Write a function expression (1) y decreases with the increase of X; (2) the image passes through the point (1, - 2)

Write a function expression (1) y decreases with the increase of X; (2) the image passes through the point (1, - 2)

Let the linear function be y = KX + B. from the first condition, we know that the slope of the straight line is K

Please write the expression of a linear function of the image passing through (2, - 1) and decreasing with the increase of X

Let y = KX + B (k)

The image of the first-order function passes through the point (- 3,2) and the value of function y decreases with the increase of the value of independent variable x

k

Write a linear function expression to satisfy the following conditions: (1) y decreases with the increase of X; (2) the image passes through the points (1, - 3)

Write a linear function expression, which satisfies (1) y decreases with the increase of X,
(2) Image passing through point (1, - 3)
Let y = KX + B
(1) If y decreases with the increase of X, then K < 0
y=-2x+b
(2) Image passing through point (1, - 3)
Bring in
-3=-2+b
be
b=-1
That is y = - 2x-1

The linear function y = - 2x + 4, when the function value is positive, the value range of X is______ ．

A function y = - 2x + 4, when the function value is positive, that is - 2x + 4 > 0, the solution is: x < 2

Make the image of function y = 2X-4, and answer the following questions according to the image: (1) when - 2 ≤ x ≤ 4, find the value range of function y; (2) when x takes what value,
To create an image

Put Y in X
Because y = 1 / 2X-4
X=（Y+4）*2
therefore
-1≤（Y+4）*2≤2
After simplification
-9/2≤Y≤-3

If the function image of function y = 2x + 1 is above the image of function y = X-1, the value range of X is obtained

2x+1-(x-1)>0
2x+1-x+1>0
x>-2

If there is only one common point between the image of function f (x) = x3-3a2x + 1 and the line y = 3, then the value range of real number a is ()
A. （-∞，+∞）B. （-∞，1）C. （-1，+∞）D. （-1，1）

If we find the first derivative, we can get f '(x) = 3x2-3a2, and substitute the two extreme points in x = A and x = - a into the function, we can get f (a) = - 2A3 + 1, f (- a) = 2A3 + 1, when a > 0, f (a) > 3 or F (- a) < 3, we can get a < 1, when a < 0, f (a) < 3 or F (- a) > 3, we can get a > 1, when a = 0, it is obviously true; then the answer is: - 1 < a < 1, so choose D

Given M & # 178; - N & # 178; = 4Mn, find the value of the algebraic formula M & # 8308; + n & # 8308; / M & # 178; n & # 178; (Mn ≠ 0)

m⁴+n⁴/m²n²=(m⁴-2m²n²+n⁴+2m²n²)/m²n²=[(m²-n²)²+2m²n²]/m²n²=[(4mn)²+2m²n²]/m²n²...

Given that the function f (x) = [x2 + (2a-2) x + 2-2a-b] ex (a, B ∈ R) is a decreasing function in the interval [- 1, 3], then the minimum value of a + B is ()
A. 4B. 2C. 32D. 23

∵ f (x) = [x2 + (2a-2) x + 2-2a-b] ex (a, B ∈ R) is a decreasing function in the interval [- 1,3], ∵ f ′ (x) = ex (x2 + 2ax-b) ＜ 0, ∵ x2 + 2ax-b ＜ 0, let g (x) = x2 + 2ax-b, ∵ f (x) = [x2 + (2a-2) x + 2-2a-b] ex (a, B ∈ R) is a decreasing function in the interval [- 1,3], ∵ g (− 1) ≤ 0g (3) ≤ 0, that is, 2A + B ≥ 1b − 6A ≥ 9, ① make a straight line in the coordinate plane 1-2a-b = 0, 9 + 6a-b = 0, they intersect at a (- 1, 3), satisfy that (a, b) is the area above point a, let a + B = t, then B = - A + T, t is the intercept of the straight line on b axis, translate the straight line, we can see that when the straight line passes through a, the minimum of T is 3-1 = 2