What Is the Relationship Between the Amplitude of Y = 50 Sinx and the Coefficient 50? << Read Less

Analyzing the Graph of y = tan x and Its Variations

We will begin with the graph of the tangent part, plotting points every bit we did for the sine and cosine functions. Recall that

[latex]\tan x=\frac{\sin x}{\cos 10}\\[/latex]

The menstruation of the tangent function is π because the graph repeats itself on intervals of kπ where chiliad is a constant. If we graph the tangent function on [latex]−\frac{\pi}{2}\\[/latex] to [latex]\frac{\pi}{2}\\[/latex], we tin see the behavior of the graph on one complete cycle. If we look at any larger interval, nosotros will see that the characteristics of the graph echo.

We can make up one's mind whether tangent is an odd or even role past using the definition of tangent.

[latex]\begin{array} � \tan(−x)=\frac{\sin(−x)}{\cos(−x)} \hfill& \text{Definition of tangent.} \\ =\frac{−\sin ten}{\cos x} \hfill& \text{Sine is an odd role, cosine is even.} \\ =−\frac{\sin ten}{\cos x} &\hfill \text{The quotient of an odd and an fifty-fifty part is odd.} \hfill \\ =−\tan 10 \hfill& \text{Definition of tangent.} \end{array}\\[/latex]

Therefore, tangent is an odd office. We can further clarify the graphical behavior of the tangent office past looking at values for some of the special angles, as listed in the table below.

x	[latex]−\frac{\pi}{2}\\[/latex]	[latex]−\frac{\pi}{iii}\\[/latex]	[latex]−\frac{\pi}{4}\\[/latex]	[latex]−\frac{\pi}{6}\\[/latex]	0	[latex]\frac{\pi}{6}\\[/latex]	[latex]\frac{\pi}{iv}\\[/latex]	[latex]\frac{\pi}{3}\\[/latex]	[latex]\frac{\pi}{2}\\[/latex]
tan (x)	undefined	[latex]−\sqrt{3}\\[/latex]	–1	[latex]−\frac{\sqrt{3}}{iii}\\[/latex]	0	[latex]\frac{\sqrt{3}}{three}\\[/latex]	i	[latex]\sqrt{three}\\[/latex]	undefined

These points will help u.s.a. describe our graph, but we demand to determine how the graph behaves where it is undefined. If nosotros look more than closely at values when [latex]\frac{\pi}{three}<10<\frac{\pi}{ii}\\[/latex], nosotros can use a table to look for a trend. Because [latex]\frac{\pi}{iii}\approx 1.05\\[/latex] and [latex]\frac{\pi}{2}\approx ane.57\\[/latex], we volition evaluate x at radian measures ane.05 < x < ane.57 equally shown in the table beneath.

x	i.3	1.5	one.55	one.56
tan x	3.6	14.1	48.1	92.half-dozen

As x approaches [latex]\frac{\pi}{2}\\[/latex], the outputs of the function go larger and larger. Because [latex]y=\tan x\\[/latex] is an odd part, we see the corresponding table of negative values in the table below.

x	−1.iii	−i.5	−1.55	−1.56
tan 10	−three.6	−14.i	−48.i	−92.half-dozen

We can see that, as x approaches [latex]−\frac{\pi}{2}\\[/latex], the outputs go smaller and smaller. Recall that there are some values of x for which cos x = 0. For case, [latex]\cos\left(\frac{\pi}{two}\right)=0\\[/latex] and [latex]\cos\left(\frac{three\pi}{2}\right)=0\\[/latex]. At these values, the tangent function is undefined, and so the graph of [latex]y=\tan 10[/latex] has discontinuities at [latex]x=\frac{\pi}{2}\\[/latex] and [latex]\frac{3\pi}{2}\\[/latex]. At these values, the graph of the tangent has vertical asymptotes. Figure i represents the graph of [latex]y=\tan x\\[/latex]. The tangent is positive from 0 to [latex]\frac{\pi}{2}\\[/latex] and from π to [latex]\frac{three\pi}{two}\\[/latex], corresponding to quadrants I and III of the unit circle.

Figure i. Graph of the tangent function

Graphing Variations of y = tan x

As with the sine and cosine functions, the tangent function can be described by a full general equation.

[latex]y=A\tan(Bx)\\[/latex]

We tin can identify horizontal and vertical stretches and compressions using values of A and B. The horizontal stretch can typically be determined from the flow of the graph. With tangent graphs, it is often necessary to determine a vertical stretch using a point on the graph.

Because there are no maximum or minimum values of a tangent office, the term aamplitude cannot be interpreted every bit information technology is for the sine and cosine functions. Instead, nosotros will use the phrase stretching/compressing factor when referring to the constant A.

A Full general Note: Features of the Graph of y = Atan(Bx)

• The stretching cistron is |A| .
• The period is [latex]P=\frac{\pi}{|B|}\\[/latex].
• The domain is all real numbers ten, where [latex]x\ne \frac{\pi}{ii|B|} + \frac{\pi}{|B|} grand\\[/latex] such that k is an integer.
• The range is (−∞, ∞).
• The asymptotes occur at [latex]ten=\frac{\pi}{2|B|} + \frac{\pi}{|B|}1000\\[/latex], where m is an integer.
• [latex]y = A \tan (Bx)\\[/latex] is an odd part.

