Given the function f (x) = asin (Wx + φ) (a > 0, | φ | π / 2, w > 0), find the analytic formula of F (x), as shown in Fig

Given the function f (x) = asin (Wx + φ) (a > 0, | φ | π / 2, w > 0), find the analytic formula of F (x), as shown in Fig

Given the function f (x) = asin (Wx + φ) (a > 0, ||||||||||||||||||||||||||||||||||||||||||||||||||||||||

The (0 / ψ) function of (0 / ψ) is known, The first highest point on the right side of the y-axis is m (2, root 2) and the first intersection point of x-axis on the right side of the origin is n (6,0). Find (1) its analytic formula and all symmetry centers (2) g (x) and image f (x) are symmetric about electricity P (4,0), and find the monotone increasing interval of G (x)

(1) According to the coordinates of the highest point: a = root 2, and from the first point n to the right of the highest point, the period of 1 / 4 is 4, so w = 2 π / 16 = π / 8. If the coordinates of m point are brought in, then we can know that ψ = π / 4;
(2) According to the symmetry, G (x) = - f (8-x), the latter is easy to obtain

It is known that the image of function f (x) = asin (Wx + φ), (a > 0, w0,0 < φ < π / 2) is symmetric with respect to point B (- π / 4,0). The shortest distance between point B and the symmetry axis of the image with function y = f (x) is π / 2, f (π / 2) = 1. Then the analytic formula of F (x) is

The shortest distance from point B to the symmetry axis of the image with function y = f (x) is π / 2
T / 4 = π / 2, t = 2 π
So w = 1
If f (- π / 4) = 0, sin (- π / 4 + φ) = 0, so - π / 4 + φ = k π
And 0

A series of corresponding values of the known function f [x] = asin [Wx + φ] + B, [a > 0, w > 0], are as follows: X ：-π/6 π/3 5π/6 4π/3 11π/6 7π/3 17π/6 Y ：-1 1 3 1 -1 1 3 1) Find the analytic formula of function f (x) according to the table (2) According to the result of [1], if the period of the function y = f (KX) (k > 0) is 2 π / 3, when x is included in the closed interval, [0, π / 3], the equation f (KX) = m has exactly two different solutions, and the value range of real number m is obtained

A series of corresponding values of the known function f [x] = asin [Wx + φ] + B, [a > 0, w > 0], are as follows:
X ：-π/6 π/3 5π/6 4π/3 11π/6 7π/3 17π/6
Y ：-1 1 3 1 -1 1 3
1) Find the analytic formula of function f (x) according to the table
(2) According to the result of [1], if the period of the function y = f (KX) (k > 0) is 2 π / 3, when x is included in the closed interval, [0, π / 3], the equation f (KX) = m has exactly two different solutions, and the value range of real number m is obtained
It can be seen from the table that the two maximum points of the function are (5 π / 6,3), (17 π / 6,3)
∴T=17π/6-5π/6=2π==>w=1
The two minimum points are (- π / 6, - 1), (11 π / 6, - 1)
∴A=(3+1)/2=2,B=(3-1)/2=1
∴f(x)=2sin(x+φ)+1
f(π/3)=2sin(π/3+φ)+1=1==> sin(π/3+φ)=0==>φ=-π/3
∴f(x)=2sin(x-π/3)+1
(2) Analysis: F (KX) = 2Sin (KX - π / 3) + 1, t = 2 π / 3
∴k=2π/T=3
∴f(x)=2sin(3x-π/3)+1
When [0, π / 3], the equation f (KX) = m has exactly two different solutions
f(0)=2sin(-π/3)+1=1-√3
f(π/3)=2sin(π-π/3)+1=1+√3
3x-π/3=π/2==>x=5π/18
ν f (x) is the maximum value 3 at x = 5 π / 18
When m ∈ [1 + √ 3,3), the equation f (KX) = m has exactly two different solutions

It is known that the minimum positive period of the function f (x) = asin (Wx + a) is 2, If x belongs to [0,1], f (x) = a has two different real roots, then find the range of A A real number and a in a function are not a number

[1,2）.

Why is the difference between y = asin (Wx + &) image panning before scaling and first scaling and then panning A. What is the practical significance of W. & and what are the determinants of the different translation and stretching times? I hope we can have a deep understanding of the lower trigonometric function,

A decides to stretch up and down, that is, to determine the maximum value
W determines the left and right stretching, that is, the period T = 2 π / W
Translate first and then stretch
y=Asin(wx+&)
y=Asin(w（x+y)+&)
y=kAsin(nwx+y+&)
And stretch and then translate
y=Asin(wx+&)
y=kAsin(nwx+&)
y=kAsin(nw（x+y)+&)
y=kAsin(nwx+ny+&)

The problem of y = asin (Wx + Q) + B How to find the symmetry center of y = asin (Wx + Q) + B?

If there is a shortcut, we can take Wx + Q as X, then we can find the symmetry center K π of sin (x), and then we can bring Wx + Q into it (Wx + q = k π)

If the image of the function f (x) = asin (Wx + 6 / π) is shifted to the left, and the image of π 6 / 6 is symmetric about the y-axis, then the value of W may be A 2 B 3 C 4 D 6

When f (x) = asin (Wx + π / 6), the image is shifted to the left by π / 6
y=Asin[w(x+π/6)+π/6]
X = 0: asin (w π / 6 + π / 6)
=Asin [(W + 1) π / 6] is the maximum
When w = 2, it accords with the meaning of the question
Option a

If the function s = asin (ω x + φ) (a > 0, ω > 0,) represents a vibration quantity with amplitude of 1 2, the frequency is 3 2 π, the initial phase is π 6, then the analytic formula of S is______ .

The function s = asin (ω x + φ) (a > 0, ω > 0,) represents a vibration quantity with amplitude of 1
2, the frequency is 3
2 π, the initial phase is π
6, so a = 1
2，T=2π
So ω = 3, φ = π
Six
So the analytic formula of the function is: S = 1
2sin（3x+π
6）．
So the answer is: S = 1
2sin（3x+π
6）．

The maximum and minimum value of y = asin (ω x + φ) + B (a > 0, ω > 0) what is the initial phase of periodic frequency phase

A -A 2pai/w w/2pai ωx+φ φ