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    Click ^ above to see question.

    How do you know for certain y = 0 is an asymptote, it says so in the mark scheme? What method do you use to test that theory.
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    (Original post by Konnichiwa)
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    Click ^ above to see question.

    How do you know for certain y = 0 is an asymptote, it says so in the mark scheme? What method do you use to test that theory.
    You find horizontal asymptotes by looking at the function's behaviour as x \to \infty. Easiest way would be to expand out the brackets and multiply the top and bottom by \dfrac{\frac{1}{x}}{\frac{1}{x}} and then see what happens as x \to \infty
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    (Original post by Konnichiwa)
    Name:  47131314cdae4e0c86ad1a8.png
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    Click ^ above to see question.

    How do you know for certain y = 0 is an asymptote, it says so in the mark scheme? What method do you use to test that theory.
    As above:
    As x tends to infinity, the denominator tends to infinty. Despite this, like the curve\ y = \frac{1}{x} it will never quite reach the x-axis, despite coming infinitesimally close.
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    If you use that method on enough, you will be able to see that rational functions always follow these rules:

    If the numerator is a higher degree than the denominator, then there is no horizontal asymptote (but an oblique one).

    If the numerator is the same degree as the denominator, then there is a horizontal asymptote at y equals the numerator's leading coefficient divided by the denominator's leading coefficient.

    If the numerator is a lower degree than the denominator, then the horizontal asymptote is at y=0.
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    (Original post by aznkid66)
    If the numerator is one degree higher than the denominator, then there is no horizontal asymptote (but an oblique one).
    Fixed that for you.
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    (Original post by Mr M)
    Fixed that for you.


    But wait, if you have f(x)/g(x), and f(x) is a higher degree than g(x), then there will be no horizontal asymptote because f(x) divided by g(x) without the remainder will not be a constant function but a polynomial with a degree greater than or equal to 1.

    So you should've edited the oblique part instead ^^
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    (Original post by aznkid66)


    But wait, if you have f(x)/g(x), and f(x) is a higher degree than g(x), then there will be no horizontal asymptote because f(x) divided by g(x) without the remainder will not be a constant function but a polynomial with a degree greater than or equal to 1.

    So you should've edited the oblique part instead ^^
    Nope, I wanted to leave the reference to oblique asymptotes there as TSR users will encounter them.
 
 
 
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