解答:解:(1)把A(1,0),B(0,-1)代入y=a(x-1)
2+k解得
,
所以經(jīng)過A點(diǎn)、B點(diǎn)的拋物線的解析式為y=-(x-1)
2,
當(dāng)x=-1時(shí),y=-(x-1)
2=-4;當(dāng)x=2時(shí),y=-(x-1)
2=-1;當(dāng)x=4時(shí),y=-(x-1)
2=-9,
所以點(diǎn)C、D、E都不在經(jīng)過點(diǎn)A和B的拋物線上;
(2)把A(1,0),C(-1,2)代入y=a(x-1)
2+k解得a=
,
所以經(jīng)過A點(diǎn)、C點(diǎn)的拋物線的解析式為y=
(x-1)
2,
當(dāng)x=2時(shí),y=(x-1)
2=1;當(dāng)x=4時(shí),y=(x-1)
2=9,
所以點(diǎn)A、點(diǎn)C、點(diǎn)D在拋物線y=(x-1)
2上;點(diǎn)E都不在經(jīng)過點(diǎn)A和C的拋物線上;
(3)把B(0,-1),C(-1,2)代入y=a(x-1)
2+k解得
,
所以經(jīng)過B點(diǎn)、C點(diǎn)的拋物線的解析式為y=(x-1)
2-2,
當(dāng)x=2時(shí),y=(x-1)
2-2=-1;當(dāng)x=4時(shí),y=(x-1)
2-2=7,
所以點(diǎn)A、點(diǎn)C、點(diǎn)D在拋物線y=(x-1)
2上;
所以這五個(gè)點(diǎn)中拋物線y=a(x-1)
2+k經(jīng)過三個(gè)點(diǎn)的拋物線有1條.
故選A.