PDA

View Full Version : all you math guys. I NEED help. asymptotes



megadeth
09-01-2008, 05:51 PM
Can someone please explain what the fuck this means? I don't know what the fuck the "lim" means...

http://img.photobucket.com/albums/v316/megadeth600606/wtf-1.jpg

skimnczach
09-01-2008, 05:58 PM
You will have a vertical asymptote when the function is undefined. There are two types of being undefined, the names of which I don't remember.

So you have f(x)=(x^2-100)/(x^2-10x)=(x-10)(x+10) / [x(x+10)]

I arrived at the second equation by simply factoring x^2-100 and x^2+10x.

Now you cancel what you have on the bottom and top, that is, the x+10 is both on the bottom and the top so they cancel out.

You are now left with f(x) = (x-10) / x.

Now where the bottom is equal to zero, you have a vertical asymptote since the function does not exist there and it is fully simplified. Since the bottom is simply x, when you have x=0 there is a vertical asymptote.

You can always graph the equation as well.

Home this helps.

ZapBulletRider
09-01-2008, 05:59 PM
in before skimnczach

hey zach, help him with his ASYMPTOPES
lol



edit: FUCK I wasn't in before zach

hobo
09-01-2008, 06:00 PM
fail zbr

skimnczach
09-01-2008, 06:00 PM
epic fail

sandsurfngbomber
09-01-2008, 06:01 PM
all i wanna say is asymptotes are a bitch!

junwin
09-01-2008, 06:01 PM
You wouldn't know your asymptote from a hole in the ground.

Joey M.
09-01-2008, 06:01 PM
You will have a vertical asymptote when the function is undefined. There are two types of being undefined, the names of which I don't remember.

So you have f(x)=(x^2-100)/(x^2-10x)=(x-10)(x+10) / [x(x+10)]

I arrived at the second equation by simply factoring x^2-100 and x^2+10x.

Now you cancel what you have on the bottom and top, that is, the x+10 is both on the bottom and the top so they cancel out.

You are now left with f(x) = (x-10) / x.

Now where the bottom is equal to zero, you have a vertical asymptote since the function does not exist there and it is fully simplified.

Home this helps.

Nice try, but it just means your mom likes it in the butt.

Kude
09-01-2008, 06:01 PM
hahaha

ZapBulletRider
09-01-2008, 06:02 PM
well there's really no way to duplicate the way Vrablic says "asymptote" in text, but zach knows what I'm talking about

skimnczach
09-01-2008, 06:03 PM
Nice try, but it just means your mom likes it in the butt.
Thanks man, you're really funny.


well there's really no way to duplicate the way Vrablic says "asymptote" in text, but zach knows what I'm talking about
:-)

megadeth
09-01-2008, 06:06 PM
but why is it (-10,2) and not just -10? where did that 2 come from?

skimnczach
09-01-2008, 06:09 PM
When you plug x=-10 into your original f(x), you'll get an undefined answer. You only get an undefined answer when the function either has a "hole" in it, or there is a vertical asymptote.

The solution sheet is saying that there is a hole at (-10,2) which means there is NO vertical asymptote. If you plug x=-10 into the simplified equation (after you cancel the x+10's), then you'll get the 2.

The ANSWER is x=0. (-10,2) is just a point which they were referencing for the "hole" situation.

EDIT: The appropriate way to solve this type problem without caring what is going on is to just factor everything, cancel things out, and check what makes the bottom equal to zero.

They're saying that in the original f(x)=(x^2-100)/(x^2+10x), the bottom: x^2+10x = 0 if x=0 or x=-10. Outlined above, we ruled out x=-10.

Icej
09-01-2008, 06:16 PM
lim means limit and it is the limit as x approaches -10. That is all I remember from calc.

megadeth
09-01-2008, 06:17 PM
Do I have to know limits to solve this style of problem?

and I never took college alg. I hope that won't hurt me.

Icej
09-01-2008, 06:18 PM
It will.

skimnczach
09-01-2008, 06:23 PM
Do I have to know limits to solve this style of problem?
What class are you in / what math classes are you planning on taking?

megadeth
09-01-2008, 06:24 PM
well, what is the "answer" im looking for. it doesn't give me what it wants... it just keeps rabling on about q(0) and shit that I don't understand AT ALL.

megadeth
09-01-2008, 06:26 PM
What class are you in / what math classes are you planning on taking?


I am in business calculus. I have never had any pre-calc class. My SAT was high enough for me to skip college algebra and go right into a Calculus class. I have always been dumb in math, and as a senior in high school I took "integrated math" which was basically just a review of everything like Algebra. I am taking statistics next semester hopefully, so we'll see...

