Asymptotes - Online Tutor, Practice Problems & Exam Prep

Understanding rational functions involves identifying vertical and horizontal asymptotes, which are crucial for graphing. Vertical asymptotes occur where the denominator equals zero, while horizontal asymptotes depend on the degrees of the numerator and denominator. If the degree of the numerator is less than that of the denominator, the horizontal asymptote is at $y=0$. If they are equal, divide the leading coefficients. Additionally, common factors can create holes in the graph, represented by open circles. Understanding these concepts is essential for accurately graphing rational functions.

Introduction to Asymptotes

Hey everyone. Whenever we worked with polynomial functions and we graphed them, we had to consider the end behavior of our graph, and this gave us great information so that we could easily graph our polynomial function. Now with rational functions, we're going to have to consider the behavior of our graph around things called asymptotes. Asymptotes are lines that our curve or our graph is going to approach and get really, really close to but not quite touch. Just that word, asymptote, might make the graphs of rational functions sound like they're going to be complicated, but here we're going to graph a rational function together and I'm going to walk you through identifying asymptotes and the basics that you need to know about them. So let's go ahead and get started.

The first thing that I want to do here is just go ahead and graph a rational function. This function \( f(x) = \frac{1}{x} \) is a really basic and common rational function that you're going to see. So let's go ahead and start with graphing the number 1 and plot those points on our graph. So if I plug in 1 to my function, that just gives me \( \frac{1}{1} \), which we know is just 1. So, plotting that point on my graph, I have the point (1,1). Then, if I take 2, this is just \( \frac{1}{2} \). So, I can also plot that on my graph at (2, \(\frac{1}{2}\)). Then, if I go to 3, this is just \( \frac{1}{3} \). So we're getting even smaller here on our graph. And if I take it all the way to 10 that's going to give me \( \frac{1}{10} \) or 0.1. So, we see that these numbers are getting progressively smaller. \( \frac{1}{100} \) is a much smaller number than 1. So what we're going to see happen on our graph is that we're going to get really, really close to the x-axis but not quite touch it. Now, I know that if I plug in negative 1, I'm going to get negative 1. So I can go ahead and plot that point on my graph at (-1, -1). Then, if I go to negative 10, I'm going to get \( \frac{-1}{10} \). So we know that it's getting smaller in that negative direction. So we're going to see a similar thing happen over here.

Now it's getting really, really close to that x-axis but not quite touching it, which is exactly what I said an asymptote is. So what I have here is an asymptote at my x-axis, which I represent using a dotted or dashed line. Now I'm going to go ahead and highlight it just so that it's easier to see here. But here I have a horizontal asymptote, because it's just a horizontal line, at \( y = 0 \). Whenever we look at this asymptote, we see that our graph is getting really close to it but not quite touching it.

And whenever we worked with the end behavior of polynomials, we described the behavior using arrow notation. We said that as \( x \) approached infinity, \( f(x) \) approached something; for this particular graph, it's approaching infinity. And as \( x \) approaches negative infinity, here \( f(x) \) is also approaching infinity. Now, we want to do the same thing with our rational function. So here, as \( x \) approaches infinity, which is on this side of my graph, I see that \( f(x) \) is instead approaching my asymptote at 0. So \( f(x) \) is approaching 0. Then, on the other side as \( x \) approaches negative infinity, I know that the same thing is happening. I'm still approaching that asymptote at 0.

Now, let's go ahead and graph the rest of our function, because we still don't know what's happening in the middle here. Now if I take a look at some smaller numbers in between 0 and 1, like, say, 0.1, that will give me \( \frac{1}{0.1} \), which is just 10. Then, if I get even smaller, even closer to 0, and I take \( \frac{1}{0.01} \), that gets even bigger at 100. So what we're going to see happen here is that our graph is going to get really high up and get really close to my y-axis but not quite touch it. And the same exact thing is going to happen over here. So, I end up having 2 different curves on the graph of this rational function.

