Sinusoidal Functions: describe transformations, sketch graph

AI Thread Summary
The discussion focuses on the transformations of the sinusoidal function and the accuracy of the graph sketch. The original attempt included incorrect elements, particularly a negative sign in front of the sine function. Participants guided the user to start from the base function and apply transformations step by step. After several revisions, the user finally produced the correct graph. The interaction highlights the importance of understanding transformations in sinusoidal functions for accurate graphing.
Evangeline101
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Homework Statement


upload_2016-8-13_1-38-49.png


Homework Equations


none

The Attempt at a Solution



-amplitude is 3
-period is 180°
-right 60°
-down 1

Rough sketch of graph:

upload_2016-8-13_1-42-40.png
I would like to know if the graph looks right, is there any improvements to be made?

Thanks :)
 
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You got it wrong. That's the graph of ##y = -3\sin(2(x-60^o))-1##.
 
wanna know why the line at 90?
 
change the sign and it is alright
 
blue_leaf77 said:
You got it wrong. That's the graph of y=−3sin(2(x−60o))−1y=−3sin⁡(2(x−60o))−1y = -3\sin(2(x-60^o))-1.

I re-did the graph:

upload_2016-8-15_2-24-35.png


Is this better?
 
Still incorrect. Start from ##f(x) = 3\sin(2(x-60^o))## which can be obtained by translating ##f(x) = 3\sin(2x)## to [left/right, it's your part to answer] by ##60^o## degrees.
 
blue_leaf77 said:
Still incorrect. Start from f(x)=3sin(2(x−60o))f(x)=3sin⁡(2(x−60o))f(x) = 3\sin(2(x-60^o)) which can be obtained by translating f(x)=3sin(2x)f(x)=3sin⁡(2x)f(x) = 3\sin(2x) to [left/right, it's your part to answer] by 60o60o60^o degrees.
Ok I have attempted the graph again:

The dotted curve is the original base function y = sinx, and I used it to help apply the transformations. The dotted lines show where the axes have been shifted.
upload_2016-8-15_20-0-46.png
Is this an improvement??
 
Nope.
Your first drawing is almost correct. It's just the negative sign in front of the sine function that needs to be removed. May be it's better to draw the functions resulting from each transformation. So, you start from ##y=\sin(2x)##, which is a harmonic function of period ##180^o## and amplitude 1. Then draw ##y=3\sin(2x)##. Followed by ##y=3\sin(2(x-60^o))## and finally ##y=3\sin(2(x-60^o))-1##. This is the only way we can spot in which step you went wrong.
 
blue_leaf77 said:
Your first drawing is almost correct. It's just the negative sign in front of the sine function that needs to be removed.

Okay, I have attempted the graph again, let's hope its an improvement at least :/

upload_2016-8-17_22-12-35.png
 
  • Like
Likes blue_leaf77
  • #10
Yes, that's the correct one.
 
  • #11
blue_leaf77 said:
Yes, that's the correct one.

Thanks for the help! I really appreciate it :biggrin:
 
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