5.2: Non-Rigid Transformations

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Roy Simpson
Cosumnes River College

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Note to the Instructor - Why Non-Rigid Transformations First?

The choice to cover non-rigid transformations (amplitude and period) before the rigid transformations is purposeful. Traditionally, when performing transformations, it is best to save the rigid transformations for the end.

Learning Objectives

Determine the amplitude of a sinusoidal function.
Determine the period of a sine, cosine, or tangent function.
Graph a sine, cosine, or tangent function having a non-standard amplitude or period.
Determine the equation of the graph of a trigonometric function.

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The material in the next few sections relies heavily on your prerequisite graphing skills from Algebra. As such, we will try our best to "backfill" some of that information here.¹ We begin by reintroducing the definition of a transformation of the graph of a function.

Definition: Transformation (of the graph of a function)

What you may not have learned in your Algebra course is that these transformations can be categorized into two distinct groups - rigid and non-rigid transformations.

Definition: Rigid Transformation

You likely remember rigid transformations as shifts. As a reminder of the basic idea of a rigid transformation, let's consider the adorable "base graph" of my dog, $ y = \mathrm{Sophie}\left( x \right) $ in Figure $ \PageIndex{ 1 } $.²

5.2 Sophie Base.png — **Figure $ \PageIndex{ 1 } $:** *The base graph of $ y = \mathrm{Sophie}\left( x \right) $*

From Algebra, we know that if we wanted to shift the graph of Sophie 2 units to the right and 3 units down, the resulting function would be\[ g(x) = \mathrm{Sophie}\left( x - 2 \right) - 3. \nonumber \]The orientation and scale of the graph of this new "function" (seen in Figure $ \PageIndex{ 2 } $) would be the same as the base graph of the "Sophie" function, but the position is different.

Sophie Shift.png — **Figure $ \PageIndex{ 2 } $:** *$ g(x) = \mathrm{Sophie}\left( x - 2 \right) - 3 $*

In general, given a function $ f(x) $, we can obtain the graph of $ g(x) = f(x + C) + D $ by subtracting $ C $ from all of the $ x $-coordinates of the graph of $ f(x) $ and adding $ D $ to all the $ y $-coordinates.³

Definition: Non-rigid Transformation

For example, if we took the base graph of our "Sophie function" from Figure $ \PageIndex{ 1 } $ and decided to scale it vertically (also known as stretching) by a factor of 2, we would need to multiply all the $ y $-coordinates of the points in the base graph by 2. From Algebra, we know this is achieved by multiplying the function by 2 to get $ h(x) = 2 \, \mathrm{Sophie}\left( x \right) $. The graph of this new image would look like the following:

Sophie Stretch.png — **Figure $ \PageIndex{ 3 } $:** *$ h(x) = 2 \, \mathrm{Sophie}\left( x \right) $*

We also learned (again, from Algebra) to combine transformations to get a new function from an old one. For example, we could take $ \mathrm{Sophie}\left( x \right) $, scale it horizontally by a factor of $ 1/2 $ (also known as compressing), reflect it about the $ y $-axis, shift it 3 units left, and shift it 4 units up. The result of all of these manipulations is the function $ k(x) = \mathrm{Sophie}\left( -2\left( x + 3 \right) \right) + 4 $, and it looks like the following image:

Sophie Multiple.png — **Figure $ \PageIndex{ 4 } $:** *$ k(x) = \mathrm{Sophie}\left( -2\left( x + 3 \right) \right) + 4 $*

In this section, we will perform these same transformations on trigonometric functions; however, because these functions are periodic, we can only capture a small section of their graphs. The process we introduce to apply these transformations and the names we define during this development will be used throughout the rest of Trigonometry and Calculus.

Our ultimate goal is to graph a trigonometric function of the form\[ f(x) = A \, \mathrm{trig}\left( Bx + C \right) + D \label{gentrig},\]where "$ \mathrm{trig} $" could be any of the trigonometric functions.⁴ The unknown values, $ A $, $ B $, $ C $, and $ D $, are called parameters.

Definition: Parameter

We will review what happens to the graph of the base trigonometric function when we adjust these parameters.

Amplitude and the Graph of $ y = A \, \mathrm{trig}( x ) $

We start our investigation of the effect of the parameters in Equation \ref{gentrig} by setting $ B = 1 $, $ C = 0 $, and $ D = 0 $. Therefore, we are interested in the effect of $ A $ on the graph of\[ f(x) = A \, \mathrm{trig}\left( x \right). \nonumber \]If $ A \gt 0 $, then from Algebra, we know that the graph of $ y = A g(x) $ is a vertical stretch (if $ A \gt 1 $) or compression (if $ 0 \lt A \lt 1 $) of the graph of $ g $; however, does this still hold in Trigonometry? The following example might lead us to the answer.