Now, whenever we consider that it's getting really close to the y-axis, that tells us that we're dealing with another asymptote. So I have an asymptote here on my y-axis, which I will again represent with a dashed line. I'm going to go ahead and highlight it again so that we can see a little bit better. And this is a vertical asymptote at the line \( x = 0 \). Now we see that our graph is approaching this asymptote from either side, and whenever we worked with polynomial functions, if we were to consider \( x \) approaching 0 and wanting to describe the behavior of our graph, we would just get a single point, in this case 0. Now, in our rational function, we see that our graph is approaching that line from either side, so that means we need to be able to describe both of those behaviors. So we're going to split the behavior into 2 different sides.

As \( x \) approaches 0 from the right side, denoted as \( x \rightarrow 0^+ \), I see that my function is getting really, really high. It's getting to be a really large number. So it's approaching infinity. Then as \( x \) approaches 0 from the left side, denoted as \( x \rightarrow 0^- \), I see that it's going the opposite direction, so \( f(x) \) is instead approaching negative infinity. Now, this is only because we're looking at 0 here where our asymptote is, and asymptotes are not always going to be at 0. They just happen to be for this function. So this could really be any number. I could consider \( x \) approaching 2, 5, or any other number from either side. It just depends on your function and where your asymptote is. And there can actually be a lot of different possibilities with asymptotes because a rational function may have no asymptotes, we may have just one asymptote like we see in this function here, or we can even have multiple asymptotes. But they're always going to be represented with these dashed lines. So if we look at function \( b \) here, I have a vertical and a horizontal asymptote kind of shifted from what we saw in our function up there. And then I could also even have multiple vertical asymptotes. It really just depends on your function. Now that we know the basics of asymptotes, let's go ahead and get to calculating them. I'll see you in the next video.

Sketch the graph of the function $f\left(x\right)=\frac{1}{x^2}$. Identify the asymptotes on the graph.

• A. Vertical Asymptote: $x=0$, Horizontal Asymptote: None
• B. Vertical Asymptote: $x=0$, Horizontal Asymptote: $y=0$
• C. Vertical Asymptote: $x=0$, Horizontal Asymptote: $y=0$
• D. Vertical Asymptote: $x=1$ , Horizontal Asymptote: $y=0$

Determining Vertical Asymptotes

Hey everyone. Now that we know what asymptotes are, we need to be able to determine where they are on our graph when given a function. So whenever I'm given a rational function, I need to be able to determine if there even is a vertical asymptote and where exactly that vertical asymptote is. Now luckily, finding vertical asymptotes is super simple and almost identical to finding the domain of our function when we set our denominator equal to 0 and solved for x. We're going to do the same thing to find vertical asymptotes, set our denominator equal to 0, and solve for x. But before we do that, we're just going to write our function in lowest terms. So we're going to do 2 things that we already know how to do in order to find our vertical asymptotes. So let's go ahead and walk through this. So like I said, to find our vertical asymptotes, we want to go ahead and put our function in lowest terms. So looking at the function that I have here, I have x+2 plus 2 times x minus 3. So to put this in lowest terms, I'm gonna go ahead and cancel my common factors, which here is x+2. So this leaves me with the function 1 over x minus 3, and now my function is in lowest terms. Now I'm simply going to set my denominator equal to 0 and solve for x to get my vertical asymptote. So if I take x minus 3 and I set it equal to 0, I'll end up with x=3. And that's my vertical asymptote, and I'm done. That's all we need to do in order to find those vertical asymptotes. Let's go ahead and look at a couple more examples and find some vertical asymptotes of different functions.

