Understanding Rational Functions and Graphing

Algebra 2, section 7-2, day 1, graphing rational functions. Start by reading here. It says a rational function has the form f of x equals p of x over q of x, where p of x and q of x are polynomials, and of course the denominator is not zero. To say that differently is to say that you're going to have things that are in standard form, numerator and denominator. So you're going to have a polynomial in the numerator, polynomial in the denominator, typically in standard form. That's a rational function. The inverse variation function, f of x equals a over x, is a rational function. So what we started the chapter with in 7.1 and discussing inverse variation, that is a rational function. Then it says, so we'll start with the parent graph then. So if I look at this one right here and they said a over x, if we let the a be 1, then that's a parent function. And this is the graph of the parent function. The name of that graph, just like a quadratic, gives you a parabola. This type of function, a rational function, gives you a hyperbola, which consists of two symmetrical parts called branches. Each one of the blue things here in the figure would be known as a branch. The domain and range are all non-zero real numbers. The way we say that, using interval notation, is to say negative infinity to zero, parenthesis, not bracket, in union with zero to infinity. That's what the domain would be. The range of this one would also be exactly that. It goes from negative infinity down here up to, but not including zero, in union with zero to infinity. So basically that's every real number except for zero, both for the domain and range. Now we're going to change that on you here in this lesson, and the asymptote will be somewhere other than the x-axis or the y-axis, and we will be labeling those. Any function of the form g of x equals a over x, again, when a is not zero, or not again, but if the denominator, if the numerator is not zero, otherwise it would just be a horizontal line at y equals zero. Anything other than that has the same asymptotes domain and range as the function that is the parent function. So we are trying to get you in the habit of having you to label the asymptotes. Here they would be x equals zero and y equals zero. Graph the function, compare the graph with the graph of the parent function. We're not going to be doing much of that comparing the graph, but we do need to be able to graph the function. And you'll notice as I do all these that I'm going to be doing these without a calculator. So this would be a fair one to put on the no calculator part of the next quiz. So plugging in, I would start with, well, it's okay to start with zero, although that's going to be undefined, isn't it? Two divided by zero is undefined. And then we could go this way, negative 1. 2 divided by negative 1 is negative 2. And then you could go with negative 2, and you're going to divide 2 by negative 2, and that's negative 1. And then if you want to pick another convenient number, how about negative 4? 2 divided by negative 4 is negative 1 half. If we go the other direction, it's going to be 1, 2, 2, 1, 4, 1 half. There's six good points that you can plot. Now it would be the case, because there is symmetry going on here, maybe you've recognized that already, that there would be a point at negative a half, negative two, negative four, excuse me. There is a point here. There is symmetry about the line y equals x. There's symmetry about the line y equals negative x as well. So then we have one, two, and two, one, and four, one half. Again, there would be one at one half comma four just as well. That does help to give us something to aim at. Make sure that neither branch crosses the asymptote, but rather just hugs up to it like so. We've had some practice at these when it came to exponential and logarithmic functions, so it's more the same in that regard. Labeling the y-axis as x equals 0, the x-axis as y equals 0. I'm going to try to move swiftly here to make sure this video isn't too terribly long. X and g of x. In fact, I'm going to pause the video just for time's sake and I'll come back when the t-chart is done. All right, we're back and I did pick an extensive amount of points on this one just to get a good handle on what the graph's going to look like. I've plotted all those points now. If you need to... Pause the video to comprehend it a little better. I would encourage you to do so. But basically, I just picked points that I felt were fairly convenient to divide negative 6 by. And I ended up getting 5 points on the branch, and that's not all of them, right? Of course, I left out 3 and negative 3, which negative 6 divided by 3 is pretty easy. But anyway, this is certainly enough points to get a handle on an accurate graph like so. I would also note that I did, while you were away... Label the asymptotes as such. Now, they're not always going to be the axes, but on these first two, they were. Okay? Moving on. So then there's translating. So please note the minus h that is to the x and then the plus k. This is a theme that keeps rearing its head. Seems like almost every chapter we have this same idea. x minus h, that's going to control the horizontal shift, and then the plus k is going to control the vertical shift. If it's positive, it's up. If it's negative, it's down. When it comes to x minus h, if it's x minus 3, then that's going to be right 3. If it's x plus 4, that's going to be left 4. That is to be interpreted as minus a negative. I know I say that every time that issue comes up. So I'm moving right on to graphing a few of these. If I do make a t-chart, I'll keep you alive with this t-chart. And I pick, you know, you don't have to put zero because it's going to be undefined again, but you pick negative one and one and then maybe like three because three-thirds is nice and then three divided by six and then this way negative three and negative six. Two wouldn't be bad, right? You know, three over two is just one and a half minus six. So it's not bad, but, you know, do you want to pick five? You know, do you want three-fourths? Do you want three-fifths? Do you want three-eighths? You know, conveniently, I picked negative three, negative six, positive three, positive six. And when I plug that in now, negative 6, that's 3 over negative 6 is negative 1 half, minus 2 would be negative 2 and a half, or as a decimal, it would be negative 2.5. Negative 3 would be negative 1, minus 1, so write 3, divided by negative 3 is negative 1, minus 2 would be negative 3. This is going to be negative 5, this is going to be 1, 3 divided by 1 is 3, minus 2 would be 1, 3, negative 1, and 6, negative 1.5. Points that we can plot on the axes. Negative 6, negative 2.5 is like here. Negative 3, negative 3. Negative 1, negative 5. Three points per branch is a pretty good minimum. 