### Video Transcript

Horizontal and Vertical Asymptotes
of a Function

In this video, we will learn how to
find the horizontal and vertical asymptotes of a function. And weβll be looking at a variety
of examples of how we can do this. Let us start by recapping the
definition of an asymptote.

An asymptote is a line which a
curve approaches and gets arbitrarily close to but does not touch it. For example, if we consider the
graph of π¦ is equal to one over π₯. We can see that it has a horizontal
asymptote at π¦ is equal to zero and a vertical asymptote at π₯ is equal to
zero.

Letβs now look at a more rigorous
definition of a vertical and horizontal asymptote. We can define a vertical asymptote
as the following. If, as π₯ approaches some constant
π, π of π₯ approaches positive or negative β, then π₯ equals π is a vertical
asymptote. Another way to think about a
vertical asymptote is any input that does not have a defined output.

We can define a horizontal
asymptote by the following. If, as π₯ approaches positive or
negative β, π of π₯ approaches some constant π, then π¦ equals π is a horizontal
asymptote. Another way in which we can think
about horizontal asymptotes is any output which cannot be reached from any input in
the functions domain.

However, we must be careful using
this line of reasoning since itβs not always the case. Sometimes the output may be able to
be reached by an input in the functions domain. And yet there can still be an
asymptote at this point. When finding horizontal asymptotes,
itβs often easier to consider what happens when π₯ approaches positive or negative
β.

When defining and finding vertical
and horizontal asymptotes, we talk a lot about inputs and outputs. And for this reason, they link
quite heavily into the domain and range of functions. If we know the domain and range of
a function, itβs often easier to find the horizontal and vertical asymptotes. And similarly, if we know the
horizontal and vertical asymptotes, itβs often easier to find the domain and range
of that function. Letβs now move on and look at an
example of how weβre able to find vertical and horizontal asymptotes.

Determine the vertical and
horizontal asymptotes of the function π of π₯ is equal to negative one plus three
over π₯ minus four over π₯ squared.

We can start by finding the
vertical asymptote of this function. Now we can find the vertical
asymptote by finding any input that does not have a defined output. When we look at our function π of
π₯, we notice that it has two rational terms. And these are three over π₯ and
negative four over π₯ squared.

Now a rational term is undefined
when its denominator is zero. So for three over π₯, this is when
π₯ is equal to zero. And for negative four over π₯
squared, this is when π₯ squared is equal to zero. And when π₯ squared is equal to
zero, this of course means that π₯ is also equal to zero. Since these two terms appear in π
of π₯, when either of these terms is undefined, π of π₯ is also undefined.

And so, therefore, we can say that
when π₯ is equal to zero, π of π₯ does not have a defined output. Hence, we have found the vertical
asymptote of π of π₯. And itβs that π₯ is equal to
zero. In order to find any horizontal
asymptotes, we need to find any value that does not appear in the range of their
function. And to do this, we can consider
what happens as π₯ approaches β.

Well, we can look at the π₯ terms
in π of π₯. We have three over π₯ and negative
four over π₯ squared. As π₯ approaches β, we have that
the denominator of both of these rational terms will get larger and larger and
larger. And so both of these rational terms
will approach zero. However, neither of these terms
will ever actually reach zero. They will just get arbitrarily
close to zero.

Therefore, when we look at π of π₯
and we have both of these rational terms approaching zero, we can see that π of π₯
will approach negative one. And we can say that π of π₯ will
get arbitrarily close to negative one without reaching negative one. Hence, weβll have a horizontal
asymptote at π¦ is equal to negative one.

Here we have found the vertical and
horizontal asymptotes of our function π of π₯. They are at π₯ is equal to zero and
π¦ is equal to negative one. And this is the solution to the
question. However, this question is a good
example of why we need to be careful using this reasoning to find horizontal
asymptotes. Since sometimes the value may
appear in the range of the function. Yet there can still be a horizontal
asymptote at this point.

We can see this by setting π of π₯
equal to negative one. We have negative one is equal to
negative one plus three over π₯ minus four over π₯ squared. We can add one to both sides of the
equation to get zero is equal to three over π₯ minus four over π₯ squared.

In our next step, we add four over
π₯ squared to both sides. Then we multiply both sides of the
equation by π₯ squared. Then we divide both sides of the
equation by three to obtain π₯ is equal to four over three. So this tells us that when π₯ is
equal to four over three, π of π₯ is equal to negative one. Hence, negative one is in the range
of π of π₯.

