Question

已知二次函数图像的最小值是-5,且图像经过点(1,3)与(-1,-3),此函数的关系式是

Answer 1

y=a(x-b)^2-5

a>0

3=a(1-b)^2-5

a(1-b)^2=8 (1)

-3=a(-1-b)^2-5

a(-1-b)^2=2 (2)

(1-b)^2/(1+b)^2=4

3b^2+10b+3=0

1 3

3 1

(b+3)(3b+1)=0

b=-3 b=-1/3

a(1-b)^2=8

b=-3 a=1/2

y=1/2(x+3)^2-5

b=-1/3 a=9/2

y=9/2(x+1/3)^2-5