### 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 by considering certain limits. Letβs start by covering the definition of an asymptote. A horizontal or vertical asymptote of a curve is a straight line such that the
distance between the curve and the line approaches zero as either the π₯- or
π¦-coordinate approaches infinity. We can see an example of an asymptote. If we consider the graph of the function π of π₯ is equal to one over π₯. This function has a vertical asymptote at π₯ equals zero and a horizontal asymptote
at π¦ equals zero.

There are various ways in which an asymptote can occur. For vertical asymptotes, one way which the asymptote can occur is if the function
tends to positive infinity from the left and negative infinity from the right. Another similar way is if the function approaches negative infinity from the left and
positive infinity from the right which is what we saw in the case of π of π₯ is
equal to one over π₯. And now, the case is if the function approaches negative infinity from both the left
and right, similarly, the curve could approach positive infinity from both the left
and right. Alternatively, it may be that only one of the left or right limit is infinite. This could be either the left or right limit tending to either positive or negative
infinity.

From this, we can conclude that π₯ equals π is a vertical asymptote. If as π₯ approaches π from either left or right, π of π₯ tends to positive or
negative infinity. Now, we can consider horizontal asymptotes. Now, these will occur when the limit as π₯ approaches either positive or negative
infinity of π of π₯ is equal to some constant. We say that π¦ equals π is a horizontal asymptote if as π₯ approaches positive or
negative infinity, π of π₯ tends to π. We can write these definitions for vertical and horizontal asymptotes in terms of
limits.

For vertical asymptote, π₯ equals π is vertical asymptote if either the limit as π₯
approaches π from above of π of π₯ is equal to positive or negative infinity or
the limit as π₯ approaches π from below of π of π₯ is equal to positive or
negative infinity. For horizontal asymptotes, we say that π¦ equals π is a horizontal asymptote, if the
limit as π₯ approaches infinity of π of π₯ is equal to π or the limit as π₯
approaches negative infinity of π of π₯ is equal to π. Now that we have covered the definition of horizontal and vertical asymptote and the
ways in which the different asymptote can occur, weβre ready to look at an
example.

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.

Letβs start by writing our function as one fraction. We find a common denominator of π₯ squared. And we can write π of π₯ as negative π₯ squared plus three π₯ minus four over π₯
squared. In order to find the vertical asymptotes, we need to find the values of π such that
the limit as π₯ approaches π from below of π of π₯ is equal to plus or minus
infinity or the limit as π₯ approaches π from above of π of π₯ is equal to plus or
minus infinity. Since our π of π₯ is irrational function, this will happen as the denominator of our
function approaches zero.

Now, the denominator of our function is simply π₯ squared, so we can say that the
vertical asymptotes will occur when π₯ squared is equal to zero. So we have found that there will be a vertical asymptote at π₯ equals zero. In order to find our horizontal asymptotes, we need to find the values of π such
that the limit as π₯ approaches infinity of π of π₯ is equal to π. Or the limit as π₯ approaches negative infinity of π of π₯ is also equal to π. So we take the limit of our function as π₯ approaches infinity. In order to find this limit, we need to multiply the top and bottom of our fraction
by one over π₯ squared. And we obtained that the limit is π₯ approaches infinity of negative one plus three
over π₯ minus four over π₯ squared all over one.

Next, weβll be using the fact that the limit as π₯ approaches positive or negative
infinity over one over π₯ is equal to zero. Therefore, three over π₯ and negative four over π₯ squared will both tend to
zero. And so we find that our limit is left equal negative one. Therefore, we find we have a horizontal asymptote at π¦ is equal to negative one. We can quickly consider the limit as π₯ approaches negative infinity of our
function. However, since the limit which we were using, thatβs the limit as π₯ approaches
positive or negative infinity of one over π₯ is equal to zero, works for both
positive and negative infinity. We will see that the limit as π₯ approaches negative infinity gives us the same
asymptote as the limit as π₯ approaches positive infinity.

