Sinusoidal Functions: describe transformations, sketch graph

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Homework Help Overview

The discussion revolves around the transformations of sinusoidal functions, specifically focusing on the graph of a sine function with given amplitude, period, horizontal shift, and vertical shift. Participants are attempting to sketch the graph based on these transformations and seek feedback on their accuracy.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants share their attempts at sketching the graph and inquire about its correctness. There are discussions about the effects of transformations, such as amplitude and shifts, on the sine function. Some participants question the placement of specific points on the graph, particularly at 90 degrees.

Discussion Status

The discussion has seen multiple attempts at graphing the function, with some participants providing corrections and suggestions for improvement. There is a back-and-forth regarding the correct transformations and the impact of negative signs on the graph. While some participants express confidence in their corrections, others continue to seek clarity on specific aspects of the graphing process.

Contextual Notes

Participants are working under the constraints of homework guidelines, which may limit the amount of direct assistance they can provide to one another. There is an emphasis on understanding the transformations rather than simply providing the correct graph.

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   Reactions: 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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