设切点(a,a³+2)
y'=3x²
斜率k=3a²
所以切线是y-(a³+2)=3a²(x-a)
y=kx过原点
-a³-2=-3a²
a³-3a²+2=0
(a-1)(a²-2a-2)=0
a=1,a=1±√2
k=3a²
k=3
k=9-6√2
k=9+6√2