So looking at my first function here I have f of x is equal to 2 over 2x plus 6. So remember the first thing we want to do is write our function in lowest terms. So let's go ahead and do that here and factor. So my numerator cannot be factored it's just a constant, it's just 2, but in my denominator, I can go ahead and pull out a greatest common factor of 2, and that leaves me with x+3 in my denominator there. So writing this in lowest terms, I'm gonna go ahead and cancel these twos, leaving me with 1 over x+3. Now I can go ahead and set my denominator equal to 0. So setting this denominator equal to 0, isolating x, I'm just going to subtract 3 on both sides, leaving me with x=−3. So that's my vertical asymptote of this function and I'm done.

Let's look at one more example here. Here I have f of x is equal to 1 over x squared minus 9. Now looking at this function, we want to go ahead and write it in lowest terms but because my numerator is just 1, that's not going to cancel with anything in my denominator. It's already in lowest terms. So I can just go ahead and take my denominator and set it equal to 0. Now solving for x here, I wanna add 9 to both sides, canceling that 9 out, leaving me with x2=9, and then I can go ahead and square root both sides to isolate x, leaving me with x=±9, which we know is just 3. So this tells me that I actually have 2 vertical asymptotes, one at x equals positive 3 and one at x equals negative 3. It's totally okay to have more than one vertical asymptote. That's gonna happen sometimes, and it's perfectly fine. Now that we know how to find vertical asymptotes, let's go ahead and get some more practice.

Based only on the vertical asymptotes, which of the following graphs could be the graph of the given function? $f\left(x\right)=\frac{x^2-4x}{x^2-x-12}$

• A.
• B.
• C.

Determining Removable Discontinuities (Holes)

Hey, everyone. Whenever we found vertical asymptotes, we would factor our function and then cancel out any common factors in order to get our function in lowest terms before setting our denominator equal to 0 and solving for x. But having that common factor actually causes something to happen in the graph of my function called a hole. And we need to be able to identify where these holes are on our graph, which we can do by simply taking our common factor and, instead of canceling it out, just setting it equal to 0 and solving for x. So here I'm going to walk you through how to find any holes in your graph starting from exactly where we do with vertical asymptotes and just taking a different path from there. So let's go ahead and get started.

Now with vertical asymptotes, like I said, we would cancel out our common factor, and then we would set our denominator equal to 0 and solve for x. So for this function here, I end up with a vertical asymptote at x equals 3, which we can see right here on our graph, a vertical asymptote at x equals 3. Now, whenever we find the holes in our graph, we're going to start at the same place. So here I have this factored function, but instead of canceling out that common factor to put it in lowest terms, I'm going to take that common factor and set it equal to 0.

So here, my common factor is x + 2 and setting that equal to 0, I then want to solve for x. So solving for x here, I end up with x equals negative 2. And that tells me that I have a hole in the graph of my function right at x equals negative 2, which I'm going to represent with an open circle on my graph. Literally, just draw a circle on the curve at that point. And that's where our hole is. So looking at this graph, now I have both a vertical asymptote and a hole for the same function. Now you might also hear holes referred to as removable discontinuities. And that's just the more technical term for it, but it's referring to the same thing.

So now that we know how to find the holes in the graph of a rational function, let's go ahead and look at some more examples here. So let's look at this first function. I have \( f(x) = \frac{x + 3}{x^2 + 4x + 3} \). So here, we want to go ahead and factor this the same way we would if we were trying to write it in lowest terms. So my numerator can't be factored here; it just stays as x + 3, but my denominator using the a c method can be factored to (x + 3)(x + 1). Now I want to take my common factor, and I don't want to cancel it. I want to take it, and I want to set it equal to 0. So here my common factor is x + 3. Setting that equal to 0, I'm going to go ahead and solve for x, which I can do by subtracting 3 on both sides. So I end up with a hole at x equals negative 3, and I'm done. That's where the hole on my graph will be.