Example $\PageIndex{1}$

Make a table of values and sketch the graph of each function. Compare your graph to the function's base graph.

$ f(x) = 3\sin\left( x \right) $
$ g(x) = \frac{1}{5} \cos\left( x \right) $
$ h(x) = \pi \tan\left( x \right) $

Solutions

Now that we know the basic shape of the sine function, we only need to make a table of the quadrantal angles.\[ \begin{array}{|c|c|c|c|c|c|}
\hline
x & 0 & \frac{\pi}{2} & \pi & \frac{3\pi}{2} & 2\pi \\[6pt] \hline
f\left( x \right) & 0 & 3 & 0 & -3 & 0 \\[6pt] \hline \end{array} \nonumber \]Plotting these points and graphing the base sine function, we get the following:

Figure $ \PageIndex{ 5 } $
$5.2 Example 01a.png$
We can immediately see that the graph of $ f(x) = 3\sin\left( x \right) $ is a vertical stretch of the base graph by a factor of $ 3 $. That is, all $ y $-values of $ f(x) = 3\sin\left( x \right) $ are three times that of the $ y $-values for $ y = \sin\left( x \right) $. Judging from this single example, it looks like the role of $ A $ in $ f(x) = A \, \mathrm{trig}\left( x \right) $ is to multiply the $ y $-values of the base trigonometric function by $ A $. However, looks can be deceiving, and this is not a proof. Let's do the second example and see if our insight holds.
Again, we use our knowledge of the basic shape of the cosine graph, along with a table of quadrantal angles.\[ \begin{array}{|c|c|c|c|c|c|}
\hline
x & 0 & \frac{\pi}{2} & \pi & \frac{3\pi}{2} & 2\pi \\[6pt] \hline
g\left( x \right) & \dfrac{1}{5} & 0 & -\dfrac{1}{5} & 0 & \dfrac{1}{5} \\[6pt] \hline \end{array} \nonumber \]Plotting these points and graphing the base cosine function, we get the following:

Figure $ \PageIndex{ 6 } $
$5.2 Example 01B.png$
As before, we see that the graph of $ g(x) = \frac{1}{5}\cos\left( x \right) $ is a vertical compression of the base graph by a factor of $ 1/5 $.⁵ That is, all $ y $-values of $ g(x) = \frac{1}{5}\cos\left( x \right) $ are one-fifth of the $ y $-values for $ y = \cos\left( x \right) $.
The tangent function always requires special consideration because it is undefined at certain quadrantal angles. Therefore, our approach from parts (a) and (b) needs to be modified. Again, this is where having a strong grasp of the base graph for the tangent function comes into play. Since the natural period of the tangent is $ \pi $, we will plot coordinates for $ x=0, \pi/4, 3\pi/4, $ and $ \pi $ (completely avoiding the domain issue by skipping over $ x = \pi/2 $).\[ \begin{array}{|c|c|c|c|c|}
\hline
x & 0 & \frac{\pi}{4} & \frac{3\pi}{4} & \pi \\[6pt] \hline
h\left( x \right) & 0 & \pi & -\pi & 0 \\[6pt] \hline \end{array} \nonumber \]Plotting these points and graphing the base tangent function, we get the following:

Figure $ \PageIndex{ 7 } $
$5.2 Example 01C.png$
Again, $ h(x) = \pi\tan\left( x \right) $ is a vertical stretch of the base graph by a factor of $ \pi $.⁶

Example $ \PageIndex{ 1 } $ verifies that our experience in Algebra still holds in Trigonometry; however, we artificially restricted the values of $ A $ to be positive. What happens if $ A \lt 0 $? Again, from Algebra, we know the graph of $ y = -g(x) $ would be a reflection of the graph of the function $ g $ about the $ x $-axis. Does this still hold in Trigonometry?

Checkpoint $\PageIndex{1}$

Make a table of values and sketch the graph of each function. Compare your graph to the function's base graph.

$ f(x) = -2 \cos\left( x \right) $
$ g(x) = -\frac{1}{2} \sin\left( x \right) $

Answer

Figure $ \PageIndex{ 8 } $
$5.2 1.1a Checkpoint.png$
The coefficient of $ -2 $ multiplies all $ y $-values of the base graph by $ -2 $. This effectively causes a vertical stretch (by a factor of $ 2 $) and a reflection about the $ x $-axis.
Figure $ \PageIndex{ 9 } $
$5.2 1.1b Checkpoint.png$
The coefficient of $ -1/2 $ multiplies all $ y $-values of the base graph by $ -1/2 $, effectively compressing the graph vertically by a factor of $ 1/2 $ and reflecting it about the $ x $-axis.

The following theorem summarizes these results.

