## Section16.6The Graphs of the Secant and Cosecant Functions

##### The Graph of the Function $y=\sec(t)$.

Recall that the secant function is the reciprocal of the cosine function. That is, $\sec(t)=\frac{1}{\cos(t)}\text{.}$ Given that, it seems like a good place to begin our understanding of the graph of $t=\sec(t)$ is to look at a graph of $y=\cos(t)\text{.}$ Such a graph appears as Figure 16.6.1. Figure 16.6.1. $y=\cos(t)$

The first thing we should note is that the period of the secant function will be the same as the period of the cosine function; that is, $2\pi\text{.}$

We should also note that $\cos(t)=0$ at $-\frac{5\pi}{2}\text{,}$ $-\frac{3\pi}{2}\text{,}$ $-\frac{\pi}{2}\text{,}$$\frac{\pi}{2}\text{,}$ $\frac{3\pi}{2}\text{,}$ and $\frac{5\pi}{2}\text{.}$ Consequently, at each of those values of $t\text{,}$ the secant function has the form $\frac{1}{0}\text{.}$ As is the case with rational functions (or any function, for that matter), at any value where the function returns a fraction of form $\frac{\text{not zero}}{\text{zero}}\text{,}$ the graph of the function approaches a vertical asymptote. Consequently the graph of $y=\sec(t)$ has a vertical asymptote everywhere $\cos(t)=0\text{.}$ The vertical asymptote are, specifically: $\,\,...,\,t=-\frac{5\pi}{2},\,t=-\frac{3\pi}{2},\,t=-\frac{\pi}{2},\,t=\frac{\pi}{2},\,t=\frac{3\pi}{2},\,t=\frac{5\pi}{2},\,...\text{.}$

The vertical asymptotes that occur over $[-\frac{5\pi}{2},\frac{5\pi}{2}]$ have been added to the graph in Figure 16.6.2. Figure 16.6.2. $y=\cos(t)$

At $-2\pi\text{,}$ $0\text{,}$ and $2\pi\text{,}$ the value of the cosine function is $1\text{,}$ so the value of the secant function is also $1\text{.}$ Over the intervals $\left(-\frac{5\pi}{2} ,-\frac{3\pi}{2}\right)\text{,}$ $\left(-\frac{\pi}{2} ,\frac{\pi}{2}\right)\text{,}$ and $\left(\frac{3\pi}{2} ,\frac{5\pi}{2}\right)$ the cosine function is always positive, so the secant function is also always positive. Consequently, at both ends of each of these intervals the secant curve will point upward along the asymptotes. These portions of the secant function have been added in Figure 16.6.3. Figure 16.6.3. $y=\sec(t)$ (thick and red) and $y=\cos(t)$ (thin and gray)

At $-\pi$ and $\pi\text{,}$ the value of the cosine function is $-1\text{,}$ so the value of the secant function is also $-1\text{.}$ Over the intervals $\left(-\frac{3\pi}{2} ,-\frac{\pi}{2}\right)$ and $\left(\frac{\pi}{2} ,\frac{3\pi}{2}\right)$ the cosine function is always negative, so the secant function is also always negative. Consequently, at both ends of each of these intervals the secant curve will point downward along the asymptotes. The secant function has been completed in Figure 16.6.4. Figure 16.6.4. $y=\sec(t)$ (thick and red) and $y=\cos(t)$ (thin and gray)
###### Example16.6.5.

Sketch two complete periods of the function $y=\frac{1}{2}\sec\left(3\left(t-\frac{\pi}{2}\right)\right)\text{.}$

Solution

We begin by observing that $\frac{1}{2}\sec\left(3\left(t-\frac{\pi}{2}\right)\right)=\frac{1}{2} \cdot \frac{1}{\cos\left(3\left(t-\frac{\pi}{2}\right)\right)}\text{.}$ We should also note that the $y$-coordinates at the local minimum points on the secant function will be $\frac{1}{2}$ and the $y$-coordinates at the local maximum points will be $-\frac{1}{2}\text{.}$ Consequently, if we want our curves to touch at the minimum and maximum points on the cosine function, the cosine function we should graph is $y=\frac{1}{2}\cos\left(3\left(t-\frac{\pi}{2}\right)\right)\text{.}$

The amplitude of the cosine function we will sketch is $\frac{1}{2}\text{.}$ Since there is no vertical shift, the mid-line will be the $x$-axis and the minimum and maximum values of $t$ will be, respectively, $-\frac{1}{2}$ and $\frac{1}{2}\text{.}$ We should note that we do not apply the word amplitude to the secant or cosecant function.

