Softballgirl372015
  • Softballgirl372015
Please explain how to graph this function...
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Softballgirl372015
  • Softballgirl372015
\[y=-1+\cos (x-\frac{ \pi }{ 4 })\]
Softballgirl372015
  • Softballgirl372015
I need help with the phase shifting. I am confused on that part.
whpalmer4
  • whpalmer4
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.

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whpalmer4
  • whpalmer4
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
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Softballgirl372015
  • Softballgirl372015
Okay, I understand that. How would we apply that concept to phase shifting?
whpalmer4
  • whpalmer4
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 \(\pi/6\): \[\left\{0,\frac{1}{2},\frac{\sqrt{3}}{2},1,\frac{\sqrt{3}}{2},\frac{1}{2},0,-\frac{1}{2},-\frac{\sqrt{3}}{2},-1,-\frac{\sqrt{3}}{2},-\frac{1}{2},0\right\}\]
whpalmer4
  • whpalmer4
Now suppose we subtract \(\pi/6\) from the argument to the \(\sin x\) function, giving us \(\sin (x-\pi/6)\): \[\left\{-\frac{1}{2},0,\frac{1}{2},\frac{\sqrt{3}}{2},1,\frac{\sqrt{3}}{2},\frac{1}{2},0,-\frac{1}{2},-\frac{\sqrt{3}}{2},-1,-\frac{\sqrt{3}}{2},-\frac{1}{2}\right\}\]
Softballgirl372015
  • Softballgirl372015
I sorta follow this, but I don't understand how it relates.
whpalmer4
  • whpalmer4
For completeness, here's \(\sin(x+\pi/6)\): \[\left\{\frac{1}{2},\frac{\sqrt{3}}{2},1,\frac{\sqrt{3}}{2},\frac{1}{2},0,-\frac{1}{2},-\frac{\sqrt{3}}{2},-1,-\frac{\sqrt{3}}{2},-\frac{1}{2},0,\frac{1}{2}\right\}\]
whpalmer4
  • whpalmer4
It's the same sequence of numbers, just a different starting point in the list.
Softballgirl372015
  • Softballgirl372015
So to solve the problem, after you shift the graph pi/4 do you kinda ignore it and continue to graph the function?
whpalmer4
  • whpalmer4
Maybe a picture will help: The blue curve is \(y = \sin x\). The other two curves are \(y = \sin(x\pm\pi/6)\)
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whpalmer4
  • whpalmer4
Perhaps the most interesting phase shift is \(y = \sin (x+\pi/2)\). I'll graph it vs. \(y = \sin x\)
whpalmer4
  • whpalmer4
Now I'm going to graph \(y=\sin x\) vs. \(y=\cos x\):
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whpalmer4
  • whpalmer4
Notice anything about the last two graphs?
Softballgirl372015
  • Softballgirl372015
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?
whpalmer4
  • whpalmer4
The last two graphs are identical! The graph of sin x + pi/2 phase shift is the same as the graph of cos x!
whpalmer4
  • whpalmer4
So, what was your question again? ;-)
Softballgirl372015
  • Softballgirl372015
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.
whpalmer4
  • whpalmer4
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?
Softballgirl372015
  • Softballgirl372015
The question was how do I graph \[y=-1+\cos (x-\frac{ \pi }{ 4 })\]
Softballgirl372015
  • Softballgirl372015
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 \[y=-1+\cos x\]
whpalmer4
  • whpalmer4
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.
whpalmer4
  • whpalmer4
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 \(x=-\pi/4\)
whpalmer4
  • whpalmer4
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 \(\cos (x-\pi/4) = \cos(\pi/4-\pi/4) = \cos(0) = 1\)
whpalmer4
  • whpalmer4
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...
Softballgirl372015
  • Softballgirl372015
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
whpalmer4
  • whpalmer4
I live for the "hey, I do understand this!" moments :-)
Softballgirl372015
  • Softballgirl372015
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. :)
whpalmer4
  • whpalmer4
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.
Softballgirl372015
  • Softballgirl372015
I totally agree!! But I think that you know everything about sine and cosine graphs! ;)
whpalmer4
  • whpalmer4
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 :-)
Softballgirl372015
  • Softballgirl372015
I feel like sometimes you learn best that way! What interactive widget did you use to make those graphs?
whpalmer4
  • whpalmer4
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)...
whpalmer4
  • whpalmer4
It's sort of the ultimate math geek Swiss Army knife :-)
whpalmer4
  • whpalmer4
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.
whpalmer4
  • whpalmer4
The mobile apps are listed here: http://products.wolframalpha.com/mobile/
Softballgirl372015
  • Softballgirl372015
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!!
whpalmer4
  • whpalmer4
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...
whpalmer4
  • whpalmer4
Okay, I guess I can let you go for now. Meet back here in 4 hours for some more? ;-)
Softballgirl372015
  • Softballgirl372015
Haha sounds like a plan! Lol ;)

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