Graphing One Menses of a Stretched or Compressed Tangent Function

Nosotros can use what we know about the backdrop of the tangent office to quickly sketch a graph of any stretched and/or compressed tangent office of the grade [latex]f(x)=A\tan(Bx)\\[/latex]. We focus on a single period of the part including the origin, because the periodic property enables us to extend the graph to the rest of the function's domain if we wish. Our limited domain is then the interval [latex](−\frac{P}{two}, \frac{P}{2})\\[/latex] and the graph has vertical asymptotes at [latex]\pm \frac{P}{two}\\[/latex] where [latex]P=\frac{\pi}{B}\\[/latex]. On [latex](−\frac{\pi}{2}, \frac{\pi}{2})\\[/latex], the graph will come up from the left asymptote at [latex]ten=−\frac{\pi}{2}\\[/latex], cross through the origin, and continue to increment every bit it approaches the right asymptote at [latex]x=\frac{\pi}{2}\\[/latex]. To brand the function arroyo the asymptotes at the correct rate, nosotros also need to fix the vertical scale by actually evaluating the office for at least one point that the graph volition pass through. For instance, we can utilise

[latex]f\left(\frac{P}{four}\right)=A \tan\left(B\frac{P}{4}\right)=A\tan\left(B\frac{\pi}{4B}\right)=A\\[/latex]

because  [latex]\tan\left(\frac{\pi}{4}\right)=ane\\[/latex].

How To: Given the function [latex]f(x)=A\tan(Bx)\\[/latex], graph i period.

1. Identify the stretching cistron, |A|.
2. Identify B and determine the period, [latex]P=\frac{\pi}{|B|}\\[/latex].
3. Draw vertical asymptotes at  [latex]x=−\frac{P}{2}\\[/latex] and [latex]ten=\frac{P}{2}\\[/latex].
4. For A > 0 , the graph approaches the left asymptote at negative output values and the right asymptote at positive output values (reverse for A < 0 ).
5. Plot reference points at [latex]\left(\frac{P}{4}\text{, }A\correct)\\[/latex] (0, 0), and ([latex]−\frac{P}{4}\\[/latex],− A), and draw the graph through these points.

Example 1: Sketching a Compressed Tangent

Sketch a graph of one menses of the function [latex]y=0.5\tan(\frac{\pi}{ii}x)\\[/latex].

Solution

First, we place A and B.

Figure 2

Because [latex]A=0.5[/latex] and [latex]B=\frac{\pi}{2}\\[/latex], nosotros can notice the stretching/compressing factor and flow. The period is [latex]\frac{\pi}{\frac{\pi}{two}}=2\\[/latex], so the asymptotes are at [latex]ten=\pm i[/latex]. At a quarter menses from the origin, we have

[latex]\begin{array}f(0.5)=0.5\tan(\frac{0.five\pi}{two}) \hfill& \\ =0.five\tan(\frac{\pi}{4}) \hfill& \\ =0.v \stop{array}\\[/latex]

This means the curve must laissez passer through the points(0.5,0.5),(0,0),and(−0.5,−0.v).The only inflection point is at the origin. Effigy shows the graph of ane menstruation of the function.

Figure 3

Try Information technology 1

Sketch a graph of [latex]f(ten)=3\tan\left(\frac{\pi}{half dozen}10\correct)\\[/latex].

Solution

Graphing 1 Flow of a Shifted Tangent Role

At present that nosotros can graph a tangent function that is stretched or compressed, we will add a vertical and/or horizontal (or stage) shift. In this case, we add together C and D to the full general course of the tangent function.

[latex]f(x)=A\tan(Bx−C)+D\\[/latex]

The graph of a transformed tangent function is unlike from the basic tangent function tan x in several ways:

A Full general Annotation: Features of the Graph of y = Atan(Bx−C)+D

• The stretching factor is |A|.
• The menstruation is [latex]\frac{\pi}{|B|}\\[/latex].
• The domain is [latex]x\ne\frac{C}{B}+\frac{\pi}{|B|}k\\[/latex], where k is an integer.
• The range is (−∞,−|A|] ∪ [|A|, ∞).
• The vertical asymptotes occur at [latex]ten=\frac{C}{B}+\frac{\pi}{2|B|}1000\\[/latex], where k is an odd integer.
• There is no amplitude.
• [latex]y=A\tan(Bx)[/latex] is an odd function considering information technology is the quotient of odd and fifty-fifty functions (sin and cosine perspectively).

How To: Given the role [latex]y=A\tan(Bx−C)+D\\[/latex], sketch the graph of one period.

1. Limited the function given in the form [latex]y=A\tan(Bx−C)+D\\[/latex].
2. Identify the stretching/compressing factor, |A|.
3. Place B and determine the period, [latex]P=\frac{\pi}{|B|}\\[/latex].
4. Place C and determine the stage shift, [latex]\frac{C}{B}\\[/latex].
5. Describe the graph of [latex]y=A\tan(Bx)\\[/latex] shifted to the right by [latex]\frac{C}{B}\\[/latex] and upwardly by D.
6. Sketch the vertical asymptotes, which occur at [latex]x=\frac{C}{B}+\frac{\pi}{2|B|}k\\[/latex], where k is an odd integer.
7. Plot any three reference points and draw the graph through these points.

Example 2: Graphing One Flow of a Shifted Tangent Function

Graph one period of the function [latex]y=−2\tan(\pi x+\pi)−ane\\[/latex].

Solution

Step 1. The function is already written in the form [latex]y=A\tan(Bx−C)+D\\[/latex].

Step 2. [latex]A=−ii[/latex], so the stretching factor is [latex]|A|=two[/latex].

Footstep three. [latex]B=\pi[/latex], so the period is [latex]P=\frac{\pi}{|B|}=\frac{\pi}{\pi}=1\\[/latex].

Pace 4. [latex]C=−\pi[/latex], and then the stage shift is [latex]\frac{C}{B}=\frac{−\pi}{\pi}=−one\\[/latex].

