asymptotes help
- Thread starter bcddd214
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I guess they have bbcode disabled?
http://91.215.159.135/images/asymtopes1.png
http://91.215.159.135/images/asymtopes1.png
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- Apr 12, 2005
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- 11,339
That is not difficult to type in.
f(x) = (x+3)/(x-5)
Okay, what is it that you want to know. There are two asymptotes, one horizontal and one veritcal.
What is your plan for identifying them?
You can also learn just a little LaTeX
\(\displaystyle f(x) = \frac{x+3}{x-5}\)
f(x) = (x+3)/(x-5)
Okay, what is it that you want to know. There are two asymptotes, one horizontal and one veritcal.
What is your plan for identifying them?
You can also learn just a little LaTeX
\(\displaystyle f(x) = \frac{x+3}{x-5}\)
What is Latex?
If it is a way to show my equations easily here, please, do tell!
I do all my stuff in word to make it easy to toss around.
Here is a new one (without Latext).
f(x)=(x^3+x^2-4x-1)/(6x^3-7x-5)
factoring look miserable for the bottom.
and, if this is the case, can I 'always cheat' by completing the square?
Can I get a $2 tutorial on plugging in the above in a manner to show it correctly via latex?
If it is a way to show my equations easily here, please, do tell!
I do all my stuff in word to make it easy to toss around.
Here is a new one (without Latext).
f(x)=(x^3+x^2-4x-1)/(6x^3-7x-5)
factoring look miserable for the bottom.
and, if this is the case, can I 'always cheat' by completing the square?
Can I get a $2 tutorial on plugging in the above in a manner to show it correctly via latex?
- Joined
- Apr 12, 2005
- Messages
- 11,339
It's easy to find. Here's one place. http://www.maths.tcd.ie/~dwilkins/LaTeXPrimer/
You'll have a spot of trouble completing the square on that cubic expression.
It has one Positive Real zero. Can you find it?
The numerator has three Real zeros. Can you find those?
Really, you must find only one. After taht, you can reduce to a quadratic and those are easy, right?
Horizontal Asymptote is an eyeball problem. QUICK - Write it down!
You'll have a spot of trouble completing the square on that cubic expression.
It has one Positive Real zero. Can you find it?
The numerator has three Real zeros. Can you find those?
Really, you must find only one. After taht, you can reduce to a quadratic and those are easy, right?
Horizontal Asymptote is an eyeball problem. QUICK - Write it down!
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- Apr 12, 2005
- Messages
- 11,339
bcddd214 said:Yikes, here is a weird one.
f(x)=(3x-1)/?(x^2-1)
As i recall, there is a trick to do with the denominator, I just forgot it....
Not tricky at all. Factor and read. You shoudl recognize a "Difference of Squares".
Okay, quit trying to get me to do all your homework. You show me the next one.
tkhunny said:bcddd214 said:Yikes, here is a weird one.
f(x)=(3x-1)/?(x^2-1)
As i recall, there is a trick to do with the denominator, I just forgot it....
Not tricky at all. Factor and read. You shoudl recognize a "Difference of Squares".
Okay, quit trying to get me to do all your homework. You show me the next one.
correct?
f(1)=(3x-1)/?(x^2-1)=(3(1)-1)/?(1^2-1)=2/0 is a VA
f(-1)=(3x-1)/?(x^2-1)=(3(-1)-1)/?(?-1?^2-1)=-4/0 is a VA
bcddd214 said:tkhunny said:bcddd214 said:Yikes, here is a weird one.
f(x)=(3x-1)/?(x^2-1)
As i recall, there is a trick to do with the denominator, I just forgot it....
Not tricky at all. Factor and read. You shoudl recognize a "Difference of Squares".
Okay, quit trying to get me to do all your homework. You show me the next one.
correct?
f(1)=(3x-1)/?(x^2-1)=(3(1)-1)/?(1^2-1)=2/0 is a VA
f(-1)=(3x-1)/?(x^2-1)=(3(-1)-1)/?(?-1?^2-1)=-4/0 is a VA
Not trying to get you to do my homework, leading step and a check of the final will be just fine.
