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CollieBean

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

  • 7 months ago
  • 7 months ago

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  1. samsterz
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    find c?

    • 7 months ago
  2. samsterz
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    tell me if 1/x^2-c is a fraction form?

    • 7 months ago
  3. CollieBean
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    Yes it is in fraction form, 1 on top and x^2-c on the bottom

    • 7 months ago
  4. samsterz
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    this is calculus y do u want to find c

    • 7 months ago
  5. CollieBean
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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....

    • 7 months ago
  6. samsterz
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    i cant find c with f(x) thingie

    • 7 months ago
  7. CollieBean
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    Okay.. Thanks anyway :)

    • 7 months ago
  8. samsterz
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    c=x^2-1/y there u go

    • 7 months ago
  9. myininaya
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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.

    • 7 months ago
  10. myininaya
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    know not no (darn it)

    • 7 months ago
  11. CollieBean
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    so c=9?

    • 7 months ago
  12. myininaya
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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)

    • 7 months ago
  13. CollieBean
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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

    • 7 months ago
  14. myininaya
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    All you really need to do is divide.

    • 7 months ago
  15. CollieBean
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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..

    • 7 months ago
  16. myininaya
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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.

    • 7 months ago
  17. myininaya
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    |dw:1379947728986:dw|

    • 7 months ago
  18. myininaya
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    how many times does x go into 3x^2?

    • 7 months ago
  19. CollieBean
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    2?

    • 7 months ago
  20. myininaya
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    3x^2/x=3xx/x What cancels?

    • 7 months ago
  21. CollieBean
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    X?

    • 7 months ago
  22. myininaya
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    So you are left with just 3x right?

    • 7 months ago
  23. myininaya
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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.

    • 7 months ago
  24. CollieBean
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    Wouldn't 3x times x-1 be 3x^2-2x?

    • 7 months ago
  25. myininaya
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    3x(x-1)=3x(x)-3x(1) =3x^2-3x

    • 7 months ago
  26. myininaya
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    |dw:1379948242687:dw|

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

    • 7 months ago
  28. myininaya
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    Nope.

    • 7 months ago
  29. myininaya
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    2-(-3)=?

    • 7 months ago
  30. CollieBean
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    Ugh the signs mess me up.. it would be 6

    • 7 months ago
  31. myininaya
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    Still no. 2-(-3)=2+3=5

    • 7 months ago
  32. myininaya
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    |dw:1379948443191:dw|

    • 7 months ago
  33. myininaya
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    |dw:1379948484905:dw|

    • 7 months ago
  34. CollieBean
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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

    • 7 months ago
  35. myininaya
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    Yes. Now you can still do x into 5x how many times? Or what is 5x/x=?

    • 7 months ago
  36. CollieBean
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    It goes in 5 times

    • 7 months ago
  37. myininaya
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    Right. So we have |dw:1379948593951:dw|

    • 7 months ago
  38. myininaya
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    Now you could finish it out, right? 5 times (x-1)

    • 7 months ago
  39. CollieBean
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    Would be 5x-5

    • 7 months ago
  40. myininaya
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    Yes so we have |dw:1379948698015:dw| Tell me what is remaining?

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

    • 7 months ago
  42. myininaya
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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).

    • 7 months ago
  43. myininaya
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    as x goes to infinity*

    • 7 months ago
  44. CollieBean
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    Ugh I'm bad at this... So my answer would be 3x+5?

    • 7 months ago
  45. myininaya
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    Yep.

    • 7 months ago
  46. myininaya
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    Do you want to try one?

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

    • 7 months ago
  48. CollieBean
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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... \[x ^{2}+1\div x ^{3}+2x ^{2}+5x+1\]

    • 7 months ago
  49. myininaya
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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.

    • 7 months ago
  50. myininaya
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    Err... My picture doesn't show the +1. Do you see everything?

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

    • 7 months ago
  52. myininaya
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    good so far

    • 7 months ago
  53. CollieBean
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    |dw:1379950178748:dw|

    • 7 months ago
  54. myininaya
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    wait

    • 7 months ago
  55. myininaya
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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.

    • 7 months ago
  56. myininaya
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    |dw:1379950632494:dw|

    • 7 months ago
  57. CollieBean
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    Wait I'm confused.. why is the x under the 5x?

    • 7 months ago
  58. myininaya
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    Because we are putting our like terms together

    • 7 months ago
  59. myininaya
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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

    • 7 months ago
  60. myininaya
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    You only combine like terms.

    • 7 months ago
  61. CollieBean
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    Oh okay gotcha so do I bring to 2x^2 down when I subtract?

    • 7 months ago
  62. myininaya
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    yep 2x^2-0=2x^2

    • 7 months ago
  63. myininaya
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    If you want you can put 0 placeholders in there |dw:1379951194918:dw|

    • 7 months ago
  64. CollieBean
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    |dw:1379951263384:dw| So like this?

    • 7 months ago
  65. myininaya
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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.

    • 7 months ago
  66. CollieBean
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    Ah okay, makes sense |dw:1379951489180:dw| is this right?

    • 7 months ago
  67. myininaya
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    is that +2 up there?

    • 7 months ago
  68. myininaya
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    also did you put a 2 under the 1(my screen is cutting the picture off)

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

    • 7 months ago
  70. myininaya
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    Tell me what you got for the remainder?

    • 7 months ago
  71. myininaya
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    That part is cut off for me.

    • 7 months ago
  72. myininaya
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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

    • 7 months ago
  73. CollieBean
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    Yeah that's what I got ^^

    • 7 months ago
  74. myininaya
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    Nice.

    • 7 months ago
  75. CollieBean
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    Yep! Thanks for the help! :D

    • 7 months ago
  76. myininaya
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    Np.:)

    • 7 months ago
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