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anonymous

  • one year ago

using complete sentences explain how to find the minimum value for each for each function and and determine which function has the smallest y value f(x) = 3x^2 + 12x + 16 And g(x) = 2sin (2x-pi) +4

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    @jim_thompson5910 I'm pretty sure the minimum point for f(x) is (-2, 4). Is that correct?

  3. jim_thompson5910
    • one year ago
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    yep the min for f is y = 4 when x = -2 ie the min f(x) = 4 occurs at the point (-2,4)

  4. anonymous
    • one year ago
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    Okay how to we find g(x)?

  5. jim_thompson5910
    • one year ago
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    when does sin(x) have a minimum?

  6. anonymous
    • one year ago
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    I don't know

  7. jim_thompson5910
    • one year ago
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    look at the unit circle what is the lowest point on it?

  8. anonymous
    • one year ago
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    (0, -1)?

  9. jim_thompson5910
    • one year ago
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    what is the corresponding angle

  10. jim_thompson5910
    • one year ago
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    I'll be right back in a few minutes

  11. anonymous
    • one year ago
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    (0, 1)? and okay

  12. jim_thompson5910
    • one year ago
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    I'm back no look at where it shows the angle theta

  13. jim_thompson5910
    • one year ago
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    what angle theta is where (0,-1) is located?

  14. anonymous
    • one year ago
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    I don't know what it is. I'm confused

  15. jim_thompson5910
    • one year ago
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    look at this http://upload.wikimedia.org/wikipedia/commons/thumb/4/4c/Unit_circle_angles_color.svg/1024px-Unit_circle_angles_color.svg.png and tell me what the angle is

  16. anonymous
    • one year ago
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    I don't know I'm lost..

  17. jim_thompson5910
    • one year ago
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    do you see the 270 degrees?

  18. anonymous
    • one year ago
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    Yes

  19. jim_thompson5910
    • one year ago
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    so that's why sin(270 degrees) = -1 or sin(3pi/2 radians) = -1

  20. jim_thompson5910
    • one year ago
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    the min occurs when sin(x) = -1, ie when x = 3pi/2 so you have to determine when 2x-pi is equal to 3pi/2

  21. jim_thompson5910
    • one year ago
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    2x-pi = 3pi/2 what is x equal to?

  22. anonymous
    • one year ago
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    0?

  23. jim_thompson5910
    • one year ago
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    you should get x = 5pi/4

  24. jim_thompson5910
    • one year ago
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    I see a shortcut though

  25. jim_thompson5910
    • one year ago
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    they just want the min y value sin( anything ) has a min of -1 so 2sin (2x-pi) +4 turns into 2*(-1) +4 when you replace all of "sin..." with the smallest it can get, which is -1

  26. anonymous
    • one year ago
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    This is confusing to me.. I'm trying to understand it but it's a lot of info.. I'm not good with trigonometry

  27. jim_thompson5910
    • one year ago
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    2*(-1) +4 turns into 2, so this is the smallest that g(x) can get

  28. jim_thompson5910
    • one year ago
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    notice on the wavy graph, the lowest points have a y coordinate of y = 2

  29. anonymous
    • one year ago
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    Yea I noticed that. So y=2 but what is x? Do we need to know x?

  30. jim_thompson5910
    • one year ago
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    they just want to know the smallest output of the function

  31. anonymous
    • one year ago
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    Okay. So g(x) has the smallest minimum?

  32. jim_thompson5910
    • one year ago
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    yes

  33. jim_thompson5910
    • one year ago
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    and you can see that on your graph you posted

  34. anonymous
    • one year ago
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    Yeah I see that now haha. Do you mind helping with one more?

  35. jim_thompson5910
    • one year ago
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    alright

  36. anonymous
    • one year ago
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    Prove: \[\sin \theta-\sin \theta \times \cos^2 \theta = \sin^3 \theta\]

  37. jim_thompson5910
    • one year ago
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    hint: factor out sin(theta)

  38. jim_thompson5910
    • one year ago
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    only alter the left side do not change the right side

  39. anonymous
    • one year ago
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    So it would be \[\cos^2 \theta = \sin^3 \theta?\]

  40. jim_thompson5910
    • one year ago
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    you should have \[\Large \sin(\theta)\left(1-\cos^2(\theta)\right) = \sin^3(\theta)\]

  41. jim_thompson5910
    • one year ago
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    then try to do something with the 1-sin^2

  42. anonymous
    • one year ago
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    How would you get 1-sin^2?

  43. jim_thompson5910
    • one year ago
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    oops

  44. jim_thompson5910
    • one year ago
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    I meant 1-cos^2

  45. anonymous
    • one year ago
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    Would you multiply |dw:1433476582158:dw|

  46. jim_thompson5910
    • one year ago
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    nope

  47. jim_thompson5910
    • one year ago
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    there's an identity you use for 1-cos^2

  48. jim_thompson5910
    • one year ago
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    hint: sin^2 + cos^2 = 1

  49. anonymous
    • one year ago
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    So 1-cos^2 would become sin^2

  50. jim_thompson5910
    • one year ago
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    yep

  51. anonymous
    • one year ago
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    Thank you so much

  52. jim_thompson5910
    • one year ago
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    you're welcome

  53. anonymous
    • one year ago
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    Wait how did you get the 1 again?

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