However, there still is an
asymptote at π¦ is equal to negative one. We can see why this is the case by
considering the graph of π of π₯. Using a graphical calculator or
some graphing software, we can see that the graph of π of π₯ would look something
like this. We can see the asymptotes at π₯ is
equal to zero and π¦ is equal to negative one. And we can see where the line of π
of π₯ crosses the asymptote at π¦ is equal to negative one and π₯ is equal to four
over three. Then we can see how π of π₯
continues to show asymptotic behaviour towards the line π¦ is equal to negative
one.

Since if we look to the right of π₯
is equal to four over three, we can see that π of π₯ is getting arbitrarily close
to π¦ is equal to negative one without actually touching that line. And this is why we must be careful
using this reasoning when finding horizontal asymptotes. In the next example, weβll see how
we can find the asymptote of a hyperbola. A hyperbola is a type of rational
function with two asymptotes.

What are the asymptotes of the
hyperbola π¦ is equal to eight over four π₯ minus three plus five over three?

We can start by finding the
vertical asymptote of this hyperbola. Weβll use the fact that a vertical
asymptote can be described as any input with no defined output. Looking at the equation of our
hyperbola, we see that we have a rational term, which is eight over four π₯ minus
three.

Now we know that any rational term
is undefined when the denominator is zero. So this is when four π₯ minus three
is equal to zero. We can rearrange this to find
π₯. It gives us that π₯ is equal to
three over four. We now have that when π₯ is equal
to three over four, this rational term of eight over four π₯ minus three is not
defined.

Hence, when we input π₯ is equal to
three over four into the equation for our hyperbola, weβll have an undefined
output. Therefore, our hyperbola will have
a vertical asymptote at π₯ is equal to three over four.

Now we can move on to find the
horizontal asymptote. Horizontal asymptotes are values
that are not in the range of the function. In order to find such values, we
can consider what happens as π₯ approaches positive or negative β.

Now the only π₯-dependent term in
our equation is eight over four π₯ minus three. Now as π₯ approaches positive or
negative β, this rational term approaches zero. And it in fact gets arbitrarily
close to zero. Therefore, if we look back at the
equation of the hyperbola, we can see that π¦ will get arbitrarily close to five
over three as π₯ approaches positive or negative β. Since the rational term in the
equation will approach zero. Hence, our hyperbola has a
horizontal asymptote at π¦ is equal to five over three. And so now weβve found the
asymptotes of our hyperbola, which completes the solution to this question.

Before we move on to our next
example, letβs quickly note that it is in fact possible for a function to have more
than one horizontal or vertical asymptote. For example, consider the function
π of π₯ is equal to one over π₯ squared minus four. We can factor the denominator of
this function to obtain one over π₯ minus two multiplied by π₯ plus two.

Now we can identify vertical
asymptotes as any input with no defined output. Since π of π₯ is a rational
function, this will occur when the denominator is equal to zero. So when π₯ minus two multiplied by
π₯ plus two is equal to zero. This gives us two solutions and,
therefore, two asymptotes. And this is at π₯ is equal to two
and π₯ is equal to negative two.

Using these asymptotes, we could
try to sketch the graph of π of π₯. However, we first need to consider
what happens to π of π₯ around the values of π₯ is equal to two and π₯ is equal to
negative two. We should consider when π₯ is less
than negative two, when π₯ is between negative two and two, and when π₯ is greater
than two.

When π₯ is less than negative two
and greater than two, π₯ squared minus four is greater than zero. Hence, π of π₯ must be
positive. And when π₯ is between negative two
and two, π₯ squared minus four is less than zero. Hence, π of π₯ is negative. Using this information, weβre able
to sketch a graph of π of π₯, something like this. And as we can see, it has two
vertical asymptotes. Finding these asymptotes really
helped us to sketch this graph. So we can see how useful asymptotes
can be for sketching graphs.

Now there are certain cases where
we must be very careful when trying to find asymptotes. And these are the cases where our
function has a factor which can be cancelled. Consider the following example.

Find the asymptotes of the function
π of π₯ is equal to π₯ plus two over π₯ squared minus four.

We would normally start by looking
for the vertical asymptotes of this function. However, if we look carefully at
our function, we notice that the denominator can be factored. Hence, we can write π of π₯ as π₯
plus two over π₯ plus two multiplied by π₯ minus two. And we notice that we can cancel a
factor of π₯ plus two.