Theorem: Vertical Scaling

The graph of\[ f(x) = A \, \mathrm{trig}\left( x \right) \nonumber \]vertically scales the graph of $ y = \mathrm{trig}\left( x \right) $ by a factor of $ \left|A\right| $.

If $ \left|A\right| \gt 1 $, it is a vertical stretch by a factor of $ \left|A\right| $.
If $ 0 \lt \left|A\right| \lt 1 $, it is a vertical compression by a factor of $ \left|A\right| $.

Additionally, if $ A \lt 0$, the graph is reflected about the horizontal axis.

In either case, the $ y $-coordinates of points on the graph of $ y = \mathrm{trig}\left( x \right) $ are multiplied by $ A $ to become the new points on the graph of $ f(x) = A \, \mathrm{trig}\left( x \right) $.

Note: Vertical Scaling Changes the Range

Except for the tangent function (whose range is all real numbers), the range of a trigonometric function will change under vertical scaling. For example, in Example $ \PageIndex{ 1b } $ the range of $ g(x) = \frac{1}{5} \cos\left( x \right) $ is $ \left[ -\frac{1}{5} , \frac{1}{5} \right] $.

When we introduced the definition of amplitude, we also mentioned that while the base sine and cosine functions have amplitudes of $ 1 $, the tangent does not have an amplitude because its range is unbounded. We now introduce the amplitude of any sinusoidal function.

Example $\PageIndex{2}$

Compute the amplitude of each function.

$ f(x) = 3\sin\left( x \right) $
$ g(x) = \frac{1}{5} \cos\left( x \right) $

Solutions

From Example $ \PageIndex{ 1a } $, we see the greatest and least values of the graph are $ 3 $ and $ -3 $, respectively. Thus, the amplitude of $ f(x) = 3\sin\left( x \right) $ is\[ \text{Amplitude} = \dfrac{1}{2}\left| 3 - (-3) \right| = \dfrac{1}{2} |6| = 3.\nonumber \]
From Example $ \PageIndex{ 1b } $, we see the greatest and least values of the graph are $ 1/5 $ and $ -1/5 $, respectively. Thus, the amplitude of $ g(x) = \frac{1}{5} \cos\left( x \right) $ is\[ \text{Amplitude} = \dfrac{1}{2}\left| \dfrac{1}{5} - \left( -\dfrac{1}{5} \right) \right| = \dfrac{1}{2} \left| \dfrac{2}{5} \right| = \dfrac{1}{5}.\nonumber \]

Example $ \PageIndex{ 2 } $ provides an observation that will allow us to state the amplitude of a sinusoidal function quickly.

Theorem: Amplitude of a Sinusoidal Function

Checkpoint $\PageIndex{2}$

State the amplitude of each function.

$ f(x) = -2 \cos\left( x \right) $
$ g(x) = -\frac{1}{2} \sin\left( x \right) $
$ h(x) = 12 \tan(x) $

Answer

$2$
$ \frac{1}{2} $
Trick question... the tangent function does not have an amplitude.

If you are using the digital version of this textbook, feel free to use the following Interactive Element to help solidify the concept of amplitude and the effect of $ A $ on the graph of $ f(x) = A \sin\left( x \right) $.

Interactive Element: Investigating the Effects of $ A $ on the Graph of $ y = A \sin\left( x \right) $

Interaction: Adjust the slider for $ A $ to see the effect on the graph of $ y = A \sin\left( x \right) $.

Period

The next parameter in Equation \ref{gentrig} is $ B $. Very technically, you should have already encountered graphs of functions of the form $ y = g(Bx) $ in Algebra, but I find that many people are confounded by what is truly happening and, very honestly, you don't see the true effect of that transformation until Trigonometry. Therefore, we are going to cover it in detail here. To determine this parameter's effects on the trigonometric function's graph, we take an example and build a table of values.

Suppose we are asked to graph $ y = \cos\left( 2x \right) $. By building a table of values using the quadrantal angles, we get the following:\[ \begin{array}{|c|c|c|c|c|c|}
\hline
x & 0 & \frac{\pi}{2} & \pi & \frac{3\pi}{2} & 2\pi \\[6pt] \hline
y = \cos\left( 2x \right) & 1 & -1 & 1 & -1 & 1 \\[6pt] \hline \end{array} \nonumber \]Graphing the base cosine curve along with these points yields Figure $ \PageIndex{ 10 } $.

5.2 Motivating Example 01.png — **Figure $ \PageIndex{ 10 } $**

The issue should be readily apparent - we need more points to plot. The quadrantal angles provide too coarse of a picture. Let's add some values to our table.\[ \begin{array}{|c|c|c|c|c|c|c|c|c|c|}
\hline
x & 0 & \frac{\pi}{4} & \frac{\pi}{2} & \frac{3\pi}{4} & \pi & \frac{5\pi}{4} & \frac{3\pi}{2} & \frac{7\pi}{4} & 2\pi \\[6pt] \hline
y = \cos\left( 2x \right) & 1 & 0 & -1 & 0 & 1 & 0 & -1 & 0 & 1 \\[6pt] \hline \end{array} \nonumber \]Plotting these additional points (and connecting them with a nice sinusoidal curve), we get the following figure.

