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UmarsBack

What is the vertex for the function? (You will want to find the axis of symmetry first.) Write as an ordered pair, leaving no spaces, such as (1,-4). y = -3x2 + 6x

  • one year ago
  • one year ago

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  1. UmarsBack
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    im so dumb omg

    • one year ago
  2. Eyad
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    use the square to convert to a vertex form ,y=a(x-h)^2+k

    • one year ago
  3. nincompoop
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    I guess we're not making sense, @UmarsBack

    • one year ago
  4. UmarsBack
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    you really arent :c idk what im suppsoed to do.

    • one year ago
  5. Eyad
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    its like : ax^2+bx+c the vertex is on the axis of symmetry: -b/2a when x= -b/2a we find y in this case -3x^2 +6x +0 gives us: a=-3 b=6 so when x=1; we get y= -3+6

    • one year ago
  6. UmarsBack
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    So what i gotta do? sorry for my stupidity.

    • one year ago
  7. nincompoop
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    okay let us do some reading ..

    • one year ago
  8. nincompoop
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    http://finedrafts.com/files/math/precal/Larson%20PreCal%208th/Larson%20Precal%20CH2.pdf

    • one year ago
  9. Eyad
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    x=-b/2a in this case -3x^2 +6x +0 gives us: a=-3 b=6 Which mean x=-6/-6= +1 ! y=-3+6= 3 Vertex = (1,3) =============

    • one year ago
  10. UmarsBack
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    Im completely clueless guys.

    • one year ago
  11. nincompoop
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    can you read pp 126-131?

    • one year ago
  12. nincompoop
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    I just want you to be acquainted with the terms and definitions

    • one year ago
  13. UmarsBack
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    Yes im reading right now. again, sorry for my stupidity.

    • one year ago
  14. UmarsBack
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    @nincompoop im doneee

    • one year ago
  15. nincompoop
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    so what did you learn?

    • one year ago
  16. nincompoop
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    eyad's information should make sense by now...

    • one year ago
  17. UmarsBack
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    Quadratic function = f(x) = ax^2 + bx + c

    • one year ago
  18. nincompoop
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    yes, that is the format of your equation at the moment. now, to help you identify the vertex of the parabola (graph of quadratic) you need to covert it into standard form of quadratic function

    • one year ago
  19. nincompoop
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    your vertex then is going to be the h and k

    • one year ago
  20. nincompoop
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    there are a few things you need to be good at like completing the square and factoring so you can solve this even mentally

    • one year ago
  21. nincompoop
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    do you know how to complete the square yet?

    • one year ago
  22. UmarsBack
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    wheres the h and k? and no :c

    • one year ago
  23. nincompoop
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    hehe this is going to be tough… so even eyad's info doesn't make sense right now?

    • one year ago
  24. UmarsBack
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    Im tryna understand >_<

    • one year ago
  25. UmarsBack
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    should i go to khan academy?

    • one year ago
  26. nincompoop
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    you can, but it's in the textbook and eyad pretty much gave the answer already

    • one year ago
  27. UmarsBack
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    you know i hate reading lmfao. i read the text book tho. im brain dead right now.

    • one year ago
  28. nincompoop
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    okay let us try this the long way so you understand the whole concept quadratic function f(x) = ax^2 + bx + c; f(x) is the same as "y" so this is what you have: y = -3x2 + 6x; this is the same as y = -3x2 + 6x + 0 (the equation you were given has + 0 omitted) this means that: a = -3 b = 6 c = 0 I need to know that you read and fully grasp this part

    • one year ago
  29. UmarsBack
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    Omfg i understand it way more now.

    • one year ago
  30. nincompoop
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    good, now recall the figures in the text book about vertex? I attached it so you can recall them

    • one year ago
  31. UmarsBack
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    Yea i recall them

    • one year ago
  32. nincompoop
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    good! we are making a progress here. keep in mind about vertex being either the lowest point or the highest point lowest point if a>0, which means the parabola opens upward highest point if a<0, which means the parabola opens downward the reason I want you to keep this in mind because the terms minimum and maximum give you the same definition in terms of the value of a recall your equation:y = -3x2 + 6x does this mean that your parabola opens upward or downward?

    • one year ago
  33. UmarsBack
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    The parabola opens upwards?

    • one year ago
  34. nincompoop
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    I want you to be certain of your answer. Only when you are confident that we are able to move on…

    • one year ago
  35. nincompoop
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    I made the figure bigger so you won't miss any detail

    • one year ago
  36. Eyad
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    y=-3x^2+6x y=ax^2+bx ^

    • one year ago
  37. UmarsBack
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    Its downward. right?

