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lilsis76

  • 2 years ago

Determine Amplitude/period/phase shift /graph function. over 1 period. Indicate xintercepts and coordinates of Hi/low points on graph. y = cos (2x - pi/3) +1 y = A ( Bx - C ) so Amplitude is 1 Period: 2pi/B--> 2pi/2 = Pi Okay is my phase shift right? Phaseshift = C/B pi ---- 3 --------------- = pi/6? 2

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  1. satellite73
    • 2 years ago
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    looks good to me i never remember the formula, so i set \(2x-\frac{\pi}{3}=0\) and solve for \(x\)

  2. satellite73
    • 2 years ago
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    this also help me remember whether the graph has moved left or right. in this case you get \(x=\frac{\pi}{6}\) so it is shifted to the right

  3. lilsis76
    • 2 years ago
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    oh!, okay so since it was positive it goes right, right? I thought if if was negative it goes right?? i am still confused on how they shift.

  4. lilsis76
    • 2 years ago
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    |dw:1349892894075:dw|

  5. satellite73
    • 2 years ago
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    think of it this way for regular old cosine, you know \(\cos(0)=1\) now you solve \(2x-\frac{\pi}{3}=0\) and you see \(x=\frac{\pi}{6}\) which means instead of \((0,1)\) on the graph, you have \((\frac{\pi}{6},1)\) on the graph, so it is shifted right

  6. satellite73
    • 2 years ago
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    hold on, i did not say what you wrote set it equal to zero \[2x-\frac{\pi}{3}=0\] \[2x=\frac{\pi}{3}\] \[x=\frac{\pi}{6}\]

  7. satellite73
    • 2 years ago
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    i meant set \[2x-\frac{\pi}{3}=0\] not \[2x-\frac{x}{3}=0\]

  8. lilsis76
    • 2 years ago
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    haha okay, i thought = to zero

  9. satellite73
    • 2 years ago
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    whew

  10. lilsis76
    • 2 years ago
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    haha sorry about that

  11. lilsis76
    • 2 years ago
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    okay so to figure the points, i start at pi/6 and move right by pi/6?

  12. satellite73
    • 2 years ago
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    lets go slow because i do not want to confuse you

  13. satellite73
    • 2 years ago
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    you know \(\cos(0)=1\) right?

  14. lilsis76
    • 2 years ago
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    okay then. well then what i know know is Amp = 1 period 2pi/b--> 2pi/3 phase shift---> 2x-pi/3=0--> 2x=pi/3 divide both by 2--> = pi/6 Goes Right And yes Cos(0)=1 means at the point zero it goes up to 1 which is the highest point

  15. satellite73
    • 2 years ago
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    right, so in this case you have to make \(x=\frac{\pi}{6}\) in order to get \(2x-\frac{\pi}{3}=0\) so for your graph, instead of \((0,1)\) you have \((\frac{\pi}{6},1)\) on the graph

  16. satellite73
    • 2 years ago
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    all you wrote is correct, here is a nice picture http://www.wolframalpha.com/input/?i=+y+%3D+cos+%282x+-+pi%2F3%29+%2B1

  17. lilsis76
    • 2 years ago
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    OH! i forgot i have to move it up 1

  18. lilsis76
    • 2 years ago
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    okay so then i have it like this: |dw:1349893756732:dw|

  19. lilsis76
    • 2 years ago
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    then i move up 1

  20. lilsis76
    • 2 years ago
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    |dw:1349893880189:dw|

  21. lilsis76
    • 2 years ago
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    @satellite73 Like this? so amplitude is 1 period is Pi Phase Shift is pi/6 I graphed the function the x intercepts: there are not any. the lowest point is (pi/2,0), and the highest point is (pi/6,2)

  22. lilsis76
    • 2 years ago
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    |dw:1349894136637:dw|

  23. lilsis76
    • 2 years ago
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    @swissgirl @marcoduuuh checked the answers in the back of my book. How did they come up with the x intercept of 2pi/3? the hight point i got is (pi/6,2) but how do i get (7pi/6, 2)? the low point of (2pi/3, 0 ) ?

