As shown in the figure, given the points a (4, m) and B (- 1, n) on the image with inverse scale function y = 8x, the line AB intersects the x-axis and y-axis at two points c and D respectively. (1) find the analytical formula of the line AB; (2) find the coordinates of two points c and D; (3) what is s △ AOC: s △ BOD?

As shown in the figure, given the points a (4, m) and B (- 1, n) on the image with inverse scale function y = 8x, the line AB intersects the x-axis and y-axis at two points c and D respectively. (1) find the analytical formula of the line AB; (2) find the coordinates of two points c and D; (3) what is s △ AOC: s △ BOD?

(1) ∵ a (4, m), B (- 1, n) on the inverse scale function y = 8x, ∵ M = 2, n = - 8, ∵ a (4, 2), B (- 1, - 8), let the analytic expression of the line AB be y = KX + B, then 2 = 4K + B − 8 = − K + B, the solution is k = 2B = − 6, ∵ the analytic expression of the function is y = 2x-6; (2) in y = 2x-6, when y = 0, x = 3, when x = 0, y = - 6, ∵ C (3, 0), D (0, - 6); (3) ∵ s △ AOC = 1 2×3×2=3，S△BOD=12×6×1=3，∴S△AOC：S△BOD=1：1．

Given the inverse proportional function y = K / x, when - 4 ≤ x ≤ - 1, - 4 ≤ y ≤ - 1, then the value of K is

4

Given the inverse proportional function y = K / x, when x = 4, y = - 2, the y value when x = 1, y = - 2 is obtained respectively, and the x value when x = - 1, y = - 3 is obtained respectively

The solution is y = K / x, x = 4, y = - 2
k=-2*4=-8
So y = - 8 / X
x=1 y=-8
y=-2,x=4
x=-1 y=8
y=-3 x=8/3.

It is known that the inverse scale function y = K / X and the first-order function y = 2x-1, and the image of the first-order function passes through (a, b) and (a + 1, B + k)
Finding the expression of inverse scale function
If there is a point a (1, c) on the line y = 2x-1, is point a on the hyperbola? Explain the reason
15 more points today

Given the inverse proportion function y = K / X and the primary function y = 2x-1, and the primary function image passes through (a, b) and (a + 1, B + k), find the expression of the inverse proportion function. If there is a point a (1, c) on the straight line y = 2x-1, is point a on the hyperbola? Explain the reason
(1) Analysis: ∵ function y = K / x, function y = 2x-1, and the graph of the first-order function passes through (a, b) and (a + 1, B + k)
∴b=2a-1
B+k=2a+2-1==>b=2a+1-k
∴1-k=-1==>k=2
The expression of inverse scale function is y = 2 / X
(2) Analysis: ∵ there is a point a (1, c) on the line y = 2x-1
∴c=2-1=1==>A(1,1)
Y=2/x==>y=2/1=2≠1
A (1,1) is not hyperbolic y = K / X

As shown in the figure, P is the inverse scale function, y = KX, a point on the image, which leads a vertical line to the x-axis and y-axis respectively through P. if s is negative = 3, then the analytical formula is______ ．

It is concluded that the area of rectangle is equal to | K |, and | K | = 3, and the image of inverse scale function is in the second and fourth quadrants. | K < 0 | k = - 3, and the analytic expression of inverse scale function is y = - 3x. So the answer is: y = - 3x

Come back as soon as possible
It is known that X1 and X2 are two real roots of the quadratic equation x ^ 2-6x + k = 0, and X1 ^ 2 * x2 ^ 2-x1-x2 = 115
(1) Find the value of K
(2) Find the value of X1 ^ 2 + x2 ^ 2 + 8
PS: X1 ^ 2 is the square of x1

1) X1 ^ 2 * x2 ^ 2 can be regarded as (x1 * x2) ^ 2, so the equation equal to 115 can be reduced to (x1 * x2) ^ 2
（X1*X2）^2-（X1+X2）=115
In the univariate quadratic equation, X1 + x2 = - B / A, X1 * x2 = C / A,
According to the meaning of the title, X1 + x2 = 6, X1 * x2 = k, into the just simplified equation, k = 11 or - 11
2) Substituting K into quadratic equation of one variable to calculate tilta (b ^ 2-4ac), it is found that k = 11 is not suitable for the problem
And X1 ^ 2 + x2 ^ 2 + 8 can be changed into (x1 + x2) ^ 2-2x1 * x2 + 8 = 36 + 22 + 8 = 66
In fact, we can also substitute K into the original quadratic equation of one variable to directly obtain X1 and X2, but this problem is obviously not very convenient. The method of 2) is much simpler than that of 2). We can also verify the two methods with each other

As shown in the figure, in △ ABC, ab = 3, AC = 5, point m is the midpoint of BC, ad is the bisector of ∠ BAC, MF ‖ ad, then the length of FC is______ ．

The image of positive scale function y = 2x intersects with the image of linear function y = - 3x + k at point P (1, m)
(1) The value of K;
(2) The area of a triangle bounded by two straight lines and the x-axis

(1) If P is brought into y = 2x, M = 2;
Then (1,2) is introduced into y = - 3x + K to get k = 5;
(2) Let the intersection of y = - 3x + 5 and X axis be a, and y = 0, then x = 5 / 3;
Then the area of triangle OAP is s = (5 / 3) * 2 * (1 / 2) = 5 / 3

A math problem in the second grade of junior high school
Let y = a1x + B1 and y = a2x = B2, then y = m (a1x + B1) + n (a2x + B2) (where M + n = 1) is the generating function of these two functions
(1) When x = 1, find the value of generating function of y = x + 1 and y = 2x;
(2) If the intersection point of the image of the function y = a1x + B1 and y = a2x + B2 is p, judge whether the point P is on the image of the generating function of the two functions, and explain the reason
Give the necessary steps

(1) The generating function is: y = m (x + 1) + n (2x), substituting x = 1 into
That is: y = 2m + 2n plus N + M = 1, so the original formula is 2
(2) Point P is proved on the generating function
Simultaneous: y = a1x + B1 and y = a2x = B2 get: the expression of X, and then substitute it into y = a1x + B1 and generating function (calculate patiently) to get the same value of Y, so P point is on generating function

Mathematical problems of first order function
It is known that the images of linear function y = KS + 1 and inverse scale function y = 6 of X all pass through point (2, m)
1. Find the analytic formula of a function
2. Find another intersection coordinate of these two function images

It is known that the images of the linear function y = KS + 1 and the inverse scale function y = 6 of X all pass through the point (2, m) 1. Find the analytic expression of the linear function 2. Find another intersection coordinate of these two function images. (1) substitute x = 2 into y = 6 / x, then y = 6 / 2 = 3, so m = 3. Substitute x = 2, y = 3 into y = KX + 1, then 3 = 2K + 1, the result is as follows