Finding horizontal and vertical asymptotes

Voiceover: 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.