【答案】
分析:(1)已知拋物線的解析式,根據(jù)頂點公式,可求出A點的坐標(biāo)(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574578350/SYS201311031011295745783027_DA/0.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574578350/SYS201311031011295745783027_DA/1.png)
)且a=1,b=4,c=0.
∵y=x
2+4x=(x+2)
2-4,∴A(-2,-4).
(2)若ABOP為菱形時,根據(jù)菱形的性質(zhì),則P點橫坐標(biāo)與A坐標(biāo)相同,然后再代入直線就可求出縱坐標(biāo),則P坐標(biāo)就求出;若ABOP為等腰梯形時,OA=BP,已知O,A坐標(biāo),可求出OA長度,設(shè)P橫坐標(biāo)為a,P在直線上,可用a表示出坐標(biāo),從而求出BP長度,OA=BP,可求出a的值,即求出P坐標(biāo).若ABOP為直角梯形時,BP與AB垂直,可求出直線BP的關(guān)系式,直線BP與直線l的交點即P點坐標(biāo).
(3)首先可以得出l的解析式.據(jù)圖分析有兩種情況可以構(gòu)成QABP為四邊形,即當(dāng)P在第二象限時和在第四象限時,當(dāng)P在第二象限時,四邊形由△AOB和△POB組成,△AOB面積確定,則△POB的面積可以求出來,由于△AOB+△POB代入到面積的不等式中可以得出x的取值范圍.同理當(dāng)P在第四象限時,△AOB+△AOP代入到面積不等式中可以得到x的取值范圍.
得AB所在直線的函數(shù)關(guān)系式是y=-2x-8,所以直線l對應(yīng)的函數(shù)關(guān)系式為y=-2x.
設(shè)點P坐標(biāo)為(x,-2x),分別討論點P在第二象限以及第四象限的值.
解答:![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574578350/SYS201311031011295745783027_DA/images2.png)
解:(1)∵y=x
2+4x=(x+2)
2-4,(1分)
∴A(-2,-4).(2分)
(2)由已知條件可求得AB所在直線的函數(shù)關(guān)系式是y=-2x-8,
所以直線l對應(yīng)的函數(shù)關(guān)系式為y=-2x.
當(dāng)四邊形ABOP是菱形時,P點橫坐標(biāo)與A點橫坐標(biāo)相同,縱坐標(biāo)與A點坐標(biāo)互為相反數(shù),四邊形ABP
1O為菱形時,P
1(-2,4);
四邊形ABOP
2為等腰梯形時,設(shè)P
2橫坐標(biāo)為a,將x=a代入y=-2x,得
P
2(a,-2a).
又∵AO=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574578350/SYS201311031011295745783027_DA/2.png)
=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574578350/SYS201311031011295745783027_DA/3.png)
,
∴P
2B=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574578350/SYS201311031011295745783027_DA/4.png)
,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574578350/SYS201311031011295745783027_DA/5.png)
=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574578350/SYS201311031011295745783027_DA/6.png)
,
整理得,5a
2+8a-4=0,
解得,a=-2(舍去),a=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574578350/SYS201311031011295745783027_DA/7.png)
,故P
2(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574578350/SYS201311031011295745783027_DA/8.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574578350/SYS201311031011295745783027_DA/9.png)
);
ABOP為直角梯形時,BP
3與AB垂直,則直線BP的解析式為y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574578350/SYS201311031011295745783027_DA/10.png)
x+b,
把B(-4,0)代入解析式得,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574578350/SYS201311031011295745783027_DA/11.png)
×(-4)+b=0,
解得b=2.
直線BP的解析式為y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574578350/SYS201311031011295745783027_DA/12.png)
x+2,
故得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574578350/SYS201311031011295745783027_DA/13.png)
,
解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574578350/SYS201311031011295745783027_DA/14.png)
,
四邊形ABP
3O為直角梯形時,P
3(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574578350/SYS201311031011295745783027_DA/15.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574578350/SYS201311031011295745783027_DA/16.png)
);
同理,當(dāng)AP
4垂直于AB時,四邊形ABOP
4為直角梯形,P
4(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574578350/SYS201311031011295745783027_DA/17.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574578350/SYS201311031011295745783027_DA/18.png)
).(6分)
(3)設(shè)點P坐標(biāo)為(x,-2x).
①當(dāng)點P在第二象限時,x<0,
△POB的面積S
△POB=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574578350/SYS201311031011295745783027_DA/19.png)
×4×(-2x)=-4x.
∵△AOB的面積S
△AOB=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574578350/SYS201311031011295745783027_DA/20.png)
×4×4=8,
∴S=S
△AOB+S
△POB=-4x+8(x<0).(8分)
∵4+6
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574578350/SYS201311031011295745783027_DA/21.png)
≤S≤6+8
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574578350/SYS201311031011295745783027_DA/22.png)
,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574578350/SYS201311031011295745783027_DA/23.png)
即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574578350/SYS201311031011295745783027_DA/24.png)
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574578350/SYS201311031011295745783027_DA/25.png)
∴x的取值范圍是
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574578350/SYS201311031011295745783027_DA/26.png)
.(9分)
②當(dāng)點P在第四象限時,x>0,過點A、P分別作x軸的垂線,垂足為A′、P′.則四邊形POA′A的面積S
POA′A=S
梯形PP′A′A-S
△PP′O=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574578350/SYS201311031011295745783027_DA/27.png)
•(x+2)-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574578350/SYS201311031011295745783027_DA/28.png)
•(2x)•x=4x+4.
∵△AA′B的面積S
△AA′B=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574578350/SYS201311031011295745783027_DA/29.png)
×4×2=4,
∴S=S
POA′A+S
△AA′B=4x+8(x>0).(10分)
∵4+6
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574578350/SYS201311031011295745783027_DA/30.png)
≤S≤6+8
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574578350/SYS201311031011295745783027_DA/31.png)
,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574578350/SYS201311031011295745783027_DA/32.png)
即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574578350/SYS201311031011295745783027_DA/33.png)
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574578350/SYS201311031011295745783027_DA/34.png)
∴x的取值范圍是
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574578350/SYS201311031011295745783027_DA/35.png)
≤x≤
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574578350/SYS201311031011295745783027_DA/36.png)
.(11分)
點評:該題首先是考查了拋物線函數(shù)的特性,要求掌握拋物線函數(shù)的特點.其次是利用不等式通過動態(tài)點的變化來加深了解拋物線曲線和一次函數(shù)的關(guān)系.