Let the image of function y = 4x + 3 be vector If a is translated to the image of y = 4x + 16, then vector A can be () A. （3，1） B. （-3，-1） C. （3，-1） D. （-3，1）

Let the image of function y = 4x + 3 be vector If a is translated to the image of y = 4x + 16, then vector A can be () A. （3，1） B. （-3，-1） C. （3，-1） D. （-3，1）

0

0

Because Y1 = 2x + 6 can be seen as y2 = 2x moving up 6 units along Y axis [i.e. coordinates of a (0,6)], 3 units left along X axis [i.e., coordinates of a (- 3,0)] and vector (1,8); (2,10); (- 2,2). So 4 is right, Not just four pairs]

Given a vector a, the origin o is moved to the point (2, - 2). If the image of the function y = 1 / X is translated by a, the image corresponding to the function analytic formula of the image is obtained according to

y=1/（x-2）-2

Given the function y = 2-x / 1 + X, translate the function image according to vector a, and get the image of y = 3 / x, then the vector a is? 1L answer makes people speechless

Let P (x, y) be any point on the function y = (2-x) / (1 + x), and the coordinate of point P after translation is p '(x', y '). If the coordinate of vector a is (h, K), then x' = x + H, y '= y + K. if p' (x ', y') is y = 3 / x, then (y + k) = 3 / (x + H), y = [3 + K (x + H)] / (x + H), (1), y = (2-x) / (1 + x). (2) according to (1), (2

Find the monotone interval of the following functions: (1) y = 1 + SiNx, X belongs to R; (2) y = - cosx, X belongs to R

Isn't it easy to see the picture at a glance?

What is the monotone increasing interval of the function f (x) = (SiNx + cosx) 2?

F (x) = (SiNx + cosx) 2 F (x) = 1 + 2sinxcosx = 1 + sin2x monotonically increasing interval, [K π, π / 4 + K π]

The minimum value of the function FX = SiNx * SiNx * SiNx + cosx * cosx is

f(x)=(1-cos²x)²+cos²x=1+cos^4x-2cos^2x+cos^x=cos^4x-cos^2x+1=((cos2x+1)/2)^2-cos^2x+1=(cos^2(2x)+1+2cos(2x))/4-(cos2x+1)/2+1=cos^2(2x)/4+3/4;=(cos4x+1)/8+3/4;=cos4x/8+7/8;∵-1≤cos4x≤1...

The minimum value of the function y = 3sinx + 2cosx is Hope to talk about the process in detail

y=3sinx+2cosx
=√(3^2+2^2)sin(x+θ)
=√13sin(x+θ)
ymin=-√13

The function f (x) = x + 2cosx in [0, π] 2] The minimum value on is______ .

∵f（x）=x+2cosx，
∴f′（x）=1-2sinx，
By F ′ (x) = 0, X ∈ [0, π
2] X = π
6，
∵f（0）=2，f（π
6）=π
6+
3，f（π
2）=π
2，
The function f (x) = x + 2cosx in [0, π]
2] The minimum value on is π
2．
So the answer is: π
2．

If 3sinx + 5cosx / 2sinx-7cosx = 1 / 11, find TaNx

3sinx+5cosx/2sinx-7cosx=1/11
11(3sinx+5cosx)=2sinx-7cosx
33sinx+55cosx=2sinx-7cosx
31sinx=-62cosx
So TaNx = SiNx / cosx = - 2