抛物线的解析
来源:学生作业帮 编辑:搜搜考试网作业帮 分类:数学作业 时间:2024/07/28 11:53:45
![](http://img.wesiedu.com/upload/3/8b/38b073621aa73362afad6ccc885ffd4d.jpg)
![抛物线的解析](/uploads/image/z/15982042-58-2.jpg?t=%E6%8A%9B%E7%89%A9%E7%BA%BF%E7%9A%84%E8%A7%A3%E6%9E%90)
解题思路: (1)先由直线AB的解析式为y=x+3,求出它与x轴的交点A、与y轴的交点B的坐标,再将A、B两点的坐标代入y=-x2+bx+c,运用待定系数法即可求出抛物线的解析式; (2)设第三象限内的点F的坐标为(m,-m2-2m+3),运用配方法求出抛物线的对称轴及顶点D的坐标,再设抛物线的对称轴与x轴交于点G,连接FG,根据S△AEF=S△AEG+S△AFG-S△EFG=3,列出关于m的方程,解方程求出m的值,进而得出点F的坐标; (3)设P点坐标为(-1,n).先由B、C两点坐标,运用勾股定理求出BC2=10,
解题过程:
![](http://img.wesiedu.com/upload/7/73/773f3d60029d44d8094b449f71ac3fa7.jpg)
![](http://img.wesiedu.com/upload/e/e3/ee3f099ae5c44b331d14e3d35b325e5a.jpg)
![](http://img.wesiedu.com/upload/f/75/f75a7de8bb0a4e2b157342163613998a.jpg)
解题过程:
![](http://img.wesiedu.com/upload/7/73/773f3d60029d44d8094b449f71ac3fa7.jpg)
![](http://img.wesiedu.com/upload/e/e3/ee3f099ae5c44b331d14e3d35b325e5a.jpg)
![](http://img.wesiedu.com/upload/f/75/f75a7de8bb0a4e2b157342163613998a.jpg)