Question

直线y=kx+3与抛物线y=ax^2-4x+1交于点A【-2,1】和点B,则B点坐标为

Answer 1

将A(-2,1)代入y=kx+3和y=ax^2-4x+1,得

1=-2k+3

1=a*(-2)^2-4*(-2)+1

解得k=1,a=-2

将y=kx+3=x+3代入y=ax^2-4x+1=-2x^2-4x+1,得

2x^2+5x+2=0

(x+2)(2x+1)=0

x=-2或x=-1/2

则另一点B的坐标为(-1/2,-1/2+3),也即B(-1/2,5/2)