Given the function f (x) = (x + 1-A) / (A-X), a ∈ R, this paper proves that the image of function y = f (x) is centrosymmetric with respect to point (a, - 1)

Given the function f (x) = (x + 1-A) / (A-X), a ∈ R, this paper proves that the image of function y = f (x) is centrosymmetric with respect to point (a, - 1)

On the problem of centrosymmetry, if a function f (x) is centrosymmetric with respect to (a, b), then f (x) + F (2a-x) = 2B (for any (x, y) point with respect to (a, b), the symmetric point is (2a-x, 2b-y), (2a-x, 2b-y) is also on f (x), so f (2a-x) = 2b-y = 2b-f (x), that is, f (x) + F (2a-x) = 2b). Substitute f (x) according to the above, if f (x) + F (2a-x) = 2B, then it is centrosymmetric

If the function f (x) satisfies f (x) + F (2a-x) = 2B for any X in the domain of definition, then the image of function y = f (x) is symmetric with respect to point (a, b). Q: given that the image of function f (x) = (x ^ + MX + m) / X is symmetric with respect to point (0,1), find the value of real number M

According to the meaning of the problem, the image of the function f (x) = (x ^ + MX + m) / X is symmetric with respect to the point (0,1)
So f (x) + F (- x) = 2
F (x) = (X & sup2; + MX + m) / x, so f (- x) = - (X & sup2; - MX + m) / X
The two formulas are summed up
f（x）+f（-x）=2m
So 2m = 2
So m = 1

Why does the function y = f (x) defined on R satisfy the condition f (x) = 2b-f (2a-x) for any X in the domain of definition, then y = f (x) is symmetric with respect to point (a, b)

The function y = f (x) defined on R satisfies the condition f (x) = 2b-f (2a-x) for any X in the domain of definition. 2a-x can be regarded as X ', that is, 2a-x = x' → x + X '= 2A. ① f (x) = 2b-f (x') → f (x) = 2b-f (x ') → f (x) + F (x') = 2B

Given that a focus of hyperbola x2a2 − y2b2 = 1 coincides with the focus of parabola y2 = 4x, and the eccentricity of the hyperbola is 5, then the asymptote equation of the hyperbola is______ ．

According to the meaning of the question, the focus coordinate of the parabola y2 = 4x is (1, 0) ∵ a focus of the hyperbola x2a2 − y2b2 = 1 coincides with the focus of the parabola y2 = 4x, ∵ C = 1 ∵ the eccentricity of the hyperbola is 5, ∵ CA = 5 ∵ a = 55 ∵ B2 = C2 − A2 = 45 ∵ B = 255 ∵ the asymptote equation of the hyperbola is y = ± Bax = ± 2x, so the answer is: y = ± 2x

（x+y）^2+3(x^2-y^2)+2(x+y)
X plus the square of Y, plus the square of x minus the square of Y multiplied by 3, plus x + y multiplied by 2, factorization, OK,

（x+y）²+3(x²-y²)+2(x+y)
=（x+y）²+3（x+y）（x-y）+2（x+y）
=（x+y）[x+y+3（x-y）+2]
=（x+y）（4x-2y+2）
=2（x+y）（2x-y+1）
Click: X & sup2; - Y & sup2; apply the square difference formula to (x + y) (X-Y)
Then extract the public (x + y) of each term, and use the law of multiplicative distribution
Finally, extract 2 from (4x-2y + 2) to get 2 (2x-y + 1)

Known: x ^ 2 + 3x + 1 = 0. Find the value of 3x ^ 3 + 7x ^ 2-3x + 6

3x^3+7x^2-3x+6=3x（x^2+3x+1）-2x^2-6x+6=-2(x^2+3x+1)+8=8

As shown in Figure 1, it is known that the parabola y = ax & # 178; + BX + 3 (a ≠ 0) intersects the x-axis at point a (1,0) and point B (- 3,0), and intersects the y-axis at point C
(1) The analytic formula of parabola
(2) Let the symmetric axis of the parabola intersect with the x-axis at point m, and ask if there is a point P on the symmetric axis. Is △ CMP an isosceles triangle
(3) As shown in Figure 2, if point E is a moving point on the parabola of the second quadrant, link be and CE, calculate the maximum area of the quadrilateral boce, and calculate the coordinates of point e at this time
My drawing is not good. It's like that

1）
Let a (1,0), B (- 3,0) generation Y = ax & # 178; + BX + 3,
a+b+3=0,
9a-3b+3=0,
The solution is a = - 1, B = - 2
The parabola is y = - X & # 178; - 2x + 3 = - (x + 1) &# 178; + 4
So the axis of symmetry is x = - 1, m (- 1,0)
From C (0,3)
In right triangle OCM, by Pythagorean theorem, CM = √ 10
Take M as the center of the circle and √ 10 as the radius to draw an arc, intersecting the axis of symmetry at point P,
In this case, MP = MC,
Two points meet the requirements, namely (- 1, √ 10), (- 1, √ 10)
Take C as the center of the circle and √ 10 as the radius to draw an arc, intersecting the axis of symmetry at point P,
In this case, CP = cm, i.e. P (- 1,6)
Make the vertical bisector of CM intersect the axis of symmetry at point P,
In this case, PC = PM,
The solution is p (- 1,5 / 3)
So there are three qualified points
 
2）
 
Let e (x, - X & # 178; - 2x + 3), where X & lt; 0, - X & # 178; - 2x + 3 & gt; 0,
Even OE,
S△BOE=(1/2)*BO*(-x²-2x+3)=（3/2）(-x²-2x+3)
S△COE=(1/2)*CO*(-x)=(-3/2)x
So quadrilateral boce area
=S△BOE+S△COE
=（3/2）(-x²-2x+3)+(-3/2)X
=（-3/2)x²-(9/2)x+9/2
When x = - 3 / 2, there is a maximum area, e (- 3 / 2,15 / 4)

The coordinate of the intersection of the parabola y = 4x Square-1 and the Y axis is, and the X axis

The intersection coordinates of Y-axis and y-axis are (0, - 1), and the intersection coordinates of x-axis and y-axis are (1 / 2,0) and (- 1 / 2,0)

When a team of players was trained, all the players moved forward at 35 km / h. Suddenly, the 1 player marched on the speed of 45 km / h, and after 10 km, she turned the car back and still rides back at 45 km / h until she joined the other players. How long time did the 1 players start from the team to meet the players again?

How to find the trajectory equation of a point with the square difference of the distance between O (0,0), a (C, 0) being constant C?

Let P (x, y) be the locus, then
OP ^ 2-pa ^ 2 = C, i.e
(x^2+y^2)-(x-c)^2-y^2=c
Simplification: 2cx-c ^ 2 = C
That is, the trajectory is x = (c + 1) / 2, which is a straight line perpendicular to the X axis