Pace 5–7. The asymptotes are at [latex]ten=−\frac{3}{2}\\[/latex] and [latex]x=−\frac{1}{2}\\[/latex] and the iii recommended reference points are (−1.25, ane), (−1,−one), and (−0.75, −3). The graph is shown in Figure 4.

Figure 4

Assay of the Solution

Note that this is a decreasing role because A < 0.

Try It 2

How would the graph in Example two expect dissimilar if we made A = 2 instead of −two ?

Solution

How To: Given the graph of a tangent function, identify horizontal and vertical stretches.

1. Find the period P from the spacing between successive vertical asymptotes or x-intercepts.
2. Write [latex]f(10)=A\tan(\frac{\pi}{P}x)[/latex].
3. Decide a user-friendly point (10, f(x)) on the given graph and use it to determine A.

Example 3: Identifying the Graph of a Stretched Tangent

Find a formula for the role graphed in Effigy 5.

Figure five

Solution

The graph has the shape of a tangent function.

Step 1. 1 bike extends from –4 to 4, and so the period is [latex]P=eight[/latex]. Since [latex]P=\frac{\pi}{|B|}\\[/latex], we take [latex]B=\frac{\pi}{P}=\frac{\pi}{8}\\[/latex].

Stride 2. The equation must take the [latex]\text{class}f(x)=A\tan(\frac{\pi}{8}x)\\[/latex].

Step iii. To find the vertical stretch A, we can utilise the point (2,2).

[latex]2=A\tan(\frac{\pi}{viii}\times2)=A\tan(\frac{\pi}{4})\\[/latex]

Considering [latex]\tan(\frac{\pi}{4})=1\\[/latex], A = ii.

This part would have a formula [latex]f(x)=2\tan(\frac{\pi}{8}x)\\[/latex].

Try It 3

Notice a formula for the function in Figure half dozen.

Figure 6

Solution

Using the Graphs of Trigonometric Functions to Solve Existent-World Problems

Many real-world scenarios represent periodic functions and may exist modeled past trigonometric functions. As an case, let'south return to the scenario from the section opener. Have you ever observed the axle formed past the rotating light on a police motorcar and wondered about the movement of the light beam itself across the wall? The periodic behavior of the distance the light shines every bit a function of time is obvious, but how practise we determine the distance? We tin can employ the tangent function .

Example four: Using Trigonometric Functions to Solve Real-World Scenarios

Suppose the function [latex]y=v\tan\left(\frac{\pi}{4}t\right)\\[/latex] marks the distance in the movement of a light beam from the top of a constabulary car across a wall where t is the time in seconds and y is the distance in feet from a point on the wall directly across from the police car.

1. Find and interpret the stretching factor and menstruation.
2. Graph on the interval [0, 5].
3. Evaluate f(one) and talk over the function's value at that input.

 Solution

1. We know from the general course of  [latex]y=A\tan(Bt)\\[/latex]  that |A| is the stretching factor and π B is the menses.
   

   

   Figure 7

   We see that the stretching factor is v. This means that the beam of light will have moved 5 ft after half the period.

   The period is [latex]\frac{\pi}{\frac{\pi}{iv}}=\frac{\pi}{i}\times \frac{4}{\pi}=4\\[/latex]. This means that every 4 seconds, the beam of lite sweeps the wall. The distance from the spot beyond from the police car grows larger as the police car approaches.

2. To graph the function, we draw an asymptote at [latex]t=2[/latex] and employ the stretching factor and period. See Figure viii.
   

   

   Figure eight

3. menstruum: [latex]f(i)=v\tan \left(\frac{\pi}{four}\left(one\correct)\right)=5\left(1\correct)=5\\[/latex]; after 1 2nd, the beam of has moved 5 ft from the spot across from the police motorcar.

Analyzing the Graphs of y = sec x and y = cscx and Their Variations

The secant was defined by the reciprocal identity [latex]\sec x=\frac{1}{\cos x}[/latex]. Discover that the function is undefined when the cosine is 0, leading to vertical asymptotes at [latex]\frac{\pi}{two}\text{, }\frac{3\pi}{2}\text{, etc}[/latex]. Considering the cosine is never more than than 1 in accented value, the secant, being the reciprocal, will never exist less than 1 in absolute value.

We can graph [latex]y=\sec x[/latex] by observing the graph of the cosine function because these two functions are reciprocals of 1 some other. See Figure 9. The graph of the cosine is shown equally a dashed orange wave so we can see the relationship. Where the graph of the cosine function decreases, the graph of the secant function increases. Where the graph of the cosine role increases, the graph of the secant function decreases. When the cosine function is zero, the secant is undefined.

The secant graph has vertical asymptotes at each value of x where the cosine graph crosses the x-axis; we show these in the graph below with dashed vertical lines, just volition not show all the asymptotes explicitly on all later graphs involving the secant and cosecant.

Note that, because cosine is an even function, secant is too an even function. That is, [latex]\sec(−x)=\sec x[/latex].

Figure 9.Graph of the secant function, [latex]f(x)=\sec x=\frac{1}{\cos x}[/latex]

As we did for the tangent function, we will again refer to the abiding |A| equally the stretching cistron, not the amplitude.

A Full general Note: Features of the Graph of y = Asec(Bx)

• The stretching cistron is |A|.
• The menstruum is [latex]\frac{two\pi}{|B|}[/latex].
• The domain is [latex]x\ne \frac{\pi}{2|B|}chiliad[/latex], where k is an odd integer.
• The range is (−∞, −|A|] ∪ [|A|, ∞).
• The vertical asymptotes occur at [latex]x=\frac{\pi}{2|B|}k [/latex], where thou is an odd integer.
• In that location is no amplitude.
• [latex]y=A\sec(Bx)[/latex] is an even part because cosine is an fifty-fifty part.

