If the image of the inverse scale function y = (m-1) x ^ 3-x ^ 2 is in the second or fourth quadrant, then the value of M is

If the image of the inverse scale function y = (m-1) x ^ 3-x ^ 2 is in the second or fourth quadrant, then the value of M is

Because it is an inverse proportional function, the coefficient of the cubic term is 0, that is, M-1 = 0, so m = 1

The second and fourth quadrant X of inverse scale function image increases with the increase of Y?
After I saw the image, I think it's not right to describe it in inverse proportion! x

It's not right
Should be:
Inverse scale function y = K / X
When k

If the image of inverse scale function y = (1-k) x passes through two or four quadrants, then the value range of K is smaller

The image with inverse scale function y = (1-k) x passes through two or four quadrants, indicating that 1-k < 0
So the range of K is k > 1

If the two branches of the image with inverse scale function y = m − 5x are in the second and fourth quadrants respectively, then the value range of M is ()
A. m＜0B. m＞0C. m＜5D. m＞5

From the meaning of the title, we can get m-5 < 0, that is, m < 5

If the image of inverse scale function y = (K-3) / X is located in the second and fourth quadrants, then the value range of K is ﹤ 0

k-3

Give two inverse scale functions (1) with the following characteristics: the image is distributed in the first three quadrants (2) the image is in each quadrant
It is known that the image of the inverse scale function y = K / X and the linear function y = MX + B intersect at P (- 2,1) and Q (1, n)
(1) The analytic formula of inverse proportion function
(2) Finding the value of n
(3) Finding the analytic expression of the first order function y = MX + B

(1) Substituting P (- 2,1) into y = K / x, we get
1=k/(-2)
∴k=-2
The analytic expression of inverse scale function is y = - 2 / X
(2) Substituting Q (1, n) into y = - 2 / X
N = - 2 / 1 = - 2
(3) Substituting P (- 2,1), q (1, - 2) into y = MX + B, we get
{-2m+b=1
m+b=-2
The solution is {M = - 1
b=-1
The analytic expression of a linear function is y = - X-1

1. The image with known function y = (K-2) x + (K + 2) passes through the origin to get k value; 2. The image with known function y = 2x + 4 passes through the point (m, 8) to get m value
It's worked out in five minutes. I'm offering 50 more

1 y = (K-2) x + (K + 2) passing through the origin, then when x = 0, y = 0, that is, K + 2 = 0, k = - 22, substituting (m, 8) into the y = 2x + 4 of a function, we can get 2m + 4 = 8, 2m = 4, and the solution is m = 2. I wish you happy and hope it will help you

It is known that the linear function y = (3-K) x + 2-k. (1) when the value of K is, the image passes through the origin? (2) when the value of K is, y decreases with the increase of X?
(3) When k is what, its image is parallel to the straight line y = - x? (4) when k is what, its image does not pass through the second quadrant?

y=(3-k)x+2-k
(1)0=2-k
k=2
(2)3-k3
(3)3-k=-1
k=4
(4)x=0 y=2-k
2-k2
3-k>0 k

The linear function y = 5kx-5k + 3, when k=___ The image passes through the origin
why?

When k = 3 / 5
Because if you let this image pass through the origin, this function is a linear function
Then - 5K + 3 should be equal to 0

The linear function y = 5kx-5k + 3, when k is more or less through the origin, when k is equal to 2, what is the intersection coordinates of the image and X axis

The first-order function y = 5kx-5k + 3, when k is over the origin, that is, when x = 0, y = - 5K + 3 = 0, the solution is k = 3 / 5
When k is equal to 2, the first-order function is y = 10x-7, let y = 0, and the solution is x = 7 / 10,
That is, the intersection coordinates of the image and the X axis are (7 / 10,0)