5.2 Motivating Example 02.png — **Figure $ \PageIndex{ 11 } $**

Before moving on, we need to mention a few things about Figure $ \PageIndex{ 11 } $. First, recall that the principal cycle of the cosine function is $ \left[ 0,2\pi \right] $. This means that standard cosine function goes through a single cycle on that interval; however, $ y=\cos(2x) $ goes through two complete cycles on the same interval. This is a beautiful way of thinking about the effect of $ B $ on the graph of $ y=A \, \mathrm{trig}\left( Bx \right) $. $ B $ tells us how many "cycles" of the graph get squeezed into the principal cycle of the base function. Another interpretation is that $ B $ tells us how many times faster the curve is "oscillating" than the base curve.⁷ In this case, $ y = \cos\left( 2x \right) $ oscillates twice as fast as $ y = \cos\left( x \right) $. This being said, we can safely conjecture that $ B $ modifies the function's period (how long it takes to complete one cycle).

The second item worth mentioning is that you will usually only be asked to graph a single cycle of a trigonometric function, and this single cycle will be based on the principal cycle. Hand-drawn (and computer-generated) graphs look crowded when more than one cycle is sketched. In general, your instructor will know you just copied a computer-generated graph of a given trigonometric function if you include several cycles.

This brings us to our third and most important observation. For any trigonometric function, we only need five special $ x $-values to obtain an accurate sketch of a single cycle (again, based on the principal cycle). For simplicity, we label these $ x_0$, $x_1$, $x_2$, $x_3$, and $ x_4 $. These are

$ x_0 $: the beginning $ x $-value
$ x_4 $: the ending $ x $-value
$ x_2 $: the value of $ x $ directly between the beginning and ending $ x $-values
$ x_1 $ and $ x_3 $: the two values of $ x $ between the beginning and middle, and middle and ending $ x $-values

As they say, a picture is worth a thousand words, so let's visualize this using $ y = \cos\left( 2x \right) $.

5.2 Motivating Example 03 Fixed.png — **Figure $ \PageIndex{ 12 } $**

Each of the $ x $-values in Figure $ \PageIndex{ 12 } $ is called a key numbers of a single graph cycle.

Definition: Key Numbers

To showcase how the key numbers help graph trigonometric functions, suppose we are asked to graph the principal cycle of the base cosine function, $ y = \cos\left( \theta \right) $. Again, the principal cycle for the cosine function is $ 0 \leq \theta \leq 2\pi $. Partitioning this interval into four equally-spaced subintervals requires taking the length of this interval ($ 2\pi $) and dividing by four to get\[ \dfrac{2\pi}{4} = \dfrac{\pi}{2}. \nonumber \]We will call the length of this subinterval the step size for this function.

Definition: Step Size

To graph, we mark the first $ x $-value in the principal cycle, which is $ x = 0 $. Hence, $ x_0 = 0 $. From there, we "mark by steps" until we reach the end of the single cycle (see the figure below).

5.2 Motivating Example 04 Fixed.png — **Figure $ \PageIndex{ 13 } $**

Since we know the basic shape of the cosine graph, we can complete the graph using these key numbers.

5.2 Motivating Example 05.png — **Figure $ \PageIndex{ 14 } $**

Before going into an example, let's amend a definition we created in the previous section. We know that the natural period of the cosine function (and sine function) is $ 2\pi $; however, what is the period of the modified function $ y = \cos\left( 2x \right) $?

Recall that a single cycle of $ y = \cos\left( \theta \right) $ occurs on $ 0 \leq \theta \leq 2\pi $. Replacing $ \theta $ with $ 2x $, we have $ 0 \leq 2x \leq 2 \pi $. Dividing all three sides of this inequality by $ 2 $, we get $ 0 \leq x \leq \pi $. That is, $ y = \cos\left( 2x \right) $ goes through a single cycle on the interval $ \left[ 0,\pi \right] $. Hence, the period of $ y = \cos\left( 2x \right) $ is $ \pi $. Generalizing this observation leads us to the following theorem.

Theorem: Period of a Trigonometric Function

Thus, the period of $ y=A\cos\left( Bx \right) $ is $ 2\pi/B $, and the period of $ y = A \tan\left( Bx \right) $ is $ \pi/B $ (remember, the natural period of the tangent is $ \pi $).

While we are at it, let's create a formula to help us compute the step size.

