Question

已知二次函数y=x2+（a+1）x+b（a,b为常数）,

Answer 1

y(3)=9+3(a+1)+b=3, ===> b=-9-3a

y-x=x^2+ax+b>=0 ===> delta=a^2-4b=a^2+36+12a=(a+6)^2 a=-6

b=9

y=x^2-5x+9=(x-5/2)^2+11/4

顶点为(5/2, 11/4)

顶点到原点的距离为√(25/4+121/16)=(√221)/4