Comparison between y=sinx and

  • Thread starter aisha
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  • #1
aisha
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Comparison between y=sinx and....

[tex] y+3=2\sin(3x+\frac{\pi}{2})[/tex]

vertical stretch by a factor of 2
horizontal compression by 1/3 no reflection in the x or y axis
30 degrees horizontal phase shift to the left
and 3 units down vertical displacement.

Can someone please check this?
 
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Answers and Replies

  • #2
whozum
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I think its shifted to the left, other than that you're ok.
 
  • #3
Aresius
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Don't you have to divide 2pi by the coefficient of (x + pi/6) [which is 3] to get the horizontal compression? It's been a while since basic functions but i'm pretty sure that's the case.

The exception to this is a tan equation where you divide pi by the coefficient instead.

Otherwise, yes it is shifted to the left. Whenever dealing with phase (x) shifts or stretch/shrink the +/- graph behaviour is reversed.
 
  • #4
BobG
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Aresius said:
Don't you have to divide 2pi by the coefficient of (x + pi/6) [which is 3] to get the horizontal compression? It's been a while since basic functions but i'm pretty sure that's the case.

The exception to this is a tan equation where you divide pi by the coefficient instead.

Otherwise, yes it is shifted to the left. Whenever dealing with phase (x) shifts or stretch/shrink the +/- graph behaviour is reversed.
[tex]2 \pi[/tex] divided by the coefficient will get you your period, yes. [tex]2 \pi[/tex] divided by 1 would give you a period of [tex]2 \pi[/tex]. [tex]2 \pi[/tex] divided by 3 would give you a period 1/3 as long, or [tex]\frac{2 \pi}{3}[/tex].

That would be what aisha meant by "compression by 1/3".

Yes the graph would shift to the left for a positive phase shift. At time zero (x=0), the angle is shifted forward for a positive phase shift (you're at 30 degrees instead of 0 degrees). That means the point where the angle (and the sine of the angle) was equal to zero had to occur prior to the time when x equaled zero (the "start" point shifted left).
 
  • #5
Aresius
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My mistake, hehe
 
  • #6
aisha
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so if this is my equation [tex] y= a sin [b(x-c)] +d [/tex]

if c is positive it shifts to the left and if negative shifts to the right?

or do you put the minus from this equation and the sign of ur question in my case (+) together and then if it is negative move to the left or positive move to the right im not sure what sign c should have.
 
  • #7
BobG
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aisha said:
so if this is my equation [tex] y= a sin [b(x-c)] +d [/tex]

if c is positive it shifts to the left and if negative shifts to the right?

or do you put the minus from this equation and the sign of ur question in my case (+) together and then if it is negative move to the left or positive move to the right im not sure what sign c should have.
Positive shifts to the left; negative to the right.
 
  • #8
aisha
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I have changed my original post is it correct now?
 
  • #9
roger
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wouldnt it be 90 degrees shift to the left ?
 
  • #10
Aresius
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No because you factor out the 3.

It looks right to me.
 

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