# How do I find the asymptotes of a curve?

The asymptotes of a curve are the lines or curves that approach a given curve very closely without actually touching it. They make curve sketching much easier. You will probably only have asymptotes that are lines in which case you will have either horizontal asymptotes or vertical asymptotes.

__Horizontal asymptotes__

To find the horizontal asymptotes, we want to see the behaviour of the curve as x tends to infinity. So we look at the order of both the numerator and denominator:

**Case 1**: If the numerator's order is greater than the denominator's (e.g if the function is f(x) = x^{3}/(2x^{2}+1) since x^{3 }has order 3 and x^{2} has order 2) then there will be no horizontal asymptotes.

**Case 2**: If the numerator's order is less than the denominator's (e.g if the function is f(x) = 1/x) then the horizontal asymptote will be y=0

**Case 3**: If the numerator's order is the same as the denominator's, e.g the function f(x) = (2x^{2 }- 4) / (x^{2} - 1), then we look at the highest order term for both sides and look at the coefficient of it. In other words, we ignore all other terms besides 2x^{2} and x^{2} so 2x^{2}/x^{2} = 2. So in this case the asymptote is y=2.

But why does this work? For the horizontal asymptotes we are effectively looking at what our curve will do as x tends to infinity. So in Case 1, when the numerator is bigger than the denominator, as x tends to infinity, y tends to infinity so there is no asymptote. In Case 2, as x tends to infinitiy, we find that y tends to zero. In case 3, as x tends to infinity, all of the smaller terms become insignificant. Let's take a closer look. In order to fully understand, we can substitute x=1000 (or any large number) to roughly see what happens.

**Case 1:** f(x) = x^{3}/(2x^{2}+1). Setting x=1000, we get f(1000)= 1000^{3}/(2(1000)^{2}+1) = 500 and if we set x= 10000 then f(x)=5000 so there is no horizontal asymptote here.

**Case 2:** f(x) = 1/x. Setting x = 1000 we get f(1000)= 0.001 which is close to zero. Similarly, f(10000) = 0.0001 so as we increase x, f(x) tends to zero. Which is why the asymtote is y=0

**Case 3:** f(x) = (2x^{2 }- 4) / (x^{2} -1). f(1000) = (2(1000)^{2 }- 4) / ((1000)^{2 } -1)= 1.999998 which is roughly 2. If we increase x even more then f(10000)= 1.99999998 which is even closer to 2. Which is why the asymptote is y=2

__Vertical asymptotes:__

Here we are effectively looking at the behaviour of the curve as f(x)=y tends to infinity. The only times there will be vertical asymptotes is when the order of the denominator of the fraction is greater than the numerator's. Since f(x) will tend to infinity, the denominator of our curve will tend to zero. For example, let's consider the curve f(x) = (2x^{2 }- 4) / (x^{2} -1)

Here we simply set the denominator x^{2 } -1=0 to find x=1 or x= -1 and these are the vertical asymptotes.