College Algebra Section 5.3 Graphs of Rational Functions

Hello, and welcome to Bay College's Online Lectures for College Algebra. I'm Jim Helmer. And in this section, 5.3, we're going to talk about the graphs of a rational functions. To utilize graphing rational functions, we have to review what we talked about in the previous section, 5.2, and we're just going to summarize asymptotes. The first thing we usually find when it comes to graphs is the domain. And that domain helps us determine what our vertical asymptotes, if there are any. A vertical asymptote is basically asking us, as x approaches some number from the left or the right, a domain restriction, what is the behavior of the function? Does it approach positive or negative infinity, one or the other? Well, we find the vertical asymptotes by essentially taking the denominator of our rational function and saying, what are the values that would make it equal to 0? Because we can't have 0 in our denominator. Those are domain restrictions. Then we reduce any factors. And any factors that reduce away are holes. We basically take the x value that reduced away from the domain, and we go ahead and plug it into the function say, well this is a value that is a domain restriction, but we cannot have it on our graph. So we write it as an ordered pair, whatever that value that reduced away. Any remaining ones are vertical asymptotes. x equals this value is a value that our graph can never cross. We always have to keep in mind that our graph could never cross a vertical asymptote. Then we look for any horizontal or oblique asymptotes. And the three scenarios that we look at is essentially the degree of each polynomial, the degree of the numerator versus the degree of the denominator. If the degree of our polynomials, the top and the bottom, are same degree, it is the ratio of their leading coefficients. That's the behavior of the graph. It's going to approach that ratio, maybe it's just going to be some number. Or it the degree of the top is less than the degree of the bottom, part of our rational function, its behavior will always approach 0 as we go to positive or negative infinity, the behavior of x at its end behavior. If it doesn't fit either of those two scenarios, maybe it fits this scenario, right here. If the degree of the numerator is one more than the degree of the denominator, then we have a linear oblique asymptote. And when we look for this behavior, essentially we just do division of the two polynomials to find a linear equation. Now, if the numerator is more than just 1 greater than the degree of the denominator, we'll actually have a nonlinear asymptote. So now, let's look at what we can do with this information and how we can apply it to the graph of a rational function. Because rational functions, at times, can look intimidating when we look at their graphs. So graphing and analyzing a rational function, these are the steps that we take in order to do that. And let me just move the board a little bit further down here. Now let R just stand for a rational function in our notes, here. So the first thing we want to do is always find the domain of a rational function. What are the values that make us divide by 0? We can't have those. So once we find our domain, then we want to reduce R to lowest terms, simplify it. It's just a fraction just like a rational number. We've got to reduce our fractions. Now, any values that don't reduce away from that denominator, as domain restrictions, become our vertical asymptotes, the value that our graph can ever cross. If some of them reduce away, they become holes. So we use that value to just plug into our reduced function to find what are the holes in our graph. What are the values that we can't have but it may look as if it would pass through that point? And holes are always written as ordered pairs-- keep that in mind. Now step four, we'll find any horizontal or oblique asymptotes if they exist. And that's using the summary before. We look at the degree of each polynomial, numerator and denominator. Then we can find if our rational function crosses a horizontal or oblique. Now keep in mind, it can never cross the vertical. But at some point, it could cross the horizontal or oblique. Because what these tell us is just its end behavior. It could cross it somewhere away from the end behavior, from the infinities. So we want to find that. We essentially just find it by setting the oblique or horizontal asymptote equal to our reduced function and solve for any values of x. Then six, this step here, we want to find any intercepts. It's good to have a few points to put on the graph so we can see its behavior as it approaches these asymptotes. And lastly, we take all this information we find, and we put it on a graph. And if we still need more to see its behavior, well, we can pick a few test points and see on what side of which asymptote that a piece of my graph may lie. So you can always pick extra points, if necessary. So let me move this out of the way, and we'll actually do an example where we can find this information. Here, we have a rational function, f of x equals 2x squared minus 5x plus 2 divided by x squared minus 4. The first thing we want to do is find that domain. Well, if I factor this, and I'll just write it over here, x plus 2, x minus 2-- that's what this would factor to. And I can see, well, the values that would make that 0 that I have to exclude are x being plus or minus 2. x cannot be these values. These are excluded, so my domain is x, such that x is plus or minus 