Substituting the coordinates of point A and point B into the functional formula of y2, we get K =-0.2 and M = 3.6, so Y2 =-0.2x+3.6, Y 1 cannot be determined.

Draw the image of y2, and then discuss with the image:

(1) When a >: 0, the opening of the quadratic function y 1 is upward, and y1>; The value range of x of y2 is: x >；; 8 or x, 4, this problem should be discussed in different situations: when a: a is greater than 0, the opening of the quadratic function is upward, and when y 1 is greater than y2, the value range of x is: x is greater than 8 or x is less than -2.

B: When a is less than 0, the opening of the quadratic function is downward, and the value range of y 1 x greater than y2 is: greater than -2X and less than 8. , 1, please answer the math function homework in detail, thank you! (5 22:20:55)

Known quadratic function y 1=ax? The image of +bx+c(a≠0) y2=kx+m(k≠0) intersects at point A (-2,4), while b (8 8,2) can make y 1 > y2 hold, and the value range of x is-