The image of the linear function Y1 = KX + B and the inverse scale function y2 = m / X intersect at two points a (- 2,1), B (1, n) 1) Find the expression of first-order function and inverse proportion function 2) Write Y1 > Y2

The image of the linear function Y1 = KX + B and the inverse scale function y2 = m / X intersect at two points a (- 2,1), B (1, n) 1) Find the expression of first-order function and inverse proportion function 2) Write Y1 > Y2

x=-2,y=1
So 1 = m / 2
m=2
Over b
n=m/1=2
Substituting AB into Y1 = KX + B
1=-2k+b
2=k+b
k=1/3,b=5/3
So Y1 = x / 3 + 5 / 3 and y2 = 2 / X
If Y1 > Y2, Y1 is above Y2
So it's - 2

As shown in the figure, it is known that the image of the inverse scale function Y1 = MX intersects the image of the primary function y2 = KX + B at two points a (- 2,1), B (a, - 2). (1) find the analytic expressions of the inverse scale function and the primary function; (2) if the image of the primary function y2 = KX + B intersects the y-axis at point C, find the area of △ AOC (o is the origin of setting); (3) find the value range of X when Y1 > Y2

(1) The image of ∵ function Y1 = MX passes through point a (- 2,1), that is, 1 = m − 2; (1 minute) ∵ M = - 2, that is, Y1 = - 2x, (2 minutes) and ∵ point B (a, - 2) on Y1 = - 2x, ∵ a = 1, ∵ B (1, - 2). (3 minutes) and ∵ function y2 = KX + B passes through two points a and B, that is, ∵ 2K + B = 1K + B = - 2. (4 minutes) of solution

The intersection of the image of inverse scale function Y1 = m / X and the image of linear function y2 = KX + B at two points a (- 2,1) B (a, - 2) is obtained
Given that the image of inverse scale function Y1 = m / X and the image of primary function y2 = KX + B intersect at two points a (- 2,1) B (a, - 2), find: (1) find the analytic formula of inverse scale function and primary function; (2) if the image of primary function y2 = KX + B intersects Y axis at point C, find the area of AOC
I'm going to have an exam

(1) The coordinate of point a is brought into Y1 to get m = - 2, so Y1 = - 2 / X;
The coordinate of point B is brought into Y1, and a = 1 is obtained;
The coordinates of points a and B are brought into Y2 to solve the equations, k = b = - 1, so y2 = - X-1;
(2) Let x = 0 in y2 = - X-1, get the coordinates of point C (0, - 1), and draw a picture to get the area s of ∈ AOC = 1

It is known that in the quadratic function y = AX2 + BX + C, the partial correspondence between the independent variable x and the function value Y in a ≠ 0 is shown in the following table
X minus two thirds minus one minus one half 0 half one half one half three
Y minus 4 / 5 minus 2 / 4 minus 9 / 4 minus 2 / 4 minus 50 / 4 seven
What is the analytic expression of the quadratic function
I forgot that my books were not at home. I had no choice but to ask for help. I forgot to hand them in tomorrow

From the meaning of the title: C = 1 and - B / 2A = 0, so B = 0,
Because when x = - 1, y = - 2, so - 2 = a * 1 + 1, so a = - 3, so y = - 3x ^ 2 + 1
Bring in other groups to test, and the analytical expression is valid