Graphing sinusoidal tangent functions

[opus]

Homework Statement

Graph $y=tan\left(x-\frac {π}{4}\right)$

Homework Equations

N/A

The Attempt at a Solution

To graph a tangent function, I first find the vertical asymptotes to set the boundaries for the graph:
To do so, set what's inside the parentheses equal to $\frac π 2$ and $-\frac π 2$ and solve. This gives me the vertical asymptotes of $\frac {3π}{4}$ and $-\frac π 4$.

Next, I divide the x-axis into 4 equal intervals:
Period= $\frac π b$ = $\frac π 1$ = π, since the b value of the function is 1.

Next, I evaluate the function at the three found x-values:
$y=tan\left(\frac π 4 - \frac π 4\right)$ = 0
$y=tan\left(\frac π 2 - \frac π 4\right)$ = $tan\left(\frac π 4\right)=1$
$y=tan\left(\frac {3π}{4} - \frac π 4\right)$ = $tan\left(\frac π 2\right)$ = UNDEFINED

Now here is my trouble- In looking at the original graph, the tangent function is shifted to the right by $\frac π 4$ units. This should also mean that the vertical asymptotes shift as well, while keeping their same interval value. So the graph of the given function is not undefined at $\frac π 2$ like a basic tangent function graph. However, I still need another acceptable x value so that I can draw the graph. But my third value is undefined.
Any ideas on where I'm going wrong?

 

[Alloymouse]

Since it's a phase shift (horizontal shift), all you need to do is shift the entirety of the graph by pi/4 in the positive x direction. Just plot it like a normal tan(x) graph, but play with the x-axis :)

 

[SammyS]

Homework Statement

Graph $y=tan\left(x-\frac {π}{4}\right)$

Homework Equations

N/A

The Attempt at a Solution

To graph a tangent function, I first find the vertical asymptotes to set the boundaries for the graph:
To do so, set what's inside the parentheses equal to $\frac π 2$ and $-\frac π 2$ and solve. This gives me the vertical asymptotes of $\frac {3π}{4}$ and $-\frac π 4$.

Next, I divide the x-axis into 4 equal intervals:
Period= $\frac π b$ = $\frac π 1$ = π, since the b value of the function is 1.

Next, I evaluate the function at the three found x-values:
$y=tan\left(\frac π 4 - \frac π 4\right)$ = 0
$y=tan\left(\frac π 2 - \frac π 4\right)$ = $tan\left(\frac π 4\right)=1$
$y=tan\left(\frac {3π}{4} - \frac π 4\right)$ = $tan\left(\frac π 2\right)$ = UNDEFINED

Now here is my trouble- In looking at the original graph, the tangent function is shifted to the right by $\frac π 4$ units. This should also mean that the vertical asymptotes shift as well, while keeping their same interval value. So the graph of the given function is not undefined at $\frac π 2$ like a basic tangent function graph. However, I still need another acceptable x value so that I can draw the graph. But my third value is undefined.
Any ideas on where I'm going wrong?

What is the value of $\ \tan (- \frac \pi 4 ) \,?\$

 

[opus]

Since it's a phase shift (horizontal shift), all you need to do is shift the entirety of the graph by pi/4 in the positive x direction. Just plot it like a normal tan(x) graph, but play with the x-axis :)

This is what I was messing around with visually, but I got a little confused with the actual values to be plotted.

What is the value of $\ \tan (- \frac \pi 4 ) \,?\$

Tangent is an odd function, so $tan\left(-x\right)$ = $~-tan\left(x\right)$
So $tan\left(-\frac π 4\right)$ = $~-tan\left(\frac π 4\right)$ = -1
I'm not putting together where this should fit in. Are we talking about the given function? Or a value down the line?

 

[Alloymouse]

Since you know where the vertical asymptotes are, just draw the tan(x) shape between those asymptotes and you'll be fine :)

Essentially your plot must reflect the values of the function tan(x), but all the x values will be +pi/4 more since that is what you must do to achieve the original tan(x) function

 

[opus]

Since you know where the vertical asymptotes are, just draw the tan(x) shape between those asymptotes and you'll be fine :)

Essentially your plot must reflect the values of the function tan(x), but all the x values will be +pi/4 more since that is what you must do to achieve the original tan(x) function

Ok that makes sense. Y values stay the same, x values shifted $\frac π 4$ units to the right, and the graph looks the same. But how would this be represented algebraicly? I get what the graph looks like, but I don't know why I'm having the hang up of an undefined x value that shouldn't be undefined due to the asymptote shift.

 

[Alloymouse]

Ok that makes sense. Y values stay the same, x values shifted $\frac π 4$ units to the right, and the graph looks the same. But how would this be represented algebraicly? I get what the graph looks like, but I don't know why I'm having the hang up of an undefined x value that shouldn't be undefined due to the asymptote shift.

Which value specifically?

Edit: looking at your post, perhaps you could work with a few more data points beyond what you have used

 

[opus]

My third x value is $y=tan\left(\frac {3π}{4} - \frac {π}{4}\right)$ which equals $tan \frac π 2$ which is undefined.