2. Now to write it in lowest forms, I also have to factor the top. Now to factor this, I notice that this coefficient isn't 1, so I'd want to use maybe the ac method. Hopefully, you remember how to do that. a times c of our quadratic, 2 times 2 is 4. What are the factors of 4 that sum to a negative 5? Well, that'd be negative 4 and negative 1, and then I can rewrite this middle term using those factors, negative 4x and negative x, and then factor it by the greatest common factor. It's a four term polynomial-- factor by grouping. And when I factor that, I get 2x minus 1 times x minus 2. And now, we can see hey, x minus 2 and x minus 2. These are a common factor that can reduce. So in lowest terms, 2x minus 1 over x plus 2 is our reduced rational function. Now the vertical asymptote would be the domain restriction that still exists. And we see there is a domain restriction that still exists, so x equal to negative 2 is the value of my vertical asymptote that my graph can never cross. Now what about the domain restriction that reduced away, that positive 2? Well, that is where we're going to have a hole. So I'm going to plug that value into my reduced function, which would be positive 2. 2 times 2 is 4 minus 1 is 3 over 2 plus 2 is 4-- 3/4. So when x is a positive 2, y is 3/4. This is the hole in my graph. I'm going to make sure when I put it on my graph, it's a nice, open circle, indicating that this value is not within our domain. A horizontal or oblique asymptote-- well, if this does have an asymptote, we just look at the degree of each polynomial. This is degree 2. This is degree 2. They have the same degree. That is a horizontal asymptote. That is the ratio of their coefficients. Well, the coefficient is 2 over 1, which is just 2. 2 over 1 is 2. So here is my horizontal-- it is a horizontal asymptote, y equals 2, the equation of a horizontal line. Then we want to find out, does my graph cross my horizontal asymptote? Well, I can find that out, essentially, by setting this equal to this value. This is a function-- f of x is the same as y in some concepts, right? So we can set this equal and solve for it. And if I do that, 2x minus 1 over x plus 2 equal to 2. Well, I can multiply both sides by the denominator, 2x minus 1 equals 2x plus 4, when I distribute the 2 to that. And we notice when I subtract 2x from both sides, I get negative 1 equals 4. Well, this isn't a true statement, but it does tell me something about my graph and that horizontal asymptote. If this is not a true statement, it tells me there's no value of x that makes it cross my horizontal asymptote. So does it cross my horizontal asymptote? Nope, not at all. So we can move on. Let's find a few extra points, maybe x-intercepts and y-intercepts. Well the x-intercept, I essentially set this equal to 0. So let's rework this problem. We make the function equal to 0, and I can find any x-intercepts. If I do that, I get 2x minus 1 equals 0. We really have to only worry about the top, because the denominator is never going to make it equal to 0. And if I solve this, I add 1 and divide by 2, I get 1/2. So I get 1/2 when y is 0, my x-intercept. And to find the y-intercept, we set x equal to 0. So if this term is 0, and this term is 0, I get negative 1 over 2. So when x is 0, I get negative 1/2. Now we have all this good information. We can take it, we can start putting it on the graph, and see what we get. Well, I know I have a vertical asymptote at negative 2. So I'm going to do a dashed line that I know my graph can never cross. That is my vertical asymptote. I also have a hole at x equals 2. y equals 3/4, so I'm going to draw a nice, open circle right there, because this is not within my domain. And then I can draw my horizontal asymptote at y equals 2. So I know, as the graph goes to positive or negative infinity, it's going to approach this horizontal asymptote. And I know it can never cross, because I tested that. And then I have a few points I can plug in. I have an x-intercept of 1/2. That'd be right here. And I have a y-intercept of 0, negative 1/2. That's right there. And now I can see, OK, I see a little bit of behavior here. I see a pattern. As this is going towards that asymptote that it can never cross, it's actually moving downward. As it moves towards this horizontal asymptote that it will not cross, I can see it's going to approach that value as x goes to infinity. Now we only have half the graph. There is some more graph over here, or at least, we should always check to see if there is. So what we want to do here is just pick an extra test point to see where is this graph? I know it's not going to cross these asymptotes, so is it going to be up here or down here? So let's just pick a value. Well, I'm going to plug in negative 4 into my reduced function. So negative 4 times 2 is negative 8 minus 1 is negative 9. And negative 4 plus 2 is negative 2. So I have negative 9/2. Well, negative 9/2-- that's just a positive value, right, which is 4 and 1/2. So I can say, OK, over here, when I chose negative 4, I got 4 and 1/2 which puts me right here. I'm up in this area that I know it's not going cross these asymptotes, but I do know its end behavior. So I know that's what it's going to do. It's going to approach that asymptote going towards positive infinity for y or going to this asymptote as x goes to negative infinity. So that is how we graph these rational functions. We take this information, and we put it all on the graph, pick a few extra points, if necessary. Now, I'd like you to attempt this on your own, so here's