f'(x)=3x²+2k=f'(0)=2f(0)=1所以切线方程为y=2(x-0)+1即y=2x+1法线与切线垂直则设法线方程为y=-x/2+c法线过(0,1);1=0+c解得c=1所以法线方程为y=-x/2+1即x+2y-1=0
