How to find horizontal and vertical asymptotes

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Finding horizontal and vertical asymptotes | Rational expressions | Algebra II

Video by Khan Academy

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We have F of X is equal to three X squared
minus 18X minus 81, over six X squared minus 54.
Now what I want to do in this video
is find the equations for the horizontal
and vertical asymptotes
and I encourage you to pause the video right now
and try to work it out on your own
before I try to work through it.
I’m assuming you’ve had a go at it.
Let’s think about each of them.
Let’s first think about the horizontal asymptote,
see if there at least is one.
The horizontal asymptote is really what is the line,
the horizontal line that F of X approaches
as the absolute value of X approaches,
as the absolute value of X approaches infinity
or you could say what does F of X approach
as X approaches infinity
and what does F of X approach
as X approaches negative infinity.
There’s a couple of ways you could think about it.
Let me just rewrite the definition of F of X
right over here.
It’s three X squared minus 18X minus 81.
All of that over six X squared minus 54.
Now there’s two ways you could think about it.
One you could say, okay,
as X as the absolute value of X
becomes larger and larger and larger,
the highest degree terms in the numerator
and the denominator are going to dominate.
What are the highest degree terms?
Well the numerator you have three X squared
and in the denominator you have six X squared.
As X approaches, as the absolute value of X
approaches infinity,
these two terms are going to dominate.
F of X is going to become approximately three X squared
over six X squared.
These other terms are going to matter less
obviously minus 54 isn’t going to grow at all
and minus 18X is going to grow much slower
than the three X squared,
the highest degree terms are going to be what dominates.
If we look at just those terms
then you could think of simplifying it in this way.
F of X is going to get closer and closer
to 3/6 or 1/2.
You could say that there’s a horizontal asymptote
at Y is equal to 1/2.
Another way we could have thought about this
if you don’t like this whole little bit
of hand wavy argument that these two terms dominate
is that we can divide the numerator and the denominator
by the highest degree or X raised to the highest power
in the numerator and the denominator.
The highest degree term is X squared in the numerator.
Let’s divide the numerator and the denominator
or I should say the highest degree term
in the numerator and the denominator is X squared.
Let’s divide both the numerator
and denominator by that.
If you multiply the numerator times one over X squared
and the denominator times one over X squared.
Notice we’re not changing the value
of the entire expression, we’re just multiplying it
times one if we assume X is not equal zero.
We get two.
In our numerator, let’s see three X squared
divided by X squared is going to be three
minus 18 over X minus 81 over X squared
and then all of that over six X squared
times one over X squared, this is going to be six
and then minus 54 over X squared.
What’s going to happen?
If you want to think in terms of
if you want to think of limits
as something approaches infinity.
If you want to say the limit as X
approaches infinity here.
What’s going to happen?
Well this, this and that are going to approach zero
so you’re going to approach 3/6 or 1/2.
Now, if you say this X approaches negative infinity,
it would be the same thing.
This, this and this approach zero
and once again you approach 1/2.
That’s the horizontal asymptote.
Y is equal to 1/2.
Let’s think about the vertical asymptotes.
Let me write that down right over here.
Let me scroll over a little bit.
Vertical asymptote or possibly asymptotes.
Vertical maybe there is more than one.
Now it might be very tempting to say,
“Okay, you hit a vertical asymptote”
“whenever the denominator equals to zero”
“which would make this rational expression undefined”
and as we’ll see for this case
that is not exactly right.
Just making the denominator equal to zero by itself
will not make a vertical asymptote.
It will definitely be a place
where the function is undefined
but by itself it does not make a vertical asymptote.
Let’s just think about this denominator right over here
so we can factor it out.
Actually let’s factor out the numerator
and the denominator.
We can rewrite this as F of X is equal to the numerator
is clearly every term is divisible by three
so let’s factor out three.
It’s going to be three times X squared
minus six X minus 27.
All of that over the denominator
each term is divisible by six.
Six times X squared minus 9
and let’s see if we can factor the numerators
and denominators out further.
This is going to be F of X is equal to three times
let’s see, two numbers, their product is negative 27,
their sum is negative six.
Negative nine and three seem to work.
You could have X minus nine times X plus three.
Just factor the numerator over the denominator.
This is the difference of squares right over here.
This would be X minus three times X plus three.
When does the denominator equal zero?
The denominator equals zero
when X is equal to positive three
or X is equal to negative three.
Now I encourage you to pause this video for a second.
Think about are both of these vertical asymptotes?
Well you might realize that the numerator
also equals zero when X is equal to negative three.
What we can do is actually simplify this a little bit
and then it becomes a little bit clear
where our vertical asymptotes are.
We could say that F of X,
we could essentially divide the numerator
and denominator by X plus three
and we just have to key,
if we want the function to be identical,
we have to keep the [caveat] that the function itself
is not defined when X is equal to negative three.
That definitely did make us divide by zero.
We have to remember that
but that will simplify the expression.
This exact same function is going to be
if we divide the numerator and denominator
by X plus three,
it’s going to be three times X minus nine
over six times X minus three
for X does not equal negative three.
Notice, this is an identical definition
to our original function
and I have to put this qualifier right over here
for X does not equal negative three
because our original function is undefined
at X equals negative three.
X equals negative three is not a part of the domain
of our original function.
If we take X plus three out of the numerator
and the denominator,
we have to remember that.
If we just put this right over here,
this wouldn’t be the same function
because this without the qualifier is defined
for X equals negative three
but we want to have the exact same function.
You’d actually have a point in discontinuity
right over here
and now we could think about the vertical asymptotes.
Now the vertical asymptotes going to be a point
that makes the denominator equals zero
but not the numerator equals zero.
X equals negative three made both equal zero.
Our vertical asymptote, I’ll do this in green
just to switch or blue.
Our vertical asymptote is going to be
at X is equal to positive three.
That’s what made the denominator equal zero
but not the numerator so let me write that.
The vertical asymptote is X is equal to three.
Using these two points of information
or I guess what we just figured out.
You can start to attempt to sketch the graph,
this by itself is not going to be enough.
You might want to also plot a few points
to see what happens I guess around the asymptotes
as we approach the two different asymptotes
but if we were to look at a graph.
Actually let’s just do it for fun here
just to complete the picture for ourselves.
The function is going to look something like this
and I’m not doing it at scales.
That’s one and this is 1/2 right over here.
Y equals 1/2 is the horizontal asymptote.
Y is equal to 1/2
and we have a vertical asymptote
that X is equal to positive three.
We have one, two …
I’m going to do that in blue.
One, two, three, once again I didn’t draw it to scale
or the X and Y’s aren’t on the same scale
but we have a vertical asymptote just like that.
Just looking at this we don’t know exactly
what the function looks like.
It could like something like this
and maybe does something like that
or it could do something like that
or it could do something like that and that
or something like that and that.
Hopefully you get the idea here
and to figure out what it does,
you would actually want to try out some points.
The other thing we want to be clear is that
the function is also not defined
at X is equal to negative three.
Let me make X equals negative three here.
One, two, three, so the function might look
and once again I haven’t tried out the points.
It could look something like this,
it could look something where we’re not defined
at negative three
and then it goes something like this
and maybe does something like that
or maybe it does something like that.
It’s not defined at negative three
and this would be an asymptote right now
so we get closer and closer
and it could go something like that
or it goes something like that.
Once again, to decide which of these it is,
you would actually want to try out a few values.
I encourage you to, after this video,
try that out on yourself
and try to figure out what the actual graph
of this looks like.

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