Invent a suitable rational function with characteristics

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SUMMARY

The discussion centers on constructing rational functions with specific asymptotic characteristics. For part (a), the function f(x) = x²/(3(x-1)) is proposed, which has a vertical asymptote at x=1 and behaves like y=x/3 for large x. In part (b), the challenge involves creating a function with vertical asymptotes at x=2 and x=4, zeroes at f(-1) and f(3), and a value of f(0) = 1. The participants emphasize the importance of understanding asymptotic behavior and the role of the denominator in defining vertical asymptotes.

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  • Familiarity with function zeroes and limits
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Glissando
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Homework Statement


For each part, invent a suitable rational function, f(x), with the characteristics given and make a sketch of your function.

a) y behaves like x/3 for large x and x=1 is a vertical asymptote

b) x=2 and x=4 are vertical asymptotes, f(-1) = 0, f(3) = 0, and f(0) = 1

Homework Equations


Zeroes, limits, infinity


The Attempt at a Solution



For part a, I've struggled to understand the question. I understand that there must be a y=x/3. I don't get the "large x"...if there's supposed to be a second x? and if x=1 is a vertical asymptote then you set the denominator equal to 0 and x is supposed to equal 1.

For part 2, I somehow need to figure out a denominator that allows x= 2 and 4.

I honestly have no idea how to start inventing a rational function ): Any help would be appreciated.

Thanks in advance!
 
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Hi Glissando! :smile:

Part (a) just means that you have y=x/3 as skew asymptote.
What must hold for a function to have y=x/3 as a skew asymptote?

What must hold in order for f to have an asymptote at x=1? Can you find such a function? (ignore the rest of the conditions for now...)
 
Hey Micro,

Thank you for your quick response (:

For x=1, then y = 0. I just made up 0 = x-1.

For y = x/3, y must equal 0 so x equals 0...(?) I'm not too sure about skewed asymptotes and the rules it follows ):

EDIT

Hey,

Thanks (; I got the first one to f(x) = x^2/(3(x-1))

I figured that for part b, f(x)= something/ (2-x)(4-x) but when I try to put something with x on top, everything works out except for f(0) = 1 ):
 
Last edited:

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