Graph sinusoidal functions: phase shift problem?

I'm quite confused how to find the formula for this problem.. 

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  • Alan
    Lv 7
    2 months ago
    Favorite 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) 

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    • Alan
      Lv 7
      2 months agoReport

      Your question says do it as a sine function . 
      This is the only correct answer using a sine function.

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  • 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

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  • 2 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.

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