It is proved that if and only if B = 0, the image of the first-order function y = KX + B passes through the origin

It is proved that if and only if B = 0, the image of the first-order function y = KX + B passes through the origin

When B = 0, y = KX + 0, when x = 0, y = 0, so when B = 0, the image of y = KX + B passes through the origin
When the image of y = KX + B passes through the origin, 0 = 0 + B B = 0
So if and only if B = 0, the image of the first-order function y = KX + B passes through the origin

Given that the graph of the function y = ax + B passes through the first, second and fourth quadrants and intersects the X axis at the point (2,0), then the solution set of the inequality a (x-1) - b > 0 about X is ()
A. x＜-1B. x＞-1C. x＞1D. x＜1

The graph of ∵ linear function y = ax + B passes through the first, second and fourth quadrants, ∵ b > 0, a < 0. Substituting (2, 0) into the analytic formula y = ax + B, we get: 0 = 2A + B, the solution is: 2A = - BBA = - 2, ∵ a (x-1) - b > 0, ∵ a (x-1) > b, ∵ a < 0, ∵ X-1 < Ba, ∵ x < 1, so we choose a

Given that the graph of the function y = ax + B passes through the first, second and fourth quadrants and intersects the X axis at the point (2,0), then the solution set of the inequality a (x-1) - b > 0 about X is ()
A. x＜-1B. x＞-1C. x＞1D. x＜1

The graph of ∵ linear function y = ax + B passes through the first, second and fourth quadrants, ∵ b > 0, a < 0. Substituting (2, 0) into the analytic formula y = ax + B, we get: 0 = 2A + B, the solution is: 2A = - BBA = - 2, ∵ a (x-1) - b > 0, ∵ a (x-1) > b, ∵ a < 0, ∵ X-1 < Ba, ∵ x < 1, so we choose a

Given that the graph of the function y = ax + B passes through the first, second and fourth quadrants and intersects the X axis at the point (2,0), then the solution set of the inequality a (x-1) - b > 0 about X is ()
A. x＜-1B. x＞-1C. x＞1D. x＜1

The graph of ∵ linear function y = ax + B passes through the first, second and fourth quadrants, ∵ b > 0, a < 0. Substituting (2, 0) into the analytic formula y = ax + B, we get: 0 = 2A + B, the solution is: 2A = - BBA = - 2, ∵ a (x-1) - b > 0, ∵ a (x-1) > b, ∵ a < 0, ∵ X-1 < Ba, ∵ x < 1, so we choose a

Given that the image of the first-order function y = ax + B passes through the first, second and fourth quadrants and intersects the X axis at the point (2,0), then the solution set of the inequality a (x + 1) + b > 0 about X is

If the image of a linear function y = ax + B passes through the first, second and fourth quadrants, the approximate image of the function can be as follows: b > 0.a0, a (x + 1) > - B, and both sides are divided by a at the same time, because a

Given that the graph of the function y = ax + B passes through the first, second and fourth quadrants and intersects the X axis at the point (2,0), then the solution set of the inequality a (x-1) - b > 0 about X is ()
A. x＜-1B. x＞-1C. x＞1D. x＜1

The graph of ∵ linear function y = ax + B passes through the first, second and fourth quadrants, ∵ b > 0, a < 0. Substituting (2, 0) into the analytic formula y = ax + B, we get: 0 = 2A + B, the solution is: 2A = - BBA = - 2, ∵ a (x-1) - b > 0, ∵ a (x-1) > b, ∵ a < 0, ∵ X-1 < Ba, ∵ x < 1, so we choose a

The graph of the linear function y = KX + B is_____ The coordinate of its intersection with the y-axis is_____
Why?

Er I'm also in the eighth grade, eh = v=
The graph of a linear function y = KX + B is a straight line, and its intersection coordinates with the Y axis are (0, b)
1. Do you use the book published by PEP? The book in Grade 8 has this concept. It is written in the "observation" and "conjecture" on page 29: the image of a linear function y = KX + B is a straight line. We call it a straight line y = KX + B. It can be obtained by translating | B | unit length of the straight line y = KX (when B is greater than 0, it moves upward; when B is less than 0, it moves downward)
Or, you can see what is written on page 25: generally, an image with a positive scale function y = KX (k is a constant, K is not equal to 0) is a straight line passing through the origin The positive proportion function is a special linear function (b = 0). From the special deduction, it is generally = = so the image of the linear function is a straight line
2. When it intersects with the Y axis, x = 0 (because it intersects on the Y axis, X must be equal to 0 What else is it...) Substituting x = 0 into y = KX + B to get y = B (this is the ordinate of that intersection), then the coordinates of the intersection of y = KX + B and Y axis are (0, b)

Linear function y = KX + B when k

If y = 0, the coordinate point is (- B / K, 0). If x = 0, the coordinate point is (0, b)
If y = 0, the coordinate point is (- B / K, 0). If x = 0, the coordinate point is (0, b)

It is known that in the linear function y = kx-3, when x is less than - 5, the value of the function is positive, then what is the solution set of the unary linear inequality kx-3 greater than 0 about x?
Such as the title

(- infinity, - 5)

Given that the images of functions Y1 = ax + 3a and y2 = KX + 5 intersect at point P, then according to the images, we can get the inequality (A-K) X

Substitute (- 1,1) into Y1 and Y2 respectively
So a = 1 / 2, k = 4
The original inequality = (A-K) x0, x > - 3
To sum up: x > - 1
No picture may be wrong, please forgive me