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algebra2sucks

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

  • 10 months ago
  • 10 months ago

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  1. myininaya
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    When is the bottom 0?

    • 10 months ago
  2. myininaya
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    For what values of x make the bottom 0?

    • 10 months ago
  3. algebra2sucks
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    not sure?

    • 10 months ago
  4. myininaya
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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.

    • 10 months ago
  5. myininaya
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    Do you know how to factor?

    • 10 months ago
  6. myininaya
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    Or even use the quadratic formula to solve a quadratic equation?

    • 10 months ago
  7. myininaya
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    You can even do completing the square.

    • 10 months ago
  8. algebra2sucks
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    so x^2+12x+27=0?

    • 10 months ago
  9. algebra2sucks
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    how do i do that?

    • 10 months ago
  10. myininaya
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    Do you have -12x or +12x?

    • 10 months ago
  11. algebra2sucks
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    its postive 12x

    • 10 months ago
  12. myininaya
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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

    • 10 months ago
  13. myininaya
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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)

    • 10 months ago
  14. myininaya
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    In other words find two factors of 27 that add up to be 12.

    • 10 months ago
  15. algebra2sucks
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    its like this |dw:1370027288868:dw|

    • 10 months ago
  16. myininaya
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    err so it is -12x?

    • 10 months ago
  17. algebra2sucks
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    noo x^2 minus 12x

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

    • 10 months ago
  19. myininaya
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    You have also said twice that we have -12x

    • 10 months ago
  20. myininaya
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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.

    • 10 months ago
  21. algebra2sucks
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    sorry

    • 10 months ago
  22. algebra2sucks
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    this is the problem were doing

    • 10 months ago
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  23. myininaya
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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

    • 10 months ago
  24. algebra2sucks
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    so which numbers do we use for x-m and x-n?

    • 10 months ago
  25. myininaya
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    Have you found factors of 27 that have product 27 and sum -12?

    • 10 months ago
  26. myininaya
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    Those are your m and n.

    • 10 months ago
  27. myininaya
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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.

    • 10 months ago
  28. myininaya
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    c is 27 and b is -12

    • 10 months ago
  29. myininaya
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    |dw:1370027980807:dw|

    • 10 months ago
  30. myininaya
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    See how I factored 27?

    • 10 months ago
  31. myininaya
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    1(27)=27 3(9)=27 -1(-27)=27 -3(-9)=27

    • 10 months ago
  32. algebra2sucks
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    Yes I see on my answes I have x= -9, x =-3 is that right?

    • 10 months ago
  33. myininaya
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    Do any of those factors above that multiplied to be 27 also add up to be -12?

    • 10 months ago
  34. algebra2sucks
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    are these numbers removable?

    • 10 months ago
  35. algebra2sucks
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    and -3?

    • 10 months ago
  36. myininaya
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    (x-3)(x-9)=0 Set both equal to 0.

    • 10 months ago
  37. myininaya
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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

    • 10 months ago
  38. myininaya
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    x-3=0 or x-9=0 I leaving solving these linear equations to you.

    • 10 months ago
  39. algebra2sucks
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    do we also use -1 or no?

    • 10 months ago
  40. myininaya
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    Where does that come from?

    • 10 months ago
  41. myininaya
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    Have you solve x-3=0 or x-9=0 yet?

    • 10 months ago
  42. algebra2sucks
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    no idk how

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

    • 10 months ago
  44. myininaya
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    To undo subtraction by three you add three.

    • 10 months ago
  45. myininaya
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    Try adding three to both sides and tell me what the result is.

    • 10 months ago
  46. myininaya
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    x-3=0 add 3 to both sides +3 +3 --------- what is the result?

    • 10 months ago
  47. myininaya
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    x-3+3=0+3

    • 10 months ago
  48. myininaya
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    what is -3+3? what is 0+3?

    • 10 months ago
  49. algebra2sucks
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    0 and 3

    • 10 months ago
  50. myininaya
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    yes so x-3+3=0+3 gives us x+0=3 Or just x=3

    • 10 months ago
  51. myininaya
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    Now you solve x-9=0

    • 10 months ago
  52. myininaya
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    What are you going to add to both sides?

    • 10 months ago
  53. algebra2sucks
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    9?

    • 10 months ago
  54. myininaya
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    yes so x=?

    • 10 months ago
  55. algebra2sucks
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    9 and 0?

    • 10 months ago
  56. algebra2sucks
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    so x=9

    • 10 months ago
  57. myininaya
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    yes

    • 10 months ago
  58. myininaya
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    Discontinuous at x=3 or x=9

    • 10 months ago
  59. myininaya
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    Now go back to the problem where you had everything factored.

    • 10 months ago
  60. myininaya
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    \[\frac{x-3}{(x-3)(x-9)}\]

    • 10 months ago
  61. myininaya
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    This was the problem where everything is factored, correct?

    • 10 months ago
  62. myininaya
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    Does anything cancel? If so that is where you have your removable discontinuity.

    • 10 months ago
  63. myininaya
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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?

    • 10 months ago
  64. myininaya
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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?

    • 10 months ago
  65. myininaya
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    This is just like matching socks.

    • 10 months ago
  66. myininaya
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    I have to go to the grocery store. Good luck.

    • 10 months ago
  67. algebra2sucks
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    Thanks for all ur help

    • 10 months ago
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