Let's find the intersection point of the line 3x-4, which is parallel to the line 3x-4

Let's find the intersection point of the line 3x-4, which is parallel to the line 3x-4

It is known that the line y = KX + B is parallel to the line y = - 3x + 4;
So k = - 3;
The intersection point with the line y = 5x-12 is on the x-axis
So the intersection point is (12 / 5,0)
-36/5+b=0;
b=36/5;
The analytical formula is y = - 3x + 36 / 5;

The image of the first order function y = KX + B passes through the image parallel to the positive scale function y = 0.5x and passes through the point (4,7). The analytic formula of the function of degree one and the coordinates of the intersection point with the coordinate axis are obtained

I'm glad to answer your question
Because it is parallel, so K is equal, so k = 0.5, because it passes through (4,7), it is substituted into 4 × 0.5 + B = 7, so B = 5, so the analytic formula of the first order function is y = 0.5x + 5
(- 10,0) and y-axis intersect at (0,5) tips: intersection with Y-axis is the intersection of (0, b) and X-axis (- k-branch, B, 0)

The image of a given function y = KX + B passes through point a (- 3, - 2) and point B (1,6). The analytic formula of the function of degree one is obtained and the image is drawn

The image of the function y = KX + B passes through points a (- 3, - 2) and B (1,6)
Replace these two points with:
-2=-3*k+b
6=k+b
So k = 2, B = 4
y=2x+4

The analytic expression of the image with known function y = KX + B is obtained by a (- 3. - 2) and B (1.6)

-3k+b=-2 ①
k+b=6 ②
② - ① 4K = 8
K=2
Substituting k = 2 into ② yields 2 + B = 6
B=4
Therefore, the analytic formula is y = 2x + 4

Given that the image of the first order function y = KX + B passes through points a (1,3) and B (- 1, - 1), then the analytic formula of this function is?

Because the image of the first order function y = KX + B passes through points a (1,3) and B (- 1, - 1), that is, the coordinates of points a and B satisfy the equation y = KX + B, i.e. 3 = K + B-1 = - K + B. to solve this system of quadratic equations, k = 2B = 1 is obtained. Therefore, the analytic formula of this function is y = 2x + 1

We know that the image of the first order function y = KX + B passes through point a (1,3) and is parallel to the straight line y = - 3x-2. (1) find the analytic formula of this linear function. (2) judge It is known that the image of the first order function y = KX + B passes through point a (1,3) and is parallel to the straight line y = - 3x-2. (1) find the analytic formula of this function. (2) judge whether the point (- 2,8) is on the image of this function?

(1) From the image of the linear function y = KX + B parallel to the line y = - 3x-2, we know that k = - 3
Then the first order function is y = - 3x + B, and the coordinates (1,3) of a are substituted,
The result is: - 3 + B = 1 and B = 4
The analytic expression of this function is y = - 3x + 4
(2) When x = - 2, - 3x (- 2) + 4 = 10 ≠ 8, so the point (- 2,8) is not on the image of this function

Given that the image of the first order function y = KX + B passes through the point (- 2,4) and is parallel to the straight line y = 3x, the analytic formula of this linear function is obtained

The analytic expression of this function is y = KX + B,
According to the meaning of the title:
-2k+b=4
K = 3, (2 points)
The solution is as follows:
b=10
K = 3, (3 points)
The analytic formula of this function is y = 3x + 10. (4 points)
So the answer is: y = 3x + 10

Given the linear function y = KX + B, the image is parallel to the straight line y = 3x-5, passing through the point (- 1,5) to find the analytic formula of the first order function fast

Because of parallelism, k = 3 goes through (- 1,5), so - 1 * 3 + B = 5B = 8, y = 3x + 8

Given that the area of the image of the first order function y = KX + B passing through a (0, a), B (- 1,2) and △ ABO is 2, the analytic formula of this function is obtained

The image of y = KX + B passes through a (0, a), B (- 1,2)
B=a
-k+b=2
The area of △ ABO is 2 (the base of triangle is Ao, the height is the ordinate of point B), a * 2 / 2 = 2, a = 2
b=2,k=4
y=4x+2

The image of the given function y = KX + B passes through the points (0,2) and (1, - 1) to find the analytic formula of this function, and to find the image and two coordinates of this function The area of a triangle enclosed by an axis

If we put the points into the function respectively, we have b = 2
k+b=-1
From the two solutions, k = - 3, B = 2
Therefore, the analytic formula is y = - 3x + 2
When it intersects the X axis, the Y coordinate is 0, that is - 3x + 2 = 0, x = 2 / 3
That is, the bottom length of the triangle is 2 / 3
When it intersects the Y axis, the X coordinate is 0 and y = 2
That is, the height of the triangle is 2
So the area of triangle is s = 1 / 2 * 2 / 3 * 2 = 2 / 3