Icej
09-01-2008, 06:29 PM
You should be alright if you listen in class and do the homework. It would have helped if you took trig in high school or college. When you get to cos and sin derivatives and shit it is a little tricky.

megadeth
09-01-2008, 06:33 PM
well the good thing is that my professor said that we aren't doing anything in trig. So sins, cos, or tans. Since its business calculus, its like geared towards business. ( go fucking figure. haha)

Icej
09-01-2008, 06:41 PM
I took business calc it was called Quantitative Methods. It was easy as fuck. Just listen in class.

jesse
09-01-2008, 06:43 PM
business calc= easy peasy lemon squeazy

Icej
09-01-2008, 06:45 PM
It is lemonezy.

megadeth
09-01-2008, 06:47 PM
yah jesse, did you take it? because I know they offer the same course at UF.

Black
09-01-2008, 06:51 PM
well, what is the "answer" im looking for. it doesn't give me what it wants... it just keeps rabling on about q(0) and shit that I don't understand AT ALL.
Find a student in a lot higer math class to explain it, they will usually have some short cuts that help. It's almost impossible to explain over a computer, but once they click in your head, they're easy. It's releaving as hell when you start getting em and they all check out.

skimnczach
09-01-2008, 06:57 PM
Find a student in a lot higer math class to explain it, they will usually have some short cuts that help. It's almost impossible to explain over a computer, but once they click in your head, they're easy. It's releaving as hell when you start getting em and they all check out.
Yeah, it is very difficult to do this over the computer. You better have meant "find a student in a lot higher math class to explain it in real life." Otherwise, I'm offended.

Black
09-01-2008, 07:00 PM
lol yeah, that's what I meant.

Not exactly "skimnczach is a dumbass, don't listen to the fucker,"
though that's apparently how it came out.

can-can
09-01-2008, 07:09 PM
limits are as X approaches A, what the Y value is. so as X gets really really really close to A from both sides, what is the y value.

asymptotes. if you have an equation, f(x). and you plug in number A for x, and the result is undefined (3/0, 5/0, etc.) then you have an asymptote or a hole. a vertical asymptote would be caused by f(5)=((x-3)/(x-5). since you cant divide by zero, X can never equal 5. But it can get reallyy really reallly close to 5. this is where limits come into play. You think, if X is really close, but a little bit less than five. then the top of that equation will be positive
(x-3)=(5-3)=2. and the bottom of that equation will be (x-5)= (a little bit less than five-5)=(a negative really small(-.0000000001, etc.) number). When you divide a positive number, by a reallyy really small negative number, you get a very big positive number. And as you get closer to 5, the bottom number stays negative, but gets even smaller (-.0000000000000000001), and as the denominator gets smaller, the whole fraction gets much bigger and closer to negative infinity. So the limit as X approaches 5 from the left of (x-3)/(x-5)= negative infinity. As X approaches 5 from the right (5.1,5.01,5.001, etc.) then the numerator is still about positive two, and the denominator is a positive very small number. So positive number divided by very very tiny number equals something like infinity.

summary:
example function: (x-3)/(x-5)

so as X approaches 5 from the left (4.9, 4.99, 4.999, etc.), the function gets closer to negative infinity. And as X approaches 5 from the right (5.1, 5.01, 5.001), the function gets closer to infinity. The function cannot equal five, because you cannont divide anything by zero or the whole universe gets fucked, so there is an invisible line at five, on the right side of the invisible line, the function gets close to infinity, on the left side, the function gets close to negative infinity, but there is never a value for x=5.

my explanation might not make much sense, but it gave me something to do other than my calc homework. I studied limits and asymptotes last year so its all pretty easy for me right now, until i forget it all when i realize that you dont actually use calc in real life.

megadeth
09-01-2008, 07:16 PM
^intense.

can-can
09-01-2008, 07:19 PM
did that make sense or help at all? its actually a pretty easy thing to explain, and its an easy concept to grasp i think, but everything gets tougher over the interwebz.

megadeth
09-01-2008, 07:22 PM
not right away, but I am going to go back and pick it apart, because I recognize all the terms, its just the first time where everything has been related in one section for me. So I am going to attempt to piece it together tomorrow. I've been doing homework all day and can't think anymore.

can-can
09-01-2008, 07:36 PM
all i talked about were vertical asymptotes. theres also horizontal and oblique(diagonal) asymptotes. same concept basically, the function always approaches an "invisible line", but never actually touches it. There are just different ways to find them. haha have fun with it.

skimnczach
09-01-2008, 10:32 PM
Not exactly "skimnczach is a dumbass, don't listen to the fucker,"
though that's apparently how it came out.
Haha, no man. I was just pulling your leg, I knew what you meant. I'm just killing myself with graduate math classes as an undergraduate right now. Things like this make me smile.

If you need help on vertical asymptotes, look at this link, mega:
http://www.purplemath.com/modules/asymtote.htm

This may help with limits, though it is a poorly designed web page:
http://www.coolmath.com/limit1.htm

You really don't need to know limits to solve this problem though.

narwhal
09-02-2008, 01:52 AM
lol On Zach beating Whit

ZapBulletRider
09-02-2008, 05:08 AM
it's funny b/c he's never on here, except magically right when somebody's asking a math question. It's like Spidey sense.

skimnczach
09-02-2008, 07:04 AM
Actually, it really was. I hadn't been on SOMB all day long and I decided to check it and randomly see a 5 minute old math post.