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algebra2sucks

  • 2 years ago

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

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  1. myininaya
    • 2 years ago
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    When is the bottom 0?

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

  3. algebra2sucks
    • 2 years ago
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    not sure?

  4. myininaya
    • 2 years ago
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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.

  5. myininaya
    • 2 years ago
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    Do you know how to factor?

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

  7. myininaya
    • 2 years ago
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    You can even do completing the square.

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

  9. algebra2sucks
    • 2 years ago
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    how do i do that?

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

  11. algebra2sucks
    • 2 years ago
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    its postive 12x

  12. myininaya
    • 2 years ago
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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

  13. myininaya
    • 2 years ago
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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)

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

  15. algebra2sucks
    • 2 years ago
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    its like this |dw:1370027288868:dw|

  16. myininaya
    • 2 years ago
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    err so it is -12x?

  17. algebra2sucks
    • 2 years ago
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    noo x^2 minus 12x

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

  19. myininaya
    • 2 years ago
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    You have also said twice that we have -12x

  20. myininaya
    • 2 years ago
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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.

  21. algebra2sucks
    • 2 years ago
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    sorry

  22. algebra2sucks
    • 2 years ago
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    this is the problem were doing

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  23. myininaya
    • 2 years ago
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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

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

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

  26. myininaya
    • 2 years ago
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    Those are your m and n.

  27. myininaya
    • 2 years ago
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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.

  28. myininaya
    • 2 years ago
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    c is 27 and b is -12

  29. myininaya
    • 2 years ago
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    |dw:1370027980807:dw|

  30. myininaya
    • 2 years ago
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    See how I factored 27?

  31. myininaya
    • 2 years ago
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    1(27)=27 3(9)=27 -1(-27)=27 -3(-9)=27

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

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

  34. algebra2sucks
    • 2 years ago
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    are these numbers removable?

  35. algebra2sucks
    • 2 years ago
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    and -3?

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

  37. myininaya
    • 2 years ago
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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

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

  39. algebra2sucks
    • 2 years ago
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    do we also use -1 or no?

  40. myininaya
    • 2 years ago
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    Where does that come from?

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

  42. algebra2sucks
    • 2 years ago
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    no idk how

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

  44. myininaya
    • 2 years ago
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    To undo subtraction by three you add three.

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

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

  47. myininaya
    • 2 years ago
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    x-3+3=0+3

  48. myininaya
    • 2 years ago
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    what is -3+3? what is 0+3?

  49. algebra2sucks
    • 2 years ago
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    0 and 3

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

  51. myininaya
    • 2 years ago
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    Now you solve x-9=0

  52. myininaya
    • 2 years ago
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    What are you going to add to both sides?

  53. algebra2sucks
    • 2 years ago
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    9?

  54. myininaya
    • 2 years ago
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    yes so x=?

  55. algebra2sucks
    • 2 years ago
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    9 and 0?

  56. algebra2sucks
    • 2 years ago
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    so x=9

  57. myininaya
    • 2 years ago
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    yes

  58. myininaya
    • 2 years ago
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    Discontinuous at x=3 or x=9

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

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

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

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

  63. myininaya
    • 2 years ago
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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?

  64. myininaya
    • 2 years ago
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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?

  65. myininaya
    • 2 years ago
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    This is just like matching socks.

  66. myininaya
    • 2 years ago
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    I have to go to the grocery store. Good luck.

  67. algebra2sucks
    • 2 years ago
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    Thanks for all ur help

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