HELLO!!! Mr. Tarrou. Ok, we just got done finding some vertical asymptotes. Let's turn our attention
to horizontal asymptotes. When you find a horizontal asymptote you need to ask yourself
and what you are trying to figure out is, "What does y approach as x approaches infinity?"
You are not concerned what happens here near the origin, or what happens near the vertical
asymptotes. I am going to use this basic function of y equals one over x as a parent function
and to help my explanation as to this concept I am coming up with here. This is undefined
at zero. I don't really care, for horizontal asymptotes, what is happening near that undefined
value of zero or even what is happening near the origin. No, I care about what is happening
as x becomes increasing small or increasingly large. Well, let's see. If x equals one half,
one divided by one half is two. That is really close to the origin and that is really close
to the origin though and I said that I really don't care about those points. X equals one
and we are at one. What about if x equals ten? Well one divided by ten is point one.
Well what if x equals a thousand? Now it is one divided by a thousand, that is .001. You
can see that as I let these x values get bigger and bigger and bigger, this function is going
to take on values that are very very close to but not exactly equal to zero. So we will
say not that one over x equals zero, but limit as x approaches infinity or negative infinity
of one over x is equal to zero. I am going to recall this fact to help find some horizontal
asymptotes now. Now there are three basic rules for finding horizontal asymptote of
rational functions. I am going to show those to you, but I am going to kind over explain
this for a minute to show you how this is working. We are going to take the equation
y equals x over x square minus four x minus five. I am going to fudge this just a little
bit just a tiny bit to make the algebra a little bit simpler for my PreCalculus students
as opposed to say maybe if you were in Calculus. I am going to say that we are going to take
this equation I want you to find the highest degree and divide or multiply by one over
x to the highest degree. Now a proper Calculus book or teacher may not like that, because
that might end up limits having a denominator of zero and you cannot divide by zero. So
they may have a little bit of an issue with that. You are supposed to divide by the highest
degree of the denominator. But, the only time when picking this highest degree and dividing
that through the top and bottom, the only time that is going to give you a denominator
of zero and cause you to divide by zero and hence have an undefined limit is when there
is no horizontal asymptote. So, it is a little bit fudging of the rules but not too bad I
hope. I am going to put this through here and get y equals x over x squared over...
let's see here... x squared over x squared minus four x over x squared, distributing
this one over x squared all the way through, minus five over x squared. I am going to simplify
this and get y equals one over x over one minus four over x minus five over x squared.
I am not using the proper limit notation because I am only teaching PreCalculus and Calculus.
A calculus teacher would want you to label all this with the proper limit notation as
you go through the problem. So I have taken our function that I claim has a horizontal
asymptote...or it will. I have looked for the highest degree and divided that through
the top and the bottom. I have shown a little bit of algebra and simplifying and now I am
going to come back, maybe not with proper notation, but revisit the idea that one over
x with this little graph as my backup, one over x will have a horizontal asymptote of
y equals zero. It will approach zero as x goes to infinity. That is what you are concerned
with when you deal with horizontal asymptotes. Well if I let x go to infinity. It is bothering
me so I am going to go ahead and write that. If I let x go to infinity, this is going to
approach zero (I just validated that with my picture), four over x is going to approach
zero, and five over x squared well that is going to approach zero that much faster. Now
I end up with something.... Let's see here. I have got zero over one minus zero minus
zero. What is zero over one? ZERO!!! Ok, so what does all this mean? That means that y
equals this equation. Now I have taken and I have divided top and bottom by x squared.
I re-simplified and now I have the limit as x approaches infinity is equal to zero. If
you were to type this into your calculator, you would have a horizontal asymptote of y
equals zero. You know what one of the rules are? One of the rules say that if the degree
of you denominator is larger than the degree of your numerator, then you are always going
to get a horizontal asymptote of y equals zero. That is because the bottom is growing
faster than the top. You put in a really huge number underneath a relatively small number,
you are always going to get a very small decimal and one that will eventually approach zero
as your x values get larger. I am not sure if I will have time to write the rules, just
remember or write it down now... If the bottom degree is larger than the top degree you always
get a horizontal asymptote of y equals zero. Ok, let's take a look at another example.
Let's take a look at the example of y equals x squared minus x minus two over two x squared
plus five x plus three. Now we are going to identify the highest degree. The numerator
has a degree of two and the denominator has a degree of two. I am going to multiply top
and bottom by one over x squared to algebraic show the work of trying to find the horizontal
asymptote. Distribute that through. Now I am going to try and save some space and simplify
as I go. Y is equal to x squared over x squared is one, minus x over x squared is one over
x, minus two times one over x squared is two over x squared. All of this over two x squared
times one over x squared is two plus five x over x squares is five over x, plus three
over x squared. So I have just made my rational equation look different and horizontal asymptotes
are again about when x approaches infinity, not close to zero. We are going to temporarily
and we are going to let x approach infinity. That means that this is going to approach
zero, this is going to approach zero, this is going to approach zero as x approaches
infinity not like x equals two or three but as you put in very huge numbers, and this
is also going to approach zero. So what is left after after I have done the limit? y
equals one over two. So this equation is of x squared minus x minus two over two x squared
plus five x plus three my have some vertical asymptotes if this denominator can equal zero,
but the horizontal asymptote is y equals one half. There is your second rule for horizontal
asymptotes of rational functions. If the degree of the numerator and the degree of the numerator,
if they are the same then your horizontal asymptote is going to be y equals whatever
the leading coefficients are. Kind of nice:) I am showing you the algebra and an idea of
limits to help explain that. If your teacher is not going to discuss limits at all I am
probably making this seem a bit more complicated than it is. So just remember, if the bottom
degree is larger like the previous example...if the bottom degree is larger then the horizontal
asymptote is y equals zero. If the degree on the top and bottom are the same like this
example, then the horizontal asymptote is y equals and then just pull out those leading
coefficients. What happens if the degree in the numerator is larger? One last example.
So we have y equals three x to the third plus two x squared minus five over x plus two.
So we are going to take the highest degree and multiply top and bottom by one over x
to that highest degree like I have in the past two examples, and multiply that through
the polynomial. Let's see what we get. So y equals three x cubed over x cubed is three,
plus two x squared over x cubed is two over x, minus five over x cubed. All of that is
over x over x cubed is one over x squared plus two over x cubed. Here is where a calculus
teacher may not be very happy with me. What happens as x goes to infinity? This will approach
zero. This will approach zero. This will approach zero. This will approach zero. So the limit
as x approaches infinity, I know the Larson Calculus book show this division of the highest
exponent, you should be using the highest exponent of the denominator, because they
don't want to show the you have a zero in the denominator. You cannot divide by zero.
But the fact that you cannot divide by zero, you can't divide by zero it is undefined.
And since the limit is undefined, you don't have a horizontal asymptote. WHOOOO!!! Alright.
So this is undefined so dot dot dot, or fancy math notation therefore there is no horizontal
asymptote. So that is the limit work to help explain these three rules. If the the numerator's degree is larger than
the denominator's degree then you have no horizontal asymptote. I have two minutes left.
If the numerator's degree is equals the denominator's degree, then it is y equals the leading coefficients
from the numerator and denominator. Just work out those leading coefficients and like my
previous example it was y equals one half. If the numerator's degree is smaller than
the denominators degree then the horizontal asymptote is always going to be y equals zero.
BAM!!! Alright. Horizontal Asymptotes.