假定切线的斜率为k,又因切线过点A(1,-3),所以切线方程为：y+3=k(x-1)即y=k(x-1)-3
y=x^2在切点(x,y)处的斜率为y'=2x,切点(x,y)又在直线y=k(x-1)-3上,所以切点(x,y)满足
y=2x(x-1)-3=2x^2-2x-3,x^2=2x^2-2x-3,
x^2-2x-3=0,(x+1)(x-3)=0解得x=-1或x=3,于是
k=2x=-2或k=2x=6,切线方程为y=-2(x-1)-3=-2x-1,y=-2x-1或y=6(x-1)-3=6x-9,y=6x-9