Similar to the secant, the cosecant is defined by the reciprocal identity [latex]\csc ten=1\sin 10[/latex]. Notice that the part is undefined when the sine is 0, leading to a vertical asymptote in the graph at 0, π, etc. Since the sine is never more 1 in absolute value, the cosecant, being the reciprocal, volition never exist less than 1 in absolute value.

We can graph [latex]y=\csc x[/latex] by observing the graph of the sine function considering these two functions are reciprocals of one another. See Figure 10. The graph of sine is shown as a dashed orange moving ridge so we can see the relationship. Where the graph of the sine part decreases, the graph of the cosecant function increases. Where the graph of the sine role increases, the graph of the cosecant function decreases.

The cosecant graph has vertical asymptotes at each value of x where the sine graph crosses the 10-axis; we show these in the graph below with dashed vertical lines.

Notation that, since sine is an odd part, the cosecant part is also an odd part. That is, [latex]\csc(−x)=−\csc x[/latex].

The graph of cosecant, which is shown in Figure x, is similar to the graph of secant.

Figure 10.The graph of the cosecant office, [latex]f(x)=\csc x=one\sin x[/latex]

A General Note: Features of the Graph of [latex]y=A\csc(Bx)

• The stretching factor is |A|.
• The menses is [latex]\frac{2\pi}{|B|}[/latex].
• The domain is [latex]ten\ne\frac{\pi}{|B|}chiliad[/latex], where grand is an integer.
• The range is ( −∞, −|A|] ∪ [|A|, ∞).
• The asymptotes occur at [latex]x=\frac{\pi}{|B|}thousand[/latex], where k is an integer.
• [latex]y=A\csc(Bx)[/latex] is an odd function because sine is an odd part.

Graphing Variations of y = sec x and y= csc x

For shifted, compressed, and/or stretched versions of the secant and cosecant functions, we can follow similar methods to those we used for tangent and cotangent. That is, we locate the vertical asymptotes and also evaluate the functions for a few points (specifically the local extrema). If we want to graph but a unmarried period, we can cull the interval for the period in more than than i way. The process for secant is very like, considering the cofunction identity means that the secant graph is the same equally the cosecant graph shifted one-half a period to the left. Vertical and stage shifts may be applied to the cosecant function in the same way equally for the secant and other functions. The equations become the following.

[latex]y=A\sec(Bx−C)+D[/latex]

[latex]y=A\csc(Bx−C)+D[/latex]

A General Note: Features of the Graph of [latex]y=A\sec(Bx−C)+D[/latex]

• The stretching factor is |A|.
• The flow is [latex]\frac{2\pi}{|B|}[/latex].
• The domain is [latex]x\ne \frac{C}{B}+\frac{\pi}{2|B|}one thousand[/latex], where chiliad is an odd integer.
• The range is (−∞, −|A|] ∪ [|A|, ∞).
• The vertical asymptotes occur at [latex]x=\frac{C}{B}+\frac{\pi}{2|B|}yard[/latex], where k is an odd integer.
• There is no aamplitude.
• [latex]y=A\sec(Bx)[/latex] is an even office considering cosine is an even function.

A General Note: Features of the Graph of [latex]y=A\csc(Bx−C)+D[/latex]

• The stretching factor is |A|.
• The period is [latex]\frac{2\pi}{|B|}[/latex].
• The domain is [latex]x\ne\frac{C}{B}+\frac{\pi}{ii|B|}k[/latex], where k is an integer.
• The range is (−∞, −|A|] ∪ [|A|, ∞).
• The vertical asymptotes occur at [latex]x=\frac{C}{B}+\frac{\pi}{|B|}grand[/latex], where k is an integer.
• There is no aamplitude.
• [latex]y=A\csc(Bx)[/latex] is an odd office because sine is an odd function.

How To: Given a function of the form [latex]y=A\sec(Bx)[/latex], graph one period.

1. Express the function given in the form [latex]y=A\sec(Bx)[/latex].
2. Identify the stretching/compressing factor, |A|.
3. Identify B and decide the catamenia, [latex]P=\frac{2\pi}{|B|}[/latex].
4. Sketch the graph of [latex]y=A\cos(Bx)[/latex].
5. Apply the reciprocal human relationship between [latex]y=\cos x[/latex] and [latex]y=\sec x[/latex] to depict the graph of [latex]y=A\sec(Bx)[/latex].
6. Sketch the asymptotes.
7. Plot any ii reference points and describe the graph through these points.

Case 6: Graphing a Variation of the Secant Function

Graph ane flow of [latex]f(x)=ii.5\sec(0.4x)[/latex].

Solution

Stride 1. The given function is already written in the full general form, [latex]y=A\sec(Bx)[/latex].
Step 2. [latex]A=2.five[/latex] and then the stretching cistron is 2.5.
Step 3. [latex]B=0.iv[/latex], then [latex]P=\frac{2\pi}{0.4}=five\pi[/latex]. The period is 5π units.
Stride 4. Sketch the graph of the part [latex]g(x)=two.v\cos(0.4x)[/latex].
Step 5. Utilize the reciprocal relationship of the cosine and secant functions to draw the cosecant office.
Steps six–vii. Sketch 2 asymptotes at [latex]x=1.25\pi[/latex] and [latex]10=3.75\pi[/latex]. We tin can utilise 2 reference points, the local minimum at (0, ii.5) and the local maximum at (2.5π, −2.5). Effigy 11 shows the graph.

Figure 11

Try Information technology 4

Graph one period of [latex]f(x)=−2.v\sec(0.4x)[/latex].

Solution

Q & A

Practise the vertical shift and stretch/compression affect the secant's range?

