Introduction

Howdy everyone! Today we will cover the topic of holes in rational functions. While we have done work with finding the zeroes and vertical asymptotes of rational functions, there are some times where both occur at the same place. In this case, the graph isn’t defined at that point, but it still exists around it.

Course and Exam Description Overview

The AP Precalculus Course and Exam Description expects you to be able to identify the amplitude, vertical shift, and period of a sinusoidal function. These attributes of a sinusoidal function are important when sketching a graph and analyzing its behavior.

The Form of a Sinusoidal Function

All sinusoidal functions can be written as $f(\theta )=a\cos (b(\theta -c))+d$ or $g(\theta )=a\sin (b(\theta -c))+d$, where:

Note that in this case $\theta$ is our input. However, you may occasionally see other inputs like $x$ or $t$; it really just depends on the problem.

If a sinusoidal function is written differently or simplified, we will need to transform it into one of the forms above. If the sinusoidal function is written in a different form and you try to evaluate the transformations as is, you may end up making mistakes, especially when it comes to the phase shift and the horizontal dilation.

Finding the Amplitude of a Sinusoidal Function

The amplitude of a sinusoidal function is the distance from the center of oscillation to either the minima or maxima of the function. The amplitude is simply the coefficient that the whole sine or cosine term is multiplied by. The amplitude of the parent sine or cosine function (the function which has no transformations done to it) is $1$. Let’s practice with two examples:

Example 1: Find the amplitude of the function $f(\theta )=4\sin \left(2\theta +\frac{3}{5}\right)+22$.

Solution: We will just find the coefficient of the sine or cosine term, in which case, is $4$. Therefore, the amplitude is $4$.

Example 2: Find the amplitude of the function $f(\theta )=\cos \left(6\theta +7\right)+1$.

Solution: Again, we will take a look at the coefficient of the sine or cosine term. The coefficient of the cosine term is $1$. Therefore, the amplitude is $1$.

NOTE: The amplitude is NOT the distance between the minimum and maximum! This is a common mistake by many students. The amplitude is the distance between the midline and either the minimum or the maximum, as the following graph of $y=2\sin \left(x+\frac{6}{7}\right)+2$ shows:

 Figure 1: A graph of $y=2\sin (x+\frac{6}{7})+2$  

The distance between the minimum and the maximum is double the amplitude of the function.

Finding the Horizontal Dilation Factor and Period of a Sinusoidal Function

To find the factor by which the function is dilated by, we will need the $b$ term of $f(\theta )=a\cos (b(\theta -c))+d$ or $g(\theta )=a\sin (b(\theta -c))+d$. Once you find the $b$ term, your horizontal scale factor is $\frac{1}{|b|}$. In order to find the $b$ term on a sinusoidal function that is not in the form above, you will need to factor the inside of the sine or cosine function. If you don’t do this, your answer will be incorrect. We will need this factor in order to find the period, or the “width” of each cycle of the sinusoidal function.

Example 3: A function $g$ is given by $g(\theta )=4\sin (7\theta +21)-4$. Let $f$ be defined by $f(\theta )=\sin \theta$. By what factor is $g$ horizontally dilated compared to $f$?

Solution: We will first need to factor the part of the function that is inside the sin to get into the form $g(\theta )=a\sin (b(\theta -c))+d$.

$g(\theta )=4\sin (7\theta +21)-4$

$g(\theta )=4\sin (7(\theta +3))-4$

In the form $g(\theta )=a\sin (b(\theta -c))+d$, the coefficient is $b=7$. The horizontal scale factor relative to $f(\theta )=\sin \theta$ is $\frac{1}{|b|}=\frac{1}{7}$. Multiplying the input by $7$ makes the sine argument change seven times as fast, so each cycle occupies one-seventh of the horizontal distance. Therefore, $g$ is horizontally compressed by a factor of $\frac{1}{7}$.

Finding the Period

To find the period, we will need to divide $2\pi$ by our $b$ term. Thus, the period is given by $\frac{2\pi }{|b|}$. The reason why $b$ is absolute in this question is because we only care about the magnitude of $b$ as the period is always positive.

Example 4: A function $g$ is given by $g(\theta )=4\sin (7\theta +21)-4$. What is the period of the graph of $g$?

Solution: We will first need to find our $b$ term. We already did this in the example above; our $b$ term is 7. We will need to divide $2\pi$ by our $b$ term to find the period. Thus, the period is $\frac{2\pi }{7}$.

Finding the Phase Shift of a Sinusoidal Function

The phase shift of a sinusoidal function is just a fancy term for the horizontal translation (this only applies to periodic functions; please don’t go around saying the phase shift of a parabola :D). To find the phase shift, write the function in the form $g(\theta )=a\sin (b(\theta -c))+d$ or $g(\theta )=a\cos (b(\theta -c))+d$. The phase shift is $c$: positive values shift right, and negative values shift left.

Example 5: A function $f$ is given by $f(\theta )=6\sin (7\theta +4\pi )+1$. Find the phase shift of $f$. 

Solution: Factor $7$ from the argument of sine to write the function in the form $f(\theta )=a\sin (b(\theta -c))+d$.

$f(\theta )=6\sin (7\theta +4\pi )+1$

$f(\theta )=6\sin \left(7\left(\theta +\frac{4\pi }{7}\right)\right)+1$

Comparing $\theta +\frac{4\pi }{7}$ with $\theta -c$, we get $c=-\frac{4\pi }{7}$. Therefore, the phase shift of $f$ is $\frac{4\pi }{7}$ units to the left. Factoring the argument does not change the vertical shift, so the outside term remains $+1$.

