# 6.2 Graphs of the other trigonometric functions  (Page 2/9)

 Page 2 / 9

## Graphing variations of y = tan x

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

$y=A\mathrm{tan}\left(Bx\right)$

We can identify horizontal and vertical stretches and compressions using values of $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}B.\text{\hspace{0.17em}}$ The horizontal stretch can typically be determined from the period 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 function, the term amplitude cannot be interpreted as it is for the sine and cosine functions. Instead, we will use the phrase stretching/compressing factor when referring to the constant $\text{\hspace{0.17em}}A.$

## Features of the graph of y = A Tan( Bx )

• The stretching factor is $\text{\hspace{0.17em}}|A|.$
• The period is $\text{\hspace{0.17em}}P=\frac{\pi }{|B|}.$
• The domain is all real numbers $\text{\hspace{0.17em}}x,$ where $\text{\hspace{0.17em}}x\ne \frac{\pi }{2|B|}+\frac{\pi }{|B|}k\text{\hspace{0.17em}}$ such that $\text{\hspace{0.17em}}k\text{\hspace{0.17em}}$ is an integer.
• The range is $\text{\hspace{0.17em}}\left(\mathrm{-\infty },\infty \right).$
• The asymptotes occur at $\text{\hspace{0.17em}}x=\frac{\pi }{2|B|}+\frac{\pi }{|B|}k,\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}k\text{\hspace{0.17em}}$ is an integer.
• $y=A\mathrm{tan}\left(Bx\right)\text{\hspace{0.17em}}$ is an odd function.

## Graphing one period of a stretched or compressed tangent function

We can use what we know about the properties of the tangent function to quickly sketch a graph of any stretched and/or compressed tangent function of the form $\text{\hspace{0.17em}}f\left(x\right)=A\mathrm{tan}\left(Bx\right).\text{\hspace{0.17em}}$ We focus on a single period    of the function 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 $\text{\hspace{0.17em}}\left(-\frac{P}{2},\frac{P}{2}\right)\text{\hspace{0.17em}}$ and the graph has vertical asymptotes at $\text{\hspace{0.17em}}±\frac{P}{2}\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}P=\frac{\pi }{B}.\text{\hspace{0.17em}}$ On $\text{\hspace{0.17em}}\left(-\frac{\pi }{2},\frac{\pi }{2}\right),\text{\hspace{0.17em}}$ the graph will come up from the left asymptote at $\text{\hspace{0.17em}}x=-\frac{\pi }{2},\text{\hspace{0.17em}}$ cross through the origin, and continue to increase as it approaches the right asymptote at $\text{\hspace{0.17em}}x=\frac{\pi }{2}.\text{\hspace{0.17em}}$ To make the function approach the asymptotes at the correct rate, we also need to set the vertical scale by actually evaluating the function for at least one point that the graph will pass through. For example, we can use

$f\left(\frac{P}{4}\right)=A\mathrm{tan}\left(B\frac{P}{4}\right)=A\mathrm{tan}\left(B\frac{\pi }{4B}\right)=A$

because $\text{\hspace{0.17em}}\mathrm{tan}\left(\frac{\pi }{4}\right)=1.$

Given the function $\text{\hspace{0.17em}}f\left(x\right)=A\mathrm{tan}\left(Bx\right),\text{\hspace{0.17em}}$ graph one period.

1. Identify the stretching factor, $\text{\hspace{0.17em}}|A|.$
2. Identify $\text{\hspace{0.17em}}B\text{\hspace{0.17em}}$ and determine the period, $\text{\hspace{0.17em}}P=\frac{\pi }{|B|}.$
3. Draw vertical asymptotes at $\text{\hspace{0.17em}}x=-\frac{P}{2}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}x=\frac{P}{2}.$
4. For $\text{\hspace{0.17em}}A>0,\text{\hspace{0.17em}}$ the graph approaches the left asymptote at negative output values and the right asymptote at positive output values (reverse for $\text{\hspace{0.17em}}A<0$ ).
5. Plot reference points at $\text{\hspace{0.17em}}\left(\frac{P}{4},A\right),\text{\hspace{0.17em}}$ $\left(0,0\right),\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(-\frac{P}{4},-A\right),\text{\hspace{0.17em}}$ and draw the graph through these points.

## Sketching a compressed tangent

Sketch a graph of one period of the function $\text{\hspace{0.17em}}y=0.5\mathrm{tan}\left(\frac{\pi }{2}x\right).$

First, we identify $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}B.$

Because $\text{\hspace{0.17em}}A=0.5\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}B=\frac{\pi }{2},\text{\hspace{0.17em}}$ we can find the stretching/compressing factor and period. The period is $\text{\hspace{0.17em}}\frac{\pi }{\frac{\pi }{2}}=2,\text{\hspace{0.17em}}$ so the asymptotes are at $\text{\hspace{0.17em}}x=±1.\text{\hspace{0.17em}}$ At a quarter period from the origin, we have

$\begin{array}{l}f\left(0.5\right)=0.5\mathrm{tan}\left(\frac{0.5\pi }{2}\right)\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=0.5\mathrm{tan}\left(\frac{\pi }{4}\right)\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=0.5\hfill \end{array}$

This means the curve must pass through the points $\text{\hspace{0.17em}}\left(0.5,0.5\right),$ $\left(0,0\right),$ and $\text{\hspace{0.17em}}\left(-0.5,-0.5\right).\text{\hspace{0.17em}}$ The only inflection point is at the origin. [link] shows the graph of one period of the function.

