∴y随x的增大而减小．
∴当x?4时，y??8?4?280?248（元）． ································································· 9分 答：有两种购买方案．第一种：买甲种水壶3个，乙种水壶7个； 第二种：买甲种水壶4个，乙种水壶6个．
其中最省钱的方案是第二种，最少费用是248元． ··························································· 10分 （其它方法可参照此答案给分）
D C 25．解：（1）BD?MF，BD⊥MF． ····················· 1分
N M 延长FM交BD于点N，
E
由题意得：△BAD≌△MAF． ∴BD?MF，?ADB??AFM． ······························ 2分 又∵?DMN??AMF， F B
A
∴?ADB??DMN??AFM??AMF?90°， ∴?DNM?90°，∴BD⊥MF． ····················································································· 3分 （2）?的度数为60°或15°（答对一个得2分） ····························································· 7分 （3）由题意得矩形PNA2A．设A2A?x，则PN?x， 在Rt△A2M2F2中，∵F2M2?FM?8， ∴A2M2?4，A2F2?43，∴AF2?43?x． ∵?PAF2?90°，?PF2A?30°，
M2 N B
A2
D M P A
F2
F
3x． ∴AP?AF2?tan30°?4?3∴PD?AD?AP?43?4?3x． 3∵NP∥AB，∴?DNP??B．
∵?D??D，∴△DPN∽△DAB． ·············································································· 9分 ∴
PNDP． ···················································································································· 10分 ?ABDAx?443?4?433x3，解得x?6?23． ··································································· 11分
∴
即A2A?6?23．
答：平移的距离是(6?23)cm． ······················································································ 12分 （其它方法可参照此答案给分）
26．解：（1）∵抛物线y?ax?x?3（a?0）的对称轴为直线x??2．
2?11??2，∴a??， 2a41∴y??x2?x?3． ··········································································································· 2分
4∴?∴D(?2，························································································································· 3分 4)． ·（2）探究一：当0?t?4时，W有最大值． ∵抛物线y??12x?x?3交x轴于A、B两点，交y轴于点C， 4y D M C P B O x ∴A(?6，0)，B(2，0)，C(0，3)，
∴OA?6，OC?3． ················································ 4分 当0?t?4时，作DM⊥y轴于M， 则DM?2，OM?4． ∵P(0，t)，
∴OP?t，MP?OM?OP?4?t． ∵S△PAD?S梯形OADM?S△AOP?S△DMP
A 111(DM?OA)?OM?OA?OP?DM?MP 222111 ?(2?6)?4??6?t??2?(4?t)
222 ?12?2t ·················································································································· 6分
?∴W?t(12?2t)??2(t?3)?18 ······················································································· 7分 ∴当t?3时，W有最大值，W最大值?18． ········································································ 8分 探究二：
存在．分三种情况：
2?90°时，作DE⊥x轴于E，则OE?2，DE?4，?DEA?90°， ①当?PDA1∴AE?OA?OE?6?2?4?DE． ∴?DAE??ADE?45°，AD?2DE?42， y D M C P1 B E O ??PDA??ADE?90°?45°?45°． ∴?PDE11∵DM⊥y轴，OA⊥y轴，
∴DM∥OA，∴?MDE??DEA?90°，
A x ?90°?45°?45°． ∴?MDP1??MDE??PDE1??DM?2，PD∴PM112DM?22．
P2 此时
OCOA32???90°， ，又因为?AOC??PDA1PDAD41?4?2?2，∴P2)． ∴Rt△ADP1∽Rt△AOC，∴OP1?OM?PM11(0，?90°时，存在点P1，使Rt△ADP∴当?PDA11∽Rt△AOC，
此时P··················································· 10分（结论1分，过程1分） 1点的坐标为（0，2）． ②当?P2AD?90°时，则?P2AO?45°， ∴P2A?PA62OA?2． ?62，∴2?OA6cos45°∵
AD42ADP2A?，∴． ?OC3OCOA∴△P2AD与△AOC不相似，此时点P2不存在． ················· 12分（结论1分，过程1分） ③当?AP3D?90°时，以AD为直径作⊙O1，则⊙O1的半径r?圆心O1到y轴的距离d?4．∵d?r，∴⊙O1与y轴相离． 不存在点P3，使?AP3D?90°．
∴综上所述，只存在一点P(0，2)使Rt△ADP与Rt△AOC相似．
································································· 14分（结论1分，过程1分） （其它方法可参照此答案给分）
AD?22， 2