Let's look at one more example here. On this side, I have \( f(x) = \frac{x^2 + 1}{x - 1} \). So let's go ahead and factor this. So my numerator can be factored into (x + 1)(x - 1) and I'm going to divide that still by x - 1 because that denominator can't be factored anymore. Now, again, I want to take my common factor, and I don't want to cancel it. I'm going to take and set it equal to 0. So here, x - 1 equals 0. And then we want to solve for x. So adding 1 to both sides leaves me with x is equal to 1, which is where the hole on my graph is at x equals 1. And we're done.

So now that we know how to find the holes in our graph, let's get a little bit more practice.

Find all vertical asymptotes and holes of each function.

$f\left(x\right)=\frac{-5x}{\left(2x-3\right)^2}$

• A. Hole(s): $x=0$ , Vertical Asymptote(s): $x=\frac32$
• B. Hole(s): $x=\frac32$ , Vertical Asymptote(s): $x=\frac32$
• C. Hole(s): $x=0$ , Vertical Asymptote(s): $x=0$
• D. Hole(s): None , Vertical Asymptote(s): $x=\frac32$

Find all vertical asymptotes and holes of each function.

$f\left(x\right)=\frac{x^2-2x}{2x^3-x^2-6x}$

• A. Hole(s): None, Vertical Asymptote(s): $x=0,x=2,x=-\frac32$
• B. Hole(s): $x=0$, Vertical Asymptote(s): $x=-\frac32$
• C. Hole(s): $x=0$, $x=2$, Vertical Asymptote(s): $x=-\frac32$
• D. Hole(s): $x=0$, $x=2$, Vertical Asymptote(s): $x=\frac32$

Find all vertical asymptotes and holes of each function.

$f\left(x\right)=\frac{x^2+10x+25}{2x^2+8x-10}$

• A. Hole(s): None, Vertical Asymptote(s): $x=-5,$ $x=1$
• B. Hole(s): $x=-5$ , Vertical Asymptote(s): $x=1$
• C. Hole(s): $x=1$ , Vertical Asymptote(s): $x=-5$
• D. Hole(s): $x=-5$ , Vertical Asymptote(s): $x=-1$

Determining Horizontal Asymptotes

Hey, everyone. When working with vertical asymptotes, we found that they affect the domain of our rational function, and we could even calculate them in a way that's really similar to finding the domain. Now with horizontal asymptotes, they're instead going to affect the range of our rational function because now we're dealing with horizontal lines that go across our y-axis. Since we don't have a way of finding the range of our function, you may be worried that this is going to get complicated. But don't worry because it just depends on only two things that we can determine by just looking at our function, and that's the degree of both our numerator, the top of our rational function, and the denominator, the bottom of our rational function. So with that in mind, let's go ahead and just jump right into finding some horizontal asymptotes.

Looking at this first function here, I have f(x)=1/x and we want to be able to determine what our horizontal asymptote is. Now, as I said, we'll look at the degree of the numerator and the degree of the denominator. Here the degree of my numerator is 0 because there's not even a variable there, and the degree of my denominator is 1. So, 0 is less than 1, and whenever the degree of your numerator is less than the degree of your denominator, you are always simply going to have a horizontal asymptote at the line y=0, which is what we can see on our graph is already happening here.

Now let's look at another function here: I have f(x)=2x+3x. So, looking at the degree of my numerator and denominator, the degree of my numerator is 1 and the degree of my denominator is also 1. Hence, 1 is equal to 1. So here, since the degree of my numerator is equal to the degree of my denominator, I actually need to consider one more thing: my leading coefficients. So, I'm going to take the leading coefficient of my numerator and then divide it by the leading coefficient of my denominator. And that will give me my horizontal asymptote y=2, which is exactly where that horizontal asymptote is on our graph.

Now, I'll discuss another function: 4xx2-xx3-5x+9. Here, the degree of our numerator is 2, and the degree of our denominator is 3. So, 2 is less than 3. Accordingly, we have a horizontal asymptote at y=0, following our original examples.

