如图,抛物线y=x 2 +bx+c与x轴交于A(-1,0)、B(3,0)两点。 (1)求该抛物线的解析式; 如图,抛物线y=x 2 +bx+c与x轴交于A(﹣1,0)和...

\u5982\u56fe\uff0c\u629b\u7269\u7ebfy=x 2 +bx+c\u4e0ex\u8f74\u4ea4\u4e8eA\uff08-1\uff0c0\uff09\u3001B\uff083\uff0c0\uff09\u4e24\u70b9\uff0c\u76f4\u7ebfL\u4e0e\u629b\u7269\u7ebf\u4ea4\u4e8eA\u3001C\u4e24\u70b9\uff0c\u5176\u4e2dC\u70b9\u7684\u6a2a

(1)\u629b\u7269\u7ebf\u7684\u89e3\u6790\u5f0fy=x 2 -2x-3\uff0c\u76f4\u7ebfAC\u7684\u51fd\u6570\u89e3\u6790\u5f0f\u662fy=-x-1\uff1b\uff082\uff09PE\u7684\u6700\u5927\u503c= \uff1b\uff083\uff09F\u70b9\u7684\u5750\u6807\u662f\uff08-3\uff0c0\uff09\uff0c\uff081\uff0c0\uff09\uff0c\uff084- \uff0c0\uff09\uff0c\uff084+ \uff0c0\uff09. \u8bd5\u9898\u5206\u6790\uff1a\uff081\uff09\u5c06A\u3001B\u7684\u5750\u6807\u4ee3\u5165\u629b\u7269\u7ebf\u4e2d\uff0c\u6613\u6c42\u51fa\u629b\u7269\u7ebf\u7684\u89e3\u6790\u5f0f\uff1b\u5c06C\u70b9\u6a2a\u5750\u6807\u4ee3\u5165\u629b\u7269\u7ebf\u7684\u89e3\u6790\u5f0f\u4e2d\uff0c\u5373\u53ef\u6c42\u51faC\u70b9\u7684\u5750\u6807\uff0c\u518d\u7531\u5f85\u5b9a\u7cfb\u6570\u6cd5\u53ef\u6c42\u51fa\u76f4\u7ebfAC\u7684\u89e3\u6790\u5f0f\uff0e\uff082\uff09PE\u7684\u957f\u5b9e\u9645\u662f\u76f4\u7ebfAC\u4e0e\u629b\u7269\u7ebf\u7684\u51fd\u6570\u503c\u7684\u5dee\uff0c\u53ef\u8bbeP\u70b9\u7684\u6a2a\u5750\u6807\u4e3ax\uff0c\u7528x\u5206\u522b\u8868\u793a\u51faP\u3001E\u7684\u7eb5\u5750\u6807\uff0c\u5373\u53ef\u5f97\u5230\u5173\u4e8ePE\u7684\u957f\u3001x\u7684\u51fd\u6570\u5173\u7cfb\u5f0f\uff0c\u6839\u636e\u6240\u5f97\u51fd\u6570\u7684\u6027\u8d28\u5373\u53ef\u6c42\u5f97PE\u7684\u6700\u5927\u503c\uff0e\uff083\uff09\u6b64\u9898\u8981\u5206\u4e24\u79cd\u60c5\u51b5\uff1a\u2460\u4ee5AC\u4e3a\u8fb9\uff0c\u2461\u4ee5AC\u4e3a\u5bf9\u89d2\u7ebf\uff0e\u786e\u5b9a\u5e73\u884c\u56db\u8fb9\u5f62\u540e\uff0c\u53ef\u76f4\u63a5\u5229\u7528\u5e73\u884c\u56db\u8fb9\u5f62\u7684\u6027\u8d28\u6c42\u51faF\u70b9\u7684\u5750\u6807\uff0e\u8bd5\u9898\u89e3\u6790\uff1a\u89e3\uff1a\uff081\uff09\u5c06A\uff08-1\uff0c0\uff09\uff0cB\uff083\uff0c0\uff09\u4ee3\u5165y=x 2 +bx+c\uff0c\u5f97b=-2\uff0cc=-3\uff1b\u2234y=x 2 -2x-3\uff0e\u5c06C\u70b9\u7684\u6a2a\u5750\u6807x=2\u4ee3\u5165y=x 2 -2x-3\uff0c\u5f97y=-3\uff0c\u2234C\uff082\uff0c-3\uff09\uff1b\u2234\u76f4\u7ebfAC\u7684\u51fd\u6570\u89e3\u6790\u5f0f\u662fy=-x-1\uff0e\uff082\uff09\u8bbeP\u70b9\u7684\u6a2a\u5750\u6807\u4e3ax\uff08-1\u2264x\u22642\uff09\uff0c\u5219P\u3001E\u7684\u5750\u6807\u5206\u522b\u4e3a\uff1aP\uff08x\uff0c-x-1\uff09\uff0cE\uff08x\uff0cx 2 -2x-3\uff09\uff1b\u2235P\u70b9\u5728E\u70b9\u7684\u4e0a\u65b9\uff0cPE=\uff08-x-1\uff09-\uff08x 2 -2x-3\uff09=-x 2 +x+2\uff0c\u2234\u5f53x= \u65f6\uff0cPE\u7684\u6700\u5927\u503c= \uff0e\uff083\uff09\u5b58\u57284\u4e2a\u8fd9\u6837\u7684\u70b9F\uff0c\u5206\u522b\u662fF1\uff081\uff0c0\uff09\uff0cF2\uff08-3\uff0c0\uff09\uff0cF3\uff084+ \uff0c0\uff09\uff0cF4\uff084- \uff0c0\uff09\uff0e\u2460\u5982\u56fe\uff0c\u8fde\u63a5C\u4e0e\u629b\u7269\u7ebf\u548cy\u8f74\u7684\u4ea4\u70b9\uff0c\u2235C\uff082\uff0c-3\uff09\uff0cG\uff080\uff0c-3\uff09\u2234CG\u2225X\u8f74\uff0c\u6b64\u65f6AF=CG=2\uff0c\u2234F\u70b9\u7684\u5750\u6807\u662f\uff08-3\uff0c0\uff09\uff1b \u2461\u5982\u56fe\uff0cAF=CG=2\uff0cA\u70b9\u7684\u5750\u6807\u4e3a\uff08-1\uff0c0\uff09\uff0c\u56e0\u6b64F\u70b9\u7684\u5750\u6807\u4e3a\uff081\uff0c0\uff09\uff1b \u2462\u5982\u56fe\uff0c\u6b64\u65f6C\uff0cG\u4e24\u70b9\u7684\u7eb5\u5750\u6807\u5173\u4e8ex\u8f74\u5bf9\u79f0\uff0c\u56e0\u6b64G\u70b9\u7684\u7eb5\u5750\u6807\u4e3a3\uff0c\u4ee3\u5165\u629b\u7269\u7ebf\u4e2d\u5373\u53ef\u5f97\u51faG\u70b9\u7684\u5750\u6807\u4e3a\uff081\u00b1 \uff0c3\uff09\uff0c\u7531\u4e8e\u76f4\u7ebfGF\u7684\u659c\u7387\u4e0e\u76f4\u7ebfAC\u7684\u76f8\u540c\uff0c\u56e0\u6b64\u53ef\u8bbe\u76f4\u7ebfGF\u7684\u89e3\u6790\u5f0f\u4e3ay=-x+h\uff0c\u5c06G\u70b9\u4ee3\u5165\u540e\u53ef\u5f97\u51fa\u76f4\u7ebf\u7684\u89e3\u6790\u5f0f\u4e3ay=-x+4+ \uff0e\u56e0\u6b64\u76f4\u7ebfGF\u4e0ex\u8f74\u7684\u4ea4\u70b9F\u7684\u5750\u6807\u4e3a\uff084+ \uff0c0\uff09\uff1b\u2463\u5982\u56fe\uff0c\u540c\u2462\u53ef\u6c42\u51faF\u7684\u5750\u6807\u4e3a\uff084- \uff0c0\uff09\uff1b \u7efc\u5408\u56db\u79cd\u60c5\u51b5\u53ef\u5f97\u51fa\uff0c\u5b58\u57284\u4e2a\u7b26\u5408\u6761\u4ef6\u7684F\u70b9

