If the image of quadratic function y = kx2-6x + 1 intersects with X axis, then the value range of K is

If the image of quadratic function y = kx2-6x + 1 intersects with X axis, then the value range of K is

The x-axis is y = 0
So KX & # 178; - 6x + 1 = 0 has a solution
K = 0, obviously
k≠0
Then △ > = 0
36-4k>=0
k

Given the function y = (k-1) x + k2-1, when the value of K is, it is a positive proportional function

Because it's a positive scale function
So K & sup2; - 1 = 0,
The solution is k = 1 or - 1,
When k = 1, the coefficient of X is 0, which is not a positive proportional function,
So k = 1

It is known that the abscissa and ordinate of the point on the image with positive scale function y = K1X are opposite to each other, and the y of inverse scale function y = K2 / X decreases with the increase of X, once
The function y = - K2 ^ 2-2k2 + K1 + 5 obtains the value of K1 through (- 2,4), and obtains the analytic expressions of inverse proportion function and first-order function
It should be the function y = - K2 ^ 2x-2k2 + K1 + 5. After (- 2,4), find the value of K1, find the analytic formula of inverse proportion function and primary function

∵ positive scale function y = K1X the abscissa and ordinate of the point on the image are opposite to each other
So y = - X
That is K1 = - 1
(2) A function y = - K2 ^ 2-2k2 + K1 + 5, why does not x appear

If y = KX + 2K + X is a positive proportional function, find the value of K

Y = KX + 2K + x = (K + 1) x + 2K is a positive proportional function
k+1≠0
k≠-1

Given the primary function y = {m + 2} x + {1-m}, if y decreases with the increase of X, and the intersection of the secondary function image and the Y axis is above the X axis, then the value range of M is

1. If y decreases with the increase of X, then the coefficient of X is negative, so m + 2

If the image of the first-order function y = - x + B passes through the positive half axis of Y axis, then the value range of B is smaller

Because the coordinate of the intersection of the image of the linear function y = - x + B and the Y axis is (0, b)
The image of the linear function y = - x + B passes through the positive half axis of the Y axis,
So b > 0

When the image of the linear function y = (m-1) x + m + 2 passes through a point on the negative half axis of Y axis, the value range of M is obtained

The image of a linear function y = (m-1) x + m + 2 passes through a point on the negative half axis of the y-axis
∴﹛m-1≠0 m≠1
m+2＜0 m＜-2
The range of M is m ＜ - 2

As for the linear function y = (3a + b) x + A-1 of X, if the intersection of the image and Y axis is below the X axis, what is the value range of a? I want to know the calculation process

Y-axis then x = 0
So y = A-1
Below the x-axis is y

Write a linear function expression with the following two conditions at the same time (just write one)______ (1) y decreases with the increase of X; (2) the image passes through points (1, - 3)

If ∵ y decreases with the increase of X, and ∵ K < 0, and ∵ a straight line crosses a point (1, - 3), then the analytical formula is y = - 3x or y = - 2x-1 or y = - X-2, etc

Write a linear function expression with the following two conditions at the same time. 1. Y decreases with the increase of X. 2?

A linear function expression with the following two conditions
y=-x-5