【答案】
分析:(1)已知拋物線與x軸的兩交點(diǎn)的橫坐標(biāo)分別是-3和1,設(shè)拋物線解析式的交點(diǎn)式y(tǒng)=a(x+3)(x-1),再配方為頂點(diǎn)式,可確定頂點(diǎn)坐標(biāo);
(2)①設(shè)AC與拋物線對(duì)稱軸的交點(diǎn)為E,先運(yùn)用待定系數(shù)法求出直線AC的解析式,求出點(diǎn)E的坐標(biāo),即可得到DE的長,然后由S
△ACD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193429807151173/SYS201311011934298071511023_DA/0.png)
×DE×OA列出方程,解方程求出a的值,即可確定拋物線的解析式;
②先運(yùn)用勾股定理的逆定理判斷出在△ACD中∠ACD=90°,利用三角函數(shù)求出tan∠DAC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193429807151173/SYS201311011934298071511023_DA/1.png)
.設(shè)y=-x
2-2x+3=-(x+1)
2+4向右平移后的拋物線解析式為y=-(x+m)
2+4,兩條拋物線交于點(diǎn)P,直線AP與y軸交于點(diǎn)F.根據(jù)正切函數(shù)的定義求出OF=1.分兩種情況進(jìn)行討論:(Ⅰ)如圖2①,F(xiàn)點(diǎn)的坐標(biāo)為(0,1),(Ⅱ)如圖2②,F(xiàn)點(diǎn)的坐標(biāo)為(0,-1).針對(duì)這兩種情況,都可以先求出點(diǎn)P的坐標(biāo),再得出m的值,進(jìn)而求出平移后拋物線的解析式.
解答:解:(1)∵拋物線y=ax
2+bx+c與x軸交于點(diǎn)A(-3,0)和點(diǎn)B(1,0),
∴拋物線解析式為y=a(x+3)(x-1)=ax
2+2ax-3a,
∵y=a(x+3)(x-1)=a(x
2+2x-3)=a(x+1)
2-4a,
∴頂點(diǎn)D的坐標(biāo)為(-1,-4a);
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193429807151173/SYS201311011934298071511023_DA/images2.png)
(2)如圖1,①設(shè)AC與拋物線對(duì)稱軸的交點(diǎn)為E.
∵拋物線y=ax
2+2ax-3a與y軸交于點(diǎn)C,
∴C點(diǎn)坐標(biāo)為(0,-3a).
設(shè)直線AC的解析式為:y=kx+t,
則:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193429807151173/SYS201311011934298071511023_DA/2.png)
,
解得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193429807151173/SYS201311011934298071511023_DA/3.png)
,
∴直線AC的解析式為:y=-ax-3a,
∴點(diǎn)E的坐標(biāo)為:(-1,-2a),
∴DE=-4a-(-2a)=-2a,
∴S
△ACD=S
△CDE+S
△ADE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193429807151173/SYS201311011934298071511023_DA/4.png)
×DE×OA=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193429807151173/SYS201311011934298071511023_DA/5.png)
×(-2a)×3=-3a,
∴-3a=3,解得a=-1,
∴拋物線的解析式為y=-x
2-2x+3;
②∵y=-x
2-2x+3,
∴頂點(diǎn)D的坐標(biāo)為(-1,4),C(0,3),
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193429807151173/SYS201311011934298071511023_DA/images7.png)
∵A(-3,0),
∴AD
2=(-1+3)
2+(4-0)
2=20,CD
2=(-1-0)
2+(4-3)
2=2,AC
2=(0+3)
2+(3-0)
2=18,
∴AD
2=CD
2+AC
2,
∴∠ACD=90°,
∴tan∠DAC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193429807151173/SYS201311011934298071511023_DA/6.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193429807151173/SYS201311011934298071511023_DA/7.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193429807151173/SYS201311011934298071511023_DA/8.png)
,
∵∠PAB=∠DAC,
∴tan∠PAB=tan∠DAC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193429807151173/SYS201311011934298071511023_DA/9.png)
.
