15.(本题满分5分)
解：设这个二次函数的解析式为y?a(x?1)?2………………………………..2分 ∵二次函数的图象过坐标原点,
∴0?a(0?1)?2
22…………………………………………………….….…3分
a?2 ………………………………………………………………………4分
∴这个二次函数的解析式是y?2(x?1)?2，
即y?2x?4x………………………………………………………………………5分 四、解答题：（本大题共3个小题，共15分） 16.（本题满分5分） 证明： ∵ AC?3,BC?6?3,DC2 CE42 ………………………………..2分 ∴ AC?BC .DCCE ………………………………………………3分又 ∠ACB=∠DCE=90°，……………………………………4分 ∴ △ACB∽△DCE．……………………………………….5分 17．（本题满分5分） 解：(1)∵AD是BC上的高，
∴∠ADB=∠ADC=90°．
∵sinB=
224，AD=12， 5∴AB=15. ??????????????????????2分 BD?AB2?AD2
?152?122?9??????????????????????????3分
∵BC=14， ∴DC=BC-BD=14-9=5． ???????????????????4分 ∴tan?DAC?5???????????????????????5分 1218．（本题满分5分） 解：如图， 连结OA. ∵ CD=10cm，
∴ OA＝5cm，????????????1分 ?AB?CD,
??AMO?90?.在Rt△AOM中， ?OM?3cm,
?AM?OA2?OM2?52?32?4??????????????????????3分
又∵ CD是直径,
AB是弦,
AB⊥CD于M ， ∴ AB＝2AM .
∴ AB＝8cm . ………………………5分 五、解答题：（本大题共2个小题，共10分）
19．(本小题满分5分)
解：如图，过C点作CD垂直于AB交BA的延长线于点D. ··············································· 1分
??CAB?120?,
??CAD?180???CAB?180??120??60?. 在Rt△CDA中，AC=300,
CD?AC?sin?CAD?300sin60??1503. ·················································································· 2分
······················································································· 3分 AD?AC?cos?CAD?300?cos60??150 在Rt△CDB中，
C ?BC?700,BD2?BC2?CD2
?BD?7002?(1503)2?650
D A B ······························································································ 4分 ?AB?BD?AD?650?150?500 ·
答：A、B两个凉亭之间的距离为500m. ··················································································· 5分
20． （本题满分5分）
开始 解：
6 7 2 6 2 7 6 2 7 6 2 7
·········································································································································· 2分 （1）P（两数字相同）=
1． ··········································································································· 3分 34． ···································································································· 5分 9（2）P（两数字和大于10）=
六、解答题：（本大题共2个小题，共11分）
21．（本题满分5分）
x?解：（1）根据题意，得y?(2400?2000?x)??8?4??，
50??22x?24x?3200．???????????????????3分 2522（2）对于y??x?24x?3200，
25即y??当x??24?150时，??????????????????.4分
?2?2?????25?150??y最大值?(2400?2000?150)?8?4???250?20?5000．
50??所以，每台彩电降价150元时，商场每天销售这种彩电的利润最大，最大利润是5000元．????????????????????????????????????5分 22．（本题满分6分）（1）直线BD和⊙O相切． ································································· 1分
证明：
∵?AEC??ODB，?AEC??ABC， ∴?ABC??ODB． ························································· 2分 ∵OD⊥BC，
∴?DBC??ODB?90°． ∴?DBC??ABC?90°． 即?DBO?90°． ∵AB是⊙O直径，
∴直线BD和⊙O相切． ···················································· 3分 （2）连接AC． ∵AB是⊙O直径， ∴?ACB?90°．
在Rt△ABC中，AB?10，BC?8， ∴AC?···································································································· 4分 AB2?BC2?6． ·
∵BD和⊙O相切，
∴?OBD?90°．
∴?ACB??OBD?90°． 由（1）得?ABC??ODB， ∴△ABC∽△ODB． ·············································································································· 5分 ∴
ACBC?． OBBD68?， 5BD∴
∴BD?20． ······························································································································ 6分 3六、解答题：（本大题共3个小题，共21分）
23. （本题满分6分）
解：以A为圆心，以a为半径作圆.延长BA交⊙A于E点，
连接ED…………………………………………………………..………..1分 ∵AB∥CD,
∴?CAB??DCA,
?DAE??CDA.
∵AC=AD,
??DCA??CDA,
??DAE??CAB.…………………………………………………..…..…..2分
在△ABC和△DAE中，
?AD?AC,???DAE??CAB, ?AE?AB.?∴△CAB≌△DAE, ……………………………………………….…….……..3分 ∴ED=BC=b……………………………………………………………………..4分 ∵BE是直径， ∴?EDB?90? 在Rt△EDB中， ED=b, BE=2a, 由勾股定理得
ED2?BD2?BE2.
∴BD?BE2?ED2?(2a)2?b2?4a2?b2.……………………………….5分
BD?cos?DBA??BE4a2?b2………………………………………………………6分 2a24. （本题满分7分）（1）①由题意知：
对称轴为直线x?1，b=1 2∴?b1?, 2a2∴a??1. …………………………………………………………………..1分
1299??(x?)??????????????????2分 ②y??x?x?22442∵y的值为正整数
∴y1?1,2y2?2 ………………………………………………………………3分
2∴?x?x?2?1或?x?x?2?2 ∴x?x?1?0或x?x?0 解，得 x1?221?51?5 ，x3?0, x4?1????????4分 ,x2?222（2）∵当a?a1时，抛物线y?ax?x?2与x轴的正半轴相交于点M（m，0） ∴0?a1m2?m?2 ①
∵当a?a2时，，抛物线y?ax?x?2与x轴的正半轴相交于点N（n，0） ∴0?a2n2?n?2 ② ∴a1?2?m?2?n?2a?，??????????????????????5分 222mn?m?2?n?2?(m?2)n2?(n?2)m2?mn2?2n2?nm2?2m2???∴a1?a2? 222222mnmnmn?mn(m?n)?2(m?n)(m?n)(mn?2m?2n)(m?n)????????6分 2222mnmn
∵点M在点N的左边，且M、N均在x轴正半轴 ∴m?0,n?0,m?n
∴mn?2m?2n?0,m?n?0,mn?0 ∴a1?a2＝
22(mn?2m?2n)(m?n)?0
m2n2∴a1?a2?????????????????????????????7分 25．（本题满分8分） 解： (1)?B(1,0),?OB?1.?OC?3?B0?C(0,?3)......................................................................................................................................1分
∵y?ax?3ax?c过B(1,0)、C(0，－3)，
2