So, therefore, we have found both the vertical and horizontal asymptotes of our
function. Itβs important to know that the function π of π₯ may cross the asymptote at some
point. We can see this if we sketch a graph of the function used in this example. Here, we can see a sketch of our function, including the horizontal and vertical
asymptotes which we found, at π₯ equals zero and π¦ equals negative one. We can see that as π₯ tends to positive infinity, the function actually crosses the
horizontal asymptote. However, it still shows asymptotic behavior since as π₯ gets larger and larger and
larger, we can see that the line of the function is getting closer and closer to the
line π¦ equals negative one. We can say that as π₯ goes to infinity, π of π₯ gets arbitrarily close to π¦ equals
negative one. Now, letβs look at another example.

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

In order to find a vertical asymptote here, we need to find the values of π such
that any limit as π₯ approaches π of π¦ is equal to positive or negative
infinity. In order to find the vertical asymptotes, we simply need to find the values of π₯
such that the denominator of π¦ is equal to zero. What this mean is that three π₯ minus four is equal to zero. Rearranging this, we find that there is a vertical asymptote at π₯ is equal to
four-thirds. In order to find the horizontal asymptotes of π¦, we need to consider the limit as π₯
goes to positive or negative infinity of π¦. In order to find the limit as π₯ approaches infinity of five π₯ plus one over three
π₯ minus four, we first multiply the numerator and denominator of the fraction by
one over π₯.

We are left with the limit as π₯ approaches infinity of five plus one over π₯ over
three minus four over π₯. Then we can use the fact that the limit as π₯ approaches infinity of one over π₯ is
equal to zero which tells us that one over π₯ and negative four over π₯ will both
tend to zero as π₯ tends to infinity. And so, therefore, we find that our limit is equal to five-thirds. Letβs quickly note that if we consider the limit as π₯ approaches negative infinity
of π¦, then we would see that this limit is also equal to five-thirds. Therefore, the solution to this question is that we have a vertical asymptote at π₯
equals four-thirds and a horizontal asymptote at π¦ equals five-thirds. Next, weβll be considering a more general case in another example.

The graph of equation π¦ is equal to ππ₯ plus π over ππ₯ plus π is a hyperbola
only if π is not on zero. In that case, what are the two asymptotes?

Letβs start by finding the vertical asymptote of this function. Vertical asymptotes occur at π₯ equals π when either the limit as π₯ approaches π
from below of π¦ is equal to double negative infinity or the limit as π₯ approaches
π from above of π¦ is equal to positive or negative infinity. Since π¦ is a rational function, this will happen at π₯ values, where the denominator
of π¦ is equal to zero. Therefore, we can find our vertical asymptotes by setting the denominator of π¦ equal
to zero. Now, we solve ππ₯ plus π is equal to zero for π₯. We obtained that thereβs a vertical asymptote as π₯ is equal to negative π over
π.

In order to find the horizontal asymptote of π¦, we need to consider the limit as π₯
approaches positive or negative infinity of π¦. In order to find this limit, we can multiply the numerator and denominator by one
over π₯. Then we use the fact that the limit as π₯ approaches infinity of one over π₯ is equal
to zero, in order to say that when we take this limit, π over π₯ and π over π₯
will both tend to zero. And this leaves us with π over π. Now, we have found our horizontal asymptote. So we have found that the two asymptotes of our hyperbola are π₯ is equal to negative
π over π and π¦ is equal to π over π. We can use this result in order to help us quickly find asymptotes of hyperbolas of
this form.

For example, if we want to find the asymptotes of the function π¦ is equal to nine π₯
minus 12 over five minus 12π₯. We can use the fact that our function of the form π¦ is equal to ππ₯ plus π over
ππ₯ plus π has asymptotes π₯ equals negative π over π and π¦ is equal to π over
π in order to quickly find our asymptotes. Writing down the values of π, π π, and π might make things a bit easier. We find that our vertical asymptote is at π₯ is equal to five over 12. And our horizontal asymptote is at negative three over four. Now, letβs quickly note that it is possible for a function to have more than one
vertical and one horizontal asymptote.