The period of both the cosine and the secant function we will sketch is $\frac{2\pi}{3}\text{.}$ They both also have a rightward shift of $\frac{\pi}{2}$ from their parent functions. Without the horizontal shift we would get two complete periods over the intervals $\left(-\frac{2\pi}{3},0\right)$ and $\left(0,\frac{2\pi}{3}\right)\text{.}$ With the shift we will two complete periods over the intervals $\left(-\frac{2\pi}{3}+\frac{\pi}{2},0+\frac{\pi}{2}\right)$ and $\left(0+\frac{\pi}{2},\frac{2\pi}{3}+\frac{\pi}{2}\right)$ which simplify to $\left(-\frac{\pi}{6},\frac{\pi}{2}\right)$ and $\left(\frac{\pi}{2},\frac{7\pi}{6}\right)\text{.}$

One-quarter of the period is $\frac{\pi}{6}\text{.}$ Beginning with $-\frac{\pi}{6}$ and repeatedly adding $\frac{\pi}{6}$ until we reach $\frac{\pi}{2}\text{,}$ we see that the critical values of $t$ over the interval $\left(-\frac{\pi}{6},\frac{\pi}{2}\right)$ are $-\frac{\pi}{6}\text{,}$ $0\text{,}$ $\frac{\pi}{6}\text{,}$ $\frac{\pi}{3}\text{,}$ and $\frac{\pi}{2}\text{.}$ Similarly, the critical values of $t$ over the interval $\left(\frac{\pi}{2},\frac{7\pi}{6}\right)$ are $\frac{\pi}{2}\text{,}$ $\frac{2\pi}{3}\text{,}$ $\frac{5\pi}{6}\text{,}$ $\pi\text{,}$ and $\frac{7\pi}{6}\text{.}$

Because we are graphing a positive-cosine function, the $y$-coordinates at the critical values of $t$ will follow the patten max, mid-line, min, mid-line, max.

The cosine function is shown in Figure 16.6.6. Figure 16.6.6. Two Periods of $y=\frac{1}{2}\cos\left(3\left(t-\frac{\pi}{2}\right)\right)$

The secant function and the cosine function will touch at the points $\left(-\frac{\pi}{6},\frac{1}{2}\right)\text{,}$ $\left(\frac{\pi}{6},-\frac{1}{2}\right)\text{,}$ $\left(\frac{\pi}{2},\frac{1}{2}\right)\text{,}$ $\left(\frac{5\pi}{6},-\frac{1}{2}\right)\text{,}$ and $\left(\frac{7\pi}{6},\frac{1}{2}\right)\text{.}$

The secant function will approach the vertical asymptotes $t=0\text{,}$ $t=\frac{\pi}{3}\text{,}$ $t=\frac{2\pi}{3}\text{,}$ and $t=\pi\text{.}$ Over the intervals $\left(0,\frac{\pi}{3}\right)$ and $\left(\frac{2\pi}{3},\pi\right)$ the secant curves will point downward along the asymptotes. Over $\left(\frac{\pi}{3},\frac{2\pi}{3}\right)$ the secant curve will point upward along the asymptotes. Over the interval $\left(-\frac{\pi}{6},0\right)$ the secant curve will point upward along the asymptote $x=0$ and over the interval $\left(\pi,\frac{7\pi}{6}\right)$ the secant curve will point upward along the asymptote $t=\pi\text{.}$

The secant function is graphed along with the cosine function in Figure 16.6.7. Figure 16.6.7. Two Periods of $y=\frac{1}{2}\sec\left(3\left(t-\frac{\pi}{2}\right)\right)$(thick and red) and $y=\frac{1}{2}\cos\left(3\left(t-\frac{\pi}{2}\right)\right)$ (thin and gray)
##### The Graph of the Function $y=\csc(t)$.