To rewrite a sine expression in cosine form without changing its value, use the identity $\sin (u)=\cos \left(u-\frac{\pi }{2}\right)$. Subtract $\frac{\pi }{2}$ from the entire argument inside the cosine, then factor the coefficient of the variable to find the new phase shift.

Example 6: A function $f$ is given by $f(\theta )=\frac{2}{3}\sin \left(\frac{7}{10}\theta -\frac{2}{5}\right)+3$. Rewrite this function in terms of cosine.

Solution: First factor $\frac{7}{10}$ from the sine argument. Since $\frac{2}{5}÷\frac{7}{10}=\frac{4}{7}$, we get $f(\theta )=\frac{2}{3}\sin \left(\frac{7}{10}\left(\theta -\frac{4}{7}\right)\right)+3$.

Apply $\sin (u)=\cos \left(u-\frac{\pi }{2}\right)$ to the entire argument:

$f(\theta )=\frac{2}{3}\cos \left(\frac{7}{10}\theta -\frac{2}{5}-\frac{\pi }{2}\right)+3$

Now factor $\frac{7}{10}$ from the cosine argument. Dividing $\frac{\pi }{2}$ by $\frac{7}{10}$ gives $\frac{5\pi }{7}$:

$f(\theta )=\frac{2}{3}\cos \left(\frac{7}{10}\left(\theta -\frac{4}{7}-\frac{5\pi }{7}\right)\right)+3$

$f(\theta )=\frac{2}{3}\cos \left(\frac{7}{10}\left(\theta -\frac{4+5\pi }{7}\right)\right)+3$

This cosine form represents the same function, with a phase shift of $\frac{4+5\pi }{7}$ units to the right.

Finding the Vertical Shift of a Sinusoidal Function

Lastly, we will cover finding the vertical shift of a sinusoidal function; that is, by how much the function is “lifted” or “lowered.” Finding the vertical shift is more straightforward; it’s usually the term added to the whole sine or cosine bit, aka the $d$ term. It’s the same way you find the vertical shift for any other function.

Example 7: A function $f$ is given by $f(x)=3\cos (4\pi t-3\pi )-2.5$. Find the vertical shift of $f$.

Solution: We will take a look at the $d$ term of this function, which in this case is $-2.5$. Because the value of $d$ is negative, this function has been shifted downward.

We will add or subtract this value to the midline of either one of the sinusoidal functions to get the new midline. Because the midline is a line, you’ve got to include $y=$. So in this case the midline is $y=-2.5$.

Finding the Above Graphically

On the AP exam, you may be expected to do everything above from both a given function as well as a graph. The process is mostly the same for the vertical shift and amplitude, but gets a bit quirky with finding the phase shift and the period; most notably, the phase shift.

Example 8: The graph of a sinusoidal function is shown below. It is known that the function below can be written as $f(\theta )=a\sin (b(\theta +c))+d$. Find values for the constants $a$, $b$, $c$, and $d$. (Note: Each vertical tick represents 1 unit, and each horizontal tick represents $\frac{\pi }{6}$ units.)

Solution: We will first start off with the amplitude of the sinusoidal function. It is known that the amplitude is the distance from the midline of the function to its extrema, or half the distance from its minima to the maxima. In this graph, we can see that the distance between the extrema is $4$. We will divide this by two to get our amplitude, which is $2$. This is our $a$ value, so $a=2$.

Next, we will find the vertical shift of the midline, which is our $d$ value. Recall that the midline or center of oscillation of the parent sinusoidal functions $f(x)=\sin x$ and $g(x)=\cos x$ are $y=0$. To find the midline of a graphed sinusoid, we will locate an extrema (doesn’t matter if it is a minimum or maximum. Whether it’s a minimum or maximum doesn’t matter) and add or subtract the amplitude to the extremum’s $y-$value to find the midline of our function. 

The maximum point we have located here is at $\left(\frac{\pi }{2},1\right)$. We also know that our amplitude is $2$ units; we will subtract this from the $y-$coordinate of our maximum. Therefore, our midline is at $y=-1$, which is a shift of 1 unit down from the midline of the parent sine and cosine functions. Thus, our value for $d$ is $-1$.

Now, we will find a value for $b$. For this, we will need to find the period of the function. The period of this function is $\frac{\pi }{2}$. But this is not the horizontal dilation factor; this is the period. Now recall that the period is equal to $\frac{2\pi }{b}$ and we will need to solve for $b$. Solving for $b$ gets us a value of $4$. 

Lastly, we will need to find a value for $c$, which is the phase shift. In this case, it is very important to take a look at how the problem wants you to express it. In this case, the problem wants us to express our parameters in terms of sine. Recall that the parent sine function will “hit" the midline at $x=0$ as it goes ; what we want to do is to find at what value of $x$ the function graphed does the same and find the distance between this new value of $x$ and where the parent function “hits” the midline. In this case, the distance shifted is $\frac{\pi }{2}$ units to the left. Thus, our $c$ value is $-\frac{\pi }{2}$.

So to wrap this all up, our values are $a=2$, $b=4$, $c=-\frac{\pi }{2}$, and $d=-1$. 

So, that concludes Topic 3.6, folks! You now know how to graph basically any sinusoidal function in existence. Stay tuned for Topic 3.7, which is where we actually apply this knowledge to real-world events.

Practice Questions