How do you find the asymptotes of an equation? What is the difference between the long division method and the

Question

How do you find the asymptotes of an equation? What is the difference between the long division method and the dominant rule method of finding asymptote?

the dominant rule method is something like (2x - 4) / (x^2 - 4) becomes 2x / x^2 when x tends to infinity but i don't know how it works?

Please try to show an example and use both was to solve.

EDIT: Please try to show an example and use both ways to solve

Answer 1

Finding assymptotes (at least vertical asymptotes) has nothing to do with either of these methods. If you are dealing with rationals, then its quite easy to find the asymptotes. Just take your numerator and divide by the denominator. If there is no remainder then there is no asymptote otherwise there are asymptotes at the roots of the denominator. Confused yet?

Say you got a question find all the asymptotes in the following rational function:

y = (x^2+2x+1)/(x^2-1)

You can simplify the question by factoring the numerator and the denominator

y = (x+1)(x+1)/(x+1)(x-1)
y = (x+1)/(x-1)

Since x+1 cannot be divided by x-1 without a remainder, there will be an asymptote at x=1 since it is the root of x-1

It is also handy to know that the function is not defined at x=-1 since if you put -1 in the original y, your denominator will turn out to be 0 and that is bad.

Answer 2

to respond to your first question, confident it particularly is fairly plenty it. the reason being that as x gets somewhat great, the dominant term will make the different words so somewhat insignificant, so it rather is the only one that concerns, particularly. subsequently, your horizontal asymptote is two/x. i've got not got time to pass by using an in intensity explaination i'm afraid, i'm hoping somebody right here can arise with the money for the exhilaration. Vertical asymptotes are merely while the denominator is 0 (because of the fact because it procedures 0, the function procedures infinity or unfavorable infinity) and the numerator isn't. subsequently, algebra will practice that x=-2 is a vertical asymptote. this question additionally has a hollow (while the two the numerator and the denominator is 0 such that f(n) produces 0/0) at x=2. To summarize, on your question: HA: 2/x VA: x=-2 no longer defined: x=2 you additionally can rapidly calculate the place the function is beneficial or unfavorable, yet i will go away that as much as you. you are able to go extra with extreme factors and maxima minima yet assuming the point of your question you won't have discovered it. desire this facilitates!

Answer 3

As X gets increasingly large Y= (2x-4)/(X^2-4) gets closer and closer to Y=2x/x^2 which is just Y=2/x
as X gets increasing large Y=2/X gets as close to zero as you want and never increases
one asymptote is therefore the straight line X=0

As X approaches zero Y gets bigger without bound and the closer to zero X is the larger is Y
another asymptote is Y=0

Answer 4

you better imagine it than fully comprehending it. In maths you have imaging it