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<span style="font-size:10.5pt"><span style="line-height:150%"><span style="font-family:Calibri"><span style="line-height:150%"><span style="font-family:宋体">如图，一元二次方程<i>x</i><sup>2</sup>+2<i>x</i>﹣3＝0的二根<i>x</i><sub>1</sub>，<i>x</i><sub>2</sub>（<i>x</i><sub>1</sub>＜<i>x</i><sub>2</sub>）是抛物线<i>y</i>＝<i>ax</i><sup>2</sup>+<i>bx</i>+<i>c</i>与<i>x</i>轴的两个交点<i>B</i>，<i>C</i>的横坐标，且此抛物线过点<i>A</i>（3，6）．</span></span></span></span></span><br />
<span style="font-size:10.5pt"><span style="line-height:150%"><span style="font-family:Calibri"><span style="line-height:150%"><span style="font-family:宋体">（1）求此二次函数的解析式；</span></span></span></span></span><br />
<span style="font-size:10.5pt"><span style="line-height:150%"><span style="font-family:Calibri"><span style="line-height:150%"><span style="font-family:宋体">（2）写出不等式<i>ax</i><sup>2</sup>+<i>bx</i>+<i>c</i>≥0的解集；</span></span></span></span></span><br />
<span style="font-size:10.5pt"><span style="line-height:150%"><span style="font-family:Calibri"><span style="line-height:150%"><span style="font-family:宋体">（3）设此抛物线的顶点为<i>P</i>，对称轴与线段<i>AC</i>相交于点<i>Q</i>，求点<i>P</i>和点<i>Q</i>的坐标；</span></span></span></span></span><br />
<span style="font-size:10.5pt"><span style="line-height:150%"><span style="font-family:Calibri"><span style="line-height:150%"><span style="font-family:宋体">（4）在<i>x</i>轴上有一动点<i>M</i>，当<i>MQ</i>+<i>MA</i>取得最小值时，求<i>M</i>点的坐标．</span></span></span></span></span><br />
                                                                                                                                                   <img height="156" src="/uploads/meth/20201226/ea5cfec6edad80808aa748a472bd280e.png" width="159" />

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<span style="font-size:10.5pt"><span style="line-height:150%"><span style="font-family:Calibri"><span style="line-height:150%"><span style="font-family:宋体">解：（1）一元二次方程<i>x</i><sup>2</sup>+2<i>x</i>﹣3＝0的二根<i>x</i><sub>1</sub>，<i>x</i><sub>2</sub>（<i>x</i><sub>1</sub>＜<i>x</i><sub>2</sub>）为：</span></span></span></span></span><br />
<span style="font-size:10.5pt"><span style="line-height:150%"><span style="font-family:Calibri"><i><span style="line-height:150%"><span style="font-family:宋体">x</span></span></i><sub><span style="line-height:150%"><span style="font-family:宋体">1</span></span></sub><span style="line-height:150%"><span style="font-family:宋体">＝﹣3，<i>x</i><sub>2</sub>＝1．</span></span></span></span></span><br />
<span style="font-size:10.5pt"><span style="line-height:150%"><span style="font-family:Calibri"><span style="line-height:150%"><span style="font-family:宋体">∴抛物线<i>y</i>＝<i>ax</i><sup>2</sup>+<i>bx</i>+<i>c</i>与<i>x</i>轴的两个交点的坐标为<i>B</i>（1，0），<i>C</i>（﹣3，0）．</span></span></span></span></span><br />
<span style="font-size:10.5pt"><span style="line-height:150%"><span style="font-family:Calibri"><span style="line-height:150%"><span style="font-family:宋体">设二次函数的解析式为<i>y</i>＝<i>a</i>（<i>x</i>+3）（<i>x</i>﹣1），</span></span></span></span></span><br />
<span style="font-size:10.5pt"><span style="line-height:150%"><span style="font-family:Calibri"><span style="line-height:150%"><span style="font-family:宋体">∵抛物线过点<i>A</i>（3，6）．</span></span></span></span></span><br />
