SOLUTION: I recently learned that in order to graph any trig function you need to factor the period out of the phase shift. Can someone explain why? I think this has something to do with the

Algebra ->  Trigonometry-basics -> SOLUTION: I recently learned that in order to graph any trig function you need to factor the period out of the phase shift. Can someone explain why? I think this has something to do with the      Log On


   

Question 1132220: I recently learned that in order to graph any trig function you need to factor the period out of the phase shift. Can someone explain why? I think this has something to do with the multiple x intercepts but I am not sure.
Thanks!
Found 2 solutions by Alan3354, greenestamps:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
I recently learned that in order to graph any trig function you need to factor the period out of the phase shift. Can someone explain why?
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That's not true.

Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


You aren't saying exactly what you mean; but it is clear what you mean.

In general, if you have a parent function y = f(x), then the graph of y = f(x-h) is shifted h units to the right, because h is the value of x that makes x-h = 0.

With trig functions, to find the phase shift, you need to find how far the graph is shifted right or left.

Given a trig function like

sin%283x-pi%29

the phase shift is the value of x that makes (3x-pi) = 0:

3x-pi+=+0 --> 3x+=+pi --> x+=+pi%2F3

The phase shift is pi/3.

In practice, we usually find that phase shift by factoring out the coefficient of x in the function definition:

sin%283x-pi%29 --> sin%283%28x-pi%2F3%29%29

This shows that the phase shift is pi/3. That is, the graph is shifted pi/3 units to the right, in exactly the same way that y+=+%28x-2%29%5E2 is the graph of y+=+x%5E2 shifted 2 units to the right.