1, 1. 3, negative 1. And 6, negative 1.5. Okay? So certainly those are... Treating or respecting the y-axis as an asymptote so we can go ahead and draw that in right like so and if you can see now They're approaching the line, but never going to reach the line X equal excuse me y equals negative 2 right so we should put that in if you use a second color That's great. I've talked about this before as well if you're using just pencil then do what I'm doing here with just the blue and draw a dashed line for the asymptote and then please do Label it with its equation, y equals negative 2. And then that other branch of the hyperbola looks like that. Okay? They did ask for the domain and range. So here we have the domain, which would, oops, old habits, interval notation, which would have it going from negative infinity to 0 in union with 0 to infinity, just like the parent function was above. The range, however, goes up to negative 2 and then is not inclusive of negative 2. So negative 8 up to negative 2 in union with negative 2 to infinity. Hopefully you see the difference there. Okay? Maybe right now you'd want to pause the video and try the next one. I'm going to keep going. Let's try, I don't really like zero, right? Negative one-fourth, one, negative one-fifth, that's worse. Go the other way, negative one, negative one-third, negative one, that would give me a negative one-third as an output. That's getting better. How about negative two? Negative two plus four is two, that's negative one-half. That's a little more convenient anyway. Negative three. That would be negative 1 over 1. That's negative 1. I like that. Negative 4 would give you a divide by 0 error. Negative 5 would be 1. And negative 6 is going to be a half. Now, in this case, I said, you know, 3 per branch is a good minimum. In this case, we really weren't afforded that unless you're going to go with something that's non-integer in your X. OK, so, you know, I don't know, negative 6 and a half or something. No, I don't think that's going to work either. So this is as best as we can get. And now we're starting to understand what the shape of this is anyway, I hope. So negative 6 1 half is here. Negative 5, 1 is here. Negative 3, negative 1. And then negative 2, negative 1 half. Maybe you can see that there's going to be a vertical asymptote at x equals negative 4. Now there's a big important concept here as it relates to that. I hope you see it graphically as to where they kind of jumped over both axes, both the axis and the asymptote at the same time at a certain point. So your branch is going to look like this for this one. And like this for this one. Okay. But it should not be ignored that, of course, the asymptote being at x equals negative 4, which I am labeling, is what would cause that denominator to be 0. That's why it's an asymptote. That's why it's not defined there. Pardon me. At x equals negative 4 because negative 4 causes a divide by 0 error in that function. Okay. So then I've labeled my vertical asymptote, my horizontal one is y equals 0, domain and range, negative infinity to negative 4, in union with negative 4 to infinity. Range would go up to 0, negative infinity up to 0, in union with then 0 through infinity. Okay, moving on to the next one. I'm going to pause the video again just for time purposes. Okay, back once again. I've made a t-chart, picked four points, two on each branch. That's about all we could do. It's kind of like the other one, and you can see that I have those points plotted as well. Someone might ask, you know, why aren't you just doing a translation? I mean, the box above has has outlined it that it's a translation from the parent graph and you absolutely could. On number four you could have graphed y equals three over x and then translated that down to. I just think that the fact that you'd have to graph it and then graph it again is probably more work than just doing a t-chart. So that's that's how I'm going to model that for you. That being said, absolutely the asymptotes are visible in the equation. This one is going to be at x equals one. One causes the denominator to be zero or again if you think of it in terms of the shift minus one. is right 1 from the y-axis. Plus 5 out here means it's going to be shifted up 5. There's going to be an asymptote here that we can draw and label that is y equals 5, and an asymptote here that is vertical that is x equals 1. Okay, and I've plotted the points already, and you should know what to do at this point. You don't have to make them super long. You know, when I had to graph lines and even parabolas, I wanted things longer. You can get yourself into trouble drawing them super long like this when it's with the asymptotes. So about the length that I'm modeling for you there is the length that I think would be appropriate for that. Domain quickly, negative infinity up to 1, and then in union with 1 to infinity, which again is really all real numbers except for 1, the range excluding 5 then. Negative infinity all the way up to 5, not including 5, and union with 5 to infinity. Okay, and that is day 1. Thanks for watching.