<span style="font-size:10.5pt"><span style="line-height:150%"><span style="font-family:Calibri"><span style="line-height:150%"><span style="font-family:宋体">∴6＝<i>a</i>（3+3）（3﹣1），解得<i>a</i>＝</span></span></span></span></span><img height="35" src="/uploads/meth/20201226/ab0e63e849b032228eb94bb9219538e1.png" width="13" /><span style="font-size:10.5pt"><span style="line-height:150%"><span style="font-family:Calibri"><span style="line-height:150%"><span style="font-family:宋体">．</span></span></span></span></span><br />
<span style="font-size:10.5pt"><span style="line-height:150%"><span style="font-family:Calibri"><span style="line-height:150%"><span style="font-family:宋体">∴二次函数的解析式为<i>y</i>＝</span></span></span></span></span><img height="35" src="/uploads/meth/20201226/2c8f86a06e6aa0d8911cb87b89877f01.png" width="13" /><span style="font-size:10.5pt"><span style="line-height:150%"><span style="font-family:Calibri"><span style="line-height:150%"><span style="font-family:宋体">（<i>x</i>+3）（<i>x</i>﹣1）＝</span></span></span></span></span><img height="35" src="/uploads/meth/20201226/15ef7b3ccbb98e700c85ba4ffb0ef90e.png" width="13" /><span style="font-size:10.5pt"><span style="line-height:150%"><span style="font-family:Calibri"><span style="line-height:150%"><span style="font-family:宋体"><i>x</i><sup>2</sup>+<i>x</i>﹣</span></span></span></span></span><img height="35" src="/uploads/meth/20201226/b6cf3f13ef0413619d52e365ea1c4c9a.png" width="13" /><span style="font-size:10.5pt"><span style="line-height:150%"><span style="font-family:Calibri"><span style="line-height:150%"><span style="font-family:宋体">．</span></span></span></span></span><br />
<span style="font-size:10.5pt"><span style="line-height:150%"><span style="font-family:Calibri"><span style="line-height:150%"><span style="font-family:宋体">（2）根据图象可知：</span></span></span></span></span><br />
<span style="font-size:10.5pt"><span style="line-height:150%"><span style="font-family:Calibri"><span style="line-height:150%"><span style="font-family:宋体">不等式<i>ax</i><sup>2</sup>+<i>bx</i>+<i>c</i>≥0的解集为：<i>x</i>≤﹣3或<i>x</i>≥1；</span></span></span></span></span><br />
<span style="font-size:10.5pt"><span style="line-height:150%"><span style="font-family:Calibri"><span style="line-height:150%"><span style="font-family:宋体">（3）由<i>y</i>＝</span></span></span></span></span><img height="35" src="/uploads/meth/20201226/113b7e0510ac7aa7ba12f816b62e1694.png" width="13" /><span style="font-size:10.5pt"><span style="line-height:150%"><span style="font-family:Calibri"><span style="line-height:150%"><span style="font-family:宋体"><i>x</i><sup>2</sup>+<i>x</i>﹣</span></span></span></span></span><img height="35" src="/uploads/meth/20201226/7296a0183a4ebefc560f71de167b3400.png" width="13" /><span style="font-size:10.5pt"><span style="line-height:150%"><span style="font-family:Calibri"><span style="line-height:150%"><span style="font-family:宋体">．</span></span></span></span></span><br />
<span style="font-size:10.5pt"><span style="line-height:150%"><span style="font-family:Calibri"><span style="line-height:150%"><span style="font-family:宋体">∴抛物线的顶点坐标为<i>P</i>（﹣1，﹣2），对称轴方程为<i>x</i>＝﹣1．</span></span></span></span></span><br />
<span style="font-size:10.5pt"><span style="line-height:150%"><span style="font-family:Calibri"><span style="line-height:150%"><span style="font-family:宋体">设直线<i>AC</i>解析式为<i>y</i>＝<i>kx</i>+<i>b</i>，</span></span></span></span></span><br />
