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CollieBean Group Title

The graph of f(x)=1/x^2-c has a vertical asymptote at x=3. Find c. PreCalculus help please

  • one year ago
  • one year ago

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  1. samsterz Group Title
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    find c?

    • one year ago
  2. samsterz Group Title
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    tell me if 1/x^2-c is a fraction form?

    • one year ago
  3. CollieBean Group Title
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    Yes it is in fraction form, 1 on top and x^2-c on the bottom

    • one year ago
  4. samsterz Group Title
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    this is calculus y do u want to find c

    • one year ago
  5. CollieBean Group Title
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    I don't know... It confuses me... This is more of an algebra question than calculus.. and I sucked at algebra....

    • one year ago
  6. samsterz Group Title
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    i cant find c with f(x) thingie

    • one year ago
  7. CollieBean Group Title
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    Okay.. Thanks anyway :)

    • one year ago
  8. samsterz Group Title
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    c=x^2-1/y there u go

    • one year ago
  9. myininaya Group Title
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    If f has a vertical asymptote at x=3, then that means the function shouldn't exist at x=3. The function doesn't exist when the fraction doesn't exist. The fraction doesn't exist when the denominator is 0 (when the bottom is 0). When is x^2-c equal to 0? And since the vertical asymptote is at x=3 then we no 3^2-c equal to 0 will give us a value of c that makes the function undefined at x=3.

    • one year ago
  10. myininaya Group Title
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    know not no (darn it)

    • one year ago
  11. CollieBean Group Title
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    so c=9?

    • one year ago
  12. myininaya Group Title
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    Yep! That would give us f(x)=1/(x^2-9) As you can see this function does not exist at x=3 (or x=-3 <---but whatever :p)

    • one year ago
  13. CollieBean Group Title
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    Awesome!! Thanks a million!!! Do you think you could help me with this one too? Find the equation of the slant asymptote for the graph of y=\[y=\frac{ 3x ^{2}+2x-3 }{ x-1 }\] @myininaya

    • one year ago
  14. myininaya Group Title
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    All you really need to do is divide.

    • one year ago
  15. CollieBean Group Title
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    I know I have to turn it into a division problem.. But after that I'm not sure... Because the long division with Xs confuses me..

    • one year ago
  16. myininaya Group Title
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    If you can do the division correctly and if the numerator is 1 degree higher than the numerator, then the slant asymptote will be the quotient.

    • one year ago
  17. myininaya Group Title
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    |dw:1379947728986:dw|

    • one year ago
  18. myininaya Group Title
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    how many times does x go into 3x^2?

    • one year ago
  19. CollieBean Group Title
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    2?

    • one year ago
  20. myininaya Group Title
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    3x^2/x=3xx/x What cancels?

    • one year ago
  21. CollieBean Group Title
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    X?

    • one year ago
  22. myininaya Group Title
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    So you are left with just 3x right?

    • one year ago
  23. myininaya Group Title
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    Since 3x^2/x=3x, then 3x is the amount of times x goes into 3x^2. So we put this up on top of that little half rectangle thing (or division problem). |dw:1379948010561:dw| Take that number we just wrote and multiply it by (x-1) to find out how much remains.

    • one year ago
  24. CollieBean Group Title
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    Wouldn't 3x times x-1 be 3x^2-2x?

    • one year ago
  25. myininaya Group Title
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    3x(x-1)=3x(x)-3x(1) =3x^2-3x

    • one year ago
  26. myininaya Group Title
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    |dw:1379948242687:dw|

    • one year ago
  27. CollieBean Group Title
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    Oh okay, then I subtract that by 3x^2+2x and get -1x don't I?

    • one year ago
  28. myininaya Group Title
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    Nope.

    • one year ago
  29. myininaya Group Title
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    2-(-3)=?

    • one year ago
  30. CollieBean Group Title
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    Ugh the signs mess me up.. it would be 6

    • one year ago
  31. myininaya Group Title
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    Still no. 2-(-3)=2+3=5

    • one year ago
  32. myininaya Group Title
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    |dw:1379948443191:dw|

    • one year ago
  33. myininaya Group Title
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    |dw:1379948484905:dw|

    • one year ago
  34. CollieBean Group Title
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    Gah! I'm multiplying when I should be adding >_< okay so now you bring down the -3 and have 5x-3

    • one year ago
  35. myininaya Group Title
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    Yes. Now you can still do x into 5x how many times? Or what is 5x/x=?

    • one year ago
  36. CollieBean Group Title
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    It goes in 5 times

    • one year ago
  37. myininaya Group Title
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    Right. So we have |dw:1379948593951:dw|

    • one year ago
  38. myininaya Group Title
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    Now you could finish it out, right? 5 times (x-1)

    • one year ago
  39. CollieBean Group Title
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    Would be 5x-5

    • one year ago
  40. myininaya Group Title
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    Yes so we have |dw:1379948698015:dw| Tell me what is remaining?

    • one year ago
  41. CollieBean Group Title
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    8 is remaining, so my answer would be 3x+5+8? Could I combine them and make it 3x+13?