Because the cosecant function and the sine function are reciprocal functions, we can gain insight into the graph of $y=\csc(t)$ by looking at a graph $y=\sin(t)\text{.}$ A graph of the sine function is shown in Figure 16.6.8. Figure 16.6.8. $y=\csc(t)$ (thick and red) and $y=\sin(t)$ (thin and gray)

The cosecant function and sine function will share the points $\left(-\frac{5\pi}{2},-1\right)\text{,}$ $\left(-\frac{3\pi}{2},1\right)\text{,}$ $\left(-\frac{\pi}{2},-1\right)\text{,}$ $\left(\frac{\pi}{2},1\right)\text{,}$ $\left(\frac{3\pi}{2},-1\right)\text{,}$ and $\left(\frac{5\pi}{2},1\right)\text{.}$

The lines $t=-2\pi\text{,}$ $t=-\pi\text{,}$ $t=0\text{,}$ $t=\pi\text{,}$ and $=2\pi$ will act as vertical asymptotes for $y=\csc(t)\text{.}$ The cosecant curves will point upward along the asymptotes where the sine function is positive and the cosecant curves will point downward along the asymptotes where the sine function is negative.

Both $y=\csc(t)$ and $y=\sin(t)$ are shown in Figure 16.6.9. Figure 16.6.9. $y=\csc(t)$ (thick and red) and $y=\sin(t)$ (thin and gray)
###### Example16.6.10.

Sketch two full periods of the function $y=-\csc\left(2\left(t+\frac{\pi}{2}\right)\right)+3\text{.}$

Solution

We'll begin by sketching two full periods of $y=-\sin\left(2\left(t+\frac{\pi}{2}\right)\right)+3\text{.}$

The amplitude of the sine function is $1\text{.}$ As a reminder, the word amplitude does not apply to the cosecant function. There is an upward shift of 3, so the mid-line is $y=3\text{.}$ The $y$-coordinates at the local minimum and maximum points will be, respectively, $2$ and $4\text{.}$

The period is $\frac{2\pi}{2}$ which simplifies to $\pi\text{.}$ There is a leftward shift of $\frac{\pi}{2}\text{,}$ so the period that would normally occur over the interval $(-\pi,0)$ will be shifted to $\left(-\frac{3\pi}{2},-\frac{\pi}{2}\right)$ and the period which would normally occur over $(\pi,0)$ will be shifted to $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\text{.}$

One-quarter of a period is $\frac{\pi}{4}\text{.}$ Beginning at $-\frac{3\pi}{2}$ and repeatedly adding $\frac{\pi}{4}$ until we reach $-\frac{\pi}{2}\text{,}$ we see that the critical values of $t$ that occur over the interval $\left(-\frac{3\pi}{2},-\frac{\pi}{2}\right)$ are $-\frac{3\pi}{2}\text{,}$ $-\frac{5\pi}{4}\text{,}$ $-\pi\text{,}$ $-\frac{3\pi}{4}\text{,}$ and $-\frac{\pi}{2}\text{.}$ Similarly, the critical values of $t$ that occur over the interval $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ are $-\frac{\pi}{2}\text{,}$ $-\frac{\pi}{4}\text{,}$ $0\text{,}$ $\frac{\pi}{4}\text{,}$and $\frac{\pi}{2}\text{.}$

Because we are graphing a negative-sine function, the $y$-coordinates will follow the pattern mid-line, min, mid-line, max, mid-line over each of the periods.

The sine function is shown in Figure 16.6.11. Figure 16.6.11. Two Periods of $y=-\sin\left(2\left(t+\frac{\pi}{2}\right)\right)+3$

The cosecant function and the sine function will share points where the points on the sine function had a value of $1$ or $-1$ before the vertical shift occurred. Those points are $\left(-\frac{5\pi}{4},2\right)\text{,}$ $\left(-\frac{3\pi}{4},4\right)\text{,}$ $\left(-\frac{\pi}{4},2\right)\text{,}$ and$\left(-\frac{\pi}{4},4\right)\text{.}$

The cosecant function will approach vertical asymptotes at values of $t$ where $-\sin\left(2\left(t+\frac{\pi}{2}\right)\right)+3=0$ which gives us $\sin\left(2\left(t+\frac{\pi}{2}\right)\right)=3\text{.}$ The vertical asymptotes are $t=-\frac{3\pi}{2}\text{,}$ $t=-\pi\text{,}$ $t=-\frac{\pi}{2}\text{,}$ $t=0\text{,}$ and $t=\frac{\pi}{2}\text{.}$

The cosecant curves will point upward along the asymptotes over the intervals $\left(-\pi,-\frac{\pi}{2}\right)$ and $\left(0,\frac{\pi}{2}\right)\text{.}$ The cosecant curves will point downward along the asymptotes over the intervals $\left(-\frac{3\pi}{2},-\pi\right)$ and $\left(-\frac{\pi}{2},0\right)\text{.}$

Both the cosecant function and the sine function are graphed in Figure 16.6.12. Figure 16.6.12. Two Periods of $y=-\csc\left(2\left(t+\frac{\pi}{2}\right)\right)+3$ (thick and red) $y=-\sin\left(2\left(t+\frac{\pi}{2}\right)\right)+3$ (thin and gray)

### ExercisesExercises

Graph as directed.

