## CollieBean Group Title The graph of f(x)=1/x^2-c has a vertical asymptote at x=3. Find c. PreCalculus help please 10 months ago 10 months ago

1. samsterz Group Title

find c?

2. samsterz Group Title

tell me if 1/x^2-c is a fraction form?

3. CollieBean Group Title

Yes it is in fraction form, 1 on top and x^2-c on the bottom

4. samsterz Group Title

this is calculus y do u want to find c

5. CollieBean Group Title

I don't know... It confuses me... This is more of an algebra question than calculus.. and I sucked at algebra....

6. samsterz Group Title

i cant find c with f(x) thingie

7. CollieBean Group Title

Okay.. Thanks anyway :)

8. samsterz Group Title

c=x^2-1/y there u go

9. myininaya Group Title

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.

10. myininaya Group Title

know not no (darn it)

11. CollieBean Group Title

so c=9?

12. myininaya Group Title

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)

13. CollieBean Group Title

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

14. myininaya Group Title

All you really need to do is divide.

15. CollieBean Group Title

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..

16. myininaya Group Title

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.

17. myininaya Group Title

|dw:1379947728986:dw|

18. myininaya Group Title

how many times does x go into 3x^2?

19. CollieBean Group Title

2?

20. myininaya Group Title

3x^2/x=3xx/x What cancels?

21. CollieBean Group Title

X?

22. myininaya Group Title

So you are left with just 3x right?

23. myininaya Group Title

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.

24. CollieBean Group Title

Wouldn't 3x times x-1 be 3x^2-2x?

25. myininaya Group Title

3x(x-1)=3x(x)-3x(1) =3x^2-3x

26. myininaya Group Title

|dw:1379948242687:dw|

27. CollieBean Group Title

Oh okay, then I subtract that by 3x^2+2x and get -1x don't I?

28. myininaya Group Title

Nope.

29. myininaya Group Title

2-(-3)=?

30. CollieBean Group Title

Ugh the signs mess me up.. it would be 6

31. myininaya Group Title

Still no. 2-(-3)=2+3=5

32. myininaya Group Title

|dw:1379948443191:dw|

33. myininaya Group Title

|dw:1379948484905:dw|

34. CollieBean Group Title

Gah! I'm multiplying when I should be adding >_< okay so now you bring down the -3 and have 5x-3

35. myininaya Group Title

Yes. Now you can still do x into 5x how many times? Or what is 5x/x=?

36. CollieBean Group Title

It goes in 5 times

37. myininaya Group Title

Right. So we have |dw:1379948593951:dw|

38. myininaya Group Title

Now you could finish it out, right? 5 times (x-1)

39. CollieBean Group Title

Would be 5x-5

40. myininaya Group Title

Yes so we have |dw:1379948698015:dw| Tell me what is remaining?

41. CollieBean Group Title

8 is remaining, so my answer would be 3x+5+8? Could I combine them and make it 3x+13?

42. myininaya Group Title

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).

43. myininaya Group Title

as x goes to infinity*

44. CollieBean Group Title

45. myininaya Group Title

Yep.

46. myininaya Group Title

Do you want to try one?

47. myininaya Group Title

Find the slant asymptote for $g(x)=\frac{x^3+2x^2+5x+1}{x^2+1}$

48. CollieBean Group Title

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... $x ^{2}+1\div x ^{3}+2x ^{2}+5x+1$

49. myininaya Group Title

$(x^3+2x^2+5x+1) \div (x^2+1)$ 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.

50. myininaya Group Title

Err... My picture doesn't show the +1. Do you see everything?

51. CollieBean Group Title

Yeah I can see all of it, yeah that's what I was trying to put.. |dw:1379950041284:dw|

52. myininaya Group Title

good so far

53. CollieBean Group Title

|dw:1379950178748:dw|

54. myininaya Group Title

wait

55. myininaya Group Title

|dw:1379950290680:dw| You might want to align your like terms so you don't forget which one you should be subtracting from.

56. myininaya Group Title

|dw:1379950632494:dw|

57. CollieBean Group Title

Wait I'm confused.. why is the x under the 5x?

58. myininaya Group Title

Because we are putting our like terms together

59. myininaya Group Title

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

60. myininaya Group Title

You only combine like terms.

61. CollieBean Group Title

Oh okay gotcha so do I bring to 2x^2 down when I subtract?

62. myininaya Group Title

yep 2x^2-0=2x^2

63. myininaya Group Title

If you want you can put 0 placeholders in there |dw:1379951194918:dw|

64. CollieBean Group Title

|dw:1379951263384:dw| So like this?

65. myininaya Group Title

|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.

66. CollieBean Group Title

Ah okay, makes sense |dw:1379951489180:dw| is this right?

67. myininaya Group Title

is that +2 up there?

68. myininaya Group Title

also did you put a 2 under the 1(my screen is cutting the picture off)

69. CollieBean Group Title

Yes it is +2 on top. No I didn't, I see why I need to though|dw:1379952081201:dw|

70. myininaya Group Title

Tell me what you got for the remainder?

71. myininaya Group Title

That part is cut off for me.

72. myininaya Group Title

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

73. CollieBean Group Title

Yeah that's what I got ^^

74. myininaya Group Title

Nice.

75. CollieBean Group Title

Yep! Thanks for the help! :D

76. myininaya Group Title

Np.:)