Theorem: Step Size Formula

The step size simplifies the process of finding the key numbers. By starting with the first key number, $ x_0 $, we add the step size to arrive at the second key number, $ x_1 $. To this, we add the step size to get the third key number, $ x_2 $. And so on.

We are overdue for some examples.

Example $\PageIndex{3}$

Sketch a single cycle of each function, stating its amplitude (if applicable), period, and step size.

$ f(x) = \cos\left( 2x \right) $
$ g(x) = -3 \sin\left( \frac{x}{3} \right) $
$ y = \frac{1}{5} \tan\left( -4x \right) $

Solutions

The process we illustrate here (and that we build on moving forward) is a bit algorithmic but very helpful in graphing trigonometric functions. It's based on the "template" trigonometric function\[ y = A \, \mathrm{trig}\left( Bx + C \right) + D. \nonumber \]In this section, $ C = D = 0 $, so we are only dealing with functions of the form\[ y = A \, \mathrm{trig}\left( Bx \right). \nonumber \]Specifically, for this problem, we have $ f(x) = 1 \cos\left( 2x \right) $. We read this from left to right.\[ \begin{array}{rcccccl}
\hline
A & = & 1 & \implies & \text{Amplitude} & = & 1 \\[6pt] \hline
B & = & 2 & \implies & \text{Period} & = & \dfrac{\text{natural period}}{B} = \dfrac{2\pi}{2} = \pi \\[6pt] \hline
& & & & \text{Step Size} & = & \dfrac{\text{Period}}{4} = \dfrac{\pi}{4} \\[6pt] \hline \end{array} \nonumber \]We can now sketch an accurate graph of a single cycle. The initial $ x $-value in the principal cycle of all trigonometric functions is $ x = 0 $. To find the next key number, we add the step size. See the figure below.

Figure $ \PageIndex{ 15 } $
$5.2.3 Example.a.1.png$
The length of this interval is $ \pi $, which matches the period, so we are on track for success. Since the initial key number is at $ x = 0 $, we now know where the $ y $-axis is, so let's draw that in.

Figure $ \PageIndex{ 16 } $
$5.2.3 Example.a.2.png$
The function we are graphing is the positive cosine, so we can fill in the basic shape of a single cycle.

Figure $ \PageIndex{ 17 } $
$5.2.3 Example.a.3.png$
Finally, we use the amplitude to label the $ y $-axis. Since the amplitude is $ 1 $, the maximum value is $ 1 $ and the minimum value of the function is $ -1 $.

Figure $ \PageIndex{ 18 } $
$5.2.3 Example.a.4.png$
I know I said that, unless asked, you should refrain from graphing more than a single cycle, however, it is nice to let your reader know that this pattern is cyclic. We showcase this by extending the graph for a short distance in both directions.

Figure $ \PageIndex{ 19 } $
$5.2.3 Example.a.5.png$
We have $ g(x) = -3 \sin\left( \frac{x}{3} \right) = -3 \sin\left( \frac{1}{3}x \right) $. Reading this from left-to-right, we get the following.\[ \begin{array}{rcccccl}
\hline
A & = & -3 & \implies & \text{Amplitude} & = & 3 \\[6pt] \hline
B & = & \dfrac{1}{3} & \implies & \text{Period} & = & \dfrac{\text{natural period}}{B} = \dfrac{2\pi}{1/3} = 6\pi \\[6pt] \hline
& & & & \text{Step Size} & = & \dfrac{\text{Period}}{4} = \dfrac{6\pi}{4} = \dfrac{3\pi}{2} \\[6pt] \hline \end{array} \nonumber \]Remembering that the initial $ x $-value for the principal cycle is $ x = 0 $, we find all of the key numbers and label the $ x $-axis.

Figure $ \PageIndex{ 20 } $
$5.2.3 Example.b.1.png$
Again, we see the length of this interval is $ \frac{12\pi}{2} = 6\pi $, which is the period. Therefore, we must be doing something right! Placing the $ y $-axis appropriately, we get the following.

Figure $ \PageIndex{ 21 } $
$5.2.3 Example.b.2.png$
Here is where paying attention to the original function is essential (technically, paying attention is always important) - the amplitude of the function is $ 3 $, but the sine function has been reflected about the $ x $-axis because $ A \lt 0 $. Therefore, we label the $ y $-axis with the maximum and minimum values of $ 3 $ and $ -3 $ but draw our sine function reflected about the $ x $-axis.