To say that differently is to say that you're going to have things that are in standard form, numerator and denominator. So you're going to have a polynomial in the numerator, polynomial in the denominator, typically in standard form. That's a rational function.

The inverse variation function, f of x equals a over x, is a rational function. So what we started the chapter with in 7.1 and discussing inverse variation, that is a rational function. Then it says, so we'll start with the parent graph then. So if I look at this one right here and they said a over x, if we let the a be 1, then that's a parent function.

And this is the graph of the parent function. The name of that graph, just like a quadratic, gives you a parabola. This type of function, a rational function, gives you a hyperbola, which consists of two symmetrical parts called branches.

Each one of the blue things here in the figure would be known as a branch. The domain and range are all non-zero real numbers. The way we say that, using interval notation, is to say negative infinity to zero, parenthesis, not bracket, in union with zero to infinity.

That's what the domain would be. The range of this one would also be exactly that. It goes from negative infinity down here up to, but not including zero, in union with zero to infinity.

So basically that's every real number except for zero, both for the domain and range. Now we're going to change that on you here in this lesson, and the asymptote will be somewhere other than the x-axis or the y-axis, and we will be labeling those. Any function of the form g of x equals a over x, again, when a is not zero, or not again, but if the denominator, if the numerator is not zero, otherwise it would just be a horizontal line at y equals zero.

Anything other than that has the same asymptotes domain and range as the function that is the parent function. So we are trying to get you in the habit of having you to label the asymptotes. Here they would be x equals zero and y equals zero. Graph the function, compare the graph with the graph of the parent function.

We're not going to be doing much of that comparing the graph, but we do need to be able to graph the function. And you'll notice as I do all these that I'm going to be doing these without a calculator. So this would be a fair one to put on the no calculator part of the next quiz.

So plugging in, I would start with, well, it's okay to start with zero, although that's going to be undefined, isn't it? Two divided by zero is undefined. And then we could go this way, negative 1. 2 divided by negative 1 is negative 2. And then you could go with negative 2, and you're going to divide 2 by negative 2, and that's negative 1. And then if you want to pick another convenient number, how about negative 4? 2 divided by negative 4 is negative 1 half.

If we go the other direction, it's going to be 1, 2, 2, 1, 4, 1 half. There's six good points that you can plot. Now it would be the case, because there is symmetry going on here, maybe you've recognized that already, that there would be a point at negative a half, negative two, negative four, excuse me. There is a point here.

There is symmetry about the line y equals x. There's symmetry about the line y equals negative x as well. So then we have one, two, and two, one, and four, one half.

Again, there would be one at one half comma four just as well. That does help to give us something to aim at. Make sure that neither branch crosses the asymptote, but rather just hugs up to it like so.

We've had some practice at these when it came to exponential and logarithmic functions, so it's more the same in that regard. Labeling the y-axis as x equals 0, the x-axis as y equals 0. I'm going to try to move swiftly here to make sure this video isn't too terribly long. X and g of x. In fact, I'm going to pause the video just for time's sake and I'll come back when the t-chart is done. All right, we're back and I did pick an extensive amount of points on this one just to get a good handle on what the graph's going to look like.

I've plotted all those points now. If you need to... Pause the video to comprehend it a little better. I would encourage you to do so.

But basically, I just picked points that I felt were fairly convenient to divide negative 6 by. And I ended up getting 5 points on the branch, and that's not all of them, right? Of course, I left out 3 and negative 3, which negative 6 divided by 3 is pretty easy.

But anyway, this is certainly enough points to get a handle on an accurate graph like so. I would also note that I did, while you were away... Label the asymptotes as such. Now, they're not always going to be the axes, but on these first two, they were. Okay?

Moving on. So then there's translating. So please note the minus h that is to the x and then the plus k. This is a theme that keeps rearing its head.

Seems like almost every chapter we have this same idea. x minus h, that's going to control the horizontal shift, and then the plus k is going to control the vertical shift. If it's positive, it's up.

If it's negative, it's down. When it comes to x minus h, if it's x minus 3, then that's going to be right 3. If it's x plus 4, that's going to be left 4. That is to be interpreted as minus a negative. I know I say that every time that issue comes up.

So I'm moving right on to graphing a few of these. If I do make a t-chart, I'll keep you alive with this t-chart. And I pick, you know, you don't have to put zero because it's going to be undefined again, but you pick negative one and one and then maybe like three because three-thirds is nice and then three divided by six and then this way negative three and negative six.

Two wouldn't be bad, right? You know, three over two is just one and a half minus six. So it's not bad, but, you know, do you want to pick five?

You know, do you want three-fourths? Do you want three-fifths? Do you want three-eighths? You know, conveniently, I picked negative three, negative six, positive three, positive six. And when I plug that in now, negative 6, that's 3 over negative 6 is negative 1 half, minus 2 would be negative 2 and a half, or as a decimal, it would be negative 2.5.

