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ali1029 Group Title

Please help! how do you write y=x^2-2x-3 in vertex and intercept form using the quadratic formula?

  • 11 months ago
  • 11 months ago

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  1. ali1029 Group Title
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    @ganeshie8 @satellite73 do you guys know how to do this? :o

    • 11 months ago
  2. Mttblink Group Title
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    I think it is y=-2x-3

    • 11 months ago
  3. Mttblink Group Title
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    Oops I meant xD y=-2x+1

    • 11 months ago
  4. ali1029 Group Title
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    is that for vertex form or intercept form? :)

    • 11 months ago
  5. Mttblink Group Title
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    Oops damn I put that in slope-intercept... Geez I suck at math xD, I don't think I can help you on vertex

    • 11 months ago
  6. ali1029 Group Title
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    hahah thanks for trying :)

    • 11 months ago
  7. ganeshie8 Group Title
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    use the quadratic formula and find the x-intercepts first

    • 11 months ago
  8. ali1029 Group Title
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    can you help me figure out how to do that? thats where im stuck :(

    • 11 months ago
  9. ganeshie8 Group Title
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    sure :) u knw the quadratic formula ?

    • 11 months ago
  10. ali1029 Group Title
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    sorta :o im just so confused because we've been using so many different forms!

    • 11 months ago
  11. ganeshie8 Group Title
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    dont wry, this is going to be very easy :) First, we find x-intercepts using the quadratic formula.

    • 11 months ago
  12. ganeshie8 Group Title
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    y=x^2-2x-3 a = ? b = ? c = ?

    • 11 months ago
  13. ganeshie8 Group Title
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    \(\large x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\)

    • 11 months ago
  14. ganeshie8 Group Title
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    you need to figure out a, b, c values. and plug them above to get the x-intercepts

    • 11 months ago
  15. ali1029 Group Title
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    so a=1 b=2 c=3? what i dont get is that it wants me to put that equation in intercept and vertex form but those have h & k :o

    • 11 months ago
  16. ganeshie8 Group Title
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    b = -2 we will get to vertex/intercept forms later. first, get the x-intercepts :)

    • 11 months ago
  17. ganeshie8 Group Title
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    put them in the formula above, wat do u get for x = ?

    • 11 months ago
  18. ali1029 Group Title
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    1 becaue 1^2 is 1? let me do that equation

    • 11 months ago
  19. ganeshie8 Group Title
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    a = 1 b = -2 c = -3

    • 11 months ago
  20. ganeshie8 Group Title
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    use above, take ur time :)

    • 11 months ago
  21. ganeshie8 Group Title
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    \(\large x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\) \(\large x = \frac{--2 \pm \sqrt{(-2)^2-4 \times 1\times (-3)}}{2(1)}\)

    • 11 months ago
  22. ali1029 Group Title
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    i got 1 ± 4 .... is that close? :o

    • 11 months ago
  23. ganeshie8 Group Title
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    \(\large x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\) \(\large x = \frac{--2 \pm \sqrt{(-2)^2-4 \times 1\times (-3)}}{2(1)}\) \(\large x = \frac{2 \pm \sqrt{4 + 12}}{2}\) \(\large x = \frac{2 \pm \sqrt{16}}{2}\)

    • 11 months ago
  24. ganeshie8 Group Title
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    thats close, just a minor mistake u did... check once

    • 11 months ago
  25. ganeshie8 Group Title
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    \(\large x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\) \(\large x = \frac{--2 \pm \sqrt{(-2)^2-4 \times 1\times (-3)}}{2(1)}\) \(\large x = \frac{2 \pm \sqrt{4 + 12}}{2}\) \(\large x = \frac{2 \pm \sqrt{16}}{2}\) \(\large x = \frac{2 \pm 4}{2}\) \(\large x = 1 \pm 2\)

    • 11 months ago
  26. ali1029 Group Title
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    ohhh i see what i did wrong! thank you :) so know what do i do?

    • 11 months ago
  27. ganeshie8 Group Title
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    that gives, \(x = 1+2 ~~~ or~~~~ x = 1-2\) \(x = 3, -1\)

    • 11 months ago
  28. ganeshie8 Group Title
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    we're done wid using quadratic formula. we got the x-intercepts : 3, -1

    • 11 months ago
  29. ganeshie8 Group Title
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    so the intercept form would be : y = (x-3)(x+1)

    • 11 months ago
  30. ganeshie8 Group Title
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    wat about vertex form ? how to get \(h\), \(k\) ... any ideas ?