Figure $ \PageIndex{ 22 } $
$5.2.3 Example.b.3.png$
The tangent function might seem like the oddball of the trigonometric functions. Still, the process of sketching a single cycle remains the same, except that the tangent does not have an amplitude. One difference you should notice with this specific example is that the coefficient on $ x $ is negative. This is where remembering the Symmetry Identities pays off. Since tangent is an odd function,\[ y = \dfrac{1}{5} \tan\left( -4x \right) = -\dfrac{1}{5} \tan\left( 4x \right). \nonumber \]Now that we have taken care of the negative argument in the function, we get to work. It is essential to remember that the natural period of the tangent function is $ \pi $.\[ \begin{array}{rcccccl}
\hline
B & = & 4 & \implies & \text{Period} & = & \dfrac{\text{natural period}}{B} = \dfrac{\pi}{4} \\[6pt] \hline
& & & & \text{Step Size} & = & \dfrac{\text{Period}}{4} = \dfrac{\pi/4}{4} = \dfrac{\pi}{16} \\[6pt] \hline \end{array} \nonumber \]Again, the initial $ x $-value in the principal cycle is $ x = 0 $. To find the next key number, we add the step size. See the figure below.

Figure $ \PageIndex{ 23 } $
$5.2.3 Example.c.1.png$
We know what the base graph of the tangent looks like, so we know the beginning and ending key numbers ($ 0 $ and $ 4\pi/16 $) are going to be the zeros of the function and the middle key number ($ 2\pi/16 $) is the vertical asymptote; however, what do we do with $ \pi/16 $ and $ 3\pi/16 $? Up to this point, we have not bothered with the coefficient of the function, $ -1/5 $. Here is where it comes into play. If we were to evaluate the given function at $ \pi/16 $, we would get\[ -\dfrac{1}{5} \tan\left( 4\left( \dfrac{\pi}{16} \right) \right) = -\dfrac{1}{5} \tan\left( \dfrac{\pi}{4} \right) = -\dfrac{1}{5} \left( 1 \right) = -\dfrac{1}{5}. \nonumber \]That is, the excess key numbers, $ \pi/16 $ and $ 3\pi/16 $, give us two convenient points to plot on the graph, and the value at these points is $ \pm A $. Sketching a tangent function that has been reflected about the $ x $-axis and plotting those two points, we get the following graph.

Figure $ \PageIndex{ 24 } $
$5.2.3 Example.c.3.png$

Did you notice I didn't bother simplifying fractions when I labeled the key numbers in Examples $ \PageIndex{ 3b } $ and $ \PageIndex{ 3c } $? This was purposeful and is a technique we will use moving forward.

Note: Changes in the Period Affect the Domain of the Tangent

While the domain of the sine and cosine functions remain all real numbers no matter the period, the tangent's domain changes when we change the period. For example, in Example $ \PageIndex{ 3c } $ the domain of the tangent does not include $ \frac{\pi}{8} $ (whereas, in the base tangent function, $ \frac{\pi}{8} $ is in the domain). This is because a change in the period of a function causes horizontal scaling (stretching or compressing along the horizontal axis). When this scaling occurs, any vertical asymptotes the function might have also move.

Checkpoint $\PageIndex{3}$

Sketch a single cycle of $ y = -\frac{1}{2} \cos\left( 7x \right) $, stating its amplitude, period, and step size.

Answer: Amplitude: $ \frac{1}{2} $
Period: $ \frac{2\pi}{7} $
Step Size: $ \frac{\pi}{14} $
$5.2.3 Checkpoint.png$

As before, the following Interactive Element (accessible through the digital version of the textbook) can help to solidify the concept of period and the effect of $ B $ on the graph of $f(x) = A \sin(B x)$.

Interactive Element: Investigating the Effects of $ B $ on the Graph of $y= A \sin(Bx)$

Interaction: Adjust the slider for $ A $ and $ B $ to see the effect on the graph of $ y = A \sin\left( B x \right) $.

Summary of Graphing Trigonometric Functions Using Non-Rigid Transformations

Summary of Graphing $ f(x) = \overline{A} \, \mathrm{trig}\left( \overline{B}x \right) $

To graph the function\[ f(x) = \overline{A} \, \mathrm{trig}\left( \overline{B}x \right): \nonumber \]

Precondition Stage

We begin by cleaning up the trigonometric function and ensuring that the coefficient of $ x $ is positive. This is done using the Symmetry Identities. Once the Symmetry Identities have been used, your new function will be of the form\[ f(x) = A \mathrm{trig}\left( Bx \right). \nonumber \]

Information Stage

We now identify the following information (in the order listed):

$ \text{Amplitude} = \left| A \right| $
$ \text{Period} = \dfrac{\text{Natural Period}}{B} $
$ \text{Step Size} = \dfrac{\text{Period}}{4} $

where the natural period is $ 2\pi $ if given the sine or cosine functions, and $ \pi $ if given the tangent function.

Graphing Stage

We then use each of these in reverse order to obtain a sketch of the trigonometric function.

