Question

已知y=x^2+2mx+n的顶点在直线y= 1/2x+1/2上

Answer 1

将(1,3)代入方程,

3=1+2m+n

2m+n=2

解析式化为y=x^2+2m+m^2-m^2+2-2m=(x-m)^2-m^2-2m+2

顶点坐标为(m,-m^2-2m+2)

将顶点代入y=1/2x+1/2

-m^2-2m+2=m/2+1/2

2m^2+5m-3=0

(2m-1)(m+3)=0

m=1/2或m=-3

n=2-2m=1或n=2-2m=8

所以解析式为y=x^2+x+1

或y=x^2-6x+8