Question

已知抛物线y=x^2-4x+k的顶点A在直线y=-4x-1上,设抛物线与x轴交于BC两点,求抛物线的顶点坐标

Answer 1

y=x²-4x+k =x² -4x + 4 -4 +k = (x-2)² + k-4

顶点A(2,k-4)

x = 2,y = -4*2 -1 = -9

k-4 = -9,k = -5

A(2,-9)