Translate the image L1 of the function y = - 2x upward by 2 units to get the line L2. If L2 intersects with X axis and Y axis at points a and B respectively, what is the area of △ AOB?

Translate the image L1 of the function y = - 2x upward by 2 units to get the line L2. If L2 intersects with X axis and Y axis at points a and B respectively, what is the area of △ AOB?

Let's look at it this way
Y = - 2x two units up give y = - 2x + 2
Let y = 0 get x = 1
Let x = 0 get y = 2
So a (1,0) B (0,2) O (0,0)
Draw a graph, and the area of the triangle is equal to 1
If you don't understand, you can talk about it again

The image of the function y = - 6x + 5 is obtained by translating the straight line y = - 6x three length units?
Is it true that if you change B to move left and right, and change K to move up and down?

No, if it's a straight line, the effect of left-right translation is the same as that of school translation
Changing K is not translation
The image of the function y = - 6x + 5 is translated up 6 length units by the straight line y = - 6x
The image of the function y = - 6x + 5 is shifted 5 / 6 length units from the straight line y = - 6x to the right

If the area of the triangle formed by the line y = 3x + B and the two coordinate axes is 6, what is B
Come on, I'm in a hurry

The intersection of line and X axis is: 0 = 3x + B, x = - B / 3
The intersection point with y axis is y = 3 * 0 + B, y = B
When b > 0, s = 1 / 2 * b * | - B / 3|
6 =b^2/6
b^2=36 b=6
When B

The area of the triangle formed by the line y = 3x + B & nbsp; and the two coordinate axes is 6, and the intersection coordinate ()
A. （0，2）B. （0，-2）（0，2）C. （0，6）D. （0，6）、（0，-6）

Let x = 0, then y = B, let y = 0, then x = - B3, ∵ straight line y = 3x + B & nbsp; and the area of triangle enclosed by two coordinate axes is 6, ∵ 12 | B | ×| - B3 | = 6, that is, B26 = 6, ∵ B = ± 6, and the intersection coordinates of Y axis are (0,6), (0, - 6)