Graphing a Tan Function

[Jason76]

$\displaystyle y = 3\tan2x$

$\displaystyle A$ (amplitude) $\displaystyle = |3| = 3$

Find $\displaystyle x$ values:

$\displaystyle y = 3\tan2x$

$\displaystyle -3 = 3\tan2x$ and $\displaystyle 3 = 3\tan 2x$

Now, looking at the one on the left:

$\displaystyle -3 = 3\tan2x$

$\displaystyle -1 = \tan2x$ How did this line become the next line? :idea: Now i did notice that the $\displaystyle \arctan$ of $\displaystyle -1$ was $\displaystyle -\dfrac{\pi}{4}$

$\displaystyle 2x = -\dfrac{\pi}{4}$

$\displaystyle -\dfrac{\pi}{8}$

Moving on to the other one:

$\displaystyle 3 = 3\tan2x$

$\displaystyle 1 = \tan2x$ How did this line become the next line? :idea: I did notice that the $\displaystyle \arctan$ of $\displaystyle 1$ was $\displaystyle \dfrac{\pi}{4}$

$\displaystyle 2x = \dfrac{\pi}{4}$

$\displaystyle \dfrac{\pi}{8}$

Now Plug in $\displaystyle x$ into the original function to get $\displaystyle y$ values (you would be plotting 2 points).

Pretty much everything else I understand. At least enough to solve the problem, but maybe not on a deeper level yet.

 

[Deleted member 4993]

$\displaystyle y = 3\tan2x$

If your objective is to graph the function - you are going at in wrong way.

You should choose the independent variable (x) - and calculate the dependent variable.

so you should choose x = 0, π/12, π/8, π/6, .... etc. and calculate y - make table and plot it

$\displaystyle A$ (amplitude) $\displaystyle = |3| = 3$

Find $\displaystyle x$ values:

$\displaystyle y = 3\tan2x$

$\displaystyle -3 = 3\tan2x$ and $\displaystyle 3 = 3\tan 2x$

Now, looking at the one on the left:

$\displaystyle -3 = 3\tan2x$

$\displaystyle -1 = \tan2x$ How did this line become the next line?

By simple arithmetic - divide both sides by 3

:idea: Now i did notice that the $\displaystyle \arctan$ of $\displaystyle -1$ was $\displaystyle -\dfrac{\pi}{4}$

$\displaystyle 2x = -\dfrac{\pi}{4}$

$\displaystyle -\dfrac{\pi}{8}$

Moving on to the other one:

$\displaystyle 3 = 3\tan2x$

$\displaystyle 1 = \tan2x$ How did this line become the next line?

Again by simple arithmetic - divide both sides by 3

:idea: I did notice that the $\displaystyle \arctan$ of $\displaystyle 1$ was $\displaystyle \dfrac{\pi}{4}$

$\displaystyle 2x = \dfrac{\pi}{4}$

$\displaystyle \dfrac{\pi}{8}$

Now Plug in $\displaystyle x$ into the original function to get $\displaystyle y$ values (you would be plotting 2 points).

Pretty much everything else I understand. At least enough to solve the problem, but maybe not on a deeper level yet.

.

 

[srmichael]

$\displaystyle y = 3\tan2x$

$\displaystyle A$ (amplitude) $\displaystyle = |3| = 3$

Find $\displaystyle x$ values:

$\displaystyle y = 3\tan2x$

$\displaystyle -3 = 3\tan2x$ and $\displaystyle 3 = 3\tan 2x$

Now, looking at the one on the left:

$\displaystyle -3 = 3\tan2x$

$\displaystyle -1 = \tan2x$ How did this line become the next line? :idea: Now i did notice that the $\displaystyle \arctan$ of $\displaystyle -1$ was $\displaystyle -\dfrac{\pi}{4}$

$\displaystyle 2x = -\dfrac{\pi}{4}$

$\displaystyle -\dfrac{\pi}{8}$

Moving on to the other one:

$\displaystyle 3 = 3\tan2x$

$\displaystyle 1 = \tan2x$ How did this line become the next line? :idea: I did notice that the $\displaystyle \arctan$ of $\displaystyle 1$ was $\displaystyle \dfrac{\pi}{4}$

$\displaystyle 2x = \dfrac{\pi}{4}$

$\displaystyle \dfrac{\pi}{8}$

Now Plug in $\displaystyle x$ into the original function to get $\displaystyle y$ values (you would be plotting 2 points).

