Given the function of degree y = kx-4, when x = 2, y = - 3. (1) find the analytic formula of the function; (2) translate the image of the function upward by 6 units, and find the analytic expression of the translated image and the coordinates of its intersection with the x-axis

Given the function of degree y = kx-4, when x = 2, y = - 3. (1) find the analytic formula of the function; (2) translate the image of the function upward by 6 units, and find the analytic expression of the translated image and the coordinates of its intersection with the x-axis

(1)-3=2k-4 k=1/2
y=x/2-4
(2) y+6=x/2-4
y=x/2-10
y=0 x=20
The coordinates of its intersection with the x-axis (20,0)

The image of the first order function y = kx-1 is shifted up K units and then passes through point a (3,2 + k) Find K If a line is parallel to the image of the function y = kx-1, and the area of the triangle obtained from the two coordinate axes is half, the functional relationship of the line is obtained

Function after translation: y = kx-1 + K,
Substitution point a (3,2 + k)
2+k=3k-1+k
K = 1
If the other line is y = x + m, then it intersects with the coordinate axis at (0, m) (- m, 0)
Area s = m ^ 2 / 2 = 1 / 2
M = 1 or M = - 1
The linear equation is y = x + 1 or y = X-1

Given the image crossing point (1,2) of the first order function y = KX + B, and its image can be obtained by the downward translation of the positive proportion function y = KX by 4 units. The analytic formula of the first order function is obtained

Substituting (1,2) into y = KX + B leads to K + B = 2,
∵ y = KX, move down 4 units to get y = KX + B,
∴b=-4，
∴k-4=2，
K = 6
The analytic formula of the first order function is y = 6x-4

The image of the first order function y = KX = B is obtained by shifting the image with the function y = 2x to the right by one unit Why is there such a rule? A rule always has corresponding reasoning If y = KX + B is moved one unit to the right, what is the analytic formula of the first order function?

The image of the function y = 2x is shifted to the right by one unit
y=2(x-1)=2x-2
This kind of topic is easy to remember
Left plus right minus
If you shift K units to the right, it is:
y=2(x-k)
Do you understand?
Move to the right, minus X
Move to the left and add to X
Move up, just add y
Move down and subtract on y
For example, y = 3x + 5 moves three units to the right
y=3(x-3)+5
That is, y = 3x-4

Given the first order function y = KX + B (K ≠ 0), if its image is shifted to the left by 3 units and then down by 5 units, If the obtained image coincides with the original image, then k =? 0? It's better to have a process, thanks

Image translation is based on the principle of "left plus right minus, top plus lower subtraction", where left plus right subtraction is on X, and top plus bottom subtraction is on the whole formula
So after translation, y = K (x + 3) + B-5 = KX + B + 3k-5
KX + B = KX + B + 3k-5, 3k-5 = 0, k = 5 / 3

Given that the image of the first order function y = KX + B passes through point a (- 3,2) and point B (0,6), the expression of this linear function is obtained

The image of the first order function y = KX + B passes through point a (- 3,2) and point B (0,6),
∴﹛2=-3k+b
6=b
∴﹛k=4/3,b=6
The expression of the first order function is y = 4 / 3x + 6

As shown in the figure, it is known that the image of function y = 6 / X and the image of function y = KX + B intersect with a (1, m), B (n, 2) two points (1) to find the analytic formula of the first order function

The image of function y = 6 / X and the image of function y = KX + B intersect two points a (1, m), B (n, 2)
∴﹛m=6
2=6/n
∴m=6,n=3
X = 1, y = 6, x = 3, y = 2 into y = KX + B
It is found that {6 = K + B
2=3k+b
K = - 2, B = 8
The analytic formula of the first order function is y = - 2x + 8

The image of the first order function y = KX + B passes through point a (3,0) and intersects with point B. if the area of △ AOB is 6, try to find out 1. Coordinates of point B 2. The analytic formula of the first order function

Set the coordinates of point B as (0, b)
that
1/2*3*|b|=6
So | B | = 4
b=±4
So the coordinates of point B (0,4) or (0, - 4)
Function analytic formula let y = KX + B bring in a B, you can find it by yourself

It is known that the image of the function y = KX + B intersects point a (- 6,0) with X axis, and intersects point B with y axis. If the area of △ AOB is 12, find the expression of function of degree one

∵ the image passes through point a (- 6, 0),
∴0=-6k+b，
That is, B = 6K,
∵ the intersection of the image and the y-axis is B (0, b),
∴S△ABO＝1
2OA•OB=12，
Namely: 1
2×6×|b|＝12，
∴|b|=4，
∴b1=4，b2=-4，
Substituting into formula ①, K1 = 2
3，k2＝−2
3，
The expression of the first order function is y = 2
3x + 4 or y = - 2
3x−4．

It is known that the image of the first order function y = KX + B intersects point a (- 6,0) with X axis, and intersects point B with y axis, and Y decreases with the increase of X. if the area of △ AOB is 12, Then the relation of this function is_________

Y decreases with the increase of X, and K is less than 0
If the area of △ AOB is 12, the passing point B (0, - 4) = > b = - 4 is obtained
Substituting y = kx-4, k = - 2 / 3 can be obtained from point a
That is, y = - 2 / 3 * x-4