anonymous
  • anonymous
The graph of f(x)=1/x^2-c has a vertical asymptote at x=3. Find c. PreCalculus help please
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
find c?
anonymous
  • anonymous
tell me if 1/x^2-c is a fraction form?
anonymous
  • anonymous
Yes it is in fraction form, 1 on top and x^2-c on the bottom

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anonymous
  • anonymous
this is calculus y do u want to find c
anonymous
  • anonymous
I don't know... It confuses me... This is more of an algebra question than calculus.. and I sucked at algebra....
anonymous
  • anonymous
i cant find c with f(x) thingie
anonymous
  • anonymous
Okay.. Thanks anyway :)
anonymous
  • anonymous
c=x^2-1/y there u go
myininaya
  • myininaya
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.
myininaya
  • myininaya
know not no (darn it)
anonymous
  • anonymous
so c=9?
myininaya
  • myininaya
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)
anonymous
  • anonymous
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
myininaya
  • myininaya
All you really need to do is divide.
anonymous
  • anonymous
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..
myininaya
  • myininaya
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.
myininaya
  • myininaya
|dw:1379947728986:dw|
myininaya
  • myininaya
how many times does x go into 3x^2?
anonymous
  • anonymous
2?
myininaya
  • myininaya
3x^2/x=3xx/x What cancels?
anonymous
  • anonymous
X?
myininaya
  • myininaya
So you are left with just 3x right?
myininaya
  • myininaya
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.
anonymous
  • anonymous
Wouldn't 3x times x-1 be 3x^2-2x?
myininaya
  • myininaya
3x(x-1)=3x(x)-3x(1) =3x^2-3x
myininaya
  • myininaya
|dw:1379948242687:dw|
anonymous
  • anonymous
Oh okay, then I subtract that by 3x^2+2x and get -1x don't I?
myininaya
  • myininaya
Nope.
myininaya
  • myininaya
2-(-3)=?
anonymous
  • anonymous
Ugh the signs mess me up.. it would be 6
myininaya
  • myininaya
Still no. 2-(-3)=2+3=5
myininaya
  • myininaya
|dw:1379948443191:dw|
myininaya
  • myininaya
|dw:1379948484905:dw|
anonymous
  • anonymous
Gah! I'm multiplying when I should be adding >_< okay so now you bring down the -3 and have 5x-3
myininaya
  • myininaya
Yes. Now you can still do x into 5x how many times? Or what is 5x/x=?
anonymous
  • anonymous
It goes in 5 times
myininaya
  • myininaya
Right. So we have |dw:1379948593951:dw|
myininaya
  • myininaya
Now you could finish it out, right? 5 times (x-1)
anonymous
  • anonymous
Would be 5x-5
myininaya
  • myininaya
Yes so we have |dw:1379948698015:dw| Tell me what is remaining?
anonymous
  • anonymous
8 is remaining, so my answer would be 3x+5+8? Could I combine them and make it 3x+13?
myininaya
  • myininaya
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).
myininaya
  • myininaya
as x goes to infinity*
anonymous
  • anonymous
Ugh I'm bad at this... So my answer would be 3x+5?
myininaya
  • myininaya
Yep.
myininaya
  • myininaya
Do you want to try one?
myininaya
  • myininaya
Find the slant asymptote for \[g(x)=\frac{x^3+2x^2+5x+1}{x^2+1}\]
anonymous
  • anonymous
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... \[x ^{2}+1\div x ^{3}+2x ^{2}+5x+1\]
myininaya
  • myininaya
\[(x^3+2x^2+5x+1) \div (x^2+1)\] 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.
myininaya
  • myininaya
Err... My picture doesn't show the +1. Do you see everything?
anonymous
  • anonymous
Yeah I can see all of it, yeah that's what I was trying to put.. |dw:1379950041284:dw|
myininaya
  • myininaya
good so far
anonymous
  • anonymous
|dw:1379950178748:dw|
myininaya
  • myininaya
wait
myininaya
  • myininaya
|dw:1379950290680:dw| You might want to align your like terms so you don't forget which one you should be subtracting from.
myininaya
  • myininaya
|dw:1379950632494:dw|
anonymous
  • anonymous
Wait I'm confused.. why is the x under the 5x?
myininaya
  • myininaya
Because we are putting our like terms together
myininaya
  • myininaya
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
myininaya
  • myininaya
You only combine like terms.
anonymous
  • anonymous
Oh okay gotcha so do I bring to 2x^2 down when I subtract?
myininaya
  • myininaya
yep 2x^2-0=2x^2
myininaya
  • myininaya
If you want you can put 0 placeholders in there |dw:1379951194918:dw|
anonymous
  • anonymous
|dw:1379951263384:dw| So like this?
myininaya
  • myininaya
|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.
anonymous
  • anonymous
Ah okay, makes sense |dw:1379951489180:dw| is this right?
myininaya
  • myininaya
is that +2 up there?
myininaya
  • myininaya
also did you put a 2 under the 1(my screen is cutting the picture off)
anonymous
  • anonymous
Yes it is +2 on top. No I didn't, I see why I need to though|dw:1379952081201:dw|
myininaya
  • myininaya
Tell me what you got for the remainder?
myininaya
  • myininaya
That part is cut off for me.
myininaya
  • myininaya
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
anonymous
  • anonymous
Yeah that's what I got ^^
myininaya
  • myininaya
Nice.
anonymous
  • anonymous
Yep! Thanks for the help! :D
myininaya
  • myininaya
Np.:)

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