It is known that the image of linear function y = KX + B passes (- 1, - 5) and intersects with the image of positive scale function y = half x at point (2, a) (1) Try to find the value of a and the analytic expression of the first-order function y = KX (2) Find the area of the triangle formed by the image of these two functions and the x-axis

It is known that the image of linear function y = KX + B passes (- 1, - 5) and intersects with the image of positive scale function y = half x at point (2, a) (1) Try to find the value of a and the analytic expression of the first-order function y = KX (2) Find the area of the triangle formed by the image of these two functions and the x-axis

Because the line y = x / 2 passes through the point (2, a), a = 1
The image of y = KX + B is solved by (- 1, - 5) and (2,1) simultaneous equations, and K = 2, B = - 3
Y = 2x-3 and X-axis intersect at the point (3 / 2,0). The height of the triangle surrounded by two function images and x-axis is the ordinate of the intersection, 1
So the area of the triangle is s = 1 / 2 times 3 / 2 times 1 = 3 / 4

Given that the image of the first-order function y = KX + B passes through points (0, - 5) and is parallel to the image of the straight line y = 12x, then the expression of the first-order function is y=______ ．

Because the image of the first-order function y = KX + B passes through the point (0, - 5) and is parallel to the image of the straight line y = 12x, then: k = 12 in y = KX + B, when x = 0, y = - 5, and substituting it into y = 12x + B, the solution is b = - 5, then the expression of the first-order function is y = 12x-5

It is known that the image of the first-order function y = KX = B is parallel to y = - x, and the expression of the first-order function is obtained through points (1,8)

∵ y = KX + B is parallel to y = - X
∴k=-1
Take (1,8) into y = - x + B to get
-1+b=8
B = 9
The analytic expression of the function is y = - x + 9 (or y = 9-x)