导数切线问题

y=x^3,y'=3x²,过(a,a³)的切线为 y = 3a²(x-a) + a³
该切线过(0,2),即2 = 3a²(0-a) + a³ = -2a³,a = -1
切线：y = 3(x+1) - 1 = 3x + 2,斜率3
y=x^2+mx+2,y'=2x+m,y'=3代入,得x=(3-m)/2,y = 17/4 - m²/4
代入切线方程：
17/4 - m²/4 = 3((3-m)/2) + 2
解得m=3