Draw a horizontal line, mark the initial key number as $ 0 $, and mark the remaining key numbers by successively adding the step size to the previous key number.
Double-check that the distance from the initial key number to the final key number matches the period, then sketch one cycle of the base trigonometric function, making sure to account for the sign of $ A $ and drawing any vertical asymptotes (if the given function is the tangent).
Draw the vertical axis at $ x = 0 $ and use the amplitude to either
- label the maximum and minimum values of the trigonometric function on the vertical axis (if the given function is sinusoidal), or
- label the remaining key points on the graph (if the given function is the tangent).

To be clear, let's do a final example. This time, however, let's practice extending the graph as well.

Example $\PageIndex{4}$

Sketch the graph of each function over the given interval.

$ g(x) = -\dfrac{1}{2} \cos\left( -\dfrac{2}{5}x \right) $ for $ -10\pi \leq x \leq 10\pi $
$ h(x) = 3 \tan\left( 6x \right) $ for $ -\frac{\pi}{12} \leq x \leq \frac{\pi}{2} $

Solutions

Precondition Stage: Since the coefficient of $ x $ is negative, we use the Symmetry Identities to rewrite the given function. Cosine is an even function. Therefore, $ \cos\left( -\theta \right) = \cos\left( \theta \right)$. Hence, $ g(x) $ can be rewritten as\[ g(x) = -\dfrac{1}{2} \cos\left( \dfrac{2}{5}x \right). \nonumber \]
Information Stage: Since the given function is sinusoidal, it has an amplitude, and its natural period is $ 2\pi $.\[ \begin{array}{rclcrcl}
\hline A & = & -\dfrac{1}{2} & \implies & \text{Amplitude} & = & \left| A \right| = \dfrac{1}{2} \\[6pt] \hline B & = & \dfrac{2}{5} & \implies & \text{Period} & = & \dfrac{\text{Natural Period}}{B} = \dfrac{2 \pi}{2/5} = 10 \pi \\[6pt] \hline \quad & \quad & \quad & \implies & \text{Step Size} & = & \dfrac{\text{Period}}{4} = \dfrac{10 \pi}{4} = \dfrac{5 \pi}{2} \\[6pt] \hline \end{array} \nonumber \]
Graphing Stage: Drawing a horizontal line (the $ x $-axis), marking the initial key number ($ x = 0 $), and marking the remaining key numbers using the step size, we get the following.

Figure $ \PageIndex{ 25 } $
$5.2.4 Example.a.1.png$
Notice that I left quite a bit of open space on the left half of the number line. This is because we will eventually extend the single cycle of our cosine function so that our graph covers the interval $ \left[ -10\pi, 10\pi \right] $. We take a moment to make sure the distance from the initial key number ($ x = 0 $) to the final key number ($ x = 10\pi $) is the same as the period. In this case, the distance is\[ \left|10\pi - 0 \right| = 10\pi = \text{Period}, \nonumber \]so we are on the right track. We now sketch a single cycle of the reflected cosine (because $ A \lt 0 $).

Figure $ \PageIndex{ 26 } $
$5.2.4 Example.a.2.png$
Finally, we draw in the $ y $-axis at $ x = 0 $, label the maximum and minimum values using the amplitude, and extend the graph to cover the interval $ \left[ -10\pi, 10\pi \right] $.

Figure $ \PageIndex{ 27 } $
$5.2.4 Example.a.3.png$
Precondition Stage: Since the coefficient of $ x $ is already positive, we do not need to clean up the trigonometric function.
Information Stage: Since $ h $ is a tangent function, it does not have an amplitude, and its natural period is $ \pi $.\[ \begin{array}{rclcrcl}
\hline B & = & 6 & \implies & \text{Period} & = & \dfrac{\text{Natural Period}}{B} = \dfrac{\pi}{6} \\[6pt] \hline \quad & \quad & \quad & \implies & \text{Step Size} & = & \dfrac{\text{Period}}{4} = \dfrac{\pi/6}{4} = \dfrac{\pi}{24} \\[6pt] \hline \end{array} \nonumber \]
Graphing Stage: Starting at $ x = 0 $ and marking the key points, we get the following number line (which has been extended because of the interval over which we are asked to graph).

Figure $ \PageIndex{ 28 } $
$5.2.4 Example.b.1 Fixed.png$
Note that the distance between the beginning key number and the ending key number is\[ \left| \dfrac{4\pi}{24} - 0 \right| = \dfrac{4 \pi}{24} = \dfrac{\pi}{6} = \text{Period}. \nonumber \]Thus, we are good to go! We now graph a single cycle of the tangent function. Remember that there is a vertical asymptote in the middle of the cycle (at $ x = 2\pi/24 $).

Figure $ \PageIndex{ 29 } $
$5.2.4 Example.b.2.png$
From here, we draw the $ y $-axis at $ x=0 $, label the points on the curve at $ x = \pi/24 $ and $ x = 3\pi/24 $, and extend the graph to cover the requested interval.

