Graphing Sinusoidal Functions: Phase Shift vs. Horizontal Shift

Section 2.3 Graphing Sinusoidal Functions: Phase Shift vs. Horizontal Shift

Let’s consider the function $g(x)=\sin \mathopen{}\left( 2x-\frac{2\pi}{3} \right)\mathclose{} \text{.}$ Using what we study in MTH 111 about graph transformations, it should be apparent that the graph of $g(x)=\sin \mathopen{}\left( 2x-\frac{2\pi}{3} \right)\mathclose{} $ can be obtained by transforming the graph of $g(x)=\sin(x)\text{.}$ (To confirm this, notice that $g(x)$ can be expressed in terms of $f(x)=\sin(x),$ as $g(x)=f \mathopen{}\left( 2x-\frac{2\pi}{3} \right)\mathclose{} \text{.}$) Since the constants “$2$” and “$\frac{2\pi}{3}$” are multiplied by and subtracted from the input variable, $x\text{,}$ what we study in MTH 111 tells us that these constants represent a horizontal stretch/compression and a horizontal shift, respectively.

It is often recommended in MTH 111 that we factor-out the horizontal stretching/compressing factor before transforming the graph, i.e., it’s often recommended that we first re-write $g(x)=\sin \mathopen{}\left( 2x-\frac{2\pi}{3} \right)\mathclose{} $ as $g(x)=\sin \mathopen{}\left( 2 \mathopen{}\left( x-\frac{\pi}{3} \right)\mathclose{}\right)\mathclose{} \text{.}$

After writing $g$ in this format, we can draw its graph by performing the following sequence of transformations of the “base function” $f(x)=\sin(x)\text{:}$

Compress horizontally by a factor of $\frac{1}{2}\text{.}$
Shift $\frac{\pi}{3}$ units to the right.

The advantage of this method is that the $y$-intercept of $f(x)=\sin(x)\text{,}$ $\left( 0,0 \right) \text{,}$ ends-up exactly where the horizontal shift suggests: when we compress the graph by a factor of $\frac{1}{2}\text{,}$ the the $y$-interceot of the graph doesn’t move since $\frac{1}{2} \cdot 0 = 0\text{;}$ then, when we shift the graph $\frac{\pi}{3}$ units to the right, the point $\left( 0,0\right) $ ends up at $\left( \frac{\pi}{3},0\right) \text{;}$ so the $y$-intercept ends up moving $\frac{\pi}{3}$ units to the right, exactly how far we shifted.

Compare this with the alternative method: we can leave $g(x)=\sin \mathopen{} \left( 2x-\frac{2\pi}{3} \right)\mathclose{} $ as-is and skip factoring-out the horizontal stretching/compressing factor, but then we need the following sequence to transform $f(x)=\sin(x)$ into the graph of $g\text{:}$

Shift $\frac{2\pi}{3}$ units to the right.
Compress horizontally by a factor of $\frac{1}{2}\text{.}$

The disadvantage of this method is that the $y$-intercept of $f(x)=\sin(x)$ doesn’t end-up where the horizontal shift suggests: when we shift the graph of $f(x)=\sin(x)$ to the right by $\frac{2\pi}{3}$ units, the $y$-intercept moves from $\left( 0,0 \right) $ to $\left( \frac{2\pi}{3},0\right) \text{;}$ then, when we compress the graph by a factor of $\frac{1}{2}\text{,}$ it moves to $\left( \frac{\pi}{3},0\right) \text{,}$ so the $y$-intercept doesn’t end up shifted $\frac{2\pi}{3}$ units to the right.

Figure 2.3.1 shows the graphs of $y=f(x)$ and $y=g(x)\text{.}$ Notice that the behavior of $y=g(x)$ at $x=\frac{\pi}{3}$ is like the behavior of $y=f(x)$ at $x=0\text{,}$ i.e., $y=g(x)$ appears to have been shifted $\frac{\pi}{3}$ units to the right. For this reason, $\frac{\pi}{3}$ is called the horzontal shift of $g(x)=\sin \mathopen{} \left( 2x-\frac{2\pi}{3}\right)\mathclose{}=\sin \mathopen{}\left( 2\mathopen{}\left( x-\frac{\pi}{3}\right)\mathclose{}\right)\mathclose{}\text{.}$

This is a graph showing both f(x) and g(x) where the function g(x) is a horizontal shift of f(x). — Figure 2.3.1. $y=g(x)$ with $f(x)=\sin(x)$

The constant $\frac{2\pi}{3}$ is given a different name, phase shift, since it can be used to determine how far “out-of-phase” a sinusoidal function is in comparison with $y=\sin(x)$ or $y=\cos(x)\text{.}$ To determine how far out-of-phase a sinusoidal function is, we can determine the ratio of the phase shift and $2\pi\text{.}$ (We use $2\pi$ because it’s the period of $y=\sin(x)$ and $y=\cos(x)\text{.}$) Since $\frac{2\pi}{3}$ is the phase shift for $g(x)=\sin \mathopen{}\left( 2x-\frac{2\pi}{3} \right)\mathclose{}\text{,}$ the graph of $y=g(x)$ is out-of-phase $\frac{2\pi/3}{2\pi}=\frac{1}{3}$ of a period. (Since this number is positive, it represents a horizontal shift to the right $\frac{1}{3}$ of a period.)

Definition 2.3.2.