    • one year ago
  42. myininaya Group Title
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    no no... -3-(-5)=-3+5=2 This means we can stop with the division since there are no x's in 2. Also this means that \[\frac{3x^2+2x-3}{x-1}=3x+5+\frac{2}{x-1}\] But as goes to infinity 2/(x-1) goes to zero which means our slant asymptote is just the quotient (3x+5).

    • one year ago
  43. myininaya Group Title
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    as x goes to infinity*

    • one year ago
  44. CollieBean Group Title
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    Ugh I'm bad at this... So my answer would be 3x+5?

    • one year ago
  45. myininaya Group Title
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    Yep.

    • one year ago
  46. myininaya Group Title
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    Do you want to try one?

    • one year ago
  47. myininaya Group Title
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    Find the slant asymptote for \[g(x)=\frac{x^3+2x^2+5x+1}{x^2+1}\]

    • one year ago
  48. CollieBean Group Title
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    Awesome, thank you sooooo much for helping me out and putting up with me this long.. It's muchly appreciated! Okay so first I would put it in division form... \[x ^{2}+1\div x ^{3}+2x ^{2}+5x+1\]

    • one year ago
  49. myininaya Group Title
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    \[(x^3+2x^2+5x+1) \div (x^2+1)\] But were you trying to write |dw:1379949456171:dw| Also you can reply using my picture. (You can click my picture in the upper right hand corner and should be able to use it) It is not any trouble for me. I'm doing this because I just want to.

    • one year ago
  50. myininaya Group Title
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    Err... My picture doesn't show the +1. Do you see everything?

    • one year ago
  51. CollieBean Group Title
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    Yeah I can see all of it, yeah that's what I was trying to put.. |dw:1379950041284:dw|

    • one year ago
  52. myininaya Group Title
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    good so far

    • one year ago
  53. CollieBean Group Title
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    |dw:1379950178748:dw|

    • one year ago
  54. myininaya Group Title
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    wait

    • one year ago
  55. myininaya Group Title
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    |dw:1379950290680:dw| You might want to align your like terms so you don't forget which one you should be subtracting from.

    • one year ago
  56. myininaya Group Title
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    |dw:1379950632494:dw|

    • one year ago
  57. CollieBean Group Title
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    Wait I'm confused.. why is the x under the 5x?

    • one year ago
  58. myininaya Group Title
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    Because we are putting our like terms together

    • one year ago
  59. myininaya Group Title
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    We are doing (x^3+2x^2+5x+1)-(x^3+x) =x^3+2x^2+5x+1-x^3-x =x^3-x^3+2x^2+5x-x+1 =(x^3-x^3)+2x^2+(5x-x)+1 =(0)+2x^2+(4x)+1 =2x^2+4x+1

    • one year ago
  60. myininaya Group Title
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    You only combine like terms.

    • one year ago
  61. CollieBean Group Title
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    Oh okay gotcha so do I bring to 2x^2 down when I subtract?

    • one year ago
  62. myininaya Group Title
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    yep 2x^2-0=2x^2

    • one year ago
  63. myininaya Group Title
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    If you want you can put 0 placeholders in there |dw:1379951194918:dw|

    • one year ago
  64. CollieBean Group Title
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    |dw:1379951263384:dw| So like this?

    • one year ago
  65. myininaya Group Title
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    |dw:1379951375025:dw| And I can't see the plus one but we also need to bring him down The reason I change your -4 to a +4 is because 5-1=4 not -4.

    • one year ago
  66. CollieBean Group Title
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    Ah okay, makes sense |dw:1379951489180:dw| is this right?

    • one year ago
  67. myininaya Group Title
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    is that +2 up there?

    • one year ago
  68. myininaya Group Title
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    also did you put a 2 under the 1(my screen is cutting the picture off)

    • one year ago
  69. CollieBean Group Title
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    Yes it is +2 on top. No I didn't, I see why I need to though|dw:1379952081201:dw|

    • one year ago
  70. myininaya Group Title
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    Tell me what you got for the remainder?

    • one year ago
  71. myininaya Group Title
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    That part is cut off for me.

    • one year ago
  72. myininaya Group Title
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    It should be 4x-1 is the remainder. Yes. And we know there are no x^2's in 4x so we are done. \[\frac{x^3+2x^2+5x+1}{x^2+1}=x+2+\frac{4x-1}{x^2+1}\] Since that R/(x^2+1) goes to 0 then y=x+2 is the slant asymptote. As long as R/D goes to O, then y=Q will be the slant asymptote. This will happen with we have the numerator is 1 degree higher than the denominator. deg(x^3+2x+5x+1)=3 deg(x^2+1)=2 3 is 1 more than 2 so there is a slant asmyptote for (x^3+2x^2+5x+1)/(x^2+1) and it is y=x+2

    • one year ago
  73. CollieBean Group Title
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    Yeah that's what I got ^^

    • one year ago
  74. myininaya Group Title
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    Nice.

    • one year ago
  75. CollieBean Group Title
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    Yep! Thanks for the help! :D

    • one year ago
  76. myininaya Group Title
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    Np.:)

    • one year ago
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