True or false :
All rational functions have some form of asymptote, whether it's slant, vertical, or horizontal.
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Proof by counterexample!
1 x 2 has no asymptote.
I'm pretty sure there has to be a variable in both the numerator and the denominator...
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No, constants are fine. For one thing, we want polynomials to be special cases of rational functions.
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I thought a legitimate rational function must contain at least one unknown in the denominator (but not necessarily a numerator). For 1 x 2 , it can be reduced to x 2 and it is equivalent to a quadratic function.
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@Margaret Zheng – No, there is no such restriction in the definition of a rational function. Look up the definition from any "legitimate" source
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@Otto Bretscher – Ok and I actually just looked it up today. I apologize for the previous comment as it contains false information. Thank you!
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@Margaret Zheng – No problem... we are all here to learn (and have fun)
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@Otto Bretscher – that's so true and I am aiming to learn a new section of a chapter every single day on Brilliant.org now!
You can have curved asymptotes, too
For example the function x^2 +1/(x-1) will have a curved asymptote at y=x^2 when x is infinitely large.
All rational functions are in the form p(x)/q(x), where p(x) and q(x) are polynomials. No asymptotes will be present if q(x) has degree 0.
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The rational function f ( x ) = x x 3 + x simplifies to f ( x ) = x 2 + 1 with a hole at the coordinate ( 0 , 1 ) , but it has no asymptotes!