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algebra2sucks Group Title

What are the points of discontinuity? Are they all removable? y = (x-3)/x^2 - 12x + 27

  • one year ago
  • one year ago

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  1. myininaya Group Title
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    When is the bottom 0?

    • one year ago
  2. myininaya Group Title
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    For what values of x make the bottom 0?

    • one year ago
  3. algebra2sucks Group Title
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    not sure?

    • one year ago
  4. myininaya Group Title
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    So you should be able to solve an equation in this form ax^2+bx+c=0 by now. I'm guessing this since this is a type of problem that occurs after learning that.

    • one year ago
  5. myininaya Group Title
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    Do you know how to factor?

    • one year ago
  6. myininaya Group Title
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    Or even use the quadratic formula to solve a quadratic equation?

    • one year ago
  7. myininaya Group Title
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    You can even do completing the square.

    • one year ago
  8. algebra2sucks Group Title
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    so x^2+12x+27=0?

    • one year ago
  9. algebra2sucks Group Title
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    how do i do that?

    • one year ago
  10. myininaya Group Title
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    Do you have -12x or +12x?

    • one year ago
  11. algebra2sucks Group Title
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    its postive 12x

    • one year ago
  12. myininaya Group Title
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    Hmmm...So the problem above is written/typed wrongly I assume. Okay You should know how to factor the polynomial x^2+12x+27

    • one year ago
  13. myininaya Group Title
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    Find two factors of c that add up to be b (you can do this when the coefficient of x^2 is 1 which in this problem it is)

    • one year ago
  14. myininaya Group Title
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    In other words find two factors of 27 that add up to be 12.

    • one year ago
  15. algebra2sucks Group Title
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    its like this |dw:1370027288868:dw|

    • one year ago
  16. myininaya Group Title
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    err so it is -12x?

    • one year ago
  17. algebra2sucks Group Title
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    noo x^2 minus 12x

    • one year ago
  18. myininaya Group Title
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    I'm really confused what problem we are doing because twice now you have written +12x

    • one year ago
  19. myininaya Group Title
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    You have also said twice that we have -12x

    • one year ago
  20. myininaya Group Title
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    Anyways, we take the bottom polynomial whatever it may be and set that equal to zero and solve for x. I will let you do it since I don't know what problem we are doing anymore.

    • one year ago
  21. algebra2sucks Group Title
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    sorry

    • one year ago
  22. algebra2sucks Group Title
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    this is the problem were doing

    • one year ago
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  23. myininaya Group Title
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    Set bottom polynomial equal to zero. \[x^2-12x+27=0\] You will notice this is in the form \[ax^2+bx+c=0\] You have learned if you cannot factor this easily you could also use completing the square or quadratic formula. The easy way to solve this quadratic is by factoring in my opinion. Since the coefficient of x squared is 1, all we have to really do is find two factors of c that add up to be b; called them m and n. So The factored form would be (x-m)(x-n)=0 Then you set both factors equal to 0 like so x-m=0 or x-n=0 Solve both for x like so x=m or x=n

    • one year ago
  24. algebra2sucks Group Title
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    so which numbers do we use for x-m and x-n?

    • one year ago
  25. myininaya Group Title
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    Have you found factors of 27 that have product 27 and sum -12?

    • one year ago
  26. myininaya Group Title
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    Those are your m and n.

    • one year ago
  27. myininaya Group Title
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    Recall this part that I said: Since the coefficient of x squared is 1, all we have to really do is find two factors of c that add up to be b; called them m and n.

    • one year ago
  28. myininaya Group Title
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    c is 27 and b is -12

    • one year ago
  29. myininaya Group Title
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    |dw:1370027980807:dw|

    • one year ago
  30. myininaya Group Title
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    See how I factored 27?

    • one year ago
  31. myininaya Group Title
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    1(27)=27 3(9)=27 -1(-27)=27 -3(-9)=27

    • one year ago
  32. algebra2sucks Group Title
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    Yes I see on my answes I have x= -9, x =-3 is that right?

    • one year ago
  33. myininaya Group Title
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    Do any of those factors above that multiplied to be 27 also add up to be -12?

    • one year ago
  34. algebra2sucks Group Title
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    are these numbers removable?

    • one year ago
  35. algebra2sucks Group Title
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    and -3?

    • one year ago
  36. myininaya Group Title
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    (x-3)(x-9)=0 Set both equal to 0.

    • one year ago
  37. myininaya Group Title
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    I put (x-m)(x-n)=0 where I meant (x+m)(x+n)=0. So you would have x+m=0 or x+n=0 Which means you would have x=-m or x=-n. But anyways we have (x-3)(x-9)=0 And you are to set both factors equal to 0 and solve for x for both equations

    • one year ago
  38. myininaya Group Title
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    x-3=0 or x-9=0 I leaving solving these linear equations to you.

    • one year ago
  39. algebra2sucks Group Title
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    do we also use -1 or no?

    • one year ago
  40. myininaya Group Title
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    Where does that come from?

    • one year ago
  41. myininaya Group Title
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    Have you solve x-3=0 or x-9=0 yet?

    • one year ago
  42. algebra2sucks Group Title
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    no idk how

    • one year ago
  43. myininaya Group Title
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    x-3=0 You can't solve this for x? You don't know how to get x by itself?

    • one year ago
  44. myininaya Group Title
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    To undo subtraction by three you add three.

    • one year ago
  45. myininaya Group Title
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    Try adding three to both sides and tell me what the result is.

    • one year ago
  46. myininaya Group Title
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    x-3=0 add 3 to both sides +3 +3 --------- what is the result?

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  47. myininaya Group Title
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    x-3+3=0+3

    • one year ago
  48. myininaya Group Title
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    what is -3+3? what is 0+3?

    • one year ago
  49. algebra2sucks Group Title
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    0 and 3

    • one year ago
  50. myininaya Group Title
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    yes so x-3+3=0+3 gives us x+0=3 Or just x=3

    • one year ago
  51. myininaya Group Title
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    Now you solve x-9=0

    • one year ago
  52. myininaya Group Title
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    What are you going to add to both sides?

    • one year ago
  53. algebra2sucks Group Title
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    9?

    • one year ago
  54. myininaya Group Title
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    yes so x=?

    • one year ago
  55. algebra2sucks Group Title
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    9 and 0?

    • one year ago
  56. algebra2sucks Group Title
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    so x=9

    • one year ago
  57. myininaya Group Title
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    yes

    • one year ago
  58. myininaya Group Title
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    Discontinuous at x=3 or x=9

    • one year ago
  59. myininaya Group Title
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    Now go back to the problem where you had everything factored.

    • one year ago
  60. myininaya Group Title
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    \[\frac{x-3}{(x-3)(x-9)}\]

    • one year ago
  61. myininaya Group Title
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    This was the problem where everything is factored, correct?

    • one year ago
  62. myininaya Group Title
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    Does anything cancel? If so that is where you have your removable discontinuity.

    • one year ago
  63. myininaya Group Title
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    x-3 is just a number so is x-9 If we had 5/(5*9) you would say the 5's cancel right? Like this 1/9 Does anything cancel? Is there a factor on top that you see on bottom?

    • one year ago
  64. myininaya Group Title
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    \[\frac{(x-3)}{(x-3)(x-9)}\] If you want you can put parenthesis around the x-3 on top, Do you see any factors that match on top and bottom?

    • one year ago
  65. myininaya Group Title
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    This is just like matching socks.

    • one year ago
  66. myininaya Group Title
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    I have to go to the grocery store. Good luck.

    • one year ago
  67. algebra2sucks Group Title
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    Thanks for all ur help

    • one year ago
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