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Babynini

  • one year ago

Vertex.

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  1. Babynini
    • one year ago
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    Write the equation in standard form then find and list the vertex, focus, and directrix of the parabola. sketch its graph showing the focus and the directrix.(note: Choose the locations of the coordinate axes and draw them on the grid and choose scales so that your graph is, if possible, about three by three inches in size.) x^2+12x+4y+44=0

  2. Babynini
    • one year ago
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    @jim_thompson5910 If you have time :)

  3. jim_thompson5910
    • one year ago
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    are you able to complete the square for the x terms?

  4. Babynini
    • one year ago
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    I'm not sure how.

  5. jim_thompson5910
    • one year ago
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    x^2+12x+_____ what goes in the blank to make that a perfect square?

  6. Babynini
    • one year ago
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    36

  7. jim_thompson5910
    • one year ago
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    yep

  8. jim_thompson5910
    • one year ago
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    so add and subtract 36 on the same side or add 36 to both sides

  9. jim_thompson5910
    • one year ago
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    then group (x^2-12x+36) and factor that to get (x-6)^2

  10. Babynini
    • one year ago
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    kks, I chose to add 36 to both sides (x-6)^2+4y+44=36

  11. Babynini
    • one year ago
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    (x-6)^2+4y=-8 (subtracted 44 from both sides)

  12. jim_thompson5910
    • one year ago
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    now isolate y

  13. Babynini
    • one year ago
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    y=-(x-6)^2/4 - 8/4

  14. jim_thompson5910
    • one year ago
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    yes

  15. jim_thompson5910
    • one year ago
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    you can simplify that further

  16. Babynini
    • one year ago
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    2 instead of 8/4 :P

  17. Babynini
    • one year ago
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    hrm the original and that one don't graph the same o.0

  18. Babynini
    • one year ago
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    http://www.wolframalpha.com/input/?i=x%5E2%2B12x%2B4y%2B44%3D0 Original

  19. Babynini
    • one year ago
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    http://www.wolframalpha.com/input/?i=Parabola+y%3D%28%28x-6%29%5E2%29%2F4%29+-2 New equation

  20. Babynini
    • one year ago
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    Would it work to do: 4y=-x^2-12x-44 y= -x^2/4 -3x -11 ?

  21. Babynini
    • one year ago
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    Or does the equation we previously made look better?

  22. Babynini
    • one year ago
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    http://www.wolframalpha.com/input/?i=Parabola+y%3D-1%2F4x%5E2+-3x-11 ....haha what.

  23. jim_thompson5910
    • one year ago
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    let me check

  24. Babynini
    • one year ago
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    kks

  25. jim_thompson5910
    • one year ago
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    oh I made a mistake, it's not -12x it's +12x

  26. jim_thompson5910
    • one year ago
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    so it should factor to (x+6)^2

  27. jim_thompson5910
    • one year ago
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    you should get \[\Large y = -\frac{1}{4}(x+6)^2 - 2\]

  28. Babynini
    • one year ago
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    Yep that's what I have :)

  29. Babynini
    • one year ago
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    that is still a different graph than the original

  30. jim_thompson5910
    • one year ago
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    I have it matching up

  31. Babynini
    • one year ago
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    oh yeah, sorry. It is. Wolfram was doing a close up of the graph and I hadn't noticed.

  32. jim_thompson5910
    • one year ago
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    you forgot the negative in http://www.wolframalpha.com/input/?i=Parabola+y%3D%28%28x-6%29%5E2%29%2F4%29+-2

  33. Babynini
    • one year ago
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    ah, sorry.

  34. Babynini
    • one year ago
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    so next is finding foci, vertex, and directrix.

  35. jim_thompson5910
    • one year ago
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    have a look at this http://www.mathwords.com/f/focus_parabola.htm

  36. Babynini
    • one year ago
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    k, so mines a vertical parabola.

  37. jim_thompson5910
    • one year ago
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    yes

  38. Babynini
    • one year ago
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    um and I use x^2=4py to find everything?

  39. jim_thompson5910
    • one year ago
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    Convert \[\Large y = -\frac{1}{4}(x+6)^2 - 2\] to 4p(y-k) = (x-h)^2 form

  40. Babynini
    • one year ago
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    would that be (y+2)/4=(x+6)^2 ??

  41. jim_thompson5910
    • one year ago
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    more like -4(y+2) = (x+6)^2

  42. jim_thompson5910
    • one year ago
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    -4(y+2) = (x+6)^2 is in the form 4p(y-k) = (x-h)^2 p = -1 h = -6 k = -2

  43. Babynini
    • one year ago
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    ahh ok

  44. jim_thompson5910
    • one year ago
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    |dw:1433470391325:dw|

  45. jim_thompson5910
    • one year ago
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    |dw:1433470408839:dw|

  46. jim_thompson5910
    • one year ago
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    |dw:1433470431794:dw|

  47. Babynini
    • one year ago
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    f: = (-6,-3)

  48. jim_thompson5910
    • one year ago
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    yes

  49. Babynini
    • one year ago
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    d: y = - 3

  50. jim_thompson5910
    • one year ago
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    no

  51. Babynini
    • one year ago
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    k-p ?

  52. Babynini
    • one year ago
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    oh sorry, -1

  53. jim_thompson5910
    • one year ago
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    yeah directrix is y = -1

  54. anonymous
    • one year ago
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    hi can i borrow jim_thompson5910

  55. anonymous
    • one year ago
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    ?

  56. Babynini
    • one year ago
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    umm so the graph looks like |dw:1433470684289:dw|

  57. Babynini
    • one year ago
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    |dw:1433470710284:dw|

  58. Babynini
    • one year ago
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    |dw:1433470757831:dw| ish lol

  59. Babynini
    • one year ago
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    the prof loves complete accuracy. so if I wanted to find the exact points that the parabola passes though do I just substitute numbers for x into the equation and find out y?

  60. jim_thompson5910
    • one year ago
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    yeah you can do that to plot points or use desmos to get an accurate graph desmos.com/calculator

  61. jim_thompson5910
    • one year ago
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    https://www.desmos.com/calculator

  62. Babynini
    • one year ago
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    well, i've got to show work. but i'll use desmos to make sure :) thanks!

  63. Babynini
    • one year ago
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    er it worked until I got to -7 and -8 values for x :o

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