∴二次函数的表达式为y??x2?2x?15
令y=0得?x2?2x?15?0 ························································································· 5分 解得x1=－3，x2=5
y ∴点B的坐标为（5，0）······························································································ 6分 D E
（3）如图，过D作DE⊥y轴，垂足为E．∴∠DEC＝∠COB＝90°， 当BC⊥CD时，∠DCE +∠BCO＝90°，
∵∠DEC＝90°，∴∠DCE +∠EDC＝90°，∴∠EDC＝∠BCO． A O B x ECED
∴△DEC∽△COB，∴ ＝ ．·············································································· 7分
OBOC由题意得：OE＝m+1，OC＝m，DE＝1，∴EC＝1．∴
11
＝ ． OBm
C ∴OB＝m，∴B的坐标为（m，0）． ·············································································· 8分 将（m，0）代入y??x2?2x?m得：－m 2+2 m + m＝0．
解得：m1＝0（舍去）， m2＝3． ··················································································· 9分
27．（本题9分） 发现：（1）小明的这个发现正确． ················································································ 1分 理由：解法一：如图一：连接AC、BC、AB，∵AC＝BC＝5 ，AB＝10
∴AC2+BC2＝AB2 ∴∠BAC＝90°， ············································· 2分
∴AB为该圆的直径． ···································································· 3分
解法二：如图二：连接AC、BC、AB．易证△AMC≌△BNC，∴∠ACM＝∠CBN．
又∵∠BCN＋∠CBN＝90°，∴∠BCN＋∠ACM＝90°，即∠BAC＝90°， · 2分 ∴AB为该圆的直径． ··········································································· 3分
N
A D
H
E
F
M
图一
图二
图三 C B
（2）如图三：易证△ADE≌△EHF，∴AD＝EH＝1． ·················································· 4分 ADDE12∵DE∥BC，∴△ADE∽△ACB，∴ ＝ ∴ ＝ ，∴BC＝8．··························· 5分
ACCB4CB∴S△ACB＝16． ·············································································································· 6分 展开图的面积6
∴该方案纸片利用率＝ ×100%＝ ×100%＝37.5% ···························· 7分
纸板的总面积16180
探究：（3） ············································································································ 9分
361 28．（本题12分）
13
3
解：（1） 2 ············································································································· 2分
2 （2）∵在△ABC中，∠C=90°，AC=BC=4．
∴∠A＝∠B＝45°，AB=42 ，∴∠ADE＋∠AED＝135°； 又∵∠DEF＝45°，∴∠BEF＋∠AED＝135°，∴∠ADE＝∠BEF；
∴△ADE∽△BEF ·········································································································· 4分 ADAE∴ ＝ ， BEBF
3x14∴ ＝ ，∴y＝－ x2+ 2 x ····································································· 5分
y33 42 －x
1418
∴y＝－ x2+ 2 x＝－ ( x－22 )2+
3333
8
∴当x＝22 时，y有最大值＝ ················································································· 6分
316
∴点F运动路程为 cm······························································································· 7分
3
A C C
F D
F
D
第28题（1）（2）图
（3）这里有三种情况：
①如图，若EF＝BF，则∠B＝∠BEF；
E B A E
第28题（3）①图
B
又∵△ADE∽△BEF，∴∠A＝∠ADE＝45° 3∴∠AED＝90°，∴AE＝DE＝ 2 ，
23
∵动点E的速度为1cm/s ，∴此时x＝ 2 s；
2②如图，若EF＝BE，则∠B＝∠EFB；
又∵△ADE∽△BEF，∴∠A＝∠AED＝45°
∴∠ADE＝90°，∴AE＝32 ， ∵动点E的速度为1cm/s ∴此时x＝32 s；
C A
第28题（3）②图
C
D
F
F
D E B
A
14
第28题（3）③图
E B
③如图，若BF＝BE，则∠FEB＝∠EFB；
又∵△ADE∽△BEF，∴∠ADE＝∠AED ∴AE＝AD＝3，
∵动点E的速度为1cm/s ∴此时x＝3s；
3
综上所述，当△BEF为等腰三角形时，x的值为 2 s或32 s或3s．
2
（注：求对一个结论得2分，求对两个结论得4分，求对三个结论得5分）
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