Next, suppose we have f(x)=2x23x2+x-1. Here the degrees for numerator and denominator are both 2. Thus, the degree of my numerator is equal to the degree of my denominator. Then, by dividing the leading coefficients, I end up with a horizontal asymptote at y=23.

And finally, a word about the graph behavior: sometimes, the graph of your rational function may actually intersect the horizontal asymptote and fully cross it, then approach it from the other side. While this may seem to contradict the typical behavior of asymptotes, it's just one of those nuances you should be aware of. It can happen, so don't be startled if you see a graph crossing a horizontal asymptote.

With this knowledge, you're now better equipped to identify horizontal asymptotes and understand their role in the behavior of rational functions. Let's get some practice!

Find the horizontal asymptote of each function.

$f\left(x\right)=\frac{-5x}{\left(2x+3\right)^2}$

• A. Horizontal Asymptote at $y=0$
• B. Horizontal Asymptote at $y=-\frac54$
• C. Horizontal Asymptote at $y=-\frac52$

Find the horizontal asymptote of each function.

$f\left(x\right)=\frac{8x^2+1}{2x^2-x-6}$

• A. Horizontal Asymptote at $y=0$
• B. Horizontal Asymptote at $y=\frac14$
• C. Horizontal Asymptote at $y=4$

Find the horizontal asymptote of each function.

$f\left(x\right)=\frac{x^2+4x}{2x^3+8x^2}$

• A. Horizontal Asymptote at $y=0$
• B. Horizontal Asymptote at $y=\frac12$
• C. Horizontal Asymptote at $y=2$

Here’s what students ask on this topic:

What are vertical asymptotes and how do you find them?

Vertical asymptotes are lines that a graph approaches but never touches or crosses. They occur where the denominator of a rational function equals zero, causing the function to be undefined. To find vertical asymptotes, follow these steps:

1. Write the function in its lowest terms by canceling any common factors.
2. Set the denominator equal to zero.
3. Solve for $x$.

For example, for the function $\frac{1}{x^2-9}$, set $x^2-9=0$ and solve to get $x=\pm 3$. Thus, the vertical asymptotes are at $x=3$ and $x=-3$.

How do you determine horizontal asymptotes of a rational function?

Horizontal asymptotes depend on the degrees of the numerator and denominator of a rational function. Follow these rules:

1. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is $y=0$.
2. If the degrees are equal, divide the leading coefficients of the numerator and denominator.
3. If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

For example, for $\frac{2x^2+3}{3x^2+x-1}$, the degrees are equal, so the horizontal asymptote is $y=\frac{2}{3}$.

What is a hole in the graph of a rational function and how do you find it?

A hole in the graph of a rational function, also known as a removable discontinuity, occurs where a common factor in the numerator and denominator cancels out. To find a hole:

1. Factor the numerator and denominator.
2. Identify and cancel the common factor.
3. Set the common factor equal to zero and solve for $x$.

For example, for $\frac{x+3}{x^2+4x+3}$, factor to get $\frac{x+3}{x+3x+1}$. The common factor $x+3$ cancels, leaving a hole at $x=-3$.

Can a rational function cross its horizontal asymptote?

Yes, a rational function can cross its horizontal asymptote. While asymptotes are lines that the graph approaches, it is possible for the graph to intersect a horizontal asymptote and then approach it from the other side. This behavior does not contradict the definition of an asymptote, as the function still approaches the asymptote as $x$ approaches infinity or negative infinity. However, this crossing typically occurs in the middle of the graph, not at the extremes. Understanding this can help you accurately graph rational functions and interpret their behavior.

How do vertical asymptotes affect the domain of a rational function?

Vertical asymptotes affect the domain of a rational function by indicating values of $x$ where the function is undefined. These occur where the denominator equals zero. To find these values, set the denominator equal to zero and solve for $x$. The solutions are excluded from the domain. For example, for $\frac{1}{x-3}$, set $x-3=0$ to get $x=3$. Thus, $x=3$ is not in the domain.