Negative 3 would be negative 1, minus 1, so write 3, divided by negative 3 is negative 1, minus 2 would be negative 3. This is going to be negative 5, this is going to be 1, 3 divided by 1 is 3, minus 2 would be 1, 3, negative 1, and 6, negative 1.5. Points that we can plot on the axes. Negative 6, negative 2.5 is like here.

Negative 3, negative 3. Negative 1, negative 5. Three points per branch is a pretty good minimum. 1, 1. 3, negative 1. And 6, negative 1.5. Okay?

So certainly those are... Treating or respecting the y-axis as an asymptote so we can go ahead and draw that in right like so and if you can see now They're approaching the line, but never going to reach the line X equal excuse me y equals negative 2 right so we should put that in if you use a second color That's great. I've talked about this before as well if you're using just pencil then do what I'm doing here with just the blue and draw a dashed line for the asymptote and then please do Label it with its equation, y equals negative 2. And then that other branch of the hyperbola looks like that. Okay? They did ask for the domain and range.

So here we have the domain, which would, oops, old habits, interval notation, which would have it going from negative infinity to 0 in union with 0 to infinity, just like the parent function was above. The range, however, goes up to negative 2 and then is not inclusive of negative 2. So negative 8 up to negative 2 in union with negative 2 to infinity. Hopefully you see the difference there. Okay?

Maybe right now you'd want to pause the video and try the next one. I'm going to keep going. Let's try, I don't really like zero, right?

Negative one-fourth, one, negative one-fifth, that's worse. Go the other way, negative one, negative one-third, negative one, that would give me a negative one-third as an output. That's getting better.

How about negative two? Negative two plus four is two, that's negative one-half. That's a little more convenient anyway. Negative three. That would be negative 1 over 1. That's negative 1. I like that.

Negative 4 would give you a divide by 0 error. Negative 5 would be 1. And negative 6 is going to be a half. Now, in this case, I said, you know, 3 per branch is a good minimum.

In this case, we really weren't afforded that unless you're going to go with something that's non-integer in your X. OK, so, you know, I don't know, negative 6 and a half or something. No, I don't think that's going to work either. So this is as best as we can get. And now we're starting to understand what the shape of this is anyway, I hope.

So negative 6 1 half is here. Negative 5, 1 is here. Negative 3, negative 1. And then negative 2, negative 1 half.

Maybe you can see that there's going to be a vertical asymptote at x equals negative 4. Now there's a big important concept here as it relates to that. I hope you see it graphically as to where they kind of jumped over both axes, both the axis and the asymptote at the same time at a certain point. So your branch is going to look like this for this one. And like this for this one. Okay.

But it should not be ignored that, of course, the asymptote being at x equals negative 4, which I am labeling, is what would cause that denominator to be 0. That's why it's an asymptote. That's why it's not defined there. Pardon me. At x equals negative 4 because negative 4 causes a divide by 0 error in that function.

Okay. So then I've labeled my vertical asymptote, my horizontal one is y equals 0, domain and range, negative infinity to negative 4, in union with negative 4 to infinity. Range would go up to 0, negative infinity up to 0, in union with then 0 through infinity. Okay, moving on to the next one.

I'm going to pause the video again just for time purposes. Okay, back once again. I've made a t-chart, picked four points, two on each branch. That's about all we could do. It's kind of like the other one, and you can see that I have those points plotted as well.

Someone might ask, you know, why aren't you just doing a translation? I mean, the box above has has outlined it that it's a translation from the parent graph and you absolutely could. On number four you could have graphed y equals three over x and then translated that down to.

I just think that the fact that you'd have to graph it and then graph it again is probably more work than just doing a t-chart. So that's that's how I'm going to model that for you. That being said, absolutely the asymptotes are visible in the equation. This one is going to be at x equals one.

One causes the denominator to be zero or again if you think of it in terms of the shift minus one. is right 1 from the y-axis. Plus 5 out here means it's going to be shifted up 5. There's going to be an asymptote here that we can draw and label that is y equals 5, and an asymptote here that is vertical that is x equals 1. Okay, and I've plotted the points already, and you should know what to do at this point.

You don't have to make them super long. You know, when I had to graph lines and even parabolas, I wanted things longer. You can get yourself into trouble drawing them super long like this when it's with the asymptotes. So about the length that I'm modeling for you there is the length that I think would be appropriate for that.

Domain quickly, negative infinity up to 1, and then in union with 1 to infinity, which again is really all real numbers except for 1, the range excluding 5 then. Negative infinity all the way up to 5, not including 5, and union with 5 to infinity. Okay, and that is day 1. Thanks for watching.

Transcript for:Understanding Rational Functions and Graphing

Transcript for:
Understanding Rational Functions and Graphing