1. We know the linear function y = (M + 3) x + M-9. (1) when m is the value, its image passes through the origin; (2) when m is the value (3) write a value of m to make his image pass through the second and fourth image limits. 2. We know the function of x y = (3K + 12) x + K + 1. (1) when the function is a linear function, can the value of K be - 4? - 2? Please explain the reason. (2) when the value of K is what, the function is a positive proportion function?

1. We know the linear function y = (M + 3) x + M-9. (1) when m is the value, its image passes through the origin; (2) when m is the value (3) write a value of m to make his image pass through the second and fourth image limits. 2. We know the function of x y = (3K + 12) x + K + 1. (1) when the function is a linear function, can the value of K be - 4? - 2? Please explain the reason. (2) when the value of K is what, the function is a positive proportion function?

(1) ∵ a linear function y = (M + 3) x + M-9;
∴m+3≠0,m-9=0
∴m≠-3,m=9
When m = 9, its image passes through the origin
(2) ∵ a linear function y = (M + 3) x + M-9, whose image passes through points (1,9);
By substituting the point (1,9) into y = (M + 3) x + M-9, we get the following result:
9=m+3+m-9
-2m=-15
m=7.5
When m = 7.5, its image passes through points (1,9);
(3) ) ∵ a linear function y = (M + 3) x + M-9, his image passes through the second and fourth image limits
∴m+3﹤0
∴m﹤-3
When m = - 4, his image passes through the second and fourth image limits
2. We know the function y = (3K + 12) x + K + 1
(1) When the function is a linear function, can the value of K be - 4? - 2?
∵ y = (3K + 12) x + K + 1
∴3K+12≠0
K=-4
When k = - 2, when the function is linear
2) When the value of K is what, the function is a positive proportional function?
∵ y = (3K + 12) x + K + 1
∴3K+12≠0,K+I=0
∴K≠-4,K=-1
When k = - 1, y = (3K + 12) x + K + 1

1. Given that the image of the first-order function y = (m-2) x + m ^ 2-6 intersects with the Y axis, the coordinate of the intersection is - 2, find the value of M
2. When m and N are values, the function y = (5m-3) x ^ (2-N) + (M + n)
(1) Is a linear function (2) is a positive proportional function

∵ the image of the linear function y = (m-2) x + M & sup2; - 6 intersects the Y axis, and the coordinate of the intersection is - 2
When x = 0, y = - 2, substituting into the equation - 2 = M & sup2; - 6; M & sup2; = 4, the solution is m = - 2 or M = 2 (x coefficient is 0, rounding off)
The linear function equation is y = - 4x-2
2. (1) is a linear function
2-N = 1 and 5m-3 ≠ 0, the solution is n = 1, m ≠ 0.6
(2) It's a positive scale function
2-N = 1 and 5m-3 ≠ 0 and M + n = 0, the solution is n = 1, M = - 1

It is known that the image of the first-order function y = (m-2) x + M & sup2; 6 intersects with the y-axis, and the coordinate of the intersection is - 2

According to the meaning of the question, I think it should be y = (m-2) x + M & sup2; - 6
∵ the image intersects the Y axis, and the coordinate of the intersection is - 2
When x = 0, y = - 2
Substituting x = 0, y = - 2 into the equation, we get
-2 = M & sup2; - 6; M & sup2; = 4, M = ± 2
∵ the coefficient (m-2) of linear function x is not 0, that is, m ≠ 2
∴m=-2