Horizontal and Phase Shifts of Sinusoidal Functions. The general form is a sine (or cosine) graph with amplitude A, midline/vertical shift M and period

MBxAxf += )sin()(

BP π2= . In this form there is no horizontal shift. If we want to include a horizontal shift, we would write MCxBAxf +−= ))(sin()( , where C is the horizontal shift. Notice how the B has been factored outside the inner parentheses first. If we distribute the B in with the C, we get a form

MBxAxf ++= )sin()( ϕ , where ϕ (phi) represents the phase shift. Frankly, telling the two – horizontal shift and phase shift – apart is very difficult. My suggestion is to adopt one form and stick with it; I always go with the first form MCxBAxf +−= ))(sin()( , where the B has been factored out. Part A: Sketching a function that includes phase/horizontal shift: Example 1: Let’s sketch 1))(2sin(2)( 6 +−= πxxf . We simply identify all the important bits: amplitude is 2, midline is 1, the period is ππ =2

2 and the horizontal shift is 6π to the right.

Graph everything first without the horizontal shift, then perform the horizontal shift. And you are done!

If you get a problem like 1)2cos(3)( 4 ++= πxxf , the 4π is the phase shift (do you see why?). Factor

the two from the inside first to get 1))(2cos(3)( 8 ++= πxxf . Now we have the horizontal shift of 8π

to the left. Then plot it. Your homework, part A: On a separate sheet of paper, give accurate sketches of these two functions. Show all guidelines and information. 1) 1))(4sin(2)( 4 −+−= πxxf 2) 2))2cos(4)( 3

2 ++= πxxf

Part B: Creating a function from a graph that includes horizontal shift. Suppose we are given a sinusoidal graph:

We can start by picking off the obvious information: the highest points are 0=y , and the lowest points are , so the midline would be the average of these: 2−=y 1−=M . Hence, the amplitude is 1=A . The period can be found by determining the distance from ‘top to top’ (or ‘bottom to bottom’). In this example, the period is 5, so the B value is 5

2π=B . It looks like a sine function, so we’ll go with that, and we have so far:

1))(sin()( 52 −−= Cxxf π

Now, the trick is to determine the C value (the horizontal shift). Pick an obvious point and plug into the function. In this example, (1,0) is on the graph and is easy to use. Then solve for C, as shown:

)(1))246.0(sin()()(246.01

)()1(26.157.1)(arcsin)1(

)1())1(sin(1)(1))1(sin(0

1))(sin()()0,1(

52

26.157.1

52

2

52

52

52

answerfinalxxfCforsolveCC

decimalsgetCsidesbothC

sidesbothtoaddCpttheinplugC

Cxxf

−+=⇒

−=→−=→

−=→

−=→

−=→

−−=→

−−=→

π

ππ

π

π

π

Homework, part B: Give the complete sinudoidal model for this graph: