If the image of quadratic function y = X2 is translated according to vector a, the image obtained has only one common point (3,1) with the image of linear function y = 2x-5, then vector a = () A. （2，0）B. （2，1）C. （3，0）D. （3，1）

If the image of quadratic function y = X2 is translated according to vector a, the image obtained has only one common point (3,1) with the image of linear function y = 2x-5, then vector a = () A. （2，0）B. （2，1）C. （3，0）D. （3，1）

Let a = (h, K), that is, translate the image of y = x2 according to the vector a = (h, K) to get the image of Y-K = (X-H) 2 (that is, translate h units to the right, and then translate K units to the upper). From the meaning of 1-k = (3-h) 2, from y − k = (x − h) 2Y = 2x − 5, we can get x2-2 (H + 1) x + H2 + K + 5 = 0, from △ = 0, we can get

Vector translation problem
After the image of function f (x) = (x ^ 2 + 2x + a) / X is translated according to vector m, the image of function y = g (x) is obtained. When y = g (x) is an odd function, the vector m is obtained

Let m = (MX, my), then any point B: (x, G (x)) on G (x) changes into a: (x-mx, G (x) - my after translation along - M (i.e., reverse direction), and point a is just on f (x), i.e., G (x) - my = f (x-mx) = ((x-mx) ^ 2 + 2 (x-mx) + a) / (x-mx) = = = = > G (x) = my + (x-mx) + 2 + A / (x-mx). [1] from y

Translation of vector
2x-y + 1 = 0; is it 2 (x-1) - (y + 1) + 1 = 0 after translation according to vector a = (1, - 1)? Why not 2 (x + 1) - (Y-1) + 1 = 0?

Find out the vector and decompose it into (1,0) + (0, - 1), that is, the sum of two vectors,
Know to shift 2x - y + 1 = 0 one unit to the right, and then one unit down
We obtain 2 (x-1) - (y + 1) + 1 = 0

Given m (2,3), n (- 3,2), P (- 1,2), translation vector a = (3,2), the coordinates of corresponding points after translation are obtained

Let the translation be divided into M '(x1, Y1), n' (X2, Y2), p '(X3, Y3)
From the topic meaning: vector mm '= vector NN' = vector PP '= vector a
Namely:
X 1-2 = 3, y 1-3 = 2; x 1 = 5, Y 1 = 5;
X 2 + 3 = 3, y 2-2 = 2; x 2 = 0, y 2 = 4;
X 3 + 1 = 3, y 3-2 = 2; x 3 = 2, y 3 = 4;
So, M '(5,5), n' (0,4), p '(2,4)

If the translation vector a of point m (3,4) is = (- 2,3) to the corresponding point n, what is the coordinate of point n?

Because the abscissa of a is - 2, that is to translate left, add left and subtract right, so 3 + 2 = 5. The ordinate of a is 3, add up and subtract down, so 4-3 = 1. So (5,1)

Given that the coordinates of a (- 1,2) B (6,1) are a '(- 3, m), B' (n, 4), then the vector a=
A(-2,3) B(2,-3) C(-3,2) D(3,-2)
Why?

Let the translation vector a = (h, K)
Then: X '= x + H, y' = y + K;
So we have: - 3 = - 1 + H, H = - 2,
4=1+k,k=3
So a = (- 2,3)
Choose a

Given the vector M = (a + 1.1), n = (a + 2.2), if (M + n) is vertical (m-n), then a =?

M + n = (2a + 3,3) M-N = (- 1, - 1) vertical has - 1 * (2a + 3) + - 1 * 3 = 0, so a = - 3

After the point m (1,3) is translated into vector a = (A1, A2), the corresponding point m (- 2,1) is obtained, then the coordinate of vector a is

The corresponding point M1 (- 2,1) is obtained by the translation of point m (1,3), that is, three units to the left and two units to the down. The corresponding vector is a = (- 3, - 2)

After the point a (- 2,3) is translated according to the vector a = (1,2), the point a ′ is obtained, then the vector → AA ′=

（—1,5）

Under the translation transformation, point a (- 1,2) becomes point a '(3, - 4), then the translation vector a=

4,-6..