_DA/36.png)
t-4 解得t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190754070841345/SYS201311011907540708413027_DA/37.png)
;
②點(diǎn)Q在點(diǎn)M左側(cè),QM=xM-xQ=3t-16,NM=NP-MP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190754070841345/SYS201311011907540708413027_DA/38.png)
t-4
則有3t-16=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190754070841345/SYS201311011907540708413027_DA/39.png)
t-4 解得t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190754070841345/SYS201311011907540708413027_DA/40.png)
但是,點(diǎn)Q的運(yùn)動(dòng)時(shí)間為(5+8)÷2=6.5秒,故將②舍去.
當(dāng)t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190754070841345/SYS201311011907540708413027_DA/41.png)
時(shí),△QMN為等腰三角形.
點(diǎn)評(píng):本題是二次函數(shù)的綜合題,其中涉及到的知識(shí)點(diǎn)有拋物線最大值的求法和動(dòng)點(diǎn)問(wèn)題等知識(shí)點(diǎn),是各地中考的熱點(diǎn)和難點(diǎn),解題時(shí)注意數(shù)形結(jié)合和分類討論等數(shù)學(xué)思想的運(yùn)用,同學(xué)們要加強(qiáng)訓(xùn)練,屬于難題.