The intersection coordinates of the image of the primary function y = kx-4 and the positive proportional function y = KX are (2, - 1) (1) write the expressions of these two functions ⑵ find the area of the triangle surrounded by the two images and the X axis (3) let the intersection of the straight line y = kx-4 and the coordinate axis be a and C respectively. If point B is on the straight line y = KX and the abscissa is four, calculate the area of the quadrilateral ABCO. (o is the coordinate origin)

The intersection coordinates of the image of the primary function y = kx-4 and the positive proportional function y = KX are (2, - 1) (1) write the expressions of these two functions ⑵ find the area of the triangle surrounded by the two images and the X axis (3) let the intersection of the straight line y = kx-4 and the coordinate axis be a and C respectively. If point B is on the straight line y = KX and the abscissa is four, calculate the area of the quadrilateral ABCO. (o is the coordinate origin)

1. Make y = - 1 and x = 2 in y = kx-4, then k = 3 / 2, y = 3 / 2X-4, and y = - 1 and x = 2 in y = KX, so k = - 1 / 2, y = - 1 / 2x 2. Y = kx-4, the images all pass through points P (2, - 1), - 1 = k * 2-4, k = 3 / 2, y = 3 / 2X-4, and the images of y = MX all pass through points P (2, - 1), - 1 = m * 2, M = - 1 / 2, y = - 1 / 2x, y = 3 / 2X-4, solution, X

The images of the positive proportional function y = KX and the primary function y = ax + B pass through point a (1,2), and the image of the primary function intersects the X axis at point B (4,0). Find the expressions of the positive proportional function and the primary function

Through the point (1, 2) of the image with the positive scale function y = KX,
Get: k = 2,
Therefore, the expression of the positive proportional function is y = 2x;
The image of the first-order function y = ax + B passes through points (1,2) and (4,0)
have to
a+b=2
4a+b=0
Solution: a = − 2
3,b=8
3,
The expression of the primary function is y = − 2
3x+8
3.

It is known that the solution set of inequality log2 (ax Λ 2-3x 6) > 2 about X is {x|xb} (1) Find the value of a and B; (2) Solve the inequality about X: (AX b) (C-X) > 0 (C is a constant)

The war of mosquito repellent incense,
(1) When a = 1, B = 2, (2) (X-2) (C-X) > 0 (X-2) (x-C) 2, the solution set is 2

If x is less than 3, the solution set of inequality ax greater than 3x + 1 is

ax > 3x + 1
So (3 - a) x < - 1
Because a < 3
So 3 - a > 0
So x < - 1 / (3 - a)

Find the solution of this equation with the image of the function: x square - 3x + 2 = 0

Let y = x square - 3x + 2,
Drawing in a plane rectangular coordinate system,
The intersection of the curve in the graph on the X axis is the solution of the original agenda

Solving inequality by function image: x-4 < 3x + 1. Then? Excuse me, I simplified this problem to 2x-5 < 0, Yes - 2x-5 < 0

Make the image, and the intersection with the X axis is (- 2.5,0)
It can be seen that the image below the Y axis corresponds to x > - 2.5