Given a sinusoidal function of the form $y=A\sin(wx-C)+k$ or $y=A\cos(wx-C)+k\text{,}$ the phase shift is $C$ and $\frac{|C|}{2\pi}$ represents the fraction of a period that the graph has been shifted (shift to the right if $C$ is positive or to the left if $C$ is negative).

Definition 2.3.3.

If we re-write the function as $y=A\sin \mathopen{}\left( w \mathopen{}\left( x-\frac{C}{w}\right)\mathclose{} \right)\mathclose{}+k$ or $y=A\cos \mathopen{} \left( w \mathopen{}\left( x-\frac{C}{w}\right)\mathclose{} \right)\mathclose{}+k\text{,}$ we can see that the horizontal shift is $\frac{C}{w}$ units (shift to the right if $\frac{C}{w}$ is positive or to the left if $\frac{C}{w}$ is negative).

Example 2.3.4.

Identify the phase shift and horizontal shift of $g(x)=\cos \mathopen{}\left( 3x-\frac{\pi}{4} \right)\mathclose{} \text{.}$

Solution.

The phase shift of $g(x)=\cos \mathopen{}\left( 3x-\frac{\pi}{4} \right)\mathclose{}$ is $\frac{\pi}{4}\text{.}$ This tells us that the graph of $y=g(x)$ is out of phase $\frac{|\pi/4|}{2\pi}=\frac{1}{8}$ of a period, i.e., compared with $y=\cos(x)\text{,}$ the graph of $g(x)=\cos(3x-\frac{\pi}{4})$ has been shifted one-eighth of a period to the right.

To find the horizontal shift, we need to factor-out $3$ from $3x-\frac{\pi}{4}\text{.}$

\begin{align*} g(x) \amp=\cos\mathopen{}\left( 3x-\frac{\pi}{4}\right)\mathclose{}\\ \amp=\cos\mathopen{}\left(3\mathopen{}\left(x-\frac{\pi}{3 \cdot 4}\right)\mathclose{} \right)\mathclose{}\\ \amp=\cos\mathopen{}\left(3\mathopen{}\left(x-\frac{\pi}{12}\right)\mathclose{}\right)\mathclose{} \end{align*}

So the horizontal shift is $\frac{\pi}{12}\text{.}$ This tells us, that compared with $y=\cos(x)\text{,}$ the graph of $g(x)=\cos(3x-\frac{\pi}{4})$ has been shifted $\frac{\pi}{12}$ to the right.

Notice that the period of $g(x)=\cos(3x-\frac{\pi}{4})$ is $2\pi \cdot \frac{1}{3}=\frac{2\pi}{3}\text{,}$ and one-eighth of $\frac{2\pi}{3}$ is $\frac{2\pi}{3} \cdot \frac{1}{8}=\frac{\pi}{12}\text{,}$ so a shift of one-eighth of a period is the same as a shift of $\frac{\pi}{12}$ units!

Example 2.3.5.

Draw a graph $q(t)=2\sin(4t+\pi)+1\text{.}$ First, find it’s amplitude, period, midline, phase shift, and horiontal shift.

Solution.

Amplitude: $|A|=|2|=2$

Period: $P=2\pi \cdot \frac{1}{|w|}=\frac{2\pi}{4}=\frac{\pi}{2}$

Midline: $y=1$

Phase shift: $-\pi$ (this tells is that the graph is out-of-phase $\frac{|-\pi|}{2\pi}=\frac{1}{2}$ of a period)

Horizontal shift: $\frac{\pi}{4}$ units to the left since:

\begin{align*} q(t) \amp=2\sin\mathopen{}\left( 4t+\pi \right)\mathclose{}+1\\ \amp=2\sin \mathopen{}\left(4 \mathopen{}\left(t+\frac{\pi}{4} \right)\mathclose{} \right)\mathclose{}+1\\ \amp=2\sin\mathopen{}\left( 4 \mathopen{}\left(t- \mathopen{}\left( -\frac{\pi}{4}\right)\mathclose{}\right)\mathclose{}\right)\mathclose{} +1 \end{align*}

Now we can draw a graph of $q(t)=2\sin(4t+\pi)+1$ by drawing a sinusoidal function with the necessary features; see Figure 2.3.6.

Exercises Exercises

Sketch Sinusoidal Graphs.

In Exercises 1–4, draw a graph of each of the following functions. List the amplitude, midline, period, phase shift, and horizontal shift.

1.

$f(x)=3\sin\mathopen{} \left( 3x-\frac{\pi}{2} \right)\mathclose{}$

2.

$g(t)=\cos(4t+\pi)+3$

3.

$m(\theta)=2 \cos \mathopen{}\left( 2\pi \theta - \pi \right)\mathclose{}+4$

4.

$n(x)=-4\sin \mathopen{}\left( \pi x+\frac{\pi}{4} \right)\mathclose{}-2$

Find the Formula.

In Exercises 5–6, find two algebraic rules (one involving sine and one involving cosine) for each of the functions graphed below.

5.

$y=p(t)$

$Graph of y=p(t). The highest points are at 2 and the lowest are at -6. On the y-axis the intercept is (0, -6). The period is pi.$

6.

$y=q(x)$

$Graph of y=q(t). The highest points are at 2 and the lowest are at -4. It has points at (0.25, 2), (0.75, -1), and (1.25, -4). The period is 2.$

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