J
JeffM
Guest
What tk means by an "eyeball" problem is that the answer is obvious once you understand a fundamental concept. Here you need to think a bit about the MEANING of vertical and horizontal asymptotes in conjunction with the MEANING of a rational function. Once you have put those two concepts firmly together in your mind, many asymptotes become very, very easy to identify. You look at them and just "see" the answer without having to do real work and spend time. They are the low hanging fruit on an exam.
What does it mean if y = f(x) has a vertical asymptote at x = v? It means that as x --> v, y --> plus or minus infinity. In other words, y approaches something that is UNDEFINED in the real number system. What is the most common reason that something is undefined in the real number system? Why because it involves division by zero. With a rational function, there is going to be a vertical asymptote at whatever values of x make the denominator equal zero. Sometimes, finding where the denominator is zero takes some work. Sometimes, it is obvious. Where does (x - 5) = 0? Why at x = 5. No work. An "eyeball" problem.
What does it mean that y = f(x) has a horizontal asymptote at y = h. It either means that y --> h as x gets very large in the positive direction or very large in the negative direction or both. So imagine for y = (x + 3) / (x - 5) what happens when x becomes very large in either direction? A trillion plus 3 divided by a trillion less 5 is not going to be very different from a trillion divided by a trillion, is it? A trillion divided by a trillion is ???? A minus trillion divided by a minus trillion is???? This too is an eyeball problem once you understand what a horizontal asymptote MEANS. Think about what happens when x is huge in the positive direction and when it is huge in the negative direction. The +3 and the -5 become negligible in comparison with x, and (x/x) = 1 in most jurisdictions.
Now consider x / (x[sup:1lvenbr8]2[/sup:1lvenbr8] + 1). No vertical asymptote because (x[sup:1lvenbr8]2[/sup:1lvenbr8] + 1) is everywhere positive. How about when x has a really tremendous magnitude, say a trillion or a minus trillion. Do you think the 1 has much influence? y will approximately equal (x / x[sup:1lvenbr8]2[/sup:1lvenbr8]) = (1 / x). What is a trillionth close to? What is minus trillionth close to? Does zero spring to mind? Another eyeball problem.
Not all asymptote problems are eyeball problems, but lots are if you remember what asymptotes mean instead of trying to memorize a lot of seemingly meaningless rules.
What does it mean if y = f(x) has a vertical asymptote at x = v? It means that as x --> v, y --> plus or minus infinity. In other words, y approaches something that is UNDEFINED in the real number system. What is the most common reason that something is undefined in the real number system? Why because it involves division by zero. With a rational function, there is going to be a vertical asymptote at whatever values of x make the denominator equal zero. Sometimes, finding where the denominator is zero takes some work. Sometimes, it is obvious. Where does (x - 5) = 0? Why at x = 5. No work. An "eyeball" problem.
What does it mean that y = f(x) has a horizontal asymptote at y = h. It either means that y --> h as x gets very large in the positive direction or very large in the negative direction or both. So imagine for y = (x + 3) / (x - 5) what happens when x becomes very large in either direction? A trillion plus 3 divided by a trillion less 5 is not going to be very different from a trillion divided by a trillion, is it? A trillion divided by a trillion is ???? A minus trillion divided by a minus trillion is???? This too is an eyeball problem once you understand what a horizontal asymptote MEANS. Think about what happens when x is huge in the positive direction and when it is huge in the negative direction. The +3 and the -5 become negligible in comparison with x, and (x/x) = 1 in most jurisdictions.
Now consider x / (x[sup:1lvenbr8]2[/sup:1lvenbr8] + 1). No vertical asymptote because (x[sup:1lvenbr8]2[/sup:1lvenbr8] + 1) is everywhere positive. How about when x has a really tremendous magnitude, say a trillion or a minus trillion. Do you think the 1 has much influence? y will approximately equal (x / x[sup:1lvenbr8]2[/sup:1lvenbr8]) = (1 / x). What is a trillionth close to? What is minus trillionth close to? Does zero spring to mind? Another eyeball problem.
Not all asymptote problems are eyeball problems, but lots are if you remember what asymptotes mean instead of trying to memorize a lot of seemingly meaningless rules.