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 ways to solve.

does the dominant term method only find the horizontal asymptote?

Answer 1

Yes, what you call the dominant rule method looks at the highest power of top and bottom. In the example you gave, as x tends to infinity clearly this tends to 0, as the higher power is on the bottom, so we can see y = 0 is an asymptote. In this case the long division method doesn't apply since the bottom is of higher degree than the top. You realise of course that there are also asymptotes where there are excluded values of x,
i.e. since the denominator can not be zero,
we cannot have x^2 - 4 = 0,
and so the values x = 2 and x = -2 are excluded.

Therefore there are vertical asymptotes at x = 2 and x = -2.

Another point special to this particular function type is that there is a factor common to top and bottom, namely x - 2, and so it is equal to
2/(x + 2)

So the graph is just the hyperbola y = 2/(x + 2) except for the one-point "hole" at x = -2. It approaches the vertical asymptote x=2 and the horizontal asymptote y=0 (the x axis)

If it had been (2x - 3)/(x^2 - 4) then you would get a branch in the bottom left-hand corner approaching the negative x axis and the negative (i.e. bottom) part of x = -2, a sort of long backward S between x = -2 and x = 2, and a branch in the top right-hand corner (looking a bit like a branch of a hyperbola but not really a hyperbola).

If the function was (x^2 - 4)/(2x - 3) then you could say that as x tends to infinity, it tends to x^2/2x which is x/2, and so the line y=(1/2)x is an asymptote. However this isn't quite right. The other method, as you suggest, is to divide and find that the expression is equal to

x/2 + 3/4 - (7/4)/(2x - 3)

As x tends to infinity, the fraction term tends to zero, and so the expression approaches x/2 + 3/4, parallel to but not quite the same as the y = x/2 asymptote. Of course x = 3/2 is a vertical asymptote.

Answer 2

To answer your first question, yes that's pretty much it. The reason is that as x gets incredibly large, the dominant term will make the other terms so incredibly insignificant, so it's the only one that matters, really. In this case, your horizontal asymptote is 2/x.

I don't have time to go through a detailed explaination I'm afraid, I hope someone here can afford the pleasure.

Vertical asymptotes are simply when the denominator is zero (because as it approaches zero, the function approaches infinity or negative infinity) and the numerator is not. In this case, algebra will show that x=-2 is a vertical asymptote.

This question also has a hole (when both the numerator and the denominator is zero such that f(n) produces 0/0) at x=2.

To summarize, for your question:

HA: 2/x
VA: x=-2
Not defined: x=2

You can also quickly calculate where the function is positive or negative, but I'll leave that up to you.

You can go further with critical points and maxima minima but assuming the level of your question you may not have learned it.

Hope this helps!

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