Given the function y = sin (1 / 2x + π / 6), how can the image of this function be transformed from the image of y = SiNx (x belongs to R)?

Given the function y = sin (1 / 2x + π / 6), how can the image of this function be transformed from the image of y = SiNx (x belongs to R)?

Firstly, the original image is shifted to the left by π / 6 unit length, and then the ordinate of each point in the image is kept unchanged, and the abscissa is expanded to twice the original image to obtain a new image

In order to get the image of function y = sin (2x + 30 degrees), how can we translate the image of function y = cos2x

Y = sin (2x + π / 6) = cos (2x + 2 π / 3) = Cos2 (x + π / 3), so we only need to shift the image of y = cos2x to the left by π / 3 units

The image of the function y = sin (2x + π / 3) A is symmetric about the origin and B is symmetric about a point (- π / 6,0) C is symmetric about y axis and D is symmetric about straight line x = π / 6

When x = - π / 6, y = 0
So, with respect to the point (- π / 6,0), symmetry
Choose B

How to draw the function image of y = 2Sin pie x (- 2? 4)?

The function y = 2Sin π x (- 2)

How to move the image of 1 / (X-2) from the function image of 1 / x

A 1 / X image is shifted two units to the right to get a 1 / (X-2) image

Let y = - 3 / X (1) draw the function image (2) use the function image to find the change range of function value y when - 3 ≤ x ≤ - 1 (3) When 3 ≤ x ≤ 6, what are the maximum and minimum values of the function

The graph of a function is a hyperbola with two branches in the second and fourth quadrants
2. From the graph, when - 3 ≤ x ≤ - 1., 1 ≤ y ≤ 3
When 3 ≤ x ≤ 6, the maximum value is - 1 / 2.. the minimum value is - 1

How does the change of B in the general formula of quadratic function affect the function image That is to say, what kind of translational motion does the parabola (A and C remain unchanged) when B is continuously increasing or decreasing? Is it a left-right translation?

Take y = ax 2 + BX + C as an example
y=ax²+bx+c
=a(x²+b/a*x)+c
=a{x²+b/a*x+[b/(2a)]²}-a*[b/(2a)]²+c
=a[x+b/(2a)]²-b²/(4a)+c
=a[x+b/(2a)]²+(4ac-b²)/(4a)
The axis of symmetry is x = - B / (2a)
Vertex coordinates: x = - B / (2a), y = (4ac-b 2) / (4a)
The symmetry axis is x = - B / (2a)
When B increases continuously, the function image shifts to the left
When b decreases continuously, the function image shifts to the right

The function image of y = - 1 / 2x 2 is drawn by point tracing method, and the changing trend of the function image is pointed out

I can't get the picture, but the trend can be counted
First of all, the image of this function must be below the x-axis
And it's monotonous
When x is greater than 0, the larger x is, the smaller y is
When x is less than 0, the larger x is, the smaller y is

Draw the following graph of the function and show the change of Y with X （1）y=-6/x(x>0) (2)y=-x²

The first is that a branch of hyperbola Z of the inverse function in the fourth quadrant increases with the increase of X in the absence of interval y
The second is that the parabola passes through the origin and is symmetric about the y-axis

Transformation of function image The image of y = f (x) and the image of y = f (- x) are changed into y = f (x + a) and y = f (- x + a) (a > 0)? A

It's all moving to the left!
Let k = x + A, when x = 0, k = a, then y = F1 (x + a) = F0 (a), when k = 0, x = - A
Then y = F1 (x + a) = F1 (0) = F0 (- a),
Just draw a picture!