Pretty much everything else I understand. At least enough to solve the problem, but maybe not on a deeper level yet.

I'm not sure I follow all that you said, but when it comes to graphing tan functions there are a few key things to be aware of that will make the graphing fairly easy.

For this function, first the new period is π/2, do you see that?

Now, lets find an asymptote, any asymptote. A quick way of doing this is to set the angle = π/2, since π/2 is an asymptote for tangent. Thus for this problem, 2x = π/2 and so x = π/4 and is an asymptote. Now, from π/4 go to the left and right in steps of π/2 which is the new period and all of these values will also be asymptotes. π/2 to the right of π/4 bringhs you to the next asymptote at 3π/4.

Then, the middle value between every asymptotes is where the graph will cross the x-axis (provided there is no vertical shift, in which case it crosses the new midline). Thus the angle bewteen π/4 and 3π/4, two of the asymptotes, is where the graph will cross the x-axis. In this case, π/2 is midway between these two points.

Now, the angle midway between the x intercept and the asymptotes is the angle that when you plug into tan2x will give you 1 or -1 depending on the angle and thus since your amplitude is 3, you go up to the value of 3 and down to -3. So between π/2 and 3π/4 is the angle 5π/8. At this x value your y value is 3. Similarly, at 3π/8 your y value is -3.

You now have 3 points between the two asymptotes which you can now connect with the familiar elongated S-looking graph that tangent delivers.

Hope that helps.

 

[Jason76]

Sorry guys

$\displaystyle 1 = \tan2x$ How did this line become the next line? :idea: I did notice that the $\displaystyle \arctan$ of $\displaystyle 1$ was $\displaystyle \dfrac{\pi}{4}$

$\displaystyle 2x = \dfrac{\pi}{4}$

Still don't get it. You said divide both sides in the top line by $\displaystyle 3$ to yield the answer in the 2nd line. Is that right? Where did the $\displaystyle 3$ come from?

 

[Deleted member 4993]

Sorry guys



Still don't get it. You said divide both sides in the top line by $\displaystyle 3$ to yield the answer in the 2nd line. Is that right? Where did the $\displaystyle 3$ come from?

If you are given

a * x = b

and asked to solve for 'x' - what would you do?

 

[Jason76]

I don't know what the teacher at school was trying to explain. I think the best way is finding the $\displaystyle x$ intercept (it has to cross somewhere). Next, find two x values, one between the $\displaystyle x$ intercept and a period, and the other between the $\displaystyle x$ intercept and a period. Next, respectively plug those $\displaystyle x$ values into the function.

Here is a graphic representation |period| |x value|| $\displaystyle x$ intercept | | x value||period|

Period of $\displaystyle \tan = \dfrac{\pi}{|b|}$

 

[srmichael]

I don't know what the teacher at school was trying to explain. I think the best way is finding the $\displaystyle x$ intercept (it has to cross somewhere). Next, find two x values, one between the $\displaystyle x$ intercept and a period, and the other between the $\displaystyle x$ intercept and a period. Next, respectively plug those $\displaystyle x$ values into the function.

Here is a graphic representation |period| |x value|| $\displaystyle x$ intercept | | x value||period|

Period of $\displaystyle \tan = \dfrac{\pi}{|b|}$

This was the crux of what I said in my earlier post.

 

[Deleted member 4993]

I don't know what the teacher at school was trying to explain. I think the best way is finding the $\displaystyle x$ intercept (it has to cross somewhere). Next, find two x values, one between the $\displaystyle x$ intercept and a period, and the other between the $\displaystyle x$ intercept and a period. Next, respectively plug those $\displaystyle x$ values into the function.

Here is a graphic representation |period| |x value|| $\displaystyle x$ intercept | | x value||period|

Period of $\displaystyle \tan = \dfrac{\pi}{|b|}$

Not necessarily true: For example-

y = cos(x) + 2

It is a very dangerous idea in mathematics to "set up rules" - without thinking thoroughly....

 

[mmm4444bot]

3 = 3 tan(2x)

1 = tan(2x)

Still don't get it. You said divide both sides in the top line by $\displaystyle 3$ to yield the answer in the 2nd line.

