【答案】
分析:(1)根據(jù)待定系數(shù)法求直線BC的解析式即可;把點(diǎn)B、C的坐標(biāo)代入二次函數(shù),利用待定系數(shù)法求函數(shù)解析式解答;
(2)根據(jù)拋物線解析式求出頂點(diǎn)D的坐標(biāo),再根據(jù)二次函數(shù)的對(duì)稱性求出點(diǎn)A的坐標(biāo),連接AD,然后求出∠ADP=∠ABC=45°,然后證明△ADP和△ABC相似,根據(jù)相似三角形對(duì)應(yīng)邊成比例列出比例式求出PD的長(zhǎng)度,從而得解;
(3)連接BD,利用勾股定理求出BD、BC的長(zhǎng)度,再求出∠CBD=90°,然后根據(jù)∠BCD與∠ACO的正切值相等可得∠BCD=∠ACO,從而得到∠OCA與∠OCD的和等于∠BCO,是45°.
解答:解:(1)設(shè)直線BC的解析式為y=kx+m,
∵點(diǎn)B(-3,0),點(diǎn)C(0,-3),
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191103171899349/SYS201311011911031718993025_DA/0.png)
,
解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191103171899349/SYS201311011911031718993025_DA/1.png)
,
所以,直線BC的解析式為y=-x-3,
∵二次函數(shù)y=-x
2+bx+c的圖象經(jīng)過(guò)點(diǎn)B(-3,0),點(diǎn)C(0,-3),
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191103171899349/SYS201311011911031718993025_DA/2.png)
,
解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191103171899349/SYS201311011911031718993025_DA/3.png)
,
∴二次函數(shù)的解析式為y=-x
2-4x-3;
(2)∵y=-x
2-4x-3=-(x+2)
2+1,
∴拋物線的頂點(diǎn)D(-2,1),對(duì)稱軸為x=-2,
∵A、B關(guān)于對(duì)稱軸對(duì)稱,點(diǎn)B(-3,0),
∴點(diǎn)A的坐標(biāo)為(-1,0),
AB=-1-(-3)=-1+3=2,
BC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191103171899349/SYS201311011911031718993025_DA/4.png)
=3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191103171899349/SYS201311011911031718993025_DA/5.png)
,
連接AD,則AD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191103171899349/SYS201311011911031718993025_DA/6.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191103171899349/SYS201311011911031718993025_DA/7.png)
,
tan∠ADP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191103171899349/SYS201311011911031718993025_DA/8.png)
=1,
∴∠ADP=45°,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191103171899349/SYS201311011911031718993025_DA/images9.png)
又∵B(-3,0),C(0,-3),
∴△OAC是等腰直角三角形,
∴∠ABC=45°,
∴∠ADP=∠ABC=45°,
又∵∠APD=∠ACB,
∴△ADP∽△ABC,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191103171899349/SYS201311011911031718993025_DA/9.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191103171899349/SYS201311011911031718993025_DA/10.png)
,
即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191103171899349/SYS201311011911031718993025_DA/11.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191103171899349/SYS201311011911031718993025_DA/12.png)
,
解得DP=3,
點(diǎn)P到x軸的距離為3-1=2,
點(diǎn)P的坐標(biāo)為(-2,-2);
(3)連接BD,∵B(-3,0),D(-2,1),
∴tan∠DBA=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191103171899349/SYS201311011911031718993025_DA/13.png)
=1,
∴∠DBA=45°,
根據(jù)勾股定理,BD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191103171899349/SYS201311011911031718993025_DA/14.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191103171899349/SYS201311011911031718993025_DA/15.png)
,
又∵∠ABC=45°,
∴∠DBC=45°×2=90°,
∴tan∠BCD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191103171899349/SYS201311011911031718993025_DA/16.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191103171899349/SYS201311011911031718993025_DA/17.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191103171899349/SYS201311011911031718993025_DA/18.png)
,
又∵tan∠OCA=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191103171899349/SYS201311011911031718993025_DA/19.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191103171899349/SYS201311011911031718993025_DA/20.png)
,
∴∠BCD=∠OCA,
∴∠OCA+∠OCD=∠BCD+∠OCD=∠OCB,
∵B(-3,0),C(0,-3),
∴△OAC是等腰直角三角形,
∴∠OCB=45°,
即∠OCA與∠OCD兩角和是45°.
點(diǎn)評(píng):本題是對(duì)二次函數(shù)的綜合考查,主要利用了待定系數(shù)法求函數(shù)解析式,二次函數(shù)的對(duì)稱性,解直角三角形,勾股定理,以及相似三角形的判定與性質(zhì),利用數(shù)據(jù)的特殊性求出等腰直角三角形得到45°角,然后找出相等的角是解題的關(guān)鍵.