Sketch a graph of $\text{\hspace{0.17em}}f\left(x\right)=3\mathrm{tan}\left(\frac{\pi }{6}x\right).$

## Graphing one period of a shifted tangent function

Now that we can graph a tangent function that is stretched or compressed, we will add a vertical and/or horizontal (or phase) shift. In this case, we add $\text{\hspace{0.17em}}C\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}D\text{\hspace{0.17em}}$ to the general form of the tangent function.

So a horizontal compression by factor of 1/2 is the same as a horizontal stretch by a factor of 2, right?
The center is at (3,4) a focus is at (3,-1), and the lenght of the major axis is 26
The center is at (3,4) a focus is at (3,-1) and the lenght of the major axis is 26 what will be the answer?
Rima
I done know
Joe
What kind of answer is that😑?
Rima
I had just woken up when i got this message
Joe
Rima
i have a question.
Abdul
how do you find the real and complex roots of a polynomial?
Abdul
@abdul with delta maybe which is b(square)-4ac=result then the 1st root -b-radical delta over 2a and the 2nd root -b+radical delta over 2a. I am not sure if this was your question but check it up
Nare
This is the actual question: Find all roots(real and complex) of the polynomial f(x)=6x^3 + x^2 - 4x + 1
Abdul
@Nare please let me know if you can solve it.
Abdul
I have a question
juweeriya
hello guys I'm new here? will you happy with me
mustapha
The average annual population increase of a pack of wolves is 25.
how do you find the period of a sine graph
Period =2π if there is a coefficient (b), just divide the coefficient by 2π to get the new period
Am
if not then how would I find it from a graph
Imani
by looking at the graph, find the distance between two consecutive maximum points (the highest points of the wave). so if the top of one wave is at point A (1,2) and the next top of the wave is at point B (6,2), then the period is 5, the difference of the x-coordinates.
Am
you could also do it with two consecutive minimum points or x-intercepts
Am
I will try that thank u
Imani
Case of Equilateral Hyperbola
ok
Zander
ok
Shella
f(x)=4x+2, find f(3)
Benetta
f(3)=4(3)+2 f(3)=14
lamoussa
14
Vedant
pre calc teacher: "Plug in Plug in...smell's good" f(x)=14
Devante
8x=40
Chris
Explain why log a x is not defined for a < 0
the sum of any two linear polynomial is what
Momo
how can are find the domain and range of a relations
the range is twice of the natural number which is the domain
Morolake
A cell phone company offers two plans for minutes. Plan A: $15 per month and$2 for every 300 texts. Plan B: $25 per month and$0.50 for every 100 texts. How many texts would you need to send per month for plan B to save you money?
6000
Robert
more than 6000
Robert
For Plan A to reach $27/month to surpass Plan B's$26.50 monthly payment, you'll need 3,000 texts which will cost an additional \$10.00. So, for the amount of texts you need to send would need to range between 1-100 texts for the 100th increment, times that by 3 for the additional amount of texts...
Gilbert
...for one text payment for 300 for Plan A. So, that means Plan A; in my opinion is for people with text messaging abilities that their fingers burn the monitor for the cell phone. While Plan B would be for loners that doesn't need their fingers to due the talking; but those texts mean more then...
Gilbert
can I see the picture
How would you find if a radical function is one to one?
how to understand calculus?
with doing calculus
SLIMANE
Thanks po.
Jenica
Hey I am new to precalculus, and wanted clarification please on what sine is as I am floored by the terms in this app? I don't mean to sound stupid but I have only completed up to college algebra.
I don't know if you are looking for a deeper answer or not, but the sine of an angle in a right triangle is the length of the opposite side to the angle in question divided by the length of the hypotenuse of said triangle.
Marco
can you give me sir tips to quickly understand precalculus. Im new too in that topic. Thanks
Jenica
if you remember sine, cosine, and tangent from geometry, all the relationships are the same but they use x y and r instead (x is adjacent, y is opposite, and r is hypotenuse).
Natalie
it is better to use unit circle than triangle .triangle is only used for acute angles but you can begin with. Download any application named"unit circle" you find in it all you need. unit circle is a circle centred at origine (0;0) with radius r= 1.
SLIMANE
What is domain
johnphilip
the standard equation of the ellipse that has vertices (0,-4)&(0,4) and foci (0, -15)&(0,15) it's standard equation is x^2 + y^2/16 =1 tell my why is it only x^2? why is there no a^2?