Question

Please explain how to graph this function...

Answer 1

\[y=-1+\cos (x-\frac{ \pi }{ 4 })\]

Answer 2

I need help with the phase shifting. I am confused on that part.

Answer 3

The phase shifting is really the same concept as translating a function to the left or right along the x-axis by adding or subtracting from the argument.

Answer 4

Think of a simple parabola \(y = x^2\). If we want to shift it 1 unit to the right, so that a feature that takes place at \(x = x_0\) on the original takes place at \(x=x_0+1\) on the translated function, we just subtract 1 from \(x\) before we square it: \(y = (x-1)^2\)

If we make a little table of x, x-1, x^2, (x-1)^2 it is easy to see this:

-2 -3 4 9
-1 -2 1 4
0 -1 0 1
1 0 1 0
2 1 4 1
3 2 9 4

Answer 5

Okay, I understand that. How would we apply that concept to phase shifting?

Answer 6

The same thing happens with phase shifts on periodic functions. Take the sin function:

Here are the values of \(\sin x\) from 0 to \(2\pi\), at intervals of \(\pi/6\):
\[\left\{0,\frac{1}{2},\frac{\sqrt{3}}{2},1,\frac{\sqrt{3}}{2},\frac{1}{2},0,-\frac{1}{2},-\frac{\sqrt{3}}{2},-1,-\frac{\sqrt{3}}{2},-\frac{1}{2},0\right\}\]

Answer 7

Now suppose we subtract \(\pi/6\) from the argument to the \(\sin x\) function, giving us \(\sin (x-\pi/6)\):

\[\left\{-\frac{1}{2},0,\frac{1}{2},\frac{\sqrt{3}}{2},1,\frac{\sqrt{3}}{2},\frac{1}{2},0,-\frac{1}{2},-\frac{\sqrt{3}}{2},-1,-\frac{\sqrt{3}}{2},-\frac{1}{2}\right\}\]

Answer 8

I sorta follow this, but I don't understand how it relates.

Answer 9

For completeness, here's \(\sin(x+\pi/6)\):
\[\left\{\frac{1}{2},\frac{\sqrt{3}}{2},1,\frac{\sqrt{3}}{2},\frac{1}{2},0,-\frac{1}{2},-\frac{\sqrt{3}}{2},-1,-\frac{\sqrt{3}}{2},-\frac{1}{2},0,\frac{1}{2}\right\}\]

Answer 10

It's the same sequence of numbers, just a different starting point in the list.

Answer 11

So to solve the problem, after you shift the graph pi/4 do you kinda ignore it and continue to graph the function?

Answer 12

Maybe a picture will help:

The blue curve is \(y = \sin x\). The other two curves are \(y = \sin(x\pm\pi/6)\)

Answer 13

Perhaps the most interesting phase shift is \(y = \sin (x+\pi/2)\). I'll graph it vs. \(y = \sin x\)

Answer 14

Now I'm going to graph \(y=\sin x\) vs. \(y=\cos x\):

Answer 15

Notice anything about the last two graphs?

Answer 16

Yea, that they stayed in the same position, except that they moved either to the left or to the right. I assume that the same thing will happen to the periodic function right?

Answer 17

The last two graphs are identical! The graph of sin x + pi/2 phase shift is the same as the graph of cos x!

Answer 18

So, what was your question again? ;-)

Answer 19

oh! I thought you were asking about the previous graphs that you posted. I see what you mean that the sin and cos grpahs are identical.

Answer 20

Periodic functions are repeating, over and over. Think of it like a continuous loop being stenciled along the axis. The phase shift just controls where in the loop you started. Does that make sense?

Answer 21

The question was how do I graph \[y=-1+\cos (x-\frac{ \pi }{ 4 })\]

Answer 22

Yea, that makes sense to me. Since the position of the loop is being changed, will the maximum and minimum values be different from the equation I posted in the previous post and \[y=-1+\cos x\]

Answer 23

Right. So you know how to graph \(y = \cos x\), I'm sure. Graphing \(y = -1 + \cos(x)\) is just about the same thing, except shifting every point up the y-axis by 1. If you'd printed a graph of \(y = \cos x\) on a piece of transparency film, you could overlay it on your paper and shift it up by 1, and that's what it would look like.

Answer 24

The phase shift is just how you align the graph along the x-axis. Because you have a negative phase shift, instead of your \(y = \cos x\) starting at \(x=0\) (where \(\cos x = 1\)), it is as if you had aligned the starting point at \(x=-\pi/4\)

Answer 25

Still exactly the same shape, just shifted to the left. You won't get the maximum value until you get to \(x = \pi/4\), where \(\cos (x-\pi/4) = \cos(\pi/4-\pi/4) = \cos(0) = 1\)

Answer 26

If you're in the habit of making a table of values to draw your graph, as I did with the parabola example, a negative phase shift is like taking the value of y from an earlier row in the table (assuming the table has -x at the top and +x at the bottom), just like it was for our parabola shifting along the x-axis...

Answer 27

Okay. I REALLY understand it now. You explained it soo much better than my teacher could have ever done! I really appreciate you helping me!! Thanks soo much!! You're a life saver! :D

Answer 28

I live for the "hey, I do understand this!" moments :-)

Answer 29

Haha!! It really helps to better understand what you are doing when you have the background info on it. Ya know what I mean? Lol. :)

Answer 30

Another thing that helps with understanding is attempting to explain the material to someone else. You quickly get an idea of what you do and don't understand, and I find it really does help build your own understanding.

Answer 31

I totally agree!! But I think that you know everything about sine and cosine graphs! ;)

Answer 32

Some of these concepts (especially with periodic functions) are very well-suited to tinkering with some sort of interactive widget that shows you exactly what happens when you fiddle with the parameters. Playing leads directly to understanding if it's the right kind of playing :-)

Answer 33

I feel like sometimes you learn best that way! What interactive widget did you use to make those graphs?

Answer 34

I often did poorly in class situations because whenever the instructor said something interesting, I wanted to go and experiment with it right then...

I did the graphs with a program called Mathematica. It's perhaps like swatting a mosquito with a sledgehammer (and priced accordingly)...

Answer 35

It's sort of the ultimate math geek Swiss Army knife :-)

Answer 36

WolframAlpha.com (made by the same people) is free, although you can't save the graphs unless you sign up for a subscription (a couple of bucks a month, I think). They also have apps for iPhone/iPad, probably Android tablets as well, that are only a couple of bucks and aimed at typical courses.

Answer 37

The mobile apps are listed here: http://products.wolframalpha.com/mobile/

Answer 38

Okay, that sounds interesting. i will have to go and check them out. But I think that I am gonna go to bed now. Lol. Its been enough math for one day. I've been at this for like 2 and a half hours!!

Answer 39

To the extent that one can spend time actually investigating new concepts instead of plowing through the drudgery of algebraic manipulations, I think computer-assisted math is a great development...

Answer 40

Okay, I guess I can let you go for now. Meet back here in 4 hours for some more? ;-)

Answer 41

Haha sounds like a plan! Lol ;)