摘要: 解(Ⅰ)当.时.抛物线为. 方程的两个根为.. ∴该抛物线与轴公共点的坐标是和. ················································ 2分 (Ⅱ)当时.抛物线为.且与轴有公共点. 对于方程.判别式≥0.有≤. ········································ 3分 ①当时.由方程.解得. 此时抛物线为与轴只有一个公共点.································· 4分 ②当时. 时.. 时.. 由已知时.该抛物线与轴有且只有一个公共点.考虑其对称轴为. 应有 即 解得. 综上.或. ················································································ 6分 (Ⅲ)对于二次函数. 由已知时.,时.. 又.∴. 于是.而.∴.即. ∴. ············································································································ 7分 ∵关于的一元二次方程的判别式 . ∴抛物线与轴有两个公共点.顶点在轴下方.····························· 8分 又该抛物线的对称轴. 由... 得. ∴. 又由已知时.,时..观察图象. 可知在范围内.该抛物线与轴有两个公共点. ············································ 10分

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