Yes. The range of f(x) = A sec(Bx − C) + D is ( −∞, −|A| + D] ∪ [|A| + D, ∞).

How To: Given a function of the course [latex]f(x)=A\sec (Bx−C)+D[/latex], graph one period.

1. Limited the role given in the course [latex]y=A\sec(Bx−C)+D[/latex].
2. Identify the stretching/compressing gene, |A|.
3. Identify B and determine the period, [latex]\frac{2\pi}{|B|}[/latex].
4. Identify C and determine the stage shift, [latex]\frac{C}{B}[/latex].
5. Depict the graph of [latex]y=A\sec(Bx)[/latex]. but shift it to the correct past [latex]\frac{C}{B}[/latex] and up by D.
6. Sketch the vertical asymptotes, which occur at [latex]ten=\frac{C}{B}+\frac{\pi}{two|B|}thou[/latex], where k is an odd integer.

Example 7: Graphing a Variation of the Secant Part

Graph one flow of [latex]y=4\sec \left(\frac{\pi}{3}x−\frac{\pi}{2}\correct)+ane[/latex].

Solution

Step one. Express the part given in the form [latex]y=iv\sec \left(\frac{\pi}{iii}ten−\frac{\pi}{2}\right)+ane[/latex].

Step ii. The stretching/compressing factor is |A| = 4.

Step iii. The period is

[latex]\begin{array} \frac{two\pi}{|B|}=\frac{two\pi}{\frac{\pi}{iii}} \hfill& \\ =\frac{ii\pi}{i}\times\frac{three}{\pi} \hfill& \\ =half-dozen \stop{array}[/latex]

Step 4. The stage shift is

[latex]\begin{array}\frac{C}{B}=\frac{\frac{\pi}{two}}{\frac{\pi}{three}} \hfill& \\ =\frac{\pi}{2} \times \frac{3}{\pi} \hfill& \\ =one.five \end{assortment}[/latex]

Pace 5. Draw the graph of [latex]y=A\sec(Bx)[/latex],but shift information technology to the correct by [latex]\frac{C}{B}=1.five[/latex] and upwards by D= six.

Footstep half-dozen. Sketch the vertical asymptotes, which occur at ten= 0, x = 3, and x = 6. There is a local minimum at (1.five, v) and a local maximum at (four.5, −iii). Figure 12 shows the graph.

Figure 12

Endeavor Information technology 5

Graph one period of [latex]f(10)=−6\sec(4x+2)−viii[/latex].

Solution

Q & A

The domain of csc ten was given to be all x such that [latex]x\ne chiliad\pi[/latex] for whatsoever integer k. Would the domain of [latex]y=A\csc(Bx−C)+D[/latex] be [latex]x\ne\frac{C+thousand\pi}{B}[/latex]?

Yes. The excluded points of the domain follow the vertical asymptotes. Their locations show the horizontal shift and compression or expansion implied by the transformation to the original part's input.

How To: Given a part of the form [latex]y=A\csc(Bx)[/latex], graph ane period.

1. Express the office given in the form [latex]y=A\csc(Bx)[/latex].
2. |A|.
3. Identify B and decide the period, [latex]P=\frac{2\pi}{|B|}[/latex].
4. Draw the graph of [latex]y=A\sin(Bx)[/latex].
5. Utilize the reciprocal relationship between [latex]y=\sin x[/latex] and [latex]y=\csc x[/latex] to describe the graph of [latex]y=A\csc(Bx) [/latex].
6. Sketch the asymptotes.
7. Plot any two reference points and draw the graph through these points.

Instance viii: Graphing a Variation of the Cosecant Function

Graph ane flow of [latex]f(x)=−3\csc(4x)[/latex].

Solution

Step 1. The given role is already written in the general grade, [latex]y=A\csc(Bx)[/latex].

Step 2. [latex]|A|=|−three|=three[/latex], so the stretching factor is 3.

Step three. [latex]B=iv\text{, so}P=\frac{2\pi}{four}=\frac{\pi}{two}[/latex].The period is [latex]\frac{\pi}{2}[/latex] units.

Stride 4. Sketch the graph of the function [latex]g(x)=−3\sin(4x)[/latex].

Stride 5. Utilize the reciprocal human relationship of the sine and cosecant functions to draw the cosecant role.

Steps 6–7. Sketch three asymptotes at [latex]x=0\text{, }x=\frac{\pi}{4}\text{, and }ten=\frac{\pi}{ii}[/latex].We can use two reference points, the local maximum at [latex]\left(\frac{\pi}{8}\text{, }−3\right)[/latex] and the local minimum at [latex]\left(\frac{iii\pi}{8}\text{, }iii\right)[/latex]. Figure 13 shows the graph.

Figure 13

Try It 6

Graph ane menses of [latex]f(x)=0.5\csc(2x)[/latex].

Solution

How To: Given a function of the class [latex]f(x)=A\csc(Bx−C)+D[/latex], graph one period.

1. Express the office given in the form [latex]y=A\csc(Bx−C)+D[/latex].
2. Identify the stretching/compressing factor, |A|.
3. Identify B and determine the period, [latex]\frac{two\pi}{|B|}[/latex].
4. Identify C and determine the phase shift, [latex]\frac{C}{B}[/latex].
5. Draw the graph of [latex]y=A\csc(Bx)[/latex] merely shift it to the right past and up by D.
6. Sketch the vertical asymptotes, which occur at [latex]x=\frac{C}{B}+\frac{\pi}{|B|}yard[/latex], where g is an integer.