a quiz. The function f of x equals 2x plus 3 over 3x squared plus 7x minus 6. The key to this one is essentially using that ac method to factor. But you have to do it in the denominator. So what I want you to do is take this 2x plus 3 over 3x squared plus 7x minus 6 and find its domain, reduce the function, if necessary, find any vertical asymptotes, find any holes, if they exist, find any horizontal or oblique asymptotes, see if it crosses your horizontal or oblique asymptote, and then find any x or y-intercepts. And finally, graph it. And if you need to, pick a few extra points. If you're not ready to do that on your own, just wait. We'll do another example and see what we get. So for this example, we have our rational function f of x equals x squared plus 3x plus 2 over x minus 1. Now to find the domain, well, q of x, my denominator, is already a linear factor. So I know that x cannot equal a positive 1. That would make this 0, undefined. Now let's see if it reduces. Well, let's just factor the top and see what we get. The top factors to x plus 1, x plus 2, the factors of positive 2 that sum to positive 3. And we notice nothing is going to cancel here. x plus 1 is not the same as x minus 1. So it's already in reduced form, and if you want, you could leave it in a factored form like this. It's going to maybe be a little bit easier to work when you're ready to graph it or find some values that you need to find. Any vertical asymptotes? Well, since nothing reduced, the vertical asymptote is essentially x not equal to 1. It can never cross x equals 1. That's my vertical asymptote. Any holes? Well, nothing reduced away, so there are no holes. And then lastly, do I have a horizontal or oblique asymptote? Well, if I look at this, here, I notice that my power of my numerator is greater than the denominator. 2 relative to 1-- well, that means I have an oblique asymptote. And we can find that oblique asymptote by doing division. And that's what I'm going to do here. I'm going to use synthetic division. So to do that, I'm going to use the 0 here, which would be 1. And I'm going to divide it into this, here. And I'm going to use the coefficients 1, 3, and 2. If you don't recall synthetic division, go out there, find a video, review it, or go to your textbook, it is in there. Bring down the 1. 1 times 1 is 1, 3 and 1 is 4, 1 times 4 is 4, and 4 and 2 is 6. Now this is a remainder. We're not concerned about remainders, because we want to see end behavior. Remainders become insignificant as the graph goes to infinity, left or right. So this is my linear equation, y equals x plus 4. So I found the oblique asymptote. And it is the line y equals x plus 4. Now does it cross this horizontal or oblique asymptote? Well, what I can do is take this equation and set it equal to this, right here. And if I do that, I get x plus 1, x plus 2, over x minus 1. But I'm going to multiply that this side, which is x plus 4 times the denominator. So I'm skipping a step here, but hopefully, you can follow along and see that this was in the denominator, I just multiplied both sides. And now I can just FOIL this out. So this is x squared plus 3x plus 2 equals x squared plus 3x minus 4. And if I subtract x squared from both sides, it goes away. Subtract 3x from both sides, it goes away. And I get 2 equals negative 4-- not a true statement. That means it does not cross my oblique asymptote, and so the answer is no, it does not cross my oblique asymptote. Let's find some x or y-intercepts. Let's find the y-intercept, first. The y-intercept is essentially where x is 0. Well if this is 0, and that's 0, and that's 0, I get 2 over negative 1, which is negative 2. When x is 0, this is negative 2. And to find an x-intercept, well, we set the whole equation equal to 0. And if we set this equal to 0, we don't have to worry about that denominator. What are the 0's of the function? And this is why I said, if you leave it in factored form, it makes it a little easier for you. Well, my 0's would be negative 1 and negative 2. So I have the 0's of negative 1, 0; negative 2, 0. Let's take this information and put it on the graph. Well, the first thing I want to graph is my vertical asymptote of x equal to 1. And then there are no holes. We have a horizontal asymptote at y equals x plus 4. So I'm going to graph that. Let's say 1, 2, 3, 4, and I know it has a slope of 1. So there's some behavior going on in this graph, here. Let's find out what it is. Let's graph our intercepts. We have a y-intercept at negative 2. We have an x-intercept at negative 1 and at negative 2. And if we look at the behavior of this, we know what it never crosses our vertical asymptote, so it's going to approach it. And we know it doesn't cross our horizontal asymptote, so this is its behavior. But what about over here or over here? Well, we're on this side of our vertical asymptote that we know it doesn't cross. What's happening over here on this side? Well, if we pick any arbitrary point, let's say I'm going to pick x equals to 2, and just plug it into the function, find its value. 2 plus 1 is 3, 2 plus 2 is 4, 3 times 4 is 12. 2 minus 1 is 1, 12 over 1-- well, that's 12. It's way up here, some value way up here. So I know, somewhere up here, it's approaching these asymptotes without crossing them. So this would be the graph. Even though a piece of it is off of the graph that I have here, I still know its behavior because of these asymptotes. And that's what this is all about. So this has been section 5.3, Analyzing Rational Functions. Thank you for watching.