###### 1.

Graph two complete periods of the function $y=-1.5\sec(2(t+\pi))-1$ after first graphing two complete periods of the function $y=-1.5\cos(2(t+\pi))-1\text{.}$

Solution

We begin with the cosine function.

The amplitude of the function is $1.5$ and that there is a downward shift by $1$ so th $y$-coordinates will all fall between $-2.5$ and $0.5\text{.}$

The period is $\frac{2\pi}{2}$ which simplifies to $\pi\text{.}$ There is a leftward shift by $\pi\text{.}$ Because the period and shift are equal, while there is technically a shift, there is no apparent shift; i.e., the curves $y=-1.5\sec(2(t+\pi))-1$ and $y=-1.5\sec(2t)-1$ lie directly atop one another. For this reason, I choose to just go ahead and graph the periods that occur over $(-\pi,0)$ and $(0,\pi)\text{.}$

One-quarter of a period is $\frac{\pi}{4}\text{.}$ Beginning with $-\pi$ and repeatedly adding $\frac{\pi}{4}$until we reach $0\text{,}$ we see that the critical values of $t$ over the interval $(-\pi,0)$ are $-\pi\text{,}$ $-\frac{3\pi}{4}\text{,}$ $-\frac{\pi}{2}\text{,}$ $-\frac{\pi}{4}\text{,}$ and $0\text{.}$ Similarly, the critical values of $t$ that occur over the interval $(0,\pi)$ are $0\text{,}$ $-\frac{\pi}{4}\text{,}$ $\frac{\pi}{2}\text{,}$ $\frac{3\pi}{4}\text{,}$ and $\pi\text{.}$

Since we are graphing a negative-cosine function, the $y$-coordinates at the critical values of $t$ with follow the pattern min, mid-line, max, mid-line, min over each of the two periods.

The function is shown in Figure 16.6.13. Figure 16.6.13. Two Periods of $y=-1.5\cos(2(t+\pi))-1$)

The secant curves and the cosine curve will intersect at the points $(-\pi,-2.5)\text{,}$ $\left(-\frac{\pi}{2},0.5\right)\text{,}$ $(0,-2.5)\text{,}$ $\left(\frac{\pi}{2},0.5\right)\text{,}$ and $(\pi,-2.5)\text{.}$

The secant function will approach vertical asymptotes a the values of $t$ where the cosine function crosses the mid-line. The lines $t=-\frac{\pi}{4}\text{,}$ $t=-\frac{\pi}{4}\text{,}$ $t=\frac{\pi}{4}\text{,}$ and $t=\frac {3\pi}{4}$ are the vertical asymptotes.

The secant curves open upward over intervals where the cosine function lies above the mid-line and they open downward over intervals where the cosine function lies below the mid-line.

The secant function and the cosine function are both shown in Figure 16.6.14. Figure 16.6.14. Two Periods of $y=-1.5\sec(2(t+\pi))-1$ (thick and red) and $y=-1.5\cos(2(t+\pi))-1$ (thin and gray)
###### 2.

Graph two complete periods of the function $y=\csc(\pi(t+1))$ after first graphing two complete periods of the function $y=\sin(\pi(t+1))\text{.}$

Solution

We begin with the sine function.

The amplitude is $1$ and there is no vertical shift, so the $y$-coordinates will fall between $-1$ and $1\text{.}$

The period is $\frac{2\pi}{\pi}$ which simplifies to $2\text{.}$ There is a leftward shift of $1\text{,}$ so the period that would normally occur over the interval $(-2,0)$ is shifted to $(-3,-1)$ and the period that would normally occur over the interval $(0,2)$ is shifted to $(-1,1)\text{.}$

One-quarter of a period is $0.5\text{.}$ Beginning with $-3$ and repeatedly adding $0.5$ until we reach $-1\text{,}$ we see that the critical values of $t$ that occur over the interval $(-3,-1)$ are $-3\text{,}$ $-2.5\text{,}$ $-2\text{,}$ $-1.5\text{,}$ and $-1\text{.}$ Similarly, the critical values of $t$ over the interval $(-1,1)$ are $-1\text{,}$ $-0.5\text{,}$ $0\text{,}$ $0.5\text{,}$ and $1\text{.}$