Instance 9: Graphing a Vertically Stretched, Horizontally Compressed, and Vertically Shifted Cosecant

Sketch a graph of [latex]y=2\csc\left(\frac{\pi}{ii}10\correct)+1[/latex]. What are the domain and range of this role?

Solution

Step 1. Express the function given in the form [latex]y=two\csc\left(\frac{\pi}{ii}x\right)+1[/latex].

Footstep 2. Identify the stretching/compressing cistron, [latex]|A|=2[/latex].

Step 3. The flow is [latex]\frac{2\pi}{|B|}=\frac{2\pi}{\frac{\pi}{2}}=\frac{2\pi}{1}\times \frac{2}{\pi}=4[/latex].

Step 4. The phase shift is [latex]\frac{0}{\frac{\pi}{two}}=0[/latex].

Step 5. Draw the graph of [latex]y=A\csc(Bx)[/latex] but shift it up [latex]D=1[/latex].

Stride 6. Sketch the vertical asymptotes, which occur at x = 0, 10 = two, x = iv.

The graph for this function is shown in Effigy 14.

Effigy 14

Assay of the Solution

The vertical asymptotes shown on the graph mark off one period of the function, and the local extrema in this interval are shown by dots. Discover how the graph of the transformed cosecant relates to the graph of [latex]f(ten)=2\sin\left(\frac{\pi}{ii}x\right)+1[/latex], shown as the orange dashed moving ridge.

Endeavour It 7

Given the graph of [latex]f(x)=ii\cos\left(\frac{\pi}{2}x\right)+1[/latex] shown in Figure xv, sketch the graph of [latex]g(x)=2\sec\left(\frac{\pi}{2}x\correct)+1[/latex] on the same axes.

Effigy xv

Solution

Analyzing the Graph of y = cot x and Its Variations

The final trigonometric function we demand to explore is cotangent. The cotangent is divers by the reciprocal identity [latex]\cot x=\frac{1}{\tan x}[/latex]. Notice that the function is undefined when the tangent function is 0, leading to a vertical asymptote in the graph at 0, π, etc. Since the output of the tangent function is all real numbers, the output of the cotangent function is also all real numbers.

We can graph [latex]y=\cot x[/latex] past observing the graph of the tangent office because these ii functions are reciprocals of one another. See Figure 16. Where the graph of the tangent function decreases, the graph of the cotangent function increases. Where the graph of the tangent role increases, the graph of the cotangent function decreases.

The cotangent graph has vertical asymptotes at each value of ten where [latex]\tan x=0[/latex]; we prove these in the graph beneath with dashed lines. Since the cotangent is the reciprocal of the tangent, [latex]\cot x[/latex] has vertical asymptotes at all values of ten where [latex]\tan x=0[/latex] , and [latex]\cot x=0[/latex] at all values of x where tan ten has its vertical asymptotes.

Figure 16. The cotangent function

A Full general Notation: Features of the Graph of y = Acot(Bx)

• The stretching factor is |A|.
• The period is [latex]P=\frac{\pi}{|B|}[/latex].
• The domain is [latex]x\ne\frac{\pi}{|B|}k[/latex], where k is an integer.
• The range is (−∞, ∞).
• The asymptotes occur at [latex]x=\frac{\pi}{|B|}k[/latex], where k is an integer.
• [latex]y=A\cot(Bx)[/latex] is an odd function.

Graphing Variations of y = cot x

We tin transform the graph of the cotangent in much the same mode as we did for the tangent. The equation becomes the following.

[latex]y=A\cot(Bx−C)+D[/latex]

A General Note: Properties of the Graph of y = Acot(Bx−C)+D

• The stretching factor is |A|.
• The menstruation is [latex]\frac{\pi}{|B|}[/latex].
• The domain is [latex]x\ne\frac{C}{B}+\frac{\pi}{|B|}k[/latex], where grand is an integer.
• The range is (−∞, −|A|] ∪ [|A|, ∞).
• The vertical asymptotes occur at [latex]ten=\frac{C}{B}+\frac{\pi}{|B|}k[/latex], where grand is an integer.
• There is no amplitude.
• [latex]y=A\cot(Bx)[/latex] is an odd function considering it is the quotient of even and odd functions (cosine and sine, respectively)

How To: Given a modified cotangent function of the form [latex]f(x)=A\cot(Bx)[/latex], graph i period.

1. Limited the office in the form [latex]f(x)=A\cot(Bx)[/latex].
2. Identify the stretching gene, |A|.
3. Identify the period, [latex]P=\frac{\pi}{|B|}[/latex].
4. Depict the graph of [latex]y=A\tan(Bx)[/latex].
5. Plot any two reference points.
6. Apply the reciprocal relationship between tangent and cotangent to depict the graph of [latex]y=A\cot(Bx)[/latex].
7. Sketch the asymptotes.

Case ten: Graphing Variations of the Cotangent Function

Determine the stretching cistron, period, and phase shift of [latex]y=three\cot(4x)[/latex], and then sketch a graph.

Solution

Step 1. Expressing the role in the form [latex]f(ten)=A\cot(Bx)[/latex] gives [latex]f(x)=3\cot(4x)[/latex].

Step two. The stretching cistron is [latex]|A|=3[/latex].

Step iii. The period is [latex]P=\frac{\pi}{four}[/latex].

Step 4. Sketch the graph of [latex]y=3\tan(4x)[/latex].

Step 5. Plot two reference points. Two such points are [latex]\left(\frac{\pi}{sixteen}\text{, }3\correct)[/latex] and [latex]\left(\frac{3\pi}{16}\text{, }−3\right)[/latex].

Step six. Use the reciprocal human relationship to draw [latex]y=3\cot(4x)[/latex].

Step 7. Sketch the asymptotes, [latex]ten=0[/latex], [latex]x=\frac{\pi}{four}[/latex].

