Finding asymptotes.

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May 11th 2010, 03:50 AM
Yehia

Finding asymptotes.

firstly, how do you get from a function like: x-1 + (1/x-3) to the original function before long division?

and secondly i know how you find the asymptotes of a funtion like:
(x^2 + 3)/(x+2)
but how do you get them if it is:
(x+2)/(x^2+3)

??

cheers and thanks very appreciated!
May 11th 2010, 06:23 AM
Anonymous1

Quote:

Originally Posted by Yehia

firstly, how do you get from a function like: x-1 + (1/x-3) to the original function before long division?

$\color{red}{ x-1 + \frac{1}{x-3} = \frac{(x-3)(x-1)+1}{x-3} = \frac{x^2-4x+4}{x-3}}$

and secondly i know how you find the asymptotes of a funtion like:
(x^2 + 3)/(x+2)
but how do you get them if it is:
(x+2)/(x^2+3)

$\color{red}{\text{No vertical asymptotes, since the denominator never = 0.}}$ $\color{red}{\text{Find horizontal asymptotes the same way you would for the other problem.}}$

??

cheers and thanks very appreciated!

aNon1
May 13th 2010, 07:43 AM
Yehia

Quote:

Originally Posted by Anonymous1

aNon1

thanks a lot!!! ok i get it now, but i still dont see how you find the asymptote...? :(
May 13th 2010, 08:24 AM
Anonymous1

Quote:

Originally Posted by Yehia

thanks a lot!!! ok i get it now, but i still dont see how you find the asymptote...? :(

For vertical asymptotes set the denominator =0 and solve for x.

For horizontal:

Quote:

To find horizontal asymptotes:

If the degree (the largest exponent) of the denominator is bigger than the degree of the numerator, the horizontal asymptote is the x-axis (y = 0).
If the degree of the numerator is bigger than the denominator, there is no horizontal asymptote.
If the degrees of the numerator and denominator are the same, the horizontal asymptote equals the leading coefficient (the coefficient of the largest exponent) of the numerator divided by the leading coefficient of the denominator