MTH 111–112 Supplement

Subsection 2.2.1 Graphing Sinusoidal Functions: Phase Shift vs. Horizontal Shift

Let's consider the function \(g(x)=\sin \left( 2x-\frac{2\pi}{3} \right) \text{.}\) Using what we study in MTH 111 about graph transformations, it should be apparent that the graph of \(g(x)=\sin \left( 2x-\frac{2\pi}{3} \right) \) 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 \(g(x)=f \left( 2x-\frac{2\pi}{3} \right) \text{.}\)) 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 \(g(x)=\sin \left( 2x-\frac{2\pi}{3} \right) \) as \(g(x)=\sin \left( 2 \left( x-\frac{\pi}{3} \right) \right) \text{.}\)

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

1. Compress horizontally by a factor of \(\frac{1}{2}\text{.}\)

2. 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 \(g(x)=\sin \left( 2x-\frac{2\pi}{3} \right) \) 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 \(g\text{:}\)

1. Shift \(\frac{2\pi}{3}\) units to the right.

2. 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 2.2.1 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 \(g(x)=\sin ( \, 2x-\frac{2\pi}{3} ) \,=\sin ( \, 2( \,x-\frac{\pi}{3} ) \, ) \,\text{.}\)

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 \(g(x)=\sin \left( 2x-\frac{2\pi}{3} \right)\text{,}\) 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.)

Example 2.2.5.

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\left( 4t+\pi \right)+1\\ \amp=2\sin \left(4 \left(t+\frac{\pi}{4} \right) \right)+1\\ \amp=2\sin\left( 4 \left(t- \left( -\frac{\pi}{4}\right)\right)\right) +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 2.2.6.