Figure $ \PageIndex{ 30 } $
$5.2.4 Example.b.3.png$

Finding Equations Given Graphs

We wrap up this section by inverting our process. Instead of graphing a function given its equation, we find the equation for a function given its graph. This process is similar to what we introduced in the summary above.

Example $\PageIndex{5}$

Find the equation of the graphed function.

Figure $ \PageIndex{ 31 } $
$5.2.5 Example.a.1.png$
Figure $ \PageIndex{ 32 } $
$5.2.5 Example.b.1.png$

Solutions

Our first task is to identify the base function. The shape of this function is sinusoidal. Moreover, because it goes through the origin, it is a sine function (although, technically, it can be written as a cosine function as well). However, it has been reflected about the $ x $-axis, so we know that $ A \lt 0 $. We can also see that the amplitude is 2. Thus, $ A = -2 $ so that\[ f(x) = -2 \sin\left( Bx \right). \nonumber \]All we need to do is identify $ B $. We know that $ B $ is associated with the period of the function, but it is not equal to the period. The relationship between $ B $ and the period is\[ \text{Period} = \dfrac{\text{Natural Period}}{B}. \nonumber \]However, the natural period of the sine function is $ 2\pi $, so\[ \text{Period} = \dfrac{2\pi}{B} \implies B = \dfrac{2 \pi}{\text{Period}}. \nonumber \]The graph shows that the function completes one full cycle every 2 units. Hence, the period is 2. Substituting that into our formula, we get\[ B = \dfrac{2 \pi}{2} = \pi. \nonumber \]Thus, our function is\[ f(x) = -2 \sin\left( \pi x \right). \nonumber \]
The base function here is the tangent. So we know we are dealing with a function of the form\[ g(x) = A \tan\left( Bx \right). \nonumber \]The value of $ A $ will be a bit trickier to get here because the tangent does not have an amplitude; however, all is not lost. We instead start by finding $ B $. As in part (a),\[ \text{Period} = \dfrac{\text{Natural Period}}{B} \nonumber \]and we know the natural period of the tangent function is $ \pi $. Therefore,\[ \text{Period} = \dfrac{\pi}{B} \implies B = \dfrac{\pi}{\text{Period}}. \nonumber \]We can see that a single cycle gets completed in 3 units; thus, the period of this function is 3. Hence,\[ B = \dfrac{\pi}{3}. \nonumber \]Now we have\[ g(x) = A \tan\left( \dfrac{\pi}{3} x \right). \nonumber \]To find the value of $ A $, we recall that any point on the graph of an equation must also satisfy the equation. In particular, because the point $ \left( \frac{3}{4}, \frac{1}{2} \right) $ lies on this tangent curve, when $ x = \frac{3}{4} $, it must be the case that $ y = \frac{1}{2} $. That is,\[ \dfrac{1}{2} = A \tan\left( \dfrac{\pi}{3} \cdot \dfrac{3}{4} \right) \implies \dfrac{1}{2} = A \tan\left( \dfrac{\pi}{4} \right) \implies \dfrac{1}{2} = A \cdot 1 \implies A = \dfrac{1}{2}. \nonumber \]Thus, the equation of this tangent function is\[ g(x) = \dfrac{1}{2} \tan\left( \dfrac{\pi}{3} x \right). \nonumber \]

Checkpoint $\PageIndex{5}$

Find the equation of the function graphed below.

5.2.5 Checkpoint.png — **Figure $ \PageIndex{ 33 } $**

Answer: $ f(x) = - \dfrac{5}{2} \cos\left( 5x \right) $

Footnotes

¹ If you need more of a review of transformations of functions than what we cover here, it's a good idea to make your instructor aware.

² Technically, this is not a function (Sophie doesn't pass the Vertical Line Test); however, using an image usually conveys the point of transformations to students more efficiently than traditional functions.

³ Again, it is imperative that you have already had total exposure to these concepts from Algebra. What we present here is not instruction but a reminder (with a cute dog).

⁴ However, we will stay with the fundamental trigonometric functions for a while until we get used to transformations of trigonometric functions.

⁵ Some authors and instructors will say, "$ g(x) $ is a compression of the graph of $ y = \cos\left( x \right) $ by a factor of $ 5 $." I have always found this language a bit confusing. Therefore, I will adhere to the convention in this textbook of stating "a vertical compression by a factor of $ \frac{1}{5} $." The fact that the word "compression" is involved gives the context that the $ y $-values of the original function are being "squished" to make them smaller than before.

⁶ Don't let the coefficient $ \pi $ fool you - it's just a number greater than $ 1 $ (approximately $ 3.142 $).

⁷ I placed "oscillating" in quotes because this interpretation doesn't work for the tangent function. That is, $ y = \tan\left( 2x \right) $ doesn't oscillate twice as fast as the base tangent function because the tangent doesn't oscillate.