y=x？ -2(m- 1)x+(m2-2m+ 1)+2m？ -2-( m2-2m+ 1)

= x？ -2(m- 1)x+(m- 1) 2+2m？ -2-( m2-2m+ 1)

= [x - (m- 1)]？ + m？ + 2m -3

For quadratic function y=ax? For +bx+c, when a is greater than zero, the function has a minimum value, which is the vertex of the function image.

For this function, [x-(m- 1)]? The time function has a minimum value, so the vertex is m- 1 and the ordinate is m? + 2m -3 .

That is to say, the vertex of the quadratic function is always at (m- 1, m? + 2m -3).

With the change of m value, the vertex position is also changing. According to the vertex coordinates, you can know the vertex coordinates y 1 (that is, m? The relationship between +2m -3) and x 1 (that is, m- 1) is: y 1=(x 1+2)? -4

So the image vertex of the original quadratic function is always on the quadratic function image. (Not a linear function, but a quadratic function. )

(2) Let the abscissas of the intersection of the image and the X axis be X 1, X2, X2 >； respectively; X 1. Then the length of the line segment is twice that of the root number 3, that is, x2-x 1 = twice that of the root number 3.

According to Vieta's theorem, the unary quadratic equation AX 2+BX+C = 0 (A is not equal to 0).

The two X 1, X2 of the equation and the coefficients a, b and c of the equation satisfy X 1+X2=-(b/a), and x1* x2 = c/a.

x 1+X2 = 2(m- 1)； X 1*X2= 2m？ -2

Further (X2-X 1)? =(X 1+X2)？ -4X 1*X2

12=4(m- 1)？ -8m？ +8

0=-4m？ -8 meters

m？ +2m=0, solve the equation to get m 1=0, m2=-2.

Bringing in the original analytical formula, we can know that when the length of the line segment cut by the quadratic function image on the X axis is twice as long as the root number 3, the resolution function is:

y = x？ +2x-2 or y = x? + 6x + 6