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anonymous

  • one year ago

Help, will fan and medal. need someone to teach me please!

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  1. anonymous
    • one year ago
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    What is the equation of the following graph?

  2. anonymous
    • one year ago
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    @LynFran can you help me with this? please

  3. anonymous
    • one year ago
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    @Robert136 can you help?

  4. anonymous
    • one year ago
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    you know the vertex so that makes it easier

  5. anonymous
    • one year ago
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    i dont know the vertex...i dont get these at all.

  6. anonymous
    • one year ago
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    owait do you need the quadractic or both the quadratic and the dotted line...

  7. anonymous
    • one year ago
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    idk..its something to do with ellipses, parabola, or hyperbola...that what the section was about

  8. anonymous
    • one year ago
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    i can help with the parabola, not sure about that line though

  9. Loser66
    • one year ago
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    What is the directrix? y=??

  10. Loser66
    • one year ago
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    What is the vertex? You need both them to write out the equation. Please, answer.

  11. anonymous
    • one year ago
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    the vertex is (1,4)

  12. anonymous
    • one year ago
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    whats the directrix.... i dont think ive learned those

  13. Loser66
    • one year ago
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    The red line below to the graph is a directrix

  14. Loser66
    • one year ago
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    if you don't know that method, ok, pick 3 points on the graph. Please.

  15. anonymous
    • one year ago
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    y = 3

  16. anonymous
    • one year ago
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    thats the directrix

  17. Loser66
    • one year ago
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    Now, what method do you want me to follow?? directrix method or standard one?

  18. anonymous
    • one year ago
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    which one do u think would be the easiest to understand?

  19. Loser66
    • one year ago
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    Depend on you. I am ok with both

  20. Loser66
    • one year ago
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    Standard one, ok?

  21. anonymous
    • one year ago
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    okay

  22. Loser66
    • one year ago
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    what is the standard form of a parabola?

  23. anonymous
    • one year ago
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    y=ax^2+bx+c

  24. Loser66
    • one year ago
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    YYYYYYYYYYYYYES

  25. Loser66
    • one year ago
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    now, 3 points Vertex (1,4) you pick other two.

  26. anonymous
    • one year ago
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    just 2 random points?

  27. Loser66
    • one year ago
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    Yes, but for convenience, you should pick the point whose coordinates are integers like (-1,5). Got what I meant?

  28. anonymous
    • one year ago
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    so the points where the red line cross over?

  29. Loser66
    • one year ago
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    no, the point on the parabola

  30. anonymous
    • one year ago
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    like (3,5)

  31. Loser66
    • one year ago
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    Yes,

  32. Loser66
    • one year ago
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    Now, starting Vertex \((\color{red}{1}, \color{blue}{4})\) Plug them into the standard form like \(\color{blue}{4}= a\color{red}{1^2}+b\color{red}{1}+c\) you have 4 = a +b +c, right? name it as (*) Do the same with other two point, what do you get?

  33. anonymous
    • one year ago
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    im lost.. im sory. my computer froze sorry about the late anwser

  34. anonymous
    • one year ago
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    @Loser66 ?

  35. anonymous
    • one year ago
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    5=a-b+c 5=5a+5b+c

  36. anonymous
    • one year ago
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    @Loser66 am I understanding it?

  37. anonymous
    • one year ago
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    @Mehek14

  38. Loser66
    • one year ago
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    5=a -b +c, yes but for (3,5) it is not right 5 = a(3^2) +b(3) +c , that is 5= 9a +3b+c

  39. Loser66
    • one year ago
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    Now, you have 3 equations, 3 unknowns a, b, c. You can solve for them a + b + c = 4 (*) a - b + c = 5 (**) 9a + 3b + c = 5 (***)

  40. Loser66
    • one year ago
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    Take (*) -(**) , what do you get?

  41. anonymous
    • one year ago
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    -1

  42. anonymous
    • one year ago
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    @Loser66

  43. anonymous
    • one year ago
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    so i have -1 left, and 9a + 3b + c = 5

  44. Loser66
    • one year ago
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    |dw:1439603548746:dw|

  45. Loser66
    • one year ago
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    Take (***) -(**)

  46. anonymous
    • one year ago
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    8a + 4b

  47. Loser66
    • one year ago
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    |dw:1439603653405:dw|

  48. Loser66
    • one year ago
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    Replace a = 1/4, b = -1/2 in (*) you get c = ??

  49. anonymous
    • one year ago
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    c=4.25

  50. Loser66
    • one year ago
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    yes, or 17/4 Now, plug back to standard form, y = ax^2 +bx+c . that is your answer.

  51. anonymous
    • one year ago
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    y=ax^2+bc+17/4?

  52. Loser66
    • one year ago
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    a = 1/4 b= -1/2 c = 17/4 Plug in \(y= ax^2+bx +c\)

  53. anonymous
    • one year ago
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    okay thank you. can i tag u if i have anymore questions? i like tht ur showing me instad of telling me

  54. misty1212
    • one year ago
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    HI!!

  55. misty1212
    • one year ago
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    there might be a somewhat easier way to get this, using two points although above method certainly works

  56. anonymous
    • one year ago
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    thank you. im sure ill run into more of these, do u mind if i tag u if i need help?

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