| 1 |
| 2 |
解:(1)当a=| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 4 |
(2)当a=0时,y2=-x,| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 4 |
| 1 |
| 2 |
②当-1≤a<0时,如图3,| 1 |
| 2 |
=2-(1+a)2;
S=S△CEF=| 1 |
| 2 |
解:(1)当a=| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 4 |
(2)当a=0时,y2=-x,| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 4 |
| 1 |
| 2 |
②当-1≤a<0时,如图3,| 1 |
| 2 |
=2-(1+a)2;
S=S△CEF=| 1 |
| 2 |
(2012·铁岭)如图所示,在平面直角坐标系中,直线OM是正比例函数y=-| 3 |
| 3 |
| 3 |
| 3 |
为t(秒).
已知:如图,在平面直角坐标系中,A、B两点分别在x轴,y轴的正半轴上,点A(6,0),∠BAO=30°.