Question

已知抛物线y=-X²;+(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

Answer 1

OA=|x1|

OB=|x2|

OC=3|m+1|

由OA²+OB²=2OC+1得

x1^2+x2^2=6|m+1|+1=(x1+x2)^2-2x1*x2

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

(1)m+1≥0时

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

m=3或1

(2)m+10

故m-3,m=-3或-5舍去

m=3时x1=-3,x2=4,|x1||x2|

所以抛物线的解析式为y=-x^2-x+6