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

- anonymous

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

- schrodinger

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- anonymous

find c?

- anonymous

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

- anonymous

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

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## More answers

- anonymous

this is calculus y do u want to find c

- anonymous

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

- anonymous

i cant find c with f(x) thingie

- anonymous

Okay.. Thanks anyway :)

- anonymous

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

- 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

know not no (darn it)

- anonymous

so c=9?

- 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

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

All you really need to do is divide.

- 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

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

|dw:1379947728986:dw|

- myininaya

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

- anonymous

2?

- myininaya

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

- anonymous

X?

- myininaya

So you are left with just 3x right?

- 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

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

- myininaya

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

- myininaya

|dw:1379948242687:dw|

- anonymous

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

- myininaya

Nope.

- myininaya

2-(-3)=?

- anonymous

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

- myininaya

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

- myininaya

|dw:1379948443191:dw|

- myininaya

|dw:1379948484905:dw|

- anonymous

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

- myininaya

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

- anonymous

It goes in 5 times

- myininaya

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

- myininaya

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

- anonymous

Would be 5x-5

- myininaya

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

- anonymous

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

- 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

as x goes to infinity*

- anonymous

Ugh I'm bad at this... So my answer would be 3x+5?

- myininaya

Yep.

- myininaya

Do you want to try one?

- myininaya

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

- 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

\[(x^3+2x^2+5x+1) \div (x^2+1)\]
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

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

- anonymous

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

- myininaya

good so far

- anonymous

|dw:1379950178748:dw|

- myininaya

wait

- 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

|dw:1379950632494:dw|

- anonymous

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

- myininaya

Because we are putting our like terms together

- 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

You only combine like terms.

- anonymous

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

- myininaya

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

- myininaya

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

- anonymous

|dw:1379951263384:dw| So like this?

- 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

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

- myininaya

is that +2 up there?

- myininaya

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

- anonymous

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

- myininaya

Tell me what you got for the remainder?

- myininaya

That part is cut off for me.

- 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

Yeah that's what I got ^^

- myininaya

Nice.

- anonymous

Yep! Thanks for the help! :D

- myininaya

Np.:)

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