 

[opus]

But the graph shifted to the right, so shouldn't the vertical asymptote of the shifted graph also be shifted to the right with the graph? That is, the asymptote for $tan\left(x\right)$ is $\frac π 2$, and the asymptote shifts with $tan\left(x- \frac π 4\right)$ to $\frac {3π}{4}$

 

[Alloymouse]

But the graph shifted to the right, so shouldn't the vertical asymptote of the shifted graph also be shifted to the right with the graph?

Yes, and tan(pi/2) is supposed to be undefined. And yes you are right, the asymptotes shift. You are staring at the correct plot.

 

[opus]

I still don't understand. Where do I get my third x value from if, by equal x-axis interval, it has to be $\frac π 2$ but this value is undefined? I see that the graph should look exactly the same as $y=tan\left(x\right)$, but shifted to the right. I also understand that the asymptotes shift with the graph. What I don't understand is that the graph now crosses $\frac π 2$, which is okay because the asymptotes shifted. But I still need a third x value to graph. When I attempt to find this third x value, it is undefined as it's $\frac π 2$, but graphically, $\frac π 2$ is okay because the asymptotes shifted. So what is my third x value?

 

[Alloymouse]

I still don't understand. Where do I get my third x value from if, by equal x-axis interval, it has to be $\frac π 2$ but this value is undefined? I see that the graph should look exactly the same as $y=tan\left(x\right)$, but shifted to the right. I also understand that the asymptotes shift with the graph. What I don't understand is that the graph now crosses $\frac π 2$, which is okay because the asymptotes shifted. But I still need a third x value to graph. When I attempt to find this third x value, it is undefined as it's $\frac π 2$, but graphically, $\frac π 2$ is okay because the asymptotes shifted. So what is my third x value?

You can take it to be positive infinity, and just draw the curve to tend towards positive infinity.

If you are just plotting points, then I recommend using smaller intervals to show the trend (why not intervals of pi/8?)

 

[Mark44]

But the graph shifted to the right, so shouldn't the vertical asymptote of the shifted graph also be shifted to the right with the graph? That is, the asymptote for $tan\left(x\right)$ is $\frac π 2$, and the asymptote shifts with $tan\left(x- \frac π 4\right)$ to $\frac {3π}{4}$

As x approaches $\pi/2$ from the left, $\tan(x)$ gets larger and larger. In terms of limits, which you might not have seen yet, this is expressed as $\lim_{x \to \pi/2^-}\tan(x) = +\infty$. (On the other side of this asymptote, the graph grows more and more negative.) Likewise, for the shifted graph, $\lim_{x \to \pi/2 + \pi/4^-}\tan(x - \pi/4) = +\infty$. Everything on the graph of $y = \tan(x)$ is shifted to the right by $\pi/4$, including the asymptotes.

BTW, the term "sinusoidal" refers to graphs that oscillate like the sine and cosine functions. The tangent function isn't sinusoidal.

 

[opus]

I understand that portion (I get the idea with the limits, as the graph gets closer and closer to the asymptote but never touches, but I haven't properly gone over them yet), but I'm just hung up on this third x coordinate thing. The coordinate itself is $tan\frac π 2$, which is always undefined, as the cosine in the denominator is equal to zero. So my third coordinate to draw my graph is undefined, even though my graph goes over it when shifted.

 

[Ray Vickson]

I understand that portion (I get the idea with the limits, as the graph gets closer and closer to the asymptote but never touches, but I haven't properly gone over them yet), but I'm just hung up on this third x coordinate thing. The coordinate itself is $tan\frac π 2$, which is always undefined, as the cosine in the denominator is equal to zero. So my third coordinate to draw my graph is undefined, even though my graph goes over it when shifted.

I don't get the source of your difficulty. Just make a small table of $\tan y$ values for some typical value of $y$, such as $y=0, \pm \pi/6, \pm \pi/4, \pm \pi/3$. Now set $y = x - \pi/4$ (that is, $x = y + \pi/4$) and so get a table of some $x$ vs. $\tan(x- \pi/4)$ values. None of those values will be infinite.

 

[SammyS]

From Post #4:

Tangent is an odd function, so $tan\left(-x\right)$ = $~-tan\left(x\right)$
So $tan\left(-\frac π 4\right)$ = $~-tan\left(\frac π 4\right)$ = -1
I'm not putting together where this should fit in. Are we talking about the given function? Or a value down the line?

Whatever works here.

Homework Statement

Graph $y=tan\left(x-\frac {π}{4}\right)$

What must the value of x be so that your result has you evaluating $\ \tan (- \frac \pi 4 ) \,?\$

 

[Mark44]

but I'm just hung up on this third x coordinate thing. The coordinate itself is $tan\frac π 2$, which is always undefined

Then just choose a different 3rd coordinate. Certainly $\tan(\frac \pi 2)$ is undefined, but $\tan(\frac \pi 2 - 0.001)$ isn't, nor is $\tan(\frac \pi 2 + 0.001)$. The first is a very large positive number, and the second is very negative.

 

[opus]

I think I figured out what I was doing wrong. For the graph, I was starting at 0, instead of starting at the left-most vertical asymptote. This resulted in my using the right-most asymptote, $\frac {3π}{4}$, as an x value to plot the graph.