The graph of a function is known to pass through points (2,1) and (- 1, - 3). (1) find the analytic expression of the function; (2) find the relation between the function and x-axis, y-axis

The graph of a function is known to pass through points (2,1) and (- 1, - 3). (1) find the analytic expression of the function; (2) find the relation between the function and x-axis, y-axis

(1) Let the function expression be y = KX + B
Substituting (2,1) (- 1, - 3)
2K+B=1
-K+B=-3
3K=4
K=4/3
B=-5/3
The expression is y = 4x / 3-5 / 3
(2) Substituting x = 0, y = - 5 / 3
The intersection point with y axis is (0, - 5 / 3)
Substituting y = 0
4X/3-5/3=0
X=5/4
The intersection point with X axis is (5 / 4,0)

The image of inverse scale function y = 2 / X and the image of primary function y = KX + B intersect at points a (m, 2) and B (2, n). The intersection of the image of primary function and Y axis is C
(1) Find the analytic expression of a function;
(2) Find the coordinates of point C;
\X05 (3) find the area of △ AOC

y=-x+3
C(0,3)
3*1/2=1.5

The graph of a first-order function intersects with x-axis and y-axis at points a (4,0) and B (0,2), respectively. The analytic expression of this first-order function is -

Suppose that the analytic formula of a function is y = KX + B
0=4k+b
2=b
The solution is b = 2, k = - 1 / 2
So the analytic expression is y = - 1 / 2x + 2

If a zero point of the function g (x) = f (log2x) - 2 ^ x (x > 0) is 2, and for any function x, y, f (x + y) = f (x) + F (y), then f (3) = if f (x) is odd, G (x) is even, and f (x) - G (x) = x ^ 2 + 2x-3, then f (x) + G (x)=

1、f(1)=4,f(3)=f(1)+f(2)=3f(1)=12
2、f(x)+g(x)=-x^2+2x+3
Step: F (x) = - f (- x), G (x) = g (- x), according to the meaning: F (- x) - G (- x) = x ^ 2-2x-3, then: - f (x) - G (x) = x ^ 2 + 2x-3

It is known that the definition domain of function f (x) = x2-4x-4 is [T-2, T-1]. For any t ∈ R, the analytic expression of the minimum value g (T) of function f (x) is obtained

If T-1 ≤ 2, that is, t ≤ 3, f (x) in [T-2, T-1, T-1 [T-2, T-1] of [T-2, T-1] is the monotone decreasing of [T-2, T-1] for [T-2, T-1] and [T-2, T-2, T-1] as [T-2, T-1] for [T-2, T-1] and [T-2, T-2, t ≤ 2, t ≤ 2, t ≤ 2, that is, t ≤ 2, t ≤ 2, t (T) is the monotone increasing of [T-2, T-2, t [T-2, T-2 [T-2, T-2, T-2, t] as [T-2, t] as [T-2] is the [t] as [T-2 [T-2 [T-2, T-2 [T-2, T-2, T-2, T-2, T-2 T + 8t ≥ 4

Let f (x) = x2-4x-4 be defined as [T-2, T-1]. For any t ∈ R, find the analytic expression of the minimum value g (T) of F (x)

The abscissa of the lowest point is - (B / 2a), that is - (4 / 2 * 1). Its value is 2. Its domain should be only 1 different. Therefore, when both [T-2] and [T-1] are greater than 2, the minimum value corresponding to X is (T-2). Substitute (T-2) into f (x), f (T-2) = (T-2) 2-4 (T-2) - 4, and arrange f (T) = t2-8t + 8. Conversely, if both [T-2] and [T-1] are less than 2, the minimum value corresponding to x should be (t-1), substitute (t-1) into f (x), F (t-1) = (t-1) 2-4 (t-1) - 4, f (T) = t2-6t + 1. When [T-2] is less than 2 and [T-1] is greater than 2, the minimum value of 2 is - 8

If the domain of function f (2-4x) is [1,4], then the domain of function f (x) is [1,4]

The domain of F (2-4x) is [1,4], that is, X is in [1,4], so 2-4x is in [- 14, - 2]. All the domains of F (x) are in [- 14, - 2]
(summary: the definition field is the value range of X, the same function, that is, the value range of the formula in brackets is the same, and only the definition field of F (x) is the same as the value range of x)

The strictly monotone increasing function in the interval (0, + ∞) is a.sinx b.tanx c.x2 × 1 / X

A. B is a periodic function. It must be wrong
Is C: (X & # 178; + 1) / x? If so, choose C

Let x belong to R, the inequality x ^ 2log2 be the bottom 4 (a + 1) / A + 2xlog2 be the bottom 2A / A + 1 + log2 be the bottom (a + 1) ^ 2 / 4A ^ 2 > 0 hold, and find the value range of A
Let ㏒ 2 ((a + 1) / a) = t
Then ㏒ 2 [4 (a + 1) / a] = 2 + T
㏒2(2a/(a＋1)) =1-t
㏒2[(a＋1）²/4a²]=2(t-1).
It can be reduced to: (2 + T) x & # 178; + 2 (1-T) x + 2 (t-1) > 0
So only 2 + T > 0 and △ 1 are needed
That is, ㏒ 2 ((a + 1) / a) > 1
(a+1)/a>2
∴0

Solution;
This kind of constant establishment question
It's a quadratic function
The coefficient of quadratic term is greater than 0
△

If log2 (- x) < x + 1 holds, the value range of X is______ ．

It can be judged by drawing that f (x) = log2 (- x) and G (x) = x + 1 intersect at (- 1,0). The former is monotonically decreasing, and the latter is monotonically increasing. Therefore, log2 (- x) < x + 1 holds only when - 1 < x < 0, so the answer is: (- 1,0)