Where did the $\displaystyle 3$ come from?

That 3 comes from pre-algebra.

 

[Jason76]

In regards to the tangent function, the asymptotes are equidistant from the x intercept within the frame of the period (the period giving boundry lines). The x intercept can be found by plugging $\displaystyle 0$ into the $\displaystyle y$ value. But two more points are needed for an accurate graph. Each respective point is equidistant between the x intercept and a respective asymptote.

Is this right? This is all based on equidistance within the period. So if the "equidistance thing" is wrong, then this is messed up.

Now there is also a way you can find stuff using inequalities like $\displaystyle 0 < x < \pi$ etc... But the way above is another way.

Not necessarily true: For example-

y = cos(x) + 2

It is a very dangerous idea in mathematics to "set up rules" - without thinking thoroughly....

This is in regards only to the tangent function. Of course, what I posted in the other post was too simplistic. Without the framework of the period, the equidistance doesn't work.

 

[Jason76]

POST EDITED

Here is the correct method of graphing a Tangent function (correct me if you see errors or confirm what I'm doing):

$\displaystyle y = A \tan(Bx - C) + D$

$\displaystyle y = \tan x$ on the interval $\displaystyle (-\dfrac{\pi}{2}, \dfrac{\pi}{2})$

$\displaystyle A = 1, B = 1, C = 0, D = 0$

Find the Vertical Asymptotes

$\displaystyle -\dfrac{\pi}{2} < x < \dfrac{\pi}{2}$

The two asymptotes are $\displaystyle -\dfrac{\pi}{2}$ and $\displaystyle \dfrac{\pi}{2}$

Find the x intercept

$\displaystyle 0 = \tan x$

$\displaystyle 0 = \tan (1) x$ - Note $\displaystyle 1 * x = x$

$\displaystyle 0 = \tan x$

$\displaystyle \arctan(0) = 0$

The $\displaystyle x$ intercept $\displaystyle = 0$


Find Two More Points to Make an Accurate Graph

We can find these points by looking at $\displaystyle A$ (Amplitude). The positive and negative versions of $\displaystyle A$ correspond to the y coordinates of the new points. $\displaystyle A = 1$ So our two $\displaystyle y$ coordinates are $\displaystyle -1$ and $\displaystyle 1$

Now, in order to find the x coordinates, we simply set up two respective equations for each respective $\displaystyle y$ value.

$\displaystyle -1 = tan x$

$\displaystyle -1 = \tan(1)x$ - Note: $\displaystyle 1 * x = x$

$\displaystyle -1 = \tan x$

$\displaystyle \arctan(-1) = -\dfrac{\pi}{4}$

$\displaystyle 1 = tan x$

$\displaystyle 1 = \tan(1)x$ - Note: $\displaystyle 1 * x = x$

$\displaystyle 1 = \tan x$

$\displaystyle \arctan(1) = \dfrac{\pi}{4}$

So our $\displaystyle x$ values are $\displaystyle -\dfrac{\pi}{4}$ and $\displaystyle \dfrac{\pi}{4}$

Our extra points are $\displaystyle (-\dfrac{\pi}{4}, -1)$ and $\displaystyle (\dfrac{\pi}{4}, 1)$

Connect the Points

Now you have your graph

 

[mmm4444bot]

Find the x intercept

0 is the [FONT=MathJax_Math]x[/FONT] intercept.

If you want x=0 to be "the" x-intercept, then you need to specify an interval, when making up exercises like this, Jason.

The graph of y=tan(x) has infinite x-intercepts; so, without specifying an interval, you may say that x=0 is "an" x-intercept.

Here's a graph of tan(x) over the interval [-4Pi,4Pi]. :cool:

 

[Jason76]

If you want x=0 to be "the" x-intercept, then you need to specify an interval, when making up exercises like this, Jason.

The graph of y=tan(x) has infinite x-intercepts; so, without specifying an interval, you may say that x=0 is "an" x-intercept.

Here's a graph of tan(x) over the interval [-4Pi,4Pi]. :cool:

View attachment 3035

The post has been edited. Please correct me again if wrong.

