Lesson Video: Horizontal and Vertical Asymptotes of a Function | Nagwa Lesson Video: Horizontal and Vertical Asymptotes of a Function | Nagwa

Lesson Video: Horizontal and Vertical Asymptotes of a Function Mathematics

In this video, we will learn how to find the horizontal and vertical asymptotes of a function.

15:03

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 we’ll 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 𝑦 is equal to one over π‘₯. We can see that it has a horizontal asymptote at 𝑦 is equal to zero and a vertical asymptote at π‘₯ is equal to zero.

Let’s 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 𝑐, 𝑓 of π‘₯ approaches positive or negative ∞, then π‘₯ equals 𝑐 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 ∞, 𝑓 of π‘₯ approaches some constant 𝑐, then 𝑦 equals 𝑐 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 it’s 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, it’s often easier to consider what happens when π‘₯ approaches positive or negative ∞.

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, it’s often easier to find the horizontal and vertical asymptotes. And similarly, if we know the horizontal and vertical asymptotes, it’s often easier to find the domain and range of that function. Let’s now move on and look at an example of how we’re able to find vertical and horizontal asymptotes.

Determine the vertical and horizontal asymptotes of the function 𝑓 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 𝑓 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 𝑓 of π‘₯, when either of these terms is undefined, 𝑓 of π‘₯ is also undefined.

And so, therefore, we can say that when π‘₯ is equal to zero, 𝑓 of π‘₯ does not have a defined output. Hence, we have found the vertical asymptote of 𝑓 of π‘₯. And it’s 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 ∞.

Well, we can look at the π‘₯ terms in 𝑓 of π‘₯. We have three over π‘₯ and negative four over π‘₯ squared. As π‘₯ approaches ∞, 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 𝑓 of π‘₯ and we have both of these rational terms approaching zero, we can see that 𝑓 of π‘₯ will approach negative one. And we can say that 𝑓 of π‘₯ will get arbitrarily close to negative one without reaching negative one. Hence, we’ll have a horizontal asymptote at 𝑦 is equal to negative one.

Here we have found the vertical and horizontal asymptotes of our function 𝑓 of π‘₯. They are at π‘₯ is equal to zero and 𝑦 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 𝑓 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, 𝑓 of π‘₯ is equal to negative one. Hence, negative one is in the range of 𝑓 of π‘₯.

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

Since if we look to the right of π‘₯ is equal to four over three, we can see that 𝑓 of π‘₯ is getting arbitrarily close to 𝑦 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, we’ll 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 𝑦 is equal to eight over four π‘₯ minus three plus five over three?

We can start by finding the vertical asymptote of this hyperbola. We’ll 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, we’ll 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 ∞.

Now the only π‘₯-dependent term in our equation is eight over four π‘₯ minus three. Now as π‘₯ approaches positive or negative ∞, 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 𝑦 will get arbitrarily close to five over three as π‘₯ approaches positive or negative ∞. Since the rational term in the equation will approach zero. Hence, our hyperbola has a horizontal asymptote at 𝑦 is equal to five over three. And so now we’ve found the asymptotes of our hyperbola, which completes the solution to this question.

Before we move on to our next example, let’s 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 𝑓 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 𝑓 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 𝑓 of π‘₯. However, we first need to consider what happens to 𝑓 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, 𝑓 of π‘₯ must be positive. And when π‘₯ is between negative two and two, π‘₯ squared minus four is less than zero. Hence, 𝑓 of π‘₯ is negative. Using this information, we’re able to sketch a graph of 𝑓 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 𝑓 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 𝑓 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, we’re slightly changing the function. After cancelling the factor, we can call the new function 𝑔 of π‘₯. We have that 𝑔 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 𝑓 of π‘₯, we would have an undefined output. Since this would give 𝑓 a denominator of zero. However, we are able to input π‘₯ is equal to negative two into 𝑔 of π‘₯.

Now it’s important to note that although these two functions differ slightly, they in fact have the same asymptotes. Therefore, we’re able to find the asymptotes of 𝑓 by finding the asymptotes of 𝑔. So let’s find these asymptotes. We can identify vertical asymptotes as any input with no defined output.

Since 𝑔 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, 𝑔 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 ∞.

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

Since 𝑔 of π‘₯ and 𝑓 of π‘₯ share the same asymptotes, in finding the vertical and horizontal asymptotes of 𝑔, we’ve found the vertical and horizontal asymptotes of 𝑓. And this completes our solution to this question.

But before we move on, let’s quickly consider how 𝑔 and 𝑓 differ with a quick sketch. Here we have sketches of 𝑓 of π‘₯ and 𝑔 of π‘₯. We can see the asymptotes at 𝑦 is equal to zero and π‘₯ is equal to two. Now the only difference between these two graphs is that 𝑓 of π‘₯ is undefined at π‘₯ is equal to negative two. And 𝑔 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, we’ll see how we can use asymptotes in order to identify the graph of a function.

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

Let’s start by finding the vertical asymptotes of 𝑓 of π‘₯. We can find vertical asymptotes by identifying any input with no defined output. Since 𝑓 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 𝑓 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 𝑦 is equal to zero. Hence, our function 𝑓 of π‘₯ must have a horizontal asymptote at 𝑦 is equal to zero.

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

Now let’s see what happens to 𝑓 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, 𝑓 of π‘₯ must also be negative. And when π‘₯ is greater than negative one, π‘₯ plus one is positive. Hence, 𝑓 of π‘₯ is also positive. And this information about 𝑓 agrees with what we’ve shown for graph c). Therefore, our solution is that the graph c) represents our function 𝑓 of π‘₯.

We’ve now covered a variety of examples of how we can find asymptotes and how useful asymptotes can be, especially when identifying or drawing graphs. Let’s 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.

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