If we know that the image of the first-order function y = KX + B passes through the point (0,1) and does not pass through the fourth quadrant, then the analytic expression of a first-order function satisfying the above conditions is______ ．

If we know that the image of the first-order function y = KX + B passes through the point (0,1) and does not pass through the fourth quadrant, then the analytic expression of a first-order function satisfying the above conditions is______ ．

∵ the image of a function y = KX + B passes through the point (0,1) and does not pass through the fourth quadrant, ∵ k > 0, b > 0, and ∵ passes through the point (0,1) ∵ B = 1, ∵ as long as the relationship satisfies the above conditions, for example, when k = 1, y = x + 1; when k = 1, y = x + 1, the answer is not unique

If we know what is the first-order analytic expression of the image of the first-order function y = KX + B passing through the point (2,0) and not passing through the second quadrant, we write two

Substituting the point (2,0) into y = KX + B, we can get: 2K + B = 0, and because it does not pass through the second quadrant, we only need k > 0
Let k = 1, then B = - 2, that is, y = X-2
Let k = 2, then B = - 4, that is, y = 2X-4

How many quadrants does the image of linear function y = 2x-5 not pass through

Second quadrant

What is the intersection coordinate of the image of the linear function y = - 2x + 1 and the image of y = 2x-3?

Solve the equations! - 2x + 1 = 2x-3 = > - 4x = - 4 = > x = 1 = > y = 2 * 1-3 = - 1
The coordinates of the intersection point are (1, - 1)

If the intersection point of the image of the linear function y = 2x and y = x + k is in the first quadrant, then the value range of K is______ ．

By solving the equations of X and y, y = 2XY = x + K, we can get x = KY = 2K. The intersection point of the graph of ∵ primary function y = 2x and y = x + k is in the first quadrant, ∵ x > 0, Y > 0, ∵ k > 02k > 0, we can get k > 0. So the answer is k > 0

What quadrant is the intersection of the image of function y = - X and the image of function y = 2x-1?

Analysis
-x=2x-1
2x-1+x=0
3x=1
x=1/3
y=-1/3
So the intersection is in the fourth quadrant

The intersection point of the image of the function y = x + A and y = - 2x-a cannot be in the fourth quadrant (explain the reason)

The coordinates of simultaneous intersection (- 2A / 3, a / 3)
The abscissa and ordinate are different signs, but one or three quadrants

If the coordinates of the intersection point of the image of the function y = 2x + B and the Y axis are (0,3), then the image of the function does not pass through the second axis______ Quadrant

∵ the intersection coordinates of the image of a function y = 2x + B and the y-axis are (0,3), ∵ B = 3. ∵ 2 > 0, ∵ the function image passes through the first and third quadrants; from b = 3 > 0, the function image intersects with the y-axis on the positive half axis

As shown in the figure, if the image of the linear function y = (a + 4) x + A + 2 is only the second quadrant, then the value of a is

If the function graph is only in the second quadrant, it must go through one, three or four
a+4＞0,a+2＜0
The solution is: a ＞ - 4, a ＜ - 2
The range of a is - 4 < a < 2

How to draw function image in junior high school mathematics?

Two points determine a straight line y = 2x-1
When x = 1, y = 1
When x = 2, y = 3
So the image goes through (1,1) (2,3) two points
I draw a plane rectangular coordinate system, mark two points, connect two points is a straight line image
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