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cowgiljl

an asymptotes problem

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The question is

fine a equation f(x) that has a vertical asymptotes of X=1and X=3 and a horizontal asymptotes of Y=1

 

A) (x-1)(x-3) / y-1 = x^2-4x+3 / x-1

 

OR

 

B) (x-1) /(x-1)(x-3) = just the oppsite of above

 

is any of these correct?

I think B is correct but right now I can't find any info on it in my book

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The equation you are looking for is

 

[math]

f(x) = \left\{ \begin{gathered}

\frac{1}{{x^2 - 4x + 3}} + 1 \hfill \\

\frac{{ - 1}}{{x^2 - 4x + 3}} + 1 \hfill \\

\end{gathered} \right.

 

[/math]

 

I hope this helps you

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Or you could try [MATH]y=e^{\frac{1}{x^2-4x+3}}[/MATH] or [MATH]y^{x^2-4x+3}=e[/MATH]

 

In fact you could replace e by any positive constant greater than unity.

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Nice equation, however there are 3 asymptotes, which yield 6 quadrants. Your equation adjusted and shown below would only fill the outer four quadrants. The inner quadrants, that is between x = 1 and x = 3 there is a couple of curves one convex and one concave. Do not asymptotically follow the asymptote lines.

 

[math]

f(x) = \left\{ \begin{gathered}

\frac{1}{{e^{x^2 - 4x + 3} }} \hfill \\

\frac{{ - 1}}{{e^{x^2 - 4x + 3} }} + 2 \hfill \\

\end{gathered} \right.

[/math]

 

Similar sort of graphs are obtained when less than unity the outer quadrant graphs did not asymptotically follow the asymptote lines but the inner quadrants did. However the most interesting value was when 0.999 was used

 

[math]

f(x) = \left\{ \begin{gathered}

\frac{1}{{(0.999)^{x^2 - 4x + 3} }} \hfill \\

\frac{{ - 1}}{{(0.999)^{x^2 - 4x + 3} }} + 2 \hfill \\

\end{gathered} \right.

[/math]

 

We actually get the asymptotes y = 1, x = 1 and x = 3, the closer we get to Unity. Very interesting!

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thank you so much because i am really having a time with this subject because the last math class (pre cal) i took over 5 years ago. and since then i decided to return and get my degree.

 

thanks again joe

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"B is certainely not a solutions as you can get f(x)=1 for x=4"

 

I was doing some rational functions today, and in class, the problem was

 

g(x) = 4(x+1)/x(x-4)

 

It has vertical asymptotes at x=0 and x=4 and horizontal asymptote at y=0.

 

The confusing part was that the asymptote was y=0 and x=-1 gives y=0.

 

I said that it's alright for the graph to pass the asymptote because the asymptote only tells the value a variable approaches as the other variable approaches infinity, but everyone unanimously said no, even my teacher.

 

My teacher then opened his book and said, something about it being a slant or oblique asymptote. However, upon reading my book, I saw oblique asymptotes occur when the numerator is a degree greater than the denominator. And I was still certain y=0 is the horizontal asymptote.

 

So is the graph discontinuous at 0? Or can graphs pass asymptotes?

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"B is certainely not a solutions as you can get f(x)=1 for x=4"

 

I was doing some rational functions today' date=' and in class, the problem was

 

g(x) = 4(x+1)/x(x-4)

 

It has vertical asymptotes at x=0 and x=4 and horizontal asymptote at y=0.

 

The confusing part was that the asymptote was y=0 and x=-1 gives y=0. [/quote']

 

I think the problem may be that it isn't really an asymptote. However, I've not come up with anything better at this point so I can't answer that properly.

 

So is the graph discontinuous at 0? Or can graphs pass asymptotes?

 

The graph is discontinous at 0. If you look at the graph as it approaches the 0 from the left, you can see it shoots off to positive infinity. From the right, it goes down towards negative infinity, and so in effect the graph 'jumps' at 0.

 

As for the other thing, I believe the definition of the asymptote has something to do with the bounding of the function over a particular domain, so it's impossible for graphs to cross them.

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Whoops, I think I mis-stated my question about discontinuity at 0. I meant to ask whether or not (-1,0) was a part of the graph or not. The question then is whether the graph is discontinuous as -1 or not.

 

We learned that for rational functions f(x) = N(x)/D(x), if the degree of the numerator is less the denominator, y=0 is the horizontal asymptote, which means that that function has a horizontal asymptote of y=0.

 

"As for the other thing, I believe the definition of the asymptote has something to do with the bounding of the function over a particular domain, so it's impossible for graphs to cross them."

 

I can see how this is true for vertical functions, since vertical functions are figured out by finding the zeroes of the denominator. If the denominator is 0, the functions is undefined. Horizontal asymptotes are figured out by comparing the degree of the numerator and denominator, but it doesn't mean that the graph would become undefined at this this point or anything.

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