Graph sinusoidal functions: phase shift problem?
I'm quite confused how to find the formula for this problem..
- AlanLv 72 months agoFavorite Answer
by looking at the graph
x = -4pi to -5pi/2 is 1/4 of period
but the period = 2pi/b
(-5pi/2 - (-4pi) ) = (1/4) *(2pi/b)
-5*pi/2 +4*pi = pi/ (2b)
divide both sides by pi
-5/2 + 4 = 1/(2b)
3/2 = (1/2b)
6b = 2
b = (1/3)
The mid-line is when the sin(bx+c) = 0
x = -4pi to -5pi/2 sine equals 0 when it equals 0 + 2pik , or pi + 2pik which combines to pik (-4pi, 2.5) if k is an integer then, 2k is an integer pi*k = (1/3)*(-4*pi) + c pi*k = (-4/3)pi + c divide by pi k = (-4/3) + c/pi h(x) = d h( -4pi) = d = 2.5 d = 2.5 now use the 2nd point(-5pi/2 , 6) h(-5pi/2) = 6 = a*(sin((1/3)x+c) ) + 2.5 so it is a max when sin((1/3)x+c) = 1 6 = a + 2.5 a = 6 -2.5 = 3.5 h(-5pi/2) = 3.5*sin(bx+c) + 2.5 this is when sin(bx+c) = 1 where sin is a maximum at pi/2 pi/2 + 2piL = (1/3)x+c = -5pi*(1/3) /2 + c pi/2 + 2piL = -5pi*(1/3) /2 + c pi/2 + 2piL = -5pi/6 + c2piL = -pi/2 -5pi/6 + c 2piL = -3pi/6 -5pi/6 + c 2piL = -8pi/6 + c 2L = -4/3 + c/pi2L = K so if k = 1 , L = 2 2*2 = -4/3 + c/pi 12/3 + 4/3 = c/pi16/3 = c/pic =16pi/3 so one solution is : h(x) = 3.5*sin( (1/3) x + (16pi/3) ) + 2.5You can verify that this equation work for both points and maps to your graph . See graph made with Desmos Graphing Calculator (an online tool)
- 2 months ago
Amplitude: 6 - (-2) = 8
Period: 19 = 2(pi)/k, k = 9.5pi
Midline: y = 2
We can use cos or sin function
Questions asks for sine
y = 4sin(9.5pi(x - 3.5)) + 2
This is not quite the correct answer since i rounded the period and probably made a mistake
- az_lenderLv 72 months ago
The period of the function is 4*(-5pi/2 - (-4pi)) = 6pi, so
h(x) = A*cos(x/3 + phi) + b.
The "b" is the midline height, that is, 2.5.
The amplitude is 6 - 2.5 = 3.5.
h(x) = (3.5)*cos(x/3 + phi) + 2.5.
Yes, the phase shift feels like the tricky part.
Suppose we move the whole curve to the right by 5pi/2, then it will have the "standard" cosine position with a max on the y axis. Therefore, the original function can be written as
h(x) = (3.5)*cos[(x + 5pi/2)/3] + 2.5, or
(3.5)*cos(x/3 + 5pi/6) + 2.5.
Notice that when x = -5/2, the expression becomes
(3.5)*cos(-5pi/6 + 5pi/6) + 2.5
= 3.5*cos(0) + 2.5 = 6, which validates our answer.