If the image of a linear function y = KX + B (K ≠ 0) passes through a point (1, - 1) and is parallel to a straight line y = 3x + 5, the analytic expression of the linear function is () and its image passes through the () quadrant

If the image of a linear function y = KX + B (K ≠ 0) passes through a point (1, - 1) and is parallel to a straight line y = 3x + 5, the analytic expression of the linear function is () and its image passes through the () quadrant

Because the image of the linear function y = KX + B (K ≠ 0) is parallel to the line y = 3x + 5
So k = 3
Another function y = KX + B passing through point (1, - 1)
Then - 1 = 3 + B
The solution is b = - 4
So the analytic expression of the function is y = 3x-4
Function over 1,3,4 quadrant

The linear L is the image of the linear function y = KX + B. fill in the blanks
As shown in the figure, the line L is the image of the linear function y = KX + B
（1）b=______ ,k=_______ ；
We need reason, we need integrity
The format should be clear, like the version of Beijing Normal University

This kind of question investigates the relationship between undetermined coefficient method and function
The idea is to substitute the coordinates of the two points into y = KX + B to obtain a system of linear equations with respect to K and B. by solving the system of equations, the values of K and B can be obtained
For example, we know that two points are (1,2), (- 3,4)
Then K + B = 2,
-3k+b=4
By solving the equations together, we can get: k = - 1 / 2, B = 5 / 2

Given that the image of the first-order function y = KX + B passes through point P (2, - 1) and point Q (- 1,5), the analytic expression of the first-order function is obtained

[-1=2k+b
5=-k+b
b=4.5 k=-2
y=-2x+9/2

1. It is known that the image intersection of the function y = - x + K and kx-4 is on the negative half axis of X axis, and the value of K is zero
2. If the area of the triangle formed by the line y = 4x-b and the two coordinate axis is 5, the value of B is constant
3. It is known that the image of a function y = 2x + K passes through the point (1,1), so its intersection with the x-axis is () and the intersection with the y-axis is the solution of the equation 2x + k = 1
If it's good, take the 25 points

1. When - x + k = kx-4,
X = (4 + k) / (1 + k), X ＜ 0, - 4 ＜ K ＜ - 1
When x = (4 + k) / (1 + k), y = 0
-x+k=(-4-k+k+k^2)/(1+k)=0
So K ^ 2-4 = 0, so k = - 2
2. When x = 0, y = - B
When y = 0, x = B / 4
So, B / 4 = 5
So B ^ 2 = 40
B = ± (2 to 10)
3. When x = 1, y = 2 + k = 1
So k = - 1, y = 2x-1
When y = 0, x = 1 / 2, so the point of intersection with X axis (1 / 2,0)
When x = 0, y = - 1, so the point of intersection with y axis (0, - 1)
The solution of 2x + k = 1 is x = 1

If the image of the Nordic function y = - KX + K ^ 2 - K passes through the origin, then K () passes through the quadrant

For example, this topic is 0 = - k * 0 + K ^ 2-k, which is transformed into the standard formula k ^ 2-k = 0 = > k (k-1) = 0. Note: the standard formula is (relation) * (relation)... = 0. In this way, we can directly get n solutions k = 0 or K = 1, and the straight line is y = 0

If the image of a linear function y = - KX + k-1 passes through the origin, then k = the line passes through the fourth quadrant

The one upstairs is wrong. The one in front is right, k = 1. If you bring in y = - KX + k-1, you get y = - x, so you should go through two or four quadrants

If the image of a linear function y = KX + 3 passes through the origin, then K=_____ This line passes through the____ Quadrant. Cause

The image of y = KX + 3 does not seem to pass through the origin (unless k = 0),
When k is greater than 0, it passes through one two three quadrants
When k is less than 0, it passes through one two four quadrants

(1) If the image of the linear function y = - KX + k-1 passes through the origin, then K=_____ This line passes through the_____ quadrant
(2) If the image of positive scale function y = KX passes through points (- 1, - 5), then the value of K is____ , image No____ Quadrant, y increases with the increase of X____

(1) K-1 = 0, k = 1, two and four quadrants
(2) K = (- 1) (- 5) = 5, one and three quadrants, increase

If the image of a linear function y = - kx-k square + 4 (K ≠ 0) only passes through two or four quadrants, then K=

∵ y = - kx-k square + 4 (K ≠ 0) image only passes through two or four quadrants
∴b=0 a＜0
The square of - K + 4 = 0, k = ± 2
∵ a＜0
∴ -k＜0 k=2
I'm glad to answer for you, smile Diye
If you don't understand this question, you can ask,

It is known that the image of the linear function y = KX + B is parallel to the image of the linear function y = - 2x through the tangent point a (0,6)
1. Find the analytic expression of the function
2. If the image of the function passes through the point P (m, 2), find the value of M
3. Find the analytic expression of the function corresponding to the op lock on the straight line
4. Calculate the area of the figure enclosed by the line y = KX + B and the line OP and the x-axis

(1) Because the two lines are parallel, so K is equal to - 2. Substituting a (0,6) into y = - 2x + B, we get: B = 6, so the analytical formula is: y = - 2x + 6 (2) substituting P (m, 2) into y = - 2x + 6, we get: 2 = - 2m + 6, and the solution is: M = - 2 (what conditions should be missing in the following problem)