When m =? The image of the function y = MX + (m-2) crosses the origin; when m -----, y decreases with the increase of X?

When m =? The image of the function y = MX + (m-2) crosses the origin; when m -----, y decreases with the increase of X?

1.m-2=0
m=2
When m = 2, the image of function y = MX + (m-2) crosses the origin
2.k=m＜0
When m < 0 -, y decreases with the increase of X

Draw the image of inverse scale function y = 12 / X by point tracing method
Ask for pictures

It is known that the image of a linear function y = (M + 2) x + m square + 2m-2 passes through the point (0,1). (1) find the function value y of this function, which increases with the increase of x value, and find M
(2) is the intersection a of the image of this function on the positive or negative half axis of the X axis?

Substituting the point (0,1), we get m square + 2m-2 = 1
So m = - 3 (rounding off) or M = 1
(2) Let y = 0, then x = - 1 / 3
So, the intersection point a of the image of this function is on the negative half of the X axis

On the image of inverse scale function y = 6x, the number of points whose coordinates are integers is ()
A. 8B. 6C. 4D. 2

In the inverse proportion function, because y = 6x, xy = 6, so (1,6), (2,3), (- 1, - 6), (- 2, - 3), (6,1), (3,2), (- 6, - 1), (- 3, - 2)

If the image of the first-order function y = 2mx + 6-2m & # passes through the origin, and the value of Y increases with the increase of X, then M=

∵ a function passes through the origin
The substitution of point (0,0) leads to
6-2m²=0
m1=√3,m2=-√3
The value of Y increases with the increase of X
∴2m>0
∴m=√3

One branch of the following inverse scale function image in the third quadrant is in the third quadrant
Y = minus 3 / X. is this case K less than 0? The sign is not on K, it is outside

Y = minus 3 / X. in this case, K is less than 0
Then k = - 3
As long as K ＜ 0, it must be in the second and fourth quadrant

If the image of a linear function y = 2mx + M2-4 passes through the origin, then the value of M is ()
A. 0b. 2C. - 2D. 2 or - 2

∵ when the image of the first-order function y = 2mx + M2-4 passes through the origin, ∵ 0 = 0 + M2-4, that is, M2 = 4, the solution is m = ± 2

As shown in the figure, it is known that the image of positive scale function y = x and inverse scale function y = 1x intersects at two points a and B. (1) the coordinates of two points a and B are obtained; (2) according to the image, the range where the value of positive scale function is larger than that of inverse scale function x is obtained

(1) According to the meaning of the problem, the coordinates of a and B satisfy the solution of the equation system y = xy = LX, X1 = 1y1 = 1, X2 = − 1Y2 = − 1, the coordinates of a and B are a (1,1), B (- 1, - 1); (2) according to the image, when - 1 < x < 0 or x > 1, the value of positive proportion function is greater than that of inverse proportion function

If the image of a linear function y = 2mx + M2-4 passes through the origin, then the value of M is ()
A. 0b. 2C. - 2D. 2 or - 2

∵ when the image of the first-order function y = 2mx + M2-4 passes through the origin, ∵ 0 = 0 + M2-4, that is, M2 = 4, the solution is m = ± 2

If the images of the first-order function y = 2x and the inverse scale function y = 2 / X pass through points a and B, point a is known to be in the third quadrant
If the coordinates of point C are (T, 0) and T > 0, the quadrilateral ABCD is a parallelogram. When the value of T is, point D is on the y-axis

A: the intersection of y = 2x and y = 2 / X is a (- 1, - 2), point B (1,2)
The slope of CD line is the same as that of AB line, which is 2
Y-0 = 2 (x-t), that is, y = 2x-2t
Let x = 0, y = - 2T
So point d coordinates are (0, - 2t)
The slope of BC line is: (2-0) / (1-T) = 2 / (1-T)
Because AD / / BC, the slope of ad is the same as that of BC
(-2+2t)/(-1-0)=2/(1-t)
The solution is: T = 2 (the contradiction between T = 0 and condition T > 0 needs to be eliminated)
So when t = 2, D is on the y-axis