Question

已知抛物y=-x^2+(m-2)x+3(m+1)交x轴于A(x1,0),B(x2,0)两点,交y轴正半轴于点C,且x1＜

Answer 1

x = 0, y = 3(m+1)

C(0, 2(m+1))

OC = 3(m+1) (1)

y = -x^2+(m-2)x+3(m+1) = -[x^2 -(m-2)x -3(m+1)]

= -(x + 3)(x -m-1)

x1 = -3, x2 = m+1 (m+1 > 0, 否则抛物线与y轴正半轴没有交点)

OA^2 = (-3)^2 = 9

OB^2 = (m+1)^2

OA^2 + OB^2 = 2OC + 1

9 + (m+1)^2 = 6(m+1) + 1

m^2 -4m +3 = 0

(m-1)(m-3) =0

m = 1或 m =3

m=1: y = -x^2 -x +6

m=3: y = -x^2 + x +12