Horizontal and Phase Shifts of Sinusoidal Functions.
The general form f ( x ) = A sin( Bx ) + M is a sine (or cosine) graph with amplitude A, midline/vertical
shift M and period P = 2Bπ . In this form there is no horizontal shift. If we want to include a horizontal
shift, we would write f ( x ) = A sin( B ( x − C )) + M , 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
f ( x) = A sin( Bx + ϕ ) + M , 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 f ( x ) = A sin( B ( x − C )) + M , where the B has been factored out.
Part A: Sketching a function that includes phase/horizontal shift:
Example 1: Let’s sketch f ( x) = 2 sin( 2( x − π6 )) + 1 . We simply identify all the important bits:
amplitude is 2, midline is 1, the period is 22π = π and the horizontal shift is π6 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 f ( x) = 3 cos(2 x + π4 ) + 1 , the π4 is the phase shift (do you see why?). Factor
the two from the inside first to get f ( x) = 3 cos(2( x + π8 )) + 1 . Now we have the horizontal shift of π8
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)
f ( x) = −2 sin(4( x + π4 )) − 1
2)
f ( x) = 4 cos(2 x +
2π
3
)) + 2
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 y = 0 , and the lowest points
are y = −2 , so the midline would be the average of these: M = −1 . Hence, the amplitude is A = 1 .
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 B = 25π . It looks like a sine function, so we’ll go with that,
and we have so far:
f ( x) = sin( 25π ( x − C )) − 1
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:
(1,0) → f ( x) = sin( 25π ( x − C )) − 1
→ 0 = sin( 25π (1 − C )) − 1
( plug in the pt )
→ 1 = sin( 25π (1 − C )) (add 1 to both sides )
→
π
2
=
2π
5
(1 − C ) (arcsin both sides )
→ 1.57 = 1.26(1 − C ) ( get decimals)
→ 11..57
26 = 1 − C → C = −0.246
( solve for C )
⇒ f ( x) = sin( 25π ( x + 0.246)) − 1 ( final answer )
Homework, part B:
Give the complete sinudoidal model for this graph: