设P（x，x2?2x?3），则Q（x，－x－1），PQ??x2?x?2．
S?APG?S?APQ?S?GPQ?当x?1(?x2?x?2)?3 ……………………9分 21时，△APG的面积最大 2?1?2此时P点的坐标为?,?2715??，S?APG的最大值为． ……………………10分 4?811．（本小题12分）解：（1）解方程x2－10x＋16＝0得x1＝2，x2＝8
∵点B在x轴的正半轴上，点C在y轴的正半轴上，且OB＜OC ∴点B的坐标为（2，0），点C的坐标为（0，8） 又∵抛物线y＝ax2＋bx＋c的对称轴是直线x＝－2 ∴由抛物线的对称性可得点A的坐标为（－6，0）
∴A、B、C三点的坐标分别是A（－6，0）、B（2，0）、C（0，8） （2）∵点C（0，8）在抛物线y＝ax2＋bx＋c的图象上
∴c＝8，将A（－6，0）、B（2，0）代入表达式y＝ax2＋bx＋8，得
??0＝36a－6b＋8
?
?0＝4a＋2b＋8?
?a＝－3
解得?8
b＝－?3
2
28
∴所求抛物线的表达式为y＝－x2－x＋8
33（3）∵AB＝8，OC＝8
1
∴S△ABC ＝×8×8=32
2
（4）依题意，AE＝m，则BE＝8－m， ∵OA＝6，OC＝8， ∴AC＝10 ∵EF∥AC ∴△BEF∽△BAC ∴
40－5mEFBEEF8－m
＝ 即＝ ∴EF＝ ACAB1084
4过点F作FG⊥AB，垂足为G，则sin∠FEG＝sin∠CAB＝
5∴
FG4440－5m＝ ∴FG＝·＝8－m EF554
11∴S＝S△BCE－S△BFE＝（8－m）×8－（8－m）（8－m）
22111
＝（8－m）（8－8＋m）＝（8－m）m＝－m2＋4m 222自变量m的取值范围是0＜m＜8 （5）存在． 理由：
111
∵S＝－m2＋4m＝－（m－4）2＋8 且－＜0，
222∴当m＝4时，S有最大值，S最大值＝8
∵m＝4，∴点E的坐标为（－2，0）
∴△BCE为等腰三角形．
12.（12分）已知：如图14，抛物线y??直线y??
323x?3与x轴交于点A，点B，与直线y??x?b相交于点B，点C，443x?b与y轴交于点E． 4（1）写出直线BC的解析式． （2）求△ABC的面积．
（3）若点M在线段AB上以每秒1个单位长度的速度从A向B运动（不与A，B重合），同时，点N在射线BC上以每秒2个单位长度的速度从B向C运动．设运动时间为t秒，请写出△MNB的面积S与t的函数关系式，并求出点M运动多少时间时，△MNB的面积最大，最大面积是多少？
解：（1）在y??32x?3中，令y?0 4C 3??x2?3?0
4?x1?2，x2??2
······································· 1分 ?A(?2，0)，B(2，0) ·又
点B在y?? y N M D O x 3x?b上 43?0???b
23b?
233·································································· 2分 ?BC的解析式为y??x? ·
4232?y??x?3?x1??1???4（2）由?，得?9
33y1??y??x???4??429???C??1，?，B(2，0)
4???x2?2 ············································· 4分 ?y?0?29 ················································································· 5分 4199··········································································· 6分 ?S△ABC??4?? ·
242（3）过点N作NP?MB于点P EO?MB ?NP∥EO
················································································· 7分 ?△BNP∽△BEO ·
BNNP ··························································································· 8分 ??BEEO?AB?4，CD?由直线y??33?3?x?可得：E?0，? 42?2?35，则BE?
22?在△BEO中，BO?2，EO?62tNP，?NP?t ··········································································· 9分 ?5352216?S?t(4?t)
25312······································································· 10分 S??t2?t(0?t?4) ·
55312··············································································· 11分 S??(t?2)2? ·
5512此抛物线开口向下，?当t?2时，S最大?
512?当点M运动2秒时，△MNB的面积达到最大，最大为． ························ 12分
5?
二次函数动点问题解答方法技巧（含例解答案）33956
