见图
y=ax^3+bx^2
(1,3)
a+b=3 (1)
y'=3ax^2-2bx
y'(1) =0
3a-2b=0 (2)
2(1)+(2)
5a=6
a=6/5
from (1)
a+b=3
6/5 +b=3
b=9/5
(a,b) = (6/5, 9/5)
y'=3ax^2-2bx
y''= 6ax -2b
y''(1) = 6a -2b = 6(6/5) - 2(9/5) = 18/5≠0
(a,b) = (6/5, 9/5)
点（1,3）为曲线y＝ax^3+bx^2的拐点
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