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UmarsBack

  • 2 years ago

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

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  1. UmarsBack
    • 2 years ago
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    im so dumb omg

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

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

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

  5. Eyad
    • 2 years ago
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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

  6. UmarsBack
    • 2 years ago
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    So what i gotta do? sorry for my stupidity.

  7. nincompoop
    • 2 years ago
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    okay let us do some reading ..

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

  9. Eyad
    • 2 years ago
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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) =============

  10. UmarsBack
    • 2 years ago
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    Im completely clueless guys.

  11. nincompoop
    • 2 years ago
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    can you read pp 126-131?

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

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

  14. UmarsBack
    • 2 years ago
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    @nincompoop im doneee

  15. nincompoop
    • 2 years ago
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    so what did you learn?

  16. nincompoop
    • 2 years ago
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    eyad's information should make sense by now...

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

  18. nincompoop
    • 2 years ago
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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

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

  20. nincompoop
    • 2 years ago
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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

  21. nincompoop
    • 2 years ago
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    do you know how to complete the square yet?

  22. UmarsBack
    • 2 years ago
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    wheres the h and k? and no :c

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

  24. UmarsBack
    • 2 years ago
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    Im tryna understand >_<

  25. UmarsBack
    • 2 years ago
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    should i go to khan academy?

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

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

  28. nincompoop
    • 2 years ago
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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

  29. UmarsBack
    • 2 years ago
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    Omfg i understand it way more now.

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

  31. UmarsBack
    • 2 years ago
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    Yea i recall them

  32. nincompoop
    • 2 years ago
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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?

  33. UmarsBack
    • 2 years ago
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    The parabola opens upwards?

  34. nincompoop
    • 2 years ago
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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…

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

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

  37. UmarsBack
    • 2 years ago
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    Its downward. right?

  38. nincompoop
    • 2 years ago
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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

  39. UmarsBack
    • 2 years ago
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    Its upward, because the value is greater than 0

  40. Eyad
    • 2 years ago
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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 .

  41. nincompoop
    • 2 years ago
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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)

  42. nincompoop
    • 2 years ago
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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

  43. UmarsBack
    • 2 years ago
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    Okay im still with you.

  44. UmarsBack
    • 2 years ago
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    Okay continue. youre the teacher here.

  45. UmarsBack
    • 2 years ago
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    ...

  46. nincompoop
    • 2 years ago
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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?

  47. UmarsBack
    • 2 years ago
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    I dont know how to complete the square :c

  48. nincompoop
    • 2 years ago
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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

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

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

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

  52. UmarsBack
    • 2 years ago
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    It equals the same thing right? 6?

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

  54. nincompoop
    • 2 years ago
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    so your multiplier is also 3

  55. UmarsBack
    • 2 years ago
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    ohh 9.

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

  57. UmarsBack
    • 2 years ago
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    Not yet. let me read it now.

  58. nincompoop
    • 2 years ago
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    okay are you done?

  59. UmarsBack
    • 2 years ago
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    Yes, i think.

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

  61. nincompoop
    • 2 years ago
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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 \[\left( \frac{ -2 }{ 2 } \right)^2=-1^2=?\]

  62. nincompoop
    • 2 years ago
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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: \[(x-1)^2\]

  63. nincompoop
    • 2 years ago
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    can you guess the value of h?

  64. UmarsBack
    • 2 years ago
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    1

  65. nincompoop
    • 2 years ago
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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)

  66. nincompoop
    • 2 years ago
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    does this look like eyad's answer?

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

  68. Eyad
    • 2 years ago
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    Yw :)

  69. nincompoop
    • 2 years ago
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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

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

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