If f (x) = 4, what is the value of X

If f (x) = 4, what is the value of X

It is 2, which is obtained by using the segmented range

How to use the image method to solve the following binary linear equations?
 

Change ① to y = - X
Let 2 be y = x + 2
The images of these two functions are drawn in rectangular coordinates
The abscissa of the intersection of images is the value of X, and the value of Y is calculated from the abscissa
Look at the picture below

When Xiao Liang uses the method of image to solve the system of quadratic equations of two variables, he makes two corresponding linear functions in the same rectangular coordinate system. The image is shown in the figure, then the system of equations solved by him is___ ．

Let the analytic formula of the straight line passing through points (0, 2) and (2, - 2) be y = KX + B, then B = 22K + B = - 2, and the solution is k = - 2b = 2. The analytic formula of the straight line is y = - 2x + 2. Let the analytic formula of the straight line passing through points (- 2, 0) and (2, - 2) be y = MX + N, then - 2m + n = 02m + n = - 2, and the solution is m = - 12n = - 1. The analytic formula of the straight line is y = - 12x-1. Therefore, the system of equations solved by others is y = - 2x + 2Y = - 12x-1

Straight line y = x + Half 1m and hyperbola y = x part m intersect a in a quadrant, X axis intersects C, AB perpendicular to X axis s triangle AOB = 1
Find the value of M and the area of triangle ABC

Y=m/X =>X*Y=m,
S△AOB=1 =>1/2X*Y=1 =>m=1
Y=X+1
Y=2/X
X = 1 (- 2 rounds; one quadrant)
Y=2
C coordinate (- 1,0)
S△ABC=1/2（1+1）*2=2

In the right triangle AOB, the angle ABO is 90 ° and the point B is on the x-axis. The point a is the intersection of the straight line y = x + m and the hyperbola y = m / X in the first quadrant. The root of the abscissa of the point a is 15-3. The point C is the intersection of the straight line y = x + m and the x-axis, and the area of the triangle AOB is 3

From y = x + m, y = m / x, we can see that x2 + mx-m = 0, and the abscissa root of point a is 15-3, we can see that M = 6, the ordinate root is 15 + 3, then B is (root 15-3,0), and because C (3,0), then BC = 6 + root 15, then the area is (3 root 15 + 3) / 2, about 7.309475019

In a right triangle, ∠ ABO = 90 °, point B is on the x-axis, and point a is the intersection of a straight line y = x + m and a hyperbola y = m / X in the first quadrant,
If the line y = x + m and the negative half axis of X axis intersect at point C, and s triangle = 3, the expressions of line and hyperbola are obtained

Because s △ ABO = 3
So the absolute value of M = 6
Because the image is in the first quadrant
So m = 6
So the straight line is y = x + 6 and the inverse proportion is y = 6 / m

Given that point P is any point except vertex on hyperbola x2a2 − y2b2 = 1, F1 and F2 are left and right focus respectively, C is half focal length, and the inscribed circle of △ pf1f2 and F1F2 are tangent to point m, then | f1m | · | F2m|=______ ．

According to the two tangent lines drawn from a point outside a circle to a circle, and the two tangent lines drawn from a point outside a circle to a circle, we can know that: | f1f1m ? = ? F1s | F1s ?, | F2m | F2m | F2t | PS | = | Pt | ① when p is in the right branch of hyperbolimage, and according to the definition of hyperbola, we can know that f1m | f1m | - ˊ

It is known that Q point is a moving point on the hyperbola x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1 (a, b > 0), which is different from two vertices. F 1 and F 2 are the left and right focal points of the hyperbola
From the bisector of the angle f1qf2 in the direction of F2, make the vertical line F2P, the perpendicular foot is p, and find the trajectory equation of point P

I'll tell you in private

Let the hyperbola x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1, two focal points F1 and F2, and point Q be any point on the hyperbola except for the vertex
If the perpendicular of the bisector of angle f1qf2 and the perpendicular foot is p, then the trajectory of point P is:

Extend f1p, cross QF2 (or its extension) with M
Then | QF1 | = | QM|
|F2M|=| |QM|-|QF2| |=| |QF1|-|QF2| |=2a
In the triangle f1f2m, OP is the median line
|OP|=|F2M |/2=a
So the trajectory of P is a circle, the center of the circle is the origin, and the radius is a
The equation is X & # 178; + Y & # 178; = A & # 178;

Given that point P is any point on the right branch of hyperbola x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1 except vertex, F1 and F2 are its focal points,
It is proved that Tana / 2 times cotb / 2 = (C-A) / (c + a)

Let AF1 = R1
AF2=r2
From the sine theorem: R1 / SINB = R2 / Sina = 2C / sin (a + b)
And R1-R2 = 2A
From the formula of the ratio of combination and separation, we can get: (R1-R2) / (SINB Sina) = 2C / sin (a + b)
2a/(sinB-sinA)=2c/sin(B+A)
2a/2cos[(B+A)/2]sin[(B-A)/2]=2c/2sin[(B+A)/2]cos[(B+A)/2 ]
a/sin[(B-A)/2]=c/sin[(B+A)/2 ]
c/a=sin[(B+A)/2]/sin[(B-A)/2]
(c-a)/(c+a)={sin[(B+A)/2]-sin[(B-A)/2]}/{sin[(B+A)/2]+sin[(B-A)/2]}
=[2cos(B/2)sin(A/2)]/[2sin(B/2)cos(A/2)]
=tanA/2*cotB/2
That is: Tana / 2 times cotb / 2 = (C-A) / (c + a)