设切线方程为：y+3=k(x-1)即y=kx-k-3将直线方程代入曲线y=x²得：x²=kx-k-3,即x²-kx+k+3=0则△=(-k)²-4(k+3)=k²-4k-12=0即(k+2)(k-6)=0解得：k=-2或6∴切线方程为：y=-2x-1或y=6x-91°若k=-...