The orange graph in Figure 17 shows [latex]y=3\tan(4x)[/latex] and the blue graph shows [latex]y=iii\cot(4x)[/latex].

Figure 17

How To: Given a modified cotangent function of the form [latex]f(10)=A\cot(Bx−C)+D[/latex], graph one menstruum.

1. Express the function in the course [latex]f(x)=A\cot(Bx−C)+D[/latex].
2. Identify the stretching factor, |A|.
3. Identify the period, [latex]P=\frac{\pi}{|B|}[/latex].
4. Identify the phase shift, [latex]\frac{C}{B}[/latex].
5. Describe the graph of [latex]y=A\tan(Bx)[/latex] shifted to the right past [latex]\frac{C}{B}[/latex] and upward by D.
6. Sketch the asymptotes [latex]10 =\frac{C}{B}+\frac{\pi}{|B|}k[/latex], where g is an integer.
7. Plot any three reference points and draw the graph through these points.

Example 11: Graphing a Modified Cotangent

Sketch a graph of one menstruum of the role [latex]f(ten)=4\cot(\frac{\pi}{viii}ten−\frac{\pi}{ii})−2[/latex].

Solution

Step 1. The role is already written in the general form [latex]f(10)=A\cot(Bx−C)+D[/latex].

Step 2. [latex]A=4[/latex], then the stretching cistron is 4.

Step 3. [latex]B=\frac{\pi}{8}[/latex], so the period is [latex]P=\frac{\pi}{|B|}=\frac{\pi}{\frac{\pi}{eight}}=eight[/latex].

Pace 4. [latex]C=\frac{\pi}{2}[/latex], then the phase shift is [latex]\frac{C}{B}=\frac{\frac{\pi}{2}}{\frac{\pi}{8}}=4[/latex].

Step v. We describe [latex]f(10)=4\tan\left(\frac{\pi}{viii}x−\frac{\pi}{2}\right)−2[/latex].

Step 6-seven. Three points we can apply to guide the graph are (6,ii), (8,−two), and (10,−6). We employ the reciprocal relationship of tangent and cotangent to draw [latex]f(x)=4\cot(\frac{\pi}{viii}x−\frac{\pi}{2})−2[/latex].

Step 8. The vertical asymptotes are [latex]10=4[/latex] and [latex]x=12[/latex].

The graph is shown in Figure 18.

Effigy xviii. I period of a modified cotangent function.

Cardinal Equations

Shifted, compressed, and/or stretched tangent function	[latex]y=A\tan(Bx−C)+D[/latex]
Shifted, compressed, and/or stretched secant role	[latex]y=A\sec(Bx−C)+D[/latex]
Shifted, compressed, and/or stretched cosecant	[latex]y=A\csc(Bx−C)+D[/latex]
Shifted, compressed, and/or stretched cotangent role	[latex]y=A\cot(Bx−C)+D[/latex]

Key Concepts

• The tangent part has period π.
• [latex]f(10)=A\tan(Bx−C)+D[/latex] is a tangent with vertical and/or horizontal stretch/pinch and shift.
• The secant and cosecant are both periodic functions with a period of2π. [latex]f(ten)=A\sec(Bx−C)+D[/latex] gives a shifted, compressed, and/or stretched secant role graph.
• [latex]f(x)=A\csc(Bx−C)+D[/latex] gives a shifted, compressed, and/or stretched cosecant role graph.
• The cotangent function has menstruum π and vertical asymptotes at 0, ±π,±2π,….
• The range of cotangent is (−∞,∞),and the function is decreasing at each signal in its range.
• The cotangent is goose egg at [latex]\ne\frac{\pi}{ii}\text{, }\ne\frac{three\pi}{2}[/latex],….
• [latex]f(x)=A\cot(Bx−C)+D[/latex] is a cotangent with vertical and/or horizontal stretch/compression and shift.
• Real-world scenarios tin can be solved using graphs of trigonometric functions.

Department Exercises

1. Explain how the graph of the sine function can be used to graph [latex]y=\csc ten[/latex].

ii. How tin can the graph of [latex]y=\cos x[/latex] be used to construct the graph of [latex]y=\sec x[/latex]?

iii. Explicate why the period of [latex]\tan 10[/latex] is equal to π.

4. Why are there no intercepts on the graph of [latex]y=\csc 10[/latex]?

5. How does the period of [latex]y=\csc ten[/latex] compare with the period of [latex]y=\sin x[/latex]?

For the following exercises, match each trigonometric function with 1 of the following graphs.

6. [latex]f(ten)=\tan 10[/latex]

vii. [latex]f(10)=\sec x[/latex]

8. [latex]f(ten)=\csc x[/latex]

ix. [latex]f(x)=\cot ten[/latex]

For the post-obit exercises, find the period and horizontal shift of each of the functions.

x. [latex]f(x)=ii\tan(4x−32)[/latex]

11. [latex]h(x)=2\sec\left(\frac{\pi}{4}(x+ane)\correct)[/latex]

12. [latex]k(10)=6\csc\left(\frac{\pi}{3}x+\pi\right)[/latex]

thirteen. If tanx = −1.5, notice tan(−10).

14. If secx = 2, find sec(−10).

15. If cscx = −5, discover csc(−10).

16. If [latex]x\sin x=ii[/latex], discover [latex](−x)\sin(−ten)[/latex].

For the following exercises, rewrite each expression such that the argumentx is positive.

17. [latex]\cot(−ten)\cos(−x)+\sin(−x)[/latex]

18. [latex]\cos(−x)+\tan(−10)\sin(−x)[/latex]

For the post-obit exercises, sketch two periods of the graph for each of the following functions. Place the stretching gene, flow, and asymptotes.