<span style="font-size:10.5pt"><span style="line-height:150%"><span style="font-family:Calibri"><span style="line-height:150%"><span style="font-family:宋体">将<i>A</i>（3，6），<i>C</i>（﹣3，0），代入解得：</span></span></span></span></span><br />
<span style="font-size:10.5pt"><span style="line-height:150%"><span style="font-family:Calibri"><i><span style="line-height:150%"><span style="font-family:宋体">k</span></span></i><span style="line-height:150%"><span style="font-family:宋体">＝1，<i>b</i>＝3，</span></span></span></span></span><br />
<span style="font-size:10.5pt"><span style="line-height:150%"><span style="font-family:Calibri"><span style="line-height:150%"><span style="font-family:宋体">直线<i>AC</i>解析式为<i>y</i>＝<i>x</i>+3．</span></span></span></span></span><br />
<span style="font-size:10.5pt"><span style="line-height:150%"><span style="font-family:Calibri"><span style="line-height:150%"><span style="font-family:宋体">将<i>x</i>＝﹣1代入，得<i>y</i>＝2．</span></span></span></span></span><br />
<span style="font-size:10.5pt"><span style="line-height:150%"><span style="font-family:Calibri"><span style="line-height:150%"><span style="font-family:宋体">∴<i>Q</i>（﹣1，2）．</span></span></span></span></span><br />
<span style="font-size:10.5pt"><span style="line-height:150%"><span style="font-family:Calibri"><span style="line-height:150%"><span style="font-family:宋体">（4）作点<i>A</i>关于<i>x</i>轴的对称点<i>A</i>′（3，﹣6），</span></span></span></span></span><br />
<img height="280" src="/uploads/meth/20201226/62b5fcc0f7864af16845205b4d09d602.png" width="222" /><br />
<span style="font-size:10.5pt"><span style="line-height:150%"><span style="font-family:Calibri"><span style="line-height:150%"><span style="font-family:宋体">连接<i>A</i>′<i>Q</i>，<i>A</i>′<i>Q</i>与<i>x</i>轴交于点<i>M</i>即为所求的点．</span></span></span></span></span><br />
<span style="font-size:10.5pt"><span style="line-height:150%"><span style="font-family:Calibri"><span style="line-height:150%"><span style="font-family:宋体">设直线<i>A</i>′<i>Q</i>的解析式为<i>y</i>＝<i>kx</i>+<i>b</i>，</span></span></span></span></span><br />
<span style="font-size:10.5pt"><span style="line-height:150%"><span style="font-family:Calibri"><span style="line-height:150%"><span style="font-family:宋体">将<i>A</i>′（3，﹣6），<i>Q</i>（﹣1，2）代入解得：</span></span></span></span></span><br />
<span style="font-size:10.5pt"><span style="line-height:150%"><span style="font-family:Calibri"><i><span style="line-height:150%"><span style="font-family:宋体">k</span></span></i><span style="line-height:150%"><span style="font-family:宋体">＝﹣2，<i>b</i>＝0．</span></span></span></span></span><br />
<span style="font-size:10.5pt"><span style="line-height:150%"><span style="font-family:Calibri"><span style="line-height:150%"><span style="font-family:宋体">∴直线<i>A</i>′<i>C</i>的解析式为<i>y</i>＝﹣2<i>x</i>．</span></span></span></span></span><br />
<span style="font-size:10.5pt"><span style="line-height:150%"><span style="font-family:Calibri"><span style="line-height:150%"><span style="font-family:宋体">令<i>x</i>＝0，则<i>y</i>＝0．</span></span></span></span></span><br />
<span style="font-size:10.5pt"><span style="line-height:150%"><span style="font-family:Calibri"><span style="line-height:150%"><span style="font-family:宋体">∴<i>M</i>（0，0）．</span></span></span></span></span>

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