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

Mathematics
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find c?
tell me if 1/x^2-c is a fraction form?
Yes it is in fraction form, 1 on top and x^2-c on the bottom

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Other answers:

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

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