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We have 2 sinusoidal functions:

$f(x)= 0,30+2\sin (3x-\frac{1}{2}\pi$) and $g(x)=0,30+2\sin(3(x-\frac{1}{2}\pi))$.

What would happen to both functions if you performed a translation of let's say $(3\pi;0,10)$?

What would happen if you multiply with reference to the y-axis (I believe in English it's called a horizontal stretch) of f(x) and g(x) with $\frac{1}{5}$?

I have to do this in a test in 2 hours but nowhere in the book do they explain transformations of already 'complete' sinusoids (of the form a+sinb (c(x-d)) ) , I understand transformations, however this is the last thing I must know, you guys on mstackexchange have helped me a lot, since I've missed every single lessson this past month I had you guys as my 'teachers' and I'm sorry for some of my boring questions.. this is the last one, after that my questions will be better , I promiss.

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What do you mean by "what would happen to a function"? What do you mean by "a translation of $(3\pi;0,10)$"? –  Gerry Myerson Sep 18 '12 at 6:13
    
What would happen basically means what would the function be after the transformation. A translation of $(3\pi;0,10)$ to a function $y=sinx$ creates $y=0,10+sin(x-3\pi)$ I hope you now understand what I mean. –  JohnPhteven Sep 18 '12 at 6:22

1 Answer 1

up vote 1 down vote accepted

I'm going to write $.3$ instead of $0,30$ as I'm more comfortable with that notation.

Translating $f(x)=.3+2\sin(3x-(\pi/2))$ by $(3\pi,.1)$, following the example given in the comments, gives you $$f(x)=.1+.3+2\sin(3(x-3\pi)-(\pi/2))=.4+2\sin(3x-(19\pi/2))$$ which you could also write as $$f(x)=.4+2\sin(3x-(3\pi/2))$$ Can you do $g(x)$ now?

I'm not sure what's meant by "multiply with reference to the $y$-axis" but a plausible interpretation yields $$(1/5)(.3+2\sin(3x-(\pi/2)))$$ which you might write as $$.06+(.4)\sin(3x-(\pi/2))$$

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Thank you, I was mainly concered by translations upon already translated functions. With multiply with reference to the y-axis, that was my fault, you understood it wrongly, but because I tried to translate it from dutch to english, but that didn't work out. But that was not my main concern, and I think you, I believe you have answered multiple questions of mine! –  JohnPhteven Sep 18 '12 at 7:11

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