Combined Transformations of Sinusoidal Functions (College Board AP® Precalculus): Revision Note

Roger B

Written by: Roger B

Reviewed by: Mark Curtis

Updated on

Combined transformations of sinusoidal functions

How do all four transformations work together?

  • The general form of a sinusoidal function with all transformations combined is

    •  f(θ)=asin(b(θ+c))+d  or  f(θ)=acos(b(θ+c))+d

  • The parameters a, b, c and d describe the following characteristics of the graph

Parameter

Characteristic

How to find it from the graph

|a|

Amplitude

maxmin2

d

Vertical shift (midline at  y=d)

max+min2

2π|b|

Period

Horizontal distance between two consecutive maxima (or minima)

c

Phase shift

Horizontal displacement from the base function's starting position

  • If a>0, the function oscillates normally

  • If a<0, the graph is reflected over the midline

    • peaks and troughs are swapped

How can the amplitude and vertical shift be found from a graph?

  • Read the maximum and minimum output values from the graph

  • The amplitude is

    • |a|=maxmin2

  • The vertical shift (midline) is

    • d=max+min2

  • For example, if a sinusoidal graph has a maximum of 7 and a minimum of 1

    • Amplitude: |a|=712=3

    • Vertical shift: d=7+12=4

      • so the midline is y=4

How can the period and the value of b be found from a graph?

  • The period is the horizontal distance between

    • two consecutive maxima

    • two consecutive minima

    • or any two consecutive corresponding points on the graph

  • Once the period is known, the value of |b| can be found using:

    • |b|=2πperiod

  • For example, if consecutive maxima occur at x=1 and x=5

    • Period: 51=4

    • |b|=2π4=π2

How can the phase shift be determined?

  • The phase shift describes the horizontal displacement of the graph

    • compared to the base sine or cosine function

  • For a cosine model (acos(b(θ+c))+d)

    • The base cosine function has its maximum at θ=0

    • If the graph's first maximum occurs at θ=h

      • then the phase shift is h units to the right

      • which means c=h

        • or equivalently, the shift is c=h

  • For a sine model (asin(b(θ+c))+d)

    • The base sine function crosses the midline (going upward) at θ=0

    • If the graph first crosses the midline going upward at θ=h

      • then the phase shift is h units to the right

      • which means c=h

        • or equivalently, the shift is c=h

  • The phase shift is often the trickiest parameter to determine

    • It helps to first decide whether to use a sine or cosine model (a question may decide this for you)

    • and then identify the appropriate reference point on the graph

Examiner Tips and Tricks

When reading a graph to determine the phase shift, be careful about the difference between the shift in the θ-value and the value of c in the equation.

If a cosine graph reaches its first maximum at θ=h, then the equation has (θh) inside the cosine, which means c=h.

  • A common error is to write c=h instead

  • Always verify your equation by substituting a known point from the graph

What is the systematic process for finding the equation from a graph?

  • Start by reading the maximum and minimum values from the graph

    • Use these to find the amplitude (|a|) and vertical shift (d)

      • Unless a question specifies otherwise, you can assume that a is positive

        • in which case a is simply equal to the amplitude

  • Next identify two consecutive maxima (or minima)

    • Use these to find the period

    • then calculate |b|

      • Unless a question specifies otherwise, you can assume that b is positive

        • in which case b is simply equal to 2πperiod

  • At this point you need to decide whether to use a sine or cosine model

    • Cosine is often convenient when a maximum or minimum is clearly visible

    • Sine is convenient when a midline crossing (going upward) is clearly visible

    • Often a question will tell you whether a sine or cosine model is to be used

  • Now you can identify the phase shift

    • by comparing the graph's reference point

    • to where the base function would normally have that feature

  • Finally write the equation

    • and verify by checking that the function matches at least one or two key points on the graph

Examiner Tips and Tricks

Remember that the same graph can be described by either a sine or cosine model. A cosine function is just a phase-shifted sine function, and vice versa

  • However on the exam, if you are given a specific form to use (e.g. h(t)=asin(b(t+c))+d), make sure you use it

    • even if a different model might feel more natural to you

Also note that the parameters a and b will almost always be positive in exam questions.

  • It is always possible to write a model using a>0 and b>0

  • The difference between that and a model with negative values for a and b can always be represented instead as an appropriate phase shift

Worked Example

Sine wave graph with peaks at (-2pi/3,4), (pi/3,4), (4pi/3,4), and troughs at (-pi/6,-2), (5pi/6,-2), (11pi/6,-2); labelled axes x and y.

The figure shows the graph of a trigonometric function  f. Which of the following could be an expression for  f(x)?

(A)  3cos (2 (xπ3))+1

(B)  3cos (2 (xπ6))+1

(C)  3sin (2 (xπ3))+1

(D)  3sin (2 (xπ6))+1

Answer:

Start by finding the amplitude and vertical shift

  • The maximum value is 4 and the minimum value is 2

|a|=4(2)2=3,    d=4+(2)2=1

So the amplitude is 3 and the midline is at  y=1

  • This is consistent with all four answer options

Next find the period and b

  • Consecutive maxima appear at x=π3 and x=4π3

period=4π3π3=π

|b|=2ππ=2

  • This is also consistent with all four options.

To distinguish between the models, determine the phase shift

  • The graph reaches a maximum at x=π3

  • For a cosine model, the maximum of the base function cosθ occurs at θ=0, so the maximum has been shifted right to x=π3

  • This means

b(x+c)=0 when x=π3

2(π3+c)=0    c=π3 

This gives option (A) as the correct answer, 3cos (2 (xπ3))+1

  • You can verify this by checking a couple of points

At x=π3:
3cos(2·0)+1=3(1)+1=4

At x=5π6:
3cos (2 (5π6π3))+1=3cos(π)+1=3+1=2

It's worth looking at why the other options are incorrect

  • Option (B) uses c=π6 instead of c=π3

    • This function would have a maximum at (π6, 4) instead of at (π3, 4)

  • Option (C) uses a sine function, 3sin (2 (xπ3))+1

    • That particular model would have a midline crossing at x=π3, not a maximum point

  • Option (D) also uses a sine function, 3sin (2 (xπ6))+1

    • That particular model would have a midline crossing at x=π6, and a maximum point at 5π12

(A)  3cos (2 (xπ3))+1

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Roger B

Author: Roger B

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Roger's teaching experience stretches all the way back to 1992, and in that time he has taught students at all levels between Year 7 and university undergraduate. Having conducted and published postgraduate research into the mathematical theory behind quantum computing, he is more than confident in dealing with mathematics at any level the exam boards might throw at you.

Mark Curtis

Reviewer: Mark Curtis

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Mark graduated twice from the University of Oxford: once in 2009 with a First in Mathematics, then again in 2013 with a PhD (DPhil) in Mathematics. He has had nine successful years as a secondary school teacher, specialising in A-Level Further Maths and running extension classes for Oxbridge Maths applicants. Alongside his teaching, he has written five internal textbooks, introduced new spiralling school curriculums and trained other Maths teachers through outreach programmes.