所以此函數(shù)若為二次函數(shù),則b2-4ac=[-(4a-1)]2-4(a-3)×4a=0,
即40a+1=0,
解得:a=-
1 |
40 |
若a=0,二次函數(shù)圖象過原點(diǎn),滿足題意.
若此函數(shù)為一次函數(shù),則a-3=0,所以a=3.
所以若關(guān)于x的函數(shù)y=(a-3)x2-(4a-1)x+4a的圖象與坐標(biāo)軸只有兩個(gè)交點(diǎn),則a=3或0或-
1 |
40 |
故選:D.
1 |
40 |
1 |
40 |
1 |
40 |
1 |
40 |
1 |
40 |