  24. phi
    • 2 years ago
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    this curve just touches the x-axis when y=0 cos (2x - pi/3) +1 = 0 cos(2x- pi/3)= -1 arc cos both sides 2x - pi/3 = pi 2x= pi+pi/3 = 4/3 pi x= 2/3 pi

  25. lilsis76
    • 2 years ago
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    how did you do that?

  26. phi
    • 2 years ago
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    you know the period is pi, so every time you increase x by pi you will get the same value as x

  27. lilsis76
    • 2 years ago
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    sorry, im new to precalculus. all this is super new to me

  28. phi
    • 2 years ago
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    the hight point i got is (pi/6,2) but how do i get (7pi/6, 2)? pi/6 + pi = pi/6 + 6pi/6 = 7pi/6

  29. phi
    • 2 years ago
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    how did you do that? are you asking how to get the lowest point at x= 2/3 pi ?

  30. lilsis76
    • 2 years ago
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    oh....okay, now is that going to be the same step for getting the HIghs on all problems like that? you get the Period and add it too the phase shift? Yes and I cant get the lowest point either, i end up getting pi/2, 0

  31. phi
    • 2 years ago
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    Let's look at How did they come up with the x intercept of 2pi/3? First, you know x-intercept means the x value where the curve touches or crosses the x-axis?

  32. lilsis76
    • 2 years ago
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    yes how did they come up with 2pi/3, because i got... pi/6 pi/3 pi/2 and wait i guess i got the 2pi/3. but okay i know and see my question. how did they get 2pi/3 an point 0, because when i graphed it by hand it shows that 2pi/3 is at 1

  33. lilsis76
    • 2 years ago
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    |dw:1349897328597:dw|

  34. lilsis76
    • 2 years ago
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    this is the graph i got, but i got pi/2 as my lowest. did i do something wrong?

  35. phi
    • 2 years ago
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    type y = cos (2x - pi/3) +1 in the google search window, it should graph it

  36. lilsis76
    • 2 years ago
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    okay im lookin at the graph

  37. lilsis76
    • 2 years ago
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    so how do i figure the lowest with that graph?

  38. phi
    • 2 years ago
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    You see the curve just touches the x-axis. This happens when y=0 we start with y = cos (2x - pi/3) +1 cos (2x - pi/3) +1 = 0 (look for where the curve crosses or touches the x-axis) use algebra to solve for x: write -1 on both sides of the = cos(2x -pi/3) +1 -1 = 0-1 simplify to get cos(2x-pi/3) = -1

  39. lilsis76
    • 2 years ago
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    cos(2x-pi/3) = -1 okay i get this like u did

  40. phi
    • 2 years ago
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    cos(2x-pi/3) = -1 take the inverse cosine of both sides 2x -pi/3 = acos(-1) type in google acos(-1)= you get pi radians (3.14159....) or 180 degrees. you now have 2x -pi/3 = pi can you finish?

  41. lilsis76
    • 2 years ago
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    i end up getting 2pi/3. and that graph on google is confusing me with different numbers

  42. phi
    • 2 years ago
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    The graph is show the curve with x in radians. the region from x=0 to 2pi (6.283...) is the region your question is asking about. you see 2 peaks in this interval. Decimals make it hard to match these spots with x= pi/6 and x= 7pi/6 but using a calculator or google we see x= 0.523... and x=3.665...

  43. phi
    • 2 years ago
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    *The graph shows the curve with x in radians.

  44. lilsis76
    • 2 years ago
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    oh....okay, i understand that now, thank you. for your help. Imma close this question and fix my homework. Ill be back later. so thank you

  45. phi
    • 2 years ago
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    OK. For graphing, I would pick a few x values and use a calculator to find the corresponding y value. Then plot that point. After a few points you can sketch in the curve.

  46. lilsis76
    • 2 years ago
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    okay, i would have to try that, I probably wont beable to do that on a quiz tomorrow. she is not allowing calculators. but if i know the steps i know i can do the question or problem

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