Because we are graphing a positive-sine function, over each period the $y$-coordinates will follow the pattern mid-line, max, mid-line, min, mid-line at the critical values of $t\text{.}$

A graph of the sine function is shown in Figure 16.6.15. Figure 16.6.15. Two Periods of $y=\sin(\pi(t+1))$

The cosecant curves and the sine curve will touch at the points $(-2.5,1)\text{,}$ $(-1.5,-1)\text{,}$ $(-0.5,1)\text{,}$ and $(0.5,-1)\text{.}$

The cosecant function will approach vertical asymptotes at the points where the sine values are $0\text{.}$ The lines $t=-3\text{,}$ $t=-2\text{,}$ $t=-1\text{,}$ $t=0\text{,}$ and $t=1$ act as vertical asymptotes for the cosecant function.

The cosecant function will open upward along the the vertical asymptotes over the intervals where the sine function is positive and it will open downward along the vertical asymptotes over the intervals where the sine function is negative.

Both the cosecant function and the sine function are shown in Figure 16.6.16. Figure 16.6.16. Two Periods of $y=\csc(\pi(t+1))$ (thick and red)$y=\sin(\pi(t+1))$ (thin and gray)
###### 3.

Graph two complete periods of the function $y=-2\csc\left(\frac{1}{2}t-\frac{\pi}{8}\right)$ after first graphing two complete periods of the function $y=-2\sin\left(\frac{1}{2}t-\frac{\pi}{8}\right)\text{.}$

Solution

We begin by graphing the sine function.

The amplitude is $2$ and there is no vertical shift, so the $y$-coordinates will all lie between $-2$ and $2\text{.}$

The period is $\frac{2\pi}{\frac{1}{2}}$ which simplifies to $4\pi\text{.}$ Before we can state the horizontal shift, we need to fact $\frac{1}{2}$ away from both $t$ and $\frac{1}{8}\text{.}$ That is done below.

\begin{equation*} -2\sin\left(\frac{1}{2}t-\frac{\pi}{8}\right)=-2\sin\left(\frac{1}{2}\left(t-\frac{\pi}{4}\right)\right) \end{equation*}

We can now see that there is a rightward shift of $\frac{\pi}{4}\text{.}$ The period that would normally occur over the interval $(-4\pi,0)$ is shifted to $\left(-\frac{15\pi}{4},\frac{\pi}{4}\right)$ and the period that would normally occur over $(0,4\pi)$is shifted to $\left(\frac{\pi}{4},\frac{17\pi}{4}\right)\text{.}$

One-quarter of the period is $\pi\text{.}$ Starting with $-\frac{15\pi}{4}$ and repeatedly adding $\pi$ until we reach $\frac{\pi}{4}\text{,}$ we see that the critical values of $t$ that occur over the interval $(-4\pi,0)$ are $-\frac{15\pi}{4}\text{,}$ $-\frac{11\pi}{4}\text{,}$ $-\frac{7\pi}{4}\text{,}$ $-\frac{3\pi}{4}\text{,}$ and $\frac{\pi}{4}\text{.}$ Similarly, the critical values of $t$ that occur over the interval $\left(\frac{\pi}{4},\frac{17\pi}{4}\right)$ are $\frac{\pi}{4}\text{,}$ $\frac{5\pi}{4}\text{,}$ $\frac{9\pi}{4}\text{,}$ $\frac{13\pi}{4}\text{,}$ and $\frac{17\pi}{4}\text{.}$

Because we are graphing a negative-sine function, over each period the $y$-coordinates will follow the pattern mid-line, min, mid-line, max, mid-line at the critical-values of $t\text{.}$

The sine function is shown in Figure 16.6.17. Figure 16.6.17. Two Periods of $y=-2\sin\left(\frac{1}{2}t-\frac{\pi}{8}\right)$

The cosecant curves and the sine curve will meet at the points $\left(-\frac{11\pi}{4},-2\right)\text{,}$ $\left(-\frac{3\pi}{4},2\right)\text{,}$ $\left(\frac{5\pi}{4},-2\right)\text{,}$ and $\left(\frac{13\pi}{4},2\right)\text{.}$

The cosecant function will approach vertical asymptotes everywhere the sine function has a value of $0\text{.}$ The vertical asymptotes are the lines $t=-\frac{15\pi}{4}\text{,}$ $t=-\frac{7\pi}{4}\text{,}$ $t=\frac{\pi}{4}\text{,}$ $t=\frac{9\pi}{4}\text{,}$ and $t=\frac{17\pi}{4}\text{.}$

The cosecant curves will open upward along the vertical asymptotes over the intervals where the sine function is positive and they will open downward along the asymptotes over intervals where the sine function is negative.

