Question

1.若将二次函数y=x2-4x+7配方为y=（x-h）2+k的形式,则y= 2.抛物线y=x2

Answer 1

1 y=(x-2)^2-4+7.

∴ y=(x-2)^2+3.

2.y=x^2-2x+2.

y=(x-1)^2-1+2.

y=(x-1)^2+1.

∴抛物线y=x^2-2x+2的顶点坐标为:(1,1).

3.令y=x^2-5x-6=0.

则 x^2-5x-6=0.

(x-6)(x+1)=0.

x-6=0,x=6,

x+1=0,x=-1.

∴抛物线与X轴的两个交点为(6,0)和(-1,0).