    • 11 months ago
  31. ali1029 Group Title
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    wow that was easier than i thought! thank you :) & no, not too sure :(

    • 11 months ago
  32. ganeshie8 Group Title
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    vertec form is also easy, \(h\) lies exactly at the middle of both the intercepts :- \(h \) = \(\large \frac{3 + -1}{2}\)

    • 11 months ago
  33. ganeshie8 Group Title
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    \(h \) = \(\large \frac{3-1}{2}\)

    • 11 months ago
  34. ganeshie8 Group Title
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    \(h \) = \(\large \frac{2}{2}\)

    • 11 months ago
  35. ali1029 Group Title
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    so h=1?

    • 11 months ago
  36. ganeshie8 Group Title
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    \(h \) = \(\large 1\)

    • 11 months ago
  37. ali1029 Group Title
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    yay :D

    • 11 months ago
  38. ganeshie8 Group Title
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    so vertex form wud be : \(y = (x-1)^2 + k\)

    • 11 months ago
  39. ganeshie8 Group Title
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    you still need to find \(k\)

    • 11 months ago
  40. ganeshie8 Group Title
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    :)

    • 11 months ago
  41. ali1029 Group Title
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    okay, so how do we find k now? :)

    • 11 months ago
  42. ganeshie8 Group Title
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    put x = h = 1, in given quadratic. it gives u the \(k\) value

    • 11 months ago
  43. ganeshie8 Group Title
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    y=x^2-2x-3 put x = 1

    • 11 months ago
  44. ganeshie8 Group Title
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    y = 1^2 - 2(1) - 3 = 1 -2 -3 = ?

    • 11 months ago
  45. ali1029 Group Title
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    -4?

    • 11 months ago
  46. ganeshie8 Group Title
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    yup ! thats ur \(k\)

    • 11 months ago
  47. ali1029 Group Title
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    y=(x−1)^2+4 would be the answer? :)

    • 11 months ago
  48. ganeshie8 Group Title
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    who ate the - sign before 4 ?

    • 11 months ago
  49. ali1029 Group Title
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    i messed up again ._. so it would be y=(x−1)2-4?

    • 11 months ago
  50. ganeshie8 Group Title
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    yes :)

    • 11 months ago
  51. ali1029 Group Title
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    thank you so much!! you're amazing!! :) if it isnt too much too ask could you help me with two more equations similar to that? :)

    • 11 months ago
  52. ali1029 Group Title
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    i need to figure out how to put y-5=-2(x+1)^2 in standard and intercept form & y=(x+2)(x-3) in standard and vertex form :o

    • 11 months ago
  53. ganeshie8 Group Title
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    wats the standard form ?

    • 11 months ago
  54. ali1029 Group Title
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    y=ax^2+bx+c?

    • 11 months ago
  55. ganeshie8 Group Title
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    y = ax^2 + bx + c this is the standard form right ?

    • 11 months ago
  56. ganeshie8 Group Title
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    oh yes... thats right !

    • 11 months ago
  57. ganeshie8 Group Title
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    y-5=-2(x+1)^2 add 5 both sides y = -2(x+1)^2 + 5

    • 11 months ago
  58. ganeshie8 Group Title
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    next, use the formula \((a+b)^2 = a^2 +b^2 + 2ab\)

    • 11 months ago
  59. ganeshie8 Group Title
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    expand right side,

    • 11 months ago
  60. ali1029 Group Title
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    would it be y=-2x+1

    • 11 months ago
  61. ganeshie8 Group Title
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    y = -2(x+1)^2 + 5 = -2(x^2 + 1^2 + 2x ) + 5 = -2(x^2 + 1 + 2x) + 5

    • 11 months ago
  62. ganeshie8 Group Title
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    simplify :)

    • 11 months ago
  63. ali1029 Group Title
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    would it be -2x^2-4x+3?

    • 11 months ago
  64. ganeshie8 Group Title
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    correct

    • 11 months ago
  65. ganeshie8 Group Title
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    standard form : y = -2x^2-4x+3

    • 11 months ago
  66. ali1029 Group Title
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    yesss :D

    • 11 months ago
  67. ganeshie8 Group Title
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    y-5=-2(x+1)^2 to put this guy in intercept form, u need to find its x-intercepts. so put y = 0 0-5=-2(x+1)^2 5/2 = (x+1)^2 \(\pm \sqrt{5/2} = x+1\) \( -1 \pm \sqrt{5/2} = x\)

    • 11 months ago
  68. ganeshie8 Group Title
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    those are our x-intercepts

    • 11 months ago
  69. ganeshie8 Group Title
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    can u put t he intercept form ?

    • 11 months ago
  70. ali1029 Group Title
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    the intercept form is y=a(x-p)(x-q)

    • 11 months ago
  71. ali1029 Group Title
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    so if the equation is −1±5/2, how do i solve the fraction part?

    • 11 months ago
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