如圖2,設(shè)y=-x
2-2x+3=-(x+1)
2+4向右平移后的拋物線解析式為y=-(x+m)
2+4,兩條拋物線交于點(diǎn)P,直線AP與y軸交于點(diǎn)F.
∵tan∠PAB=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193429807151173/SYS201311011934298071511023_DA/10.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193429807151173/SYS201311011934298071511023_DA/11.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193429807151173/SYS201311011934298071511023_DA/12.png)
,
∴OF=1,則F點(diǎn)的坐標(biāo)為(0,1)或(0,-1).
分兩種情況:
(Ⅰ)如圖2①,當(dāng)F點(diǎn)的坐標(biāo)為(0,1)時(shí),易求直線AF的解析式為y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193429807151173/SYS201311011934298071511023_DA/13.png)
x+1,
由
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193429807151173/SYS201311011934298071511023_DA/14.png)
,解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193429807151173/SYS201311011934298071511023_DA/15.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193429807151173/SYS201311011934298071511023_DA/16.png)
(舍去),
∴P點(diǎn)坐標(biāo)為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193429807151173/SYS201311011934298071511023_DA/17.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193429807151173/SYS201311011934298071511023_DA/18.png)
),
將P點(diǎn)坐標(biāo)(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193429807151173/SYS201311011934298071511023_DA/19.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193429807151173/SYS201311011934298071511023_DA/20.png)
)代入y=-(x+m)
2+4,
得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193429807151173/SYS201311011934298071511023_DA/21.png)
=-(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193429807151173/SYS201311011934298071511023_DA/22.png)
+m)
2+4,
解得m
1=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193429807151173/SYS201311011934298071511023_DA/23.png)
,m
2=1(舍去),
∴平移后拋物線的解析式為y=-(x-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193429807151173/SYS201311011934298071511023_DA/24.png)
)
2+4;
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193429807151173/SYS201311011934298071511023_DA/images27.png)
(Ⅱ)如圖2②,當(dāng)F點(diǎn)的坐標(biāo)為(0,-1)時(shí),易求直線AF的解析式為y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193429807151173/SYS201311011934298071511023_DA/25.png)
x-1,
由
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193429807151173/SYS201311011934298071511023_DA/26.png)
,解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193429807151173/SYS201311011934298071511023_DA/27.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193429807151173/SYS201311011934298071511023_DA/28.png)
(舍去),
∴P點(diǎn)坐標(biāo)為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193429807151173/SYS201311011934298071511023_DA/29.png)
,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193429807151173/SYS201311011934298071511023_DA/30.png)
),
將P點(diǎn)坐標(biāo)(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193429807151173/SYS201311011934298071511023_DA/31.png)
,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193429807151173/SYS201311011934298071511023_DA/32.png)
)代入y=-(x+m)
2+4,
得-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193429807151173/SYS201311011934298071511023_DA/33.png)
=-(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193429807151173/SYS201311011934298071511023_DA/34.png)
+m)
2+4,
解得m
1=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193429807151173/SYS201311011934298071511023_DA/35.png)
,m
2=1(舍去),
∴平移后拋物線的解析式為y=-(x-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193429807151173/SYS201311011934298071511023_DA/36.png)
)
2+4;
綜上可知,平移后拋物線的解析式為y=-(x-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193429807151173/SYS201311011934298071511023_DA/37.png)
)
2+4或y=-(x-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193429807151173/SYS201311011934298071511023_DA/38.png)
)
2+4.
點(diǎn)評(píng):此題是二次函數(shù)的綜合題,考查了待定系數(shù)法求函數(shù)的解析式,二次函數(shù)的性質(zhì),勾股定理的逆定理,三角函數(shù)的定義,三角形的面積、兩函數(shù)交點(diǎn)坐標(biāo)的求法,函數(shù)平移的規(guī)律等知識(shí),綜合性較強(qiáng),有一定難度,解題的關(guān)鍵是方程思想、數(shù)形結(jié)合思想與分類討論思想的應(yīng)用.