解：（1）由直角三角形纸板的两直角边的长为1和2，
2)(21)． 知A，C两点的坐标分别为(1，，，设直线AC所对应的函数关系式为y?kx?b． ···················································· 2分
?k?b?2，?k??1，有?解得?
2k?b?1．b?3．??所以，直线AC所对应的函数关系式为y??x?3． ··········································· 4分 （2）①点M到x轴距离h与线段BH的长总相等．
1)， 因为点C的坐标为(2，y 所以，直线OC所对应的函数关系式为y?又因为点P在直线AC上，
3?a)． 所以可设点P的坐标为(a，1x． 2A P I C N M II O G K B H F E （第24题答图）
过点M作x轴的垂线，设垂足为点K，则有MK?h． 因为点M在直线OC上，所以有M(2h，h)． ················ 6分
x 因为纸板为平行移动，故有EF∥OB，即EF∥GH．
又EF?PF，所以PH?GH．
法一：故Rt△MKG∽Rt△PHG∽Rt△PFE，
GKGHEF1???． 从而有
MKPHPF21111得GK?MK?h，GH?PH?(3?a)．
222213所以OG?OK?GK?2h?h?h．
2213又有OG?OH?GH?a?(3?a)?(a?1)． ····················································· 8分
2233所以h?(a?1)，得h?a?1，而BH?OH?OB?a?1，
22从而总有h?BH． ································································································ 10分
GHEF1??． 法二：故Rt△PHG∽Rt△PFE，可得
PHPF211PH?(3?a)． 2213所以OG?OH?GH?a?(3?a)?(a?1)．
22故GH??3?故G点坐标为?(a?1)，0?．
?2?设直线PG所对应的函数关系式为y?cx?d，
?3?a?ca?d，?c?2?则有?解得 3?0?c(a?1)?d．?d?3?3a??2所以，直线PG所对的函数关系式为y?2x?(3?3a)． ····································· 8分 将点M的坐标代入，可得h?4h?(3?3a)．解得h?a?1．
而BH?OH?OB?a?1，从而总有h?BH． ···················································· 10分
?1?②由①知，点M的坐标为(2a?2，a?1)，点N的坐标为?a，a?．
?2?S?S△ONH?S△ONG?111113a?3NH?OH?OG?h??a?a???(a?1) 22222221331?3?3·································································· 12分 ??a2?a????a???． ·
2242?2?8当a?33时，S有最大值，最大值为． 28?33?S取最大值时点P的坐标为?，?． ·································································· 14分
22??[2008福建省南平市]25．（14分）如图，平面直角坐标系中有一矩形纸片OABC，O为原点，点A，C分别在x轴，y轴上，点B坐标为(m，2)（其中m?0），在BC边上选取适当的点E和点F，将△OCE沿OE翻折，得到△OGE；再将△ABF沿AF翻折，恰好使点B与点G重合，得到△AGF，且?OGA?90?．
（1）求m的值；
（2）求过点O，G，A的抛物线的解析式和对称轴； （3）在抛物线的对称轴上是否存在点P，使得△OPG是等腰三角形？若不存在，请说明理．．．由；若存在，直接答出所有满足条件的点P的坐标（不要求写出求解过程）． ．．．．
?b4ac?b2?b【提示：抛物线y?ax?bx?c(a?0)的对称轴是x??，顶点坐标是??，?】
2a4a2a??2
25．（1）解法一：?B(m，2)，
由题意可知AG?AB?2，OG?OC?2，OA?m ······································· 2分
??OGA?90?，?OG2?AG2?OA2 ······································································ 3分 ?2?2?m2．又?m?0，?m?2 ········································································· 4分
解法二：?B(m，2)，
由题意可知AG?AB?2，OG?OC?2，OA?m ······································· 2分
??OGA?90?，??GOA??GAO?45? ································································ 3分
?m?OA?OG2??2 ········································································ 4分
cos?GOAcos45?（2）解法一：过G作直线GH?x轴于H，
，． ·则OH?1，HG?1，故G(11)··········································································· 5分 0)， 又由（1）知A(2，设过O，G，A三点的抛物线解析式为y?ax2?bx?c ····································· 6分 ?抛物线过原点，?c?0．
?a?b?1?a??1又?抛物线过G，A两点，?? 解得?