 

[mmm4444bot]

$\displaystyle y = \tan x$ on the interval $\displaystyle (-\dfrac{\pi}{2}, \dfrac{\pi}{2})$

Find the x intercept

$\displaystyle \arctan (1) = 0$ so $\displaystyle 0$ is the $\displaystyle x$ intercept.

arctan(1) does not equal 0

Did you intend to type arctan(0) = 0 ?

To find x-intercept, set y equal to 0, and solve for x...

 

[Jason76]

arctan(1) does not equal 0

Did you intend to type arctan(0) = 0 ?

To find x-intercept, set y equal to 0, and solve for x...

My mistake. I mean the $\displaystyle \arctan(0) = 0$

 

[mmm4444bot]

POST EDITED

$\displaystyle y = \tan x$ on the interval $\displaystyle (-\dfrac{\pi}{2}, \dfrac{\pi}{2})$

Ah, you have now restricted the domain of tan(x), for your "work in progress", so as to trap a single x-intercept. That's a very interesting choice, to employ an open interval versus the closed option. Was this choice because your project is designed for an audience who has no idea what the graph of tan(x) might look like?

I'm not willing to participate in this project, unless you keep me in the loop. :cool:

 

[Jason76]

Ah, you have now restricted the domain of tan(x), for your "work in progress", so as to trap a single x-intercept. That's a very interesting choice, to employ an open interval versus the closed option. Was this choice because your project is designed for an audience who has no idea what the graph of tan(x) might look like?

I'm not willing to participate in this project, unless you keep me in the loop. :cool:

Well, from I've studied at school, it's hard to graph a tangent function unless you trap one $\displaystyle x$ intercept. Is there another way to graph a tangent function?

Also, of course I had to use an "open option" because there are vertical asymptotes, or did I get something wrong? Asymptotes don't land on a value, only approach it.

 

[Jason76]

Actually though, you don't need $\displaystyle \arctan$ to solve the problem. You simply solve for the asymptotes first. Next you find the x intercept by finding the midway point between the two asymptotes. Later you find the two extra points by find the respective midway point between the $\displaystyle x$ intercept and the respective asymptote.

It's all about "dividing by $\displaystyle 2$" and can often be much simpler than doing a lot of algebra (which is the case in more complex functions). But whether you do the $\displaystyle \arctan$ or the "Midpoint" way, you get the same answer. In my class, we do the "Midpoint" way.

For instance,

Given: $\displaystyle y = \tan x$ and

Asymptotes of $\displaystyle \pm \dfrac{\pi}{2}$

Find the x intercept.

Halfway between $\displaystyle \pm \dfrac{\pi}{2}$ is $\displaystyle 0$. So $\displaystyle 0$ is the $\displaystyle x$ intercept. In other words $\displaystyle \dfrac{-pi/2 + \pi/2}{2} = 0$

Find the extra points.

First look at the amplitude $\displaystyle A$. Since it's $\displaystyle 1$ We know that the respective $\displaystyle y$ coordinates will be $\displaystyle -1$ and $\displaystyle 1$

Now we can find the respective $\displaystyle x$ coordinates by dividing the intervals between the $\displaystyle x$ intercept and a respective asymptote:

$\displaystyle \dfrac{-\pi/2}{2} = -\dfrac{\pi}{4}$ and $\displaystyle \dfrac{\pi/2}{2} = \dfrac{\pi}{4}$

So now we found the extra points to be: $\displaystyle ( -\dfrac{\pi}{4}, -1)$ and $\displaystyle ( \dfrac{\pi}{4}, 1)$

Next, just connect the dots and graph.

 

[mmm4444bot]

Actually though, you don't need $\displaystyle \arctan$ to [graph y=tan(x)]. You simply solve for the asymptotes first. Next you find the x intercept by finding the midway point between the two asymptotes.

Okay -- I'm assuming (because you did not answer my last question) that your instructions are for somebody who does not know what the graph of tan(x) looks like.

For a person who does not know what the graph of tan(x) looks like, what process exactly do you have in mind for them to simply solve for the asymptotes first? In other words, how do you envision a person going about that?

 

[mmm4444bot]

I had to use an "open option" because there are vertical asymptotes

What do you think would happen to the graph, were you to graph tan(x) on the interval [-Pi/2,Pi/2] instead of (-Pi/2,Pi/2)?