(1)这个正比例函数的解析式为.·················································· (1分) 这个反比例函数的解析式为.························································ (2分) (2)因为点在的图象上,所以,则点.······ (3分) 设一次函数解析式为. 因为的图象是由平移得到的,所以,即. 又因为的图象过点,所以,解得, 一次函数的解析式为.························································· (5分) (3)因为的图象交轴于点,所以的坐标为. 设二次函数的解析式为. 解得 这个二次函数的解析式为.·········································· (8分)
(4)交轴于点,点的坐标是, 如图, 假设存在点,使. 四边形的顶点只能在轴上方,, . ,.···························································· (10分) 在二次函数的图象上,. 解得或. 当时,点与点重合,这时不是四边形,故舍去, 点的坐标为. |