?4a?2b?0?b?2··················································································· 8分 ?所求抛物线为y??x2?2x
它的对称轴为x?1． ······························································································· 9分
解法二：过G作直线GH?x轴于H，
，． ·则OH?1，HG?1，故G(11)··········································································· 5分
又由（1）知A(2，······ 6分 0)，?点A，O关于直线l对称，?点G为抛物线的顶点 于是可设过O，G，A三点的抛物线解析式为y?a(x?1)2?1
0)，?0?a(0?1)2?1，解得a??1 ?抛物线过点O(0，······················································· 8分 ?所求抛物线为y?(?1)(x?1)2?1??x2?2x ·
它的对称轴为x?1． ······························································································· 9分 （3）答：存在 ········································································································ 10分 满足条件的点P有(1，（每空1分） ········ 14分 ?1)，(110)，(1，，?2)，(11，?2)．
[2008年福建省宁德市]26．（本题满分14分）如图1，在Rt△ABC中，∠C＝90°，BC＝8厘米，点D在AC上，CD＝3厘米．点P、Q分别由A、C两点同时出发，点P沿AC方向向点C匀速移动，速度为每秒k厘米，行完AC全程用时8秒；点Q沿CB方向向点B匀速移动，速度为每秒1厘米．设运动的时间为x秒?0＜x＜8?，△DCQ的面积为y1平方厘米，△PCQ的面积为y2平方厘米．
⑴求y1与x的函数关系，并在图2中画出y1的图象；
⑵如图2，y2的图象是抛物线的一部分，其顶点坐标是（4，12），求点P的速度及AC的长； ⑶在图2中，点G是x轴正半轴上一点（0＜OG＜6＝，过G作EF垂直于x轴，分别交y1、y2于点E、F．
①说出线段EF的长在图1中所表示的实际意义； ②当0＜x＜６时，求线段EF长的最大值． 解： y
A 12
10
8 P ↓ 6
4 D
2 126.解：⑴∵S?DCQ??CQ?CD，CD＝3，CQ＝x， y 2C Q→ B O 2 G 4 6 8 10 x 3图1 12∴y1?x．
F 图2 210 图象如图所示．
8 1⑵方法一：S?PCQ??CQ?CP，CP＝8k－xk，CQ＝x， 6 2114 E ∴y2???8k?kx??x??kx2?4kx． 222 ∵抛物线顶点坐标是（4，12），
1O 2 G 4 6 8 10 x ∴?k?42?4k?4?12． 2 解得k?3． 23厘米，AC＝12厘米． 2方法二：观察图象知，当x=4时，△PCQ面积为12． 此时PC＝AC－AP＝8k－4k＝4k，CQ＝4．
14k?43?12．解得k?． ∴由S?PCQ??CQ?CP，得
2223则点P的速度每秒厘米，AC＝12厘米．
2则点P的速度每秒
方法三：设y2的图象所在抛物线的解析式是y?ax2?bx?c．
∵图象过（0，0），（4，12），（8，0），
3?a??，?c?0，?4??∴?16a?4b?c?12， 解得 ?b?6， ?64a?8b?c?0.?c?0.???3∴y2??x2?6x． ①
41∵S?PCQ??CQ?CP，CP＝8k－xk，CQ＝x，
21∴y2??kx2?4kx． ②
23比较①②得k?.
23则点P的速度每秒厘米，AC＝12厘米．
2⑶①观察图象，知
线段的长EF＝y2－y1，表示△PCQ与△DCQ的面积差（或△PDQ面积）．
31?33?3②由⑵得 y2??x2?6x.（方法二，y2???8??x??x??x2?6x）
42?22?4∵EF＝y2－y1，
3339∴EF＝?x2?6x?x??x2?x，
4242∵二次项系数小于０，
27∴在0＜x＜6范围，当x?3时，EF?最大．
4（2008苏州）初三数学课本上，用“描点法”画二次函数y?ax?bx?c的图象时，列了如下表格：
2x y ? ? ?2 1?6 2?1 ?4 0 1 2 ? ? 1?2 2?2 1?2 22根据表格上的信息回答问题：该二次函数y?ax?bx?c在x?3时，y? —4 ．