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algebra2sucks

  • one year ago

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

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  1. myininaya
    • one year ago
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    When is the bottom 0?

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

  3. algebra2sucks
    • one year ago
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    not sure?

  4. myininaya
    • one year 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
    • one year ago
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    Do you know how to factor?

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

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

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

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

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

  11. algebra2sucks
    • one year ago
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    its postive 12x

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

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

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

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

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

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

  20. myininaya
    • one year 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
    • one year ago
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    sorry

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

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  23. myininaya
    • one year 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
    • one year ago
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    so which numbers do we use for x-m and x-n?

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

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

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

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

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

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

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

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

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

  35. algebra2sucks
    • one year ago
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    and -3?

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

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

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

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

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

  42. algebra2sucks
    • one year ago
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    no idk how

  43. myininaya
    • one year 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
    • one year ago
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    To undo subtraction by three you add three.

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

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

  47. myininaya
    • one year ago
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    x-3+3=0+3

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

  49. algebra2sucks
    • one year ago
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    0 and 3

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

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

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

  53. algebra2sucks
    • one year ago
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    9?

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

  55. algebra2sucks
    • one year ago
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    9 and 0?

  56. algebra2sucks
    • one year ago
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    so x=9

  57. myininaya
    • one year ago
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    yes

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

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

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

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

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

  63. myininaya
    • one year 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
    • one year 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
    • one year ago
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    This is just like matching socks.

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

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

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