解:(1)∵拋物線三角形系數(shù)為[-1,b,0],
∴拋物線解析式為y=-x
2+bx=-(x-
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)
2+
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,
∴頂點(diǎn)坐標(biāo)為(
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,
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),
令y=0,則-x
2+bx=0,
解得x
1=0,x
2=b,
∴與x軸的交點(diǎn)為(0,0),(b,0),
∵“拋物線三角形”是等腰直角三角形,
∴
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=
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|b|,
∴b
2=2b或b
2=-2b,
∵b=0時(shí),拋物線與x軸只有一個(gè)交點(diǎn)(0,0),
∴b=0不符合題意,
∴b=2或b=-2,
故b的值為2或-2;
(2)如圖,∵四邊形ABCD是矩形,
∴OA=OB,
由拋物線的對(duì)稱性,OB=AB,
∴OA=OB=AB,
∴△AOB是等邊三角形,
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∵拋物線三角形系數(shù)為[-2,2m,0],
∴拋物線解析式為y=-2x
2+2mx=-2(x-
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)
2+
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,
∴頂點(diǎn)B的坐標(biāo)為(
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,
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),
令y=0,則-2x
2+2mx=0,
解得x
1=0,x
2=m,
∴與x軸的交點(diǎn)為(0,0),(m,0),
∴AO=m,
∴
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=
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m,
解得m=
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,
∴點(diǎn)A(
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,0),B(
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,
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),
∵四邊形ABCD是以原點(diǎn)O為對(duì)稱中心的矩形,
∴點(diǎn)C(-
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,0),D(-
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,-
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),
設(shè)過(guò)O、C、D三個(gè)點(diǎn)的拋物線為y=ax
2+bx(a≠0),
則
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,
解得
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,
所以,過(guò)O、C、D三個(gè)點(diǎn)的拋物線為y=2x
2+2
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x.
分析:(1)把拋物線三角形系數(shù)代入拋物線,令y=0求出點(diǎn)A的坐標(biāo),再求出頂點(diǎn)坐標(biāo),然后根據(jù)等腰直角三角形的斜邊上的高線等于斜邊的一半列出方程求解即可得到b的值;
(2)根據(jù)矩形的對(duì)角線互相平分且相等可得OA=OB,再根據(jù)拋物線的對(duì)稱性可得OB=AB,從而判定△AOB是等邊三角形,然后拋物線三角形系數(shù)代入拋物線,令y=0求出點(diǎn)A的坐標(biāo),再求出頂點(diǎn)坐標(biāo),過(guò)然后根據(jù)等邊三角形的高等于邊長(zhǎng)的
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列出方程求出m的值,從而得到點(diǎn)A、B的坐標(biāo),再根據(jù)關(guān)于原點(diǎn)對(duì)稱的點(diǎn)的橫坐標(biāo)與縱坐標(biāo)都互為相反數(shù)求出C、D的坐標(biāo),再利用待定系數(shù)法求二次函數(shù)解析式解答即可.
點(diǎn)評(píng):本題是二次函數(shù)綜合題型,主要考查了等腰直角三角形的性質(zhì),等邊三角形的判定與性質(zhì),矩形的對(duì)角線互相平分且相等的性質(zhì),待定系數(shù)法求二次函數(shù)解析式,讀懂題目信息,理解“拋物線三角形”的定義是解題的關(guān)鍵.