anyone super good with math & would like to help me?

please? thanks a lot in advance

ok so... this is the question:

find all horizontal and vertical asymptotes of the function.

f(x) = x / x - 1

ok so i’m pretty sure i understand what a vertical asymptote is. would a vertical be 1? and is that all i would need to write? and i have no idea how to find a horizontal asymptote.

thanks again for any help!

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um....i think this may well be the first math question here i don’t know. sorry, to advanced for me. no idea whats going on there, and i have a feeling i would just screw you up if i tried to help.

haha well thanks anyway. apparently everyone knows what an asymptote is just not how to figure it out

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meh, i had to look it up. lol.

the answers 12. yep, no doubt about it.

lol i was expecting something easy like (8x3)5+7-(3+2)=

are you askin the INTEGRAL .. or the DERIVATIVE of the function ..?

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Cid Highwind wrote:
Moderator record hun, ..yada yada yada ..I was talking about, dearest.

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satanic double post

Etain wrote:

Cid Highwind wrote:
Moderator record hun, ..yada yada yada ..I was talking about, dearest.

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jesus

Etain wrote:

Cid Highwind wrote:
Moderator record hun, ..yada yada yada ..I was talking about, dearest.

Cid - don’t call me “hun” or “dearest” thanks.

What is this “asymptotes” you speak of?

Assuming you mean f(x) = x/(x-1), and not f(x) = (x/x)-1;

When x = 1, y= infinity, Asymptote at x = 1

When y = 1, x is impossible, Asymptote at y = 1

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SoB Admin

doesnt it come under “Calculus” instead ?

Etain wrote:

Cid Highwind wrote:
Moderator record hun, ..yada yada yada ..I was talking about, dearest.

Cid - don’t call me “hun” or “dearest” thanks.

It’s not calculus. It’s about manipulating and understanding functions. As long as you know what an Asymptote is, you can work it out in your head anyway.

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SoB Admin

Yeah that’s basic Algebra there.

Calculous would be: Find the First and Second Derivatives and Anti-Derivative of f(x) = X^3 + 4x^2 - 5x

First Derivative would be f'(x) = 3x^2 + 8x - 5

Second Derivative would be f”(x) = 6x + 8

And the Anti-Derivative which is a lil more difficult would be F(x) = 1/4x^4 + 4/3X^3 - 5/2X^2 + C

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If anybody wants to actually solve my calculous problem BTW, please show your work and don’t just steal the answers that I gave. I want to know what level you guys are at.

THE KING HAS RETURNED!

The vt = 1.4

horizontal asymp would be 4

Next question

“Man.. It feels good when alot of losers are around. "

hushbaby wrote: please? thanks a lot in advance

ok so... this is the question:

find all horizontal and vertical asymptotes of the function.

f(x) = x / x - 1

ok so i’m pretty sure i understand what a vertical asymptote is. would a vertical be 1? and is that all i would need to write? and i have no idea how to find a horizontal asymptote.

thanks again for any help!

I’m going to do this MY way.

Okay you take the denominator which is (X-1) and set it equal to O, that means X=1, so at x=1 there is an asymptote, because the denominator can never equal 0

X=-999 (nearing -infinity), Y=.999
X=-2, Y=.66
X=-1, Y=.25
X=0, Y=0
X=1, Y= Undefined
X=2, Y=2
X=3, Y=1.5
X=4, Y=1.33
x=999 (nearing infinity), Y=1.001

Y approaches but never hits at 1, so there is a horizontal assymptote at Y=1

X approaches but never hits 1, so there is a vertical assymptote at X=1

Footnote: My method could be speed up conciderably with the use of “limits” which you learn in calculous.

THE KING HAS RETURNED!

As for you derev

FD f'(x)=3x^2+8x-5,AD F(x)=1/4x^4 + 4/3X^3-5/2X^2+C ,SD f”(x)=6x+8

“Man.. It feels good when alot of losers are around. "

I honestly don’t understand this stuff, but why would you need to know all of this stuff to have a good education? Would it really help you out when your an adult?

Math makes my head hurt. That’s why I don’t pay attention to it in school.

Rise of Shinra