If the image of the linear function y = KX + B passes through point a (0,2) B (2,0), then its analytic formula is

If the image of the linear function y = KX + B passes through point a (0,2) B (2,0), then its analytic formula is

2=b
0=2k+b
2k+2=0
k=-1
therefore
The analytic formula is y = - x + 2

Given that the abscissa of the intersection point between the image of the first order function y = KX + B and the X axis is - 4, then the solution of the univariate linear equation KX + B = 0 is A. X = - 4 b.x = 4 c.x = O D. cannot be determined

Option a
kx+b=0
So y = 0
If and only if x = - 4, y = 0

When x = 1, y = - 2, and the ordinate of the intersection point of its image and Y axis is - 5, then its analytic formula is () A y=3x+5 B y=-3x-5 C -3x+5 D 3x-5

Answer: D
Explain that "the ordinate of the intersection point of its image and Y axis is - 5", that is, when the image passes through (0, - 5) points and x = 1, y = - 2, that is, passing (1, - 2), and bringing the two points into the equation system - 2 = k × 1 + B, - 5 = 0 × K + B, that is, B = - 5, k = 3, and the equation is y = 3x-5

As shown in the figure, the image of the first order function y = KX + B and the inverse scale function y = - 8 The image of X intersects a and B, and the abscissa of point a and the ordinate of point B are - 2 (1) The relation of the first degree function; (2) Find the area of △ AOB

(1) Let a (x1, Y1), B (X2, Y2), then X1 = - 2, y2 = - 2,
Replace X1 = y2 = - 2 with y = - 8, respectively
X is Y1 = x2 = 4,
∴A（-2，4），B（4，-2）．
By substituting a (- 2,4) and B (4, - 2) into y = KX + B respectively, the
4＝−2k+b
−2＝4k+b
The solution
k＝−1
b＝2
The analytic formula of the first order function is y = - x + 2
(2) As shown in the figure,
∵ y = - x + 2 and the intersection of Y axis is C (0, 2)
∴OC=2，
∴S△AOB=S△AOC+S△BOC
=1
2×OC×|x1|+1
2×OC×|x2|
=1
2×2×2+1
2×2×4=6．

As shown in the figure, the image with the first order function y = KX + B and the image with the inverse scale function y = m / X intersect at two points a and B (1) using the conditions in the image, the sum of the inverse scale function is obtained As shown in the figure, the image of the first order function y = KX + B and the image of the inverse scale function y = m / X intersect at two points a and B (1) using the conditions in the image, the analytic expressions of the inverse scale function and the first order function are obtained (2) Find the area of △ AOB (3) According to the image, write out the value range of the first order function greater than the value of X of the inverse scale function

1) ∵ point a (- 2,1), B (1, a) is the point where the image intersects with the first order function y = KX + B and the inverse scale function y = m / X
The point a (- 2,1) and B (1, a) are on the first order function y = KX + B and the inverse proportional function y = m / X
 by substituting the point a (- 2,1) into the inverse proportional function y = m / x, the solution is m = - 2
The analytic expression of the inverse proportional function y = m / X is y = - 2 / X
 substituting point B (1, a) into the inverse proportional function y = - 2 / x, the solution is n = - 2
The coordinates of point B are (1, - 2)
And ∵ point a (- 2,1), B (1, - 2) on the function y = KX + B
 replace the point a (- 2,1), B (1, - 2) into the function of degree y = KX + B,
The solution of {1 = - 2K + B-2 = 1K + B leads to {k = - 1, B = - 1
The analytic formula of the first order function y = KX + B is y = - X-1, and that of the inverse proportional function y = m / X is y = - 2 / X
(2) The area of △ AOB = 3 / 2
(3) It can be seen from the image that when x ＜ - 2 or 0 ＜ x ＜ 1, the value of the first-order function is greater than that of the inverse proportional function

The image of the first order function y = KX + B passes through points (3, - 3) and (- 1,5), and finds the value of K in B

If (3, - 3) and (- 1,5) are brought into the first order function y = KX + B, then
-3=3k+b
5= -k+b
Solving two equations and two unknowns
Using the elimination method, K or B are eliminated
K = - 2
b= 3

The image passing through point a (- 3, - 2) and point B (1,6) with known function y = KX + B (1) . find a function expression of degree one (2) Find out the area of the triangle formed by the function image and the coordinate axis

(1) Substituting a and B into the function, we get - 2 = - 3K + B, 6 = K + B
The solution is k = 2, B = 4, so the expression is y = 2x + 4
(2) The intersection point of function and x-axis and y-axis is (- 2,0) (0,4)
So the triangle area is s = 1 / 2 * 2 * 4 = 4

It is known that the graph of the first order function y = KX + B is a straight line passing through two points a (0, - 4) and B (2, - 3) (1) Find the analytic formula of line ab; (2) The line AB is shifted to the left by 6 units, and the analytical formula of the translated line is obtained (3) Translate the line AB upward by 6 units to find the distance from the origin to the translated line

(1) ∵ the line AB: y = KX + B passes through a (0, - 4), B (2, - 3),
∴b=-4，-3=2k-4，
∴k=1
2，
The analytic formula of line AB is y = 1
2x-4；
(2) ∵ straight line AB: y = 1
2X-4 intersects the x-axis with the point E (8, 0),
﹣ shift the line AB to the left by 6 units and then pass through the point F (2,0),
Let the analytic formula of the line after the line AB is shifted to the left by 6 units is y = 1
2x+n，
∴0=1
2×2+n，
∴n=-1，
The analytic formula of the straight line after translating the line AB to the left by 6 units is y = 1
2x-1；
(3) Translate the line AB upward by 6 units to get the line CD: y = 1
2X-4 + 6, i.e. y = 1
2x+2，
∵ the intersection point of line CD and X, Y axis is C (- 4,0), D (0,2)
∴CD=
OC2+OD2=
22+42=2
5     
The distance between the straight line CD and the origin is 2 × 4
Two
5=4
Five
5．

As shown in the figure, the image of the first order function y = KX + B passes through two points a and B If the straight line a, B and X axis intersect with point C, find the area of triangle ABC

You are totally wrong in this problem. You can't do it
First of all, the image of the function y = KX + B passes through two points a and B. what are a and B here? The code of the point? Coordinates?
Secondly, if a and B represent two points, then the straight line a, B and X axis intersect with point C, then the three points of ABC should be on a straight line. How can we get a triangle?

As shown in the figure, the image of the given function y = x + 1 intersects point a with the Y axis, and the image of the first order function y = KX + B passes through point B (0, - 1), and The images of x-axis and y = x + 1 intersect at points c and D, respectively (1) If the abscissa of point D is 1, calculate the area of the quadrilateral AOCD (that is, the area of the shadow part in the figure); (2) Under the condition of sub question (1), is there such a point P on the Y axis such that the triangle with points P, B and D as its vertices is an isosceles triangle. If it exists, find out the coordinates of point p; if not, explain the reason (3) If the intersection D of the image of the first order function y = KX + B and the image of the function y = x + 1 is always in the first quadrant, then the value range of the coefficient K is___ .

(1)  the abscissa of point D is 1, and the point D is on the image of y = x + 1, ᙽ D (1,2),  the analytic formula of straight line BD is y = 3x-1,  a (0, 1), C (13, 0), (s quadrilateral AOCD = s △ AOD + s △ cod = 12 × 1 × 1 + 12 × 13 × 2 = 56; (2) when DP = dB, let P (0, y),