The cosecant and sine functions are both shown in Figure 16.6.18. Figure 16.6.18. Two Periods of $y=-2\csc\left(\frac{1}{2}t-\frac{\pi}{8}\right)$ (thick and red) and $y=-2\sin\left(\frac{1}{2}t-\frac{\pi}{8}\right)$ (thin and gray)
###### 4.

Graph two complete periods of the function $y=3\sec\left(\frac{5\pi}{2}t+5\pi\right)+4$ after first graphing two complete periods of the function $y=3\cos\left(\frac{5\pi}{2}t+5\pi\right)+4\text{.}$

Solution

We begin with the cosine function.

The amplitude is $3$ and there is a upward shit of $4\text{,}$ so the $y$-coordinates will all fall between $1$ and $7\text{.}$

The period is derived below.

\begin{align*} \frac{2\pi}{\frac{5\pi}{2}}\amp=\frac{2\pi}{1} \cdot \frac{2}{5\pi}\\ \amp=\frac{4}{5} \end{align*}

Before stating the horizontal shift, we need to factor $\frac{5\pi}{2}$ away from both $t$ and $5\pi\text{.}$ This is done below.

\begin{equation*} 3\sec\left(\frac{5\pi}{2}t+5\pi\right)+4=3\sec\left(\frac{5\pi}{2}(t+2)\right)+4 \end{equation*}

So there is leftward shift of 2. The standard period that would normally occur over $\left(\frac{4}{5},\frac{8}{5}\right)$ is shifted to $\left(-\frac{6}{5},-\frac{2}{5}\right)\text{.}$ The standard period that would normally occur over $\left(\frac{8}{5},\frac{12}{5}\right)$ is shifted to $\left(-\frac{2}{5},\frac{2}{5}\right)\text{.}$

One-quarter of the period is $\frac{1}{5}\text{.}$ Starting with $-\frac{6}{5}$ and repeatedly adding $\frac{1}{5}$ until we reach $-\frac{2}{5}\text{,}$ we see that the critical values of $t$ that occur over the interval $\left(-\frac{6}{5},-\frac{2}{5}\right)$ are $-\frac{6}{5}\text{,}$ $-1\text{,}$ $-\frac{4}{5}\text{,}$ $-\frac{3}{5}\text{,}$ and $-\frac{2}{5}\text{.}$ Similarly, the critical values of $t$ that occur over the interval $\left(-\frac{2}{5},\frac{2}{5}\right)$ are $-\frac{2}{5}\text{,}$ $-\frac{1}{5}\text{,}$ $0\text{,}$ $\frac{1}{5}\text{,}$ and$\frac{2}{5}\text{.}$

Because we are graphing a positive-cosine function, over each period the $y$-coordinates will follow the patten max, mid-line, min, mid-line, min at the critical values of $t\text{.}$

The cosine function is graphed in Figure 16.6.19. Figure 16.6.19. Two Periods of $y=3\cos\left(\frac{5\pi}{2}t+5\pi\right)+4$

The secant and cosine functions will share the points $\left(-\frac{6}{5},7\right)\text{,}$ $\left(-\frac{4}{5},1\right)\text{,}$ $\left(-\frac{2}{5},7\right)\text{,}$ $(0,1)\text{,}$ and $\left(\frac{2}{5},7\right)\text{.}$

The secant function will approach vertical asymptotes everywhere the cosine function crosses the mid-line. The vertical asymptotes are the lines $t=-1\text{,}$ $t=-\frac{3}{5}\text{,}$ $t=-\frac{1}{5}\text{,}$ and $t=\frac{1}{5}\text{.}$

The secant curves will open upward along the vertical asymptotes over intervals where the secant function lies above the mid-line. The secant curve will open downward along the vertical asymptotes over intervals where the cosine function lies below the mid-line.

Both the secant function ad the cosine function are graphed in Figure 16.6.20. Figure 16.6.20. Two Periods of $y=3\sec\left(\frac{5\pi}{2}t+5\pi\right)+4$ (thick and red) and $y=3\cos\left(\frac{5\pi}{2}t+5\pi\right)+4$ (thin and gray)