\u89e3\uff1a\uff081\uff09\u2235\u629b\u7269\u7ebfy=x 2 +bx+c\u4e0ex\u8f74\u4ea4\u4e8eA\uff08\ufe631\uff0c0\uff09\u548cB\uff083\uff0c0\uff09\u4e24\u70b9\uff0c\u2234 \uff0c\u89e3\u5f97\uff1a \u3002\u2234\u629b\u7269\u7ebf\u89e3\u6790\u5f0f\u4e3a\uff1ay=x 2 \ufe632x\ufe633\uff1b\uff082\uff09\u8054\u7acb\u5f97\uff1a \uff0c\u89e3\u5f97\uff1a \uff0c \u3002\u2234D\uff084\uff0c5\uff09\u3002\u5bf9\u4e8e\u76f4\u7ebfy=x+1\uff0c\u5f53x=0\u65f6\uff0cy=1\uff0c\u2234F\uff080\uff0c1\uff09\u3002\u5bf9\u4e8ey=x 2 \ufe632x\ufe633\uff0c\u5f53x=0\u65f6\uff0cy=\ufe633\uff0c\u2234E\uff080\uff0c\ufe633\uff09\u3002\u2234EF=4\u3002\u8fc7\u70b9D\u4f5cDM\u22a5y\u8f74\u4e8e\u70b9M\uff0c \u2234S \u25b3 DEF = EF?DM=8\u3002 \u8bd5\u9898\u5206\u6790\uff1a\uff081\uff09\u5229\u7528\u5f85\u5b9a\u7cfb\u6570\u6cd5\u6c42\u4e8c\u6b21\u51fd\u6570\u89e3\u6790\u5f0f\u5373\u53ef\u3002\uff082\uff09\u9996\u5148\u6c42\u51fa\u76f4\u7ebf\u4e0e\u4e8c\u6b21\u51fd\u6570\u7684\u4ea4\u70b9\u5750\u6807\u8fdb\u800c\u5f97\u51faE\uff0cF\u70b9\u5750\u6807\uff0c\u5373\u53ef\u5f97\u51fa\u25b3DEF\u7684\u9762\u79ef\u3002\u3000

解:①将点A(-1,0), B(3,0)代入y=x 2 +bx+c中得:
 

②解:设p点的纵坐标为t

③存在
∵点A,B关于抛物线的对称轴对称∴连接BC与对称轴的交点Q就能使 周长最小
设直线BC解析式为:y=kx+b,又∵C(0,-3) B(3,0)

∴y=x-3
又∵抛物线对称轴为x=1
∴Q(1,-2)



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