19. [latex]f(x)=two\tan(4x−32)[/latex]

twenty. [latex]h(ten)=2\sec\left(\frac{\pi}{iv}\left(ten+ane\correct)\right)[/latex]

21. [latex]m(x)=six\csc\left(\frac{\pi}{3}x+\pi\right)[/latex]

22. [latex]j(x)=\tan\left(\frac{\pi}{2}x\right)[/latex]

23. [latex]p(x)=\tan\left(10−\frac{\pi}{two}\right)[/latex]

24. [latex]f(x)=4\tan(x)[/latex]

25. [latex]f(x)=\tan\left(x+\frac{\pi}{4}\right)[/latex]

26. [latex]f(x)=\pi\tan\left(\pi x−\pi\correct)−\pi[/latex]

27. [latex]f(x)=2\csc(x)[/latex]

28. [latex]f(x)=−\frac{i}{4}\csc(10)[/latex]

29. [latex]f(10)=4\sec(3x)[/latex]

30. [latex]f(x)=−3\cot(2x)[/latex]

31. [latex]f(x)=7\sec(5x)[/latex]

32. [latex]f(x)=\frac{9}{10}\csc(\pi x)[/latex]

33. [latex]f(x)=two\csc \left(10+\frac{\pi}{iv}\correct)−1[/latex]

34. [latex]f(x)=−\sec \left(x−\frac{\pi}{3}\right)−2[/latex]

35. [latex]f(x)=\frac{7}{five}\csc \left(x−\frac{\pi}{iv}\right)[/latex]

36. [latex]f(x)=five\left(\cot\left(x+\frac{\pi}{2}\correct)−3\right)[/latex]

For the following exercises, find and graph 2 periods of the periodic role with the given stretching factor, |A|, menstruation, and phase shift.

37. A tangent curve, [latex]A=1[/latex], period of [latex]\frac{\pi}{3}[/latex]; and phase shift [latex](h\text{,}k)=\left(\frac{\pi}{iv}\text{,}2\right)[/latex]

38. A tangent curve, [latex]A=−2[/latex], menstruation of [latex]\frac{\pi}{4}[/latex], and phase shift [latex](h\text{,}1000)=\left(−\frac{\pi}{4}\text{,}−2\right)[/latex]

For the post-obit exercises, find an equation for the graph of each office.

39.

40.

41.

42.

43.

44.

45.

47.

49.

50. Graph [latex]f(x)=one+\sec^{2}(x)−\tan^{ii}(10)[/latex]. What is the role shown in the graph?

51. [latex]f(x)=\sec(0.001x)[/latex]

52. [latex]f(x)=\cot(100\pi x)[/latex]

53. [latex]f(ten)=\sin^{2}x+\cos^{ii}x[/latex]

54. The function [latex]f(x)=20\tan\left(\frac{\pi}{10}ten\right)[/latex] marks the distance in the movement of a light beam from a police motorcar across a wall for time 10, in seconds, and distance [latex]f(10)[/latex], in feet.

a. Graph on the interval[0,5].
b. Find and interpret the stretching gene, menstruation, and asymptote.
c. Evaluate f(1) and f(ii.5) and discuss the office's values at those inputs.

55. Standing on the shore of a lake, a fisherman sights a boat far in the distance to his left. Let x, measured in radians, be the angle formed by the line of sight to the send and a line due north from his position. Presume due due north is 0 and 10 is measured negative to the left and positive to the correct. (See Figure 19.) The boat travels from westward to due east and, ignoring the curvature of the Globe, the distance [latex]d(x)[/latex], in kilometers, from the fisherman to the gunkhole is given by the function [latex]d(x)=ane.5\sec(ten)[/latex].

a. What is a reasonable domain for [latex]d(10)[/latex]?
b. Graph d(x) on this domain.
c. Discover and discuss the significant of any vertical asymptotes on the graph of [latex]d(ten)[/latex].
d. Calculate and translate [latex]d(−\frac{\pi}{3})[/latex]. Circular to the second decimal place.
e. Summate and interpret [latex]d(\frac{\pi}{6})[/latex]. Circular to the second decimal place.
f. What is the minimum altitude between the fisherman and the boat? When does this occur?

Figure xix

56. A light amplification by stimulated emission of radiation rangefinder is locked on a comet approaching Earth. The distance [latex]g(x)[/latex], in kilometers, of the comet after x days, for x in the interval 0 to xxx days, is given by [latex]g(x)=250,000\csc(\frac{\pi}{thirty}ten)[/latex].

a. Graph [latex]1000(x)[/latex] on the interval [0,35].
b. Evaluate [latex]m(5)[/latex] and interpret the data.
c. What is the minimum altitude between the comet and Globe? When does this occur? To which abiding in the equation does this stand for?
d. Find and talk over the meaning of any vertical asymptotes.

57. A video camera is focused on a rocket on a launching pad ii miles from the photographic camera. The angle of summit from the footing to the rocket afterwards x seconds is [latex]\frac{\pi}{120}10[/latex].

a. Write a role expressing the altitude [latex]h(10)[/latex], in miles, of the rocket above the footing after x seconds. Ignore the curvature of the Earth.
b. Graph [latex]h(x)[/latex] on the interval (0,60).
c. Evaluate and interpret the values [latex]h(0)[/latex] and [latex]h(thirty)[/latex].
d. What happens to the values of [latex]h(10)[/latex] equally x approaches lx seconds? Interpret the pregnant of this in terms of the problem.

felicianomishought.blogspot.com

Source: https://courses.lumenlearning.com/precalctwo/chapter/graphs-of-the-other-trigonometric-functions/

Share this post