It is known that the functions Y1 = kx-2 and y2 = - 3x + B are relative to point a Given that the functions Y1 = kx-2 and y2 = - 3x + B intersect at point a (2, - 1) (1) Find the value of K and B and draw the image of two functions in the same coordinate system (2) Using the image to find out: when x takes what value, there are: ① Y1 The first question is the second and the third

It is known that the functions Y1 = kx-2 and y2 = - 3x + B are relative to point a Given that the functions Y1 = kx-2 and y2 = - 3x + B intersect at point a (2, - 1) (1) Find the value of K and B and draw the image of two functions in the same coordinate system (2) Using the image to find out: when x takes what value, there are: ① Y1 The first question is the second and the third

(1) Substituting the coordinates of point a into Y1, we can get 2k-2 = - 1, that is, k = 12; substituting the coordinates of point a into Y2, we can get: - 6 + B = - 1, that is, B = 5; the analytic expressions of the two functions are: Y1 = 12x-2, y2 = - 3x + 5; as shown in the figure; (2) from the image, we can see: ① when x < 2, Y1 < Y2; ② when x ≥ 2, Y1 ≥ Y2; 3) ∵ straight line Y1