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FP1: Asymptote #2

y = x/(x-1)

I can clearly see that x = 1 is an asymptote.

However, in the mark scheme I find that y = 1 is also another asymptote equation, but how do I find that out?

Is there a video tutorial somewhere to find these asymptotes?
Reply 1
Original post by Konnichiwa
y = x/(x-1)

I can clearly see that x = 1 is an asymptote.

However, in the mark scheme I find that y = 1 is also another asymptote equation, but how do I find that out?

Is there a video tutorial somewhere to find these asymptotes?


For x(x1)\frac{x}{(x-1)} to equal 1 x must equal x-1 this is not possible, though as limx\displaystyle\lim_{x\to \infty} you can get close
(edited 11 years ago)
Also, don't forget your horizontal asymptote rules:

1) If the degree of the numerator is the same as the degree of the denominator, the ratio of the leading coefficients will be your horizontal asymptote.
2) If the degree of the numerator exceeds the degree of the denominator, no horizontal asymptote exists.
3) If the degree of the denominator exceeds the degree of the numerator, the horizontal asymptote is y = 0.

In your case, we have situation number '1' and so you have the leading coefficient as 1.

Another way is building on Joostans point. Dividing through by the highest power and calculating the limit will give your HA.
Reply 3
Original post by claret_n_blue
Also, don't forget your horizontal asymptote rules:

1) If the degree of the numerator is the same as the degree of the denominator, the ratio of the leading coefficients will be your horizontal asymptote.
2) If the degree of the numerator exceeds the degree of the denominator, no horizontal asymptote exists.
3) If the degree of the denominator exceeds the degree of the numerator, the horizontal asymptote is y = 0.

In your case, we have situation number '1' and so you have the leading coefficient as 1.

Another way is building on Joostans point. Dividing through by the highest power and calculating the limit will give your HA.


What do you mean by 'degree'? Sorry for my ignorance but I have to self-study FP1 and I find it difficult.

I don't understand what the degree of the numerator/denominator means.
Original post by Konnichiwa
y = x/(x-1)

I can clearly see that x = 1 is an asymptote.

However, in the mark scheme I find that y = 1 is also another asymptote equation, but how do I find that out?

Is there a video tutorial somewhere to find these asymptotes?


Divide x by x-1 using your preferred method.

You will then see why y=1 is a horizontal asymptote.

Spoiler

Original post by Konnichiwa
What do you mean by 'degree'? Sorry for my ignorance but I have to self-study FP1 and I find it difficult.

I don't understand what the degree of the numerator/denominator means.


The degree (or order) of the polynomial is given by the highest power. I.e if you have a function x2+3x+4 x^2 + 3x + 4 then the highest power here is '2' and so you have a function of degree (or order) 2. If you have a function x46x3+x8+3x6 x^4 - 6x ^3 + x^8 + 3x - 6 , then even though all the powers aren't in an ascending or descending order, the highest power here is 8 and so you have a function of degree 8.
Reply 6
Original post by claret_n_blue
The degree (or order) of the polynomial is given by the highest power. I.e if you have a function x2+3x+4 x^2 + 3x + 4 then the highest power here is '2' and so you have a function of degree (or order) 2. If you have a function x46x3+x8+3x6 x^4 - 6x ^3 + x^8 + 3x - 6 , then even though all the powers aren't in an ascending or descending order, the highest power here is 8 and so you have a function of degree 8.

Thanks very much! This will be useful. :smile:

Edit: Thumbed you up. :smile:
(edited 11 years ago)

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