f(x)=x^3+ax^2+(a-3)x
f'(x)=3x^2+2ax+(a-3)
又f‘(x)是偶函数
∴f'(-x)=f(x)
3x^2+2ax+(a-3)=3x^2-2ax+(a-3)
2ax=-2ax
4ax=0
a=0
f(x)=x^3-3x
f(2)=2^3-3*2=2
f'(x)=3x^2-3
f'(2)=3*2^2-3=9
所以
y=f(x)=(2,f(2))处的切线方程为y=9x-16