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FP1: Asymptotes

47131314cdae4e0c86ad1a8.png
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.
Original post by Konnichiwa
47131314cdae4e0c86ad1a8.png
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→∞x \to \infty. Easiest way would be to expand out the brackets and multiply the top and bottom by 1x1x\dfrac{\frac{1}{x}}{\frac{1}{x}} and then see what happens as x→∞x \to \infty
Reply 2
Avatar for joostan
joostan
13
Original post by Konnichiwa
47131314cdae4e0c86ad1a8.png
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=1x\ y = \frac{1}{x} it will never quite reach the x-axis, despite coming infinitesimally close.
(edited 11 years ago)
Reply 3
Avatar for aznkid66
aznkid66
1
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.
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.
Reply 5
Avatar for aznkid66
aznkid66
1
Original post by Mr M
Fixed that for you.


:colondollar:

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 ^^
Original post by aznkid66
:colondollar:

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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