If the distance between the vertex A and the positive half axis of the graph a and B is the maximum distance on the positive half axis of the graph A. 32a−12aB. 32a+12aC. 62a−12aD. 62a+12a

If the distance between the vertex A and the positive half axis of the graph a and B is the maximum distance on the positive half axis of the graph A. 32a−12aB. 32a+12aC. 62a−12aD. 62a+12a

According to the symmetry, we can get OC ⊥ AB, ∵ △ AOB as isosceles right triangle, ab = a, ∵ od = 12ab = 12a, in RT △ BCD, BC = a, BD = 12a, according to Pythagorean theorem: CD = 32a, then OC = od + DC = 12a + 32A
Make the image of function y = 2 to the power of 1-2x
It's not necessary to sketch the process
Set & nbsp; y = 2 ^ (1-2x)
Deformation: y = & nbsp; (1 / 2) ^ (2x-1)
That is & nbsp; & nbsp; y = (1 / 4) ^ (x-1 / 2)
 
First, make the image C with y = (1 / 4) ^ x, and shift C to the right by 1 / 2 units
Is y = (1 / 4) ^ (x-1 / 2), that is y = 2 ^ (1-2x) image & nbsp;
 
Given the vertex a (2,3), B (- 1, - 1), BC ‖ Y axis of △ ABC, O is the coordinate origin. If OC vector ‖ AB vector, find the coordinate of point C
If △ ABC is a right triangle, calculate the coordinate of point C
If BC = 7 and C is in the second quadrant, find the value of Tana
(1) Because OC vector ‖ AB vector, KAB = 4 / 3, so the coordinates of point C are (3,4)
(2) There are two possible coordinates of point C (- 1,3) or (- 1,2)
(3) Because BC ∥ Y axis, BC = 7, C in the second quadrant, the coordinates of point C can be calculated as (- 1,6)
There is no picture
What is the graph of the x power of E
It is known that the three vertices a, B, C and O of △ ABC are a point in the plane, which satisfies: vector AB + vector ob + vector OC = 0. If the real number λ satisfies: vector AB + vector AC + λ vector OA = 0, then the value of λ is:
Take the midpoint of BC as M,
So the vector ob + OC = 2om
∵ vector AB + vector ob + vector OC = 0 vector
The vector AB + 2 vector om = 0 vector
‖ vector AB = - 2 vector OM
So OM / / AB
Vector AB + vector AC = 2AM
Vector AB + vector AC + λ vector OA = 0
The vector is 0A + λ
2AM=λAO
A, m and o are collinear
If there are contradictions, check the input
How to draw the image of y = a ^ (x-1) + 1 to explain X-1 power
Given that the coordinates of the three vertices of the triangle ABC are a (- 1,0), B (1,2), C (0, c), and the vector AB is perpendicular to the vector BC, then the value of C is
What is the specific process of seeking truth
AB=(1-(-1),2-0)=(2,2)
BC=(0-1,c-2)=(-1,c-2)
Because ab ⊥ CD
Then AB * CD = 0
2*(-1)+2*(c-2)=0
-2+2c-4=0
2c=6
C=3
How to draw the image of (x + 1) power of y = 2 from the image of x power of F (x) = 2
The image with F (x) = 2 to the x power is shifted one unit to the left to get the image with y = 2 to the (x + 1) power
If f (x) = x ^ 2, the image moves one unit to the left
Move the image with F (x) = 2 to the x power one unit to the left in the coordinate axis
It's too much trouble to draw the picture~~~
One side ab of △ ABC is fixed, and the vertex C moves on a fixed line L parallel to ab. suppose that the vertical center of △ ABC is in the triangle, the trajectory equation of the vertical center is obtained
Establish a rectangular coordinate system, put AB on the X axis, and be symmetrical about the origin, let a (a, 0), B (- A, 0), because C moves on y = B, let C point coordinate be (x1, b), then the equation of AC can be obtained from the "two-point formula", which is sorted out as: y = BX / (x1 + 2) + 2B / (x1 + 2),
Let BD be perpendicular to AC at D, CE be perpendicular to ab (x axis) at e, BD and CE intersect at P, P is perpendicular
Because BD is perpendicular to AC, the slope k of the straight line BD satisfies K × B / (x1 + 2) = - 1, and K = - (x1 + 2) / B is solved. Then according to the coordinates of point B, the BD equation is obtained, and the equation of CE is x = x1. The two equations are connected, and x 1 is eliminated to get y = (B + 1 - x square) / B, so it is a parabola
Semicircle centered at the midpoint of ab
If f (x) = x (1 / 2 to the power of - 1 + 1 / 2), the parity of F (x) is judged
F (x) = x (1 / 2 to X-1 + 1 / 2) 2 to X-1 is a whole
F (x) is an even function,
f(-x)=-x[1/(2^(-x)-1)+1/2]
=-x( 2^x/(1-2^x) + 1/2 )
=x( 2^x/(2^x-1) - 1/2 )
f(x)-f(-x)=x( (1-2^x)/(2^x-1) + 1/2 - (-1/2) )
=x(-1+1)=0
So f (x) = f (- x)