三次曲线相切
=>
设切点(t,t^3)
=>
斜率k=(t^3-0)/(t-1)
又k=f'(t)=3t^2
=>
t^3/(t-1)=3t^2
=>
t=1.5(t=0舍)
=>
k=6.75
=>
直线y-0=6.75(x-1)
整理得
y=6.75x-6.75
与抛物线相切
=>
相切处抛物线斜率=6.75
切点（x,6.75x-6.75）
=>
6.75=f'(x)=2ax+15/4
(=>
6.75x=2ax^2+15/4x)
且
6.75x-6.75=ax^2+15/4x-9
=>
6.75=ax^2+9且3=2ax
=>
x=-1.5