12. 解：(1)如图：，；(2)　 (b，a) ；

(3)由(2)得，D(1,-3) 关于直线l的对称点的坐标为(-3,1)，连接E交直线l于点Q，此时点Q到D、E两点的距离之和最小　

设过(-3,1) 、E(-1,-4)的设直线的解析式为，则

，∴，∴．由得 ，∴所求Q点的坐标为(，)

11. 解：由图象可知，点在直线上，．解得．直线的解析式为．令，可得．直线与轴的交点坐标为．令，可得．直线与轴的交点坐标为．

10. 解：(1)由直角三角形纸板的两直角边的长为1和2，

知两点的坐标分别为．

设直线所对应的函数关系式为．···························································· 2分

有解得

所以，直线所对应的函数关系式为．····················································· 4分

(2)①点到轴距离与线段的长总相等．

因为点的坐标为，

所以，直线所对应的函数关系式为．

又因为点在直线上，

所以可设点的坐标为．

过点作轴的垂线，设垂足为点，则有．

因为点在直线上，所以有．······················· 6分

因为纸板为平行移动，故有，即．

又，所以．

法一：故，

从而有．

得，．

所以．

又有．························································ 8分

所以，得，而，

从而总有．····································································································· 10分

法二：故，可得．

故．

所以．

故点坐标为．

设直线所对应的函数关系式为，

则有解得

所以，直线所对的函数关系式为．·············································· 8分

将点的坐标代入，可得．解得．

而，从而总有．························································· 10分

②由①知，点的坐标为，点的坐标为．

．····································································· 12分

当时，有最大值，最大值为．

取最大值时点的坐标为．　　 14分

9.

解：(1)······························· (3分)

(2)由题意，可得：．

．·········································································································· (5分)

当时，．

造这片林的总费用需45 000元．········································································· (8分)

8.

解：(1)；

(2)，

(天)

答：乙队单独完成这项工程要60天.

(3)(天)

答：图中的值是28.

7. 解：(1)设与之间的关系为一次函数，其函数表达式为··················· 1分

将，代入上式得，

　 解得

········································································································· 4分

验证：当时，，符合一次函数；

当时，，也符合一次函数．

可用一次函数表示其变化规律，

而不用反比例函数、二次函数表示其变化规律．···························································· 5分

与之间的关系是一次函数，其函数表达式为································ 6分

(2)当时，由可得

即货车行驶到处时油箱内余油16升．········································································ 8分

(3)方法不唯一，如：

方法一：由(1)得，货车行驶中每小时耗油20升，····················································· 9分

设在处至少加油升，货车才能到达地．

依题意得，，···························································· 11分

解得，(升)··································································································· 12分

方法二：由(1)得，货车行驶中每小时耗油20升，····················································· 9分

汽车行驶18千米的耗油量：(升)

之间路程为：(千米)

汽车行驶282千米的耗油量：

(升)······························································································· 11分

(升)················································································ 12分

方法三：由(1)得，货车行驶中每小时耗油20升，····················································· 9分

设在处加油升，货车才能到达地．

依题意得，，

解得，············································································································· 11分

在处至少加油升，货车才能到达地．·························································· 12分

6. 解：(1)特征数为的一次函数为，

，

．

(2)抛物线与轴的交点为，

与轴的交点为．

若，则，；

若，则，．

当时，满足题设条件．

此时抛物线为．

它与轴的交点为，

与轴的交点为，

一次函数为或，

特征数为或．

5. 解⑴y＝(63－55)x+(40－35)(500－x)……………3分

＝2x+2500。即y＝2x+2500(0≤x≤500)，………………4分

⑵由题意，得55x+35(500－x)≤20000，………………6分

解这个不等式，得x≤125，………………………………7分

∴当x＝125时,y_最大值＝3×12+2500＝2875(元),…………9分

∴该商场购进A、B两种品牌的饮料分别为125箱、375箱时，能获得最大利润2875元.………………………………………………………………10分

4. 解：设这个一次函数的解析式为y=kx+b.

则解得，函数的解析式为y=-2x+3.

由题意，得得，所以使函数为正值的x的范围为

3. 解：(1)，所以不能在60天内售完这些椪柑，

　　　 (千克)

　　　 即60天后还有库存5000千克，总毛利润为

　　　 W=；

　 (2)

　　 要在2月份售完这些椪柑，售价x必须满足不等式

　　 解得

　　 所以要在2月份售完这些椪柑，销售价最高可定为1.4元/千克。