For example, letβs consider the function π of π₯ is equal to one over π₯ squared
minus four. We notice that the denominator of our function is a difference of two squares. And the function can, therefore, be written as one over π₯ plus two multiplied by π₯
minus two. Now in order to find our vertical asymptotes, we need to find the values of π₯ which
will send our functions to infinity. In order to do this, we set the denominator of the function equal to zero. This will give us two asymptotes, one at π₯ is equal to negative two, which comes
from π₯ plus two being equals zero, and the other at π₯ is equal to two, which will
come from π₯ minus two being equal to zero.

If we take the limit of our function as π₯ goes to infinity, we can see that the
denominator of this fraction will go to infinity as π₯ goes to infinity. Therefore, the value of this limit is simply zero which gives us a horizontal
asymptote at π¦ is equal to zero, knowing the values of these three asymptotes will
help us to sketch our graph. This is what the graph of our function will look like. As we can see, it has two vertical asymptotes and one horizontal. In the next example, weβll be considering a case which we must be very careful of,
which is when we have a factor within our rational function which can be
cancelled. Letβs now look at the next example.

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

Now, weβd normally start by finding the vertical asymptote by setting the denominator
of our function equal to zero. However, we cannot do this immediately with this function. Letβs, instead, start by factorizing the denominator of π of π₯. We can spot that we have a factor of π₯ plus two in both the numerator and
denominator. And so we can cancel the factor in both the numerator and denominator here. However, we have to be careful since, in doing this, we will be slightly changing our
function, since the original π of π₯ is not defined as the value of π₯ is equal to
negative two. However, our new function, which we can called π of π π₯, is. However, the asymptotes of π and π will still be the same.

Now in order to find the asymptotes of π, we simply need to find the asymptotes of
π. In order to find the vertical asymptotes, we set the denominator of π equal to
zero. This gives us a vertical asymptote at π₯ equals two. In order to find the horizontal asymptotes, we need to find the limit of π as π₯
goes to infinity. So thatβs the limit as π₯ goes to infinity of one over π₯ minus two. Now, as π₯ goes to infinity, π₯ minus two will also go to infinity. And since π₯ minus two is in the denominator of the function here, that means that
this limit will tend to zero. Therefore, we have a horizontal asymptote at π¦ equals zero. Letβs quickly sketch the graphs of π of π₯ and π of π₯, so that we can see how
these two functions differ, despite sharing the same asymptotes.

As we can see from the sketches with the graph of π of π₯ on the left and π of π₯
on the right, the two functions look identical. The only difference is that, on the graph of π of π₯, the point which has an π₯
value of negative two is undefined. Had we had tried to find the vertical asymptotes without cancelling the factor of π₯
plus two, then our denominator will be π₯ plus two multiplied by π₯ minus two. And in finding the vertical asymptote, we would have set this denominator equal to
zero. And we would have said that there was two vertical asymptotes, one at π₯ equals
negative two and one at π₯ equals two. However, as we can see from our graph, thereβs no asymptote at π₯ equals negative
two. All that we have is an undefined point. So, therefore, itβs very important to check that thereβs no factors in our function
which can be cancelled before we start finding asymptotes.

Now, we have seen a variety of examples of how we can find asymptotes and how useful
asymptotes can be, especially when identifying or drawing graphs. We will now revisit some of the key points of this video. Key points. In order to find vertical asymptotes, we need to identify points which give a zero
denominator. However, we must be careful to check if the rational function can simplify. In order to find horizontal asymptotes, we need to consider the limit of the function
as π₯ tends to positive and negative Infinity. Asymptotes of a function can be useful to help us identify or sketch the graph of the
function.