However, we must be careful there
since, in doing this, weβre slightly changing the function. After cancelling the factor, we can
call the new function π of π₯. We have that π of π₯ is equal to
one over π₯ minus two. We can see how these two functions
differ slightly by considering the domains of these functions.

We can see that if we inputted π₯
is equal to negative two into π of π₯, we would have an undefined output. Since this would give π a
denominator of zero. However, we are able to input π₯ is
equal to negative two into π of π₯.

Now itβs important to note that
although these two functions differ slightly, they in fact have the same
asymptotes. Therefore, weβre able to find the
asymptotes of π by finding the asymptotes of π. So letβs find these asymptotes. We can identify vertical asymptotes
as any input with no defined output.

Since π of π₯ is a rational
function, this happens when the denominator is equal to zero or when π₯ minus two is
equal to zero. Rearranging this, we have π₯ is
equal to two. Hence, π of π₯ has a vertical
asymptote at π₯ is equal to two. We can identify a horizontal
asymptote as any value that is not in the range of the function. We can find such values by
considering what happens as π₯ approaches positive or negative β.

We can see that as π₯ gets very
large in either the positive or negative direction that the denominator of π of π₯
gets very large in either the positive or negative direction. Therefore, π of π₯ will get closer
and closer to zero. So we have found that π of π₯ has
a horizontal asymptote at π¦ is equal to zero.

Since π of π₯ and π of π₯ share
the same asymptotes, in finding the vertical and horizontal asymptotes of π, weβve
found the vertical and horizontal asymptotes of π. And this completes our solution to
this question.

But before we move on, letβs
quickly consider how π and π differ with a quick sketch. Here we have sketches of π of π₯
and π of π₯. We can see the asymptotes at π¦ is
equal to zero and π₯ is equal to two. Now the only difference between
these two graphs is that π of π₯ is undefined at π₯ is equal to negative two. And π of π₯ is defined at π₯ is
equal to negative two. And despite this, we can see that
the two graphs still have the same asymptotes. In our final example, weβll see how
we can use asymptotes in order to identify the graph of a function.

Which of the following graphs
represents π of π₯ is equal to one over π₯ plus one?

Letβs start by finding the vertical
asymptotes of π of π₯. We can find vertical asymptotes by
identifying any input with no defined output. Since π of π₯ is a rational
function, this occurs when its denominator is zero, so when π₯ plus one is equal to
zero. Rearranging, we can see that this
is when π₯ is equal to negative one.

Here we can deduce that π of π₯
has a vertical asymptote at π₯ is equal to negative one. c) and d) are the only
graphs with vertical asymptotes at π₯ is equal to negative one. Therefore, we can eliminate options
a) and b). When we look at the graphs of c)
and d), we can see that they both have a horizontal asymptote at π¦ is equal to
zero. Hence, our function π of π₯ must
have a horizontal asymptote at π¦ is equal to zero.

Now letβs see how graphs c) and d)
differ. For graph c), we can see that when
π₯ is less than negative one, π of π₯ is negative. And when π₯ is greater than
negative one, π of π₯ is positive. However, for graph d), when π₯ is
less than negative one, π of π₯ is positive. And when π₯ is greater than
negative one, π of π₯ is negative.

Now letβs see what happens to π of
π₯ given in the question when π₯ is less than negative one and when π₯ is greater
than negative one. We have that when π₯ is less than
negative one, π₯ plus one is negative. Therefore, π of π₯ must also be
negative. And when π₯ is greater than
negative one, π₯ plus one is positive. Hence, π of π₯ is also
positive. And this information about π
agrees with what weβve shown for graph c). Therefore, our solution is that the
graph c) represents our function π of π₯.

Weβve now covered a variety of
examples of how we can find asymptotes and how useful asymptotes can be, especially
when identifying or drawing graphs. Letβs now recap some key points of
the video.

Key Points. To find the vertical asymptotes of
a function, we need to identify any point that would lead to a denominator of
zero. But be careful if the function can
be simplified. To find the horizontal asymptotes
of a rational function, we need to identify any value that the function cannot
take. However, we must be careful here as
the function may be able to take the value at the asymptote as we saw in the first
example. We can use the asymptotes to help
us identify the domain and range of a function. And finally, we can use the
asymptotes to help us sketch and identify the graph of a function.