    • one year ago
  38. nincompoop
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    with your equation y = -3x2 + 6x I mentioned this: a = -3 b = 6 c = 0 and I also mentioned this: lowest point if a>0, which means the parabola opens upward highest point if a<0, which means the parabola opens downward ask yourself this question: is the value of a less than or greater than zero? greater than zero gives you an upward parabola less than zero gives you a downward parabola

    • one year ago
  39. UmarsBack
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    Its upward, because the value is greater than 0

    • one year ago
  40. Eyad
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    Btw @nincompoop ,After you finish you may Compile all your comments and post it in a one post ,IT would be a good tutorial Including the example .

    • one year ago
  41. nincompoop
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    awesome! now you seem more confident. now, let us move on understanding the standard form of a quadratic function I attached some information for you to re-read (since you said you read it earlier)

    • one year ago
  42. nincompoop
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    keep in mind that the values of your a, b, and c have not changed. they are still a = -3 b = 6 c = 0

    • one year ago
  43. UmarsBack
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    Okay im still with you.

    • one year ago
  44. UmarsBack
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    Okay continue. youre the teacher here.

    • one year ago
  45. UmarsBack
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    ...

    • one year ago
  46. nincompoop
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    now, we need to try to re-arrange your y = -3x^2 + 6x into the standard form of quadratic function, we can achieve it by completing the square, which for now will look like: factor 3 out of x-terms \[y=−3(x^2-2x)+0\] I included the value of c for the purpose of showing you what it would look like, but since our c is zero we can ignore it. do you know how to proceed or know how to complete the square?

    • one year ago
  47. UmarsBack
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    I dont know how to complete the square :c

    • one year ago
  48. nincompoop
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    it is now the time to learn one way is to divide the value of b by 2, and then square it but things could get more complicated than that so you will need to read and familiarize yourself http://www.mathsisfun.com/algebra/completing-square.html or this http://www.purplemath.com/modules/sqrquad.htm

    • one year ago
  49. UmarsBack
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    6/2= 3 2(3) ? like that?

    • one year ago
  50. nincompoop
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    yes, remember that you have to square it after you divide it by 2

    • one year ago
  51. nincompoop
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    so \[\left( \frac{ b }{ 2 } \right)^2\] \[\left( \frac{ 6 }{ 2 } \right)^2=3^2=?\]

    • one year ago
  52. UmarsBack
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    It equals the same thing right? 6?

    • one year ago
  53. nincompoop
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    \[3^2=3\times3=?\] to square is to multiply the multiplicand by itself twice

    • one year ago
  54. nincompoop
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    so your multiplier is also 3

    • one year ago
  55. UmarsBack
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    ohh 9.

    • one year ago
  56. nincompoop
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    did you read the contents of the links regarding completing the square I provided earlier?

    • one year ago
  57. UmarsBack
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    Not yet. let me read it now.

    • one year ago
  58. nincompoop
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    okay are you done?

    • one year ago
  59. UmarsBack
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    Yes, i think.

    • one year ago
  60. UmarsBack
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    x2 + bx + (b/2)2 = (x+b/2)2 < thats how you complete the square?

    • one year ago
  61. nincompoop
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    so this is where we left off: \[y=−3(x^2−2x)+0\] we need to complete the square of what is in the parenthesis our b is -2 \[\left( \frac{ -2 }{ 2 } \right)^2=-1^2=?\]

    • one year ago
  62. nincompoop
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    WAKE UP!!! 1 multiplied by itself is 1 this is what we will yield: \[y=-3(x^2-2x+1-1)\] After adding and subtracting 1 within the parentheses, you must now regroup the terms to form a perfect square trinomial. The -1 can be removed from inside the parentheses; however, because of the -3 outside of the parentheses, you must multiply -1 by -3 this is what it would look like: \[y=-3(x^2-2x+1)-3(-1)\] -3(-1) = 3 if you are still with me, we just solved the value of k this confirms one of @eyad's solution earlier k being 3 next step is to factor what is in the parenthesis \[(x^2-2x+1) \] turns into: \[(x-1)^2\]

    • one year ago
  63. nincompoop
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    can you guess the value of h?

    • one year ago
  64. UmarsBack
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    1

    • one year ago
  65. nincompoop
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    that is super awesome! our re-written quadratic function form looks like this: \[y=-3(x^2-1)^2+3\] \[y=a(x^2-h)^2+k, where: a \neq0\] vertex: (h,k) (-(-1), 3) or (1,3)

    • one year ago
  66. nincompoop
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    does this look like eyad's answer?

    • one year ago
  67. UmarsBack
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    Hahaha wo i cant believe i understand this now. it looks like eyads answer. thanks alot nin omfg.

    • one year ago
  68. Eyad
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    Yw :)

    • one year ago
  69. nincompoop
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    btw a few notes to consider: \[-1^2 = -(1)^2=-1\] earlier, I should have typed \[(-1)^2=1\] \[-1\times-1=1\] negative multiplied by a negative yields a positive

    • one year ago
  70. Eyad
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    A good teacher can inspire hope, ignite the imagination, and instill a love of learning.

    • one year ago
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