Rational Functions

本站添加: 2026-05-15

[00:00:00]Let's talk about rational functions and  some of the common characteristics that they have. We're going to start by sketching  the graph of our reciprocal function, f of x equals one over x. We graphed this  a long time ago when talking about those toolkit functions, but I am going to go  ahead and make a table of values just to help me remember what this shape  looks like and plot some points.

[00:00:25]If I plug in negative two for x, I get negative  one-half. If I plug in negative one for x, I get one over negative one, or negative  one. If I plug in zero, well, one over zero is undefined. Zero is actually not in the  domain of this function. Then plug in one—one divided by one is one. Plug in two—one-half.  So I'm going to go ahead and plot those points.

[00:00:58]I've got the point, let's see, negative two,  negative one-half, negative one, negative one, one, one, two, one-half. When I connect the  dots, what's happening with this graph is as the x-values are getting larger, the y-values  are approaching zero. We're getting fractions with magnitudes that are smaller and smaller. The  same thing is happening with negative values. As I plug in negative numbers with larger and larger  magnitudes, those y-values are approaching zero.

[00:01:33]In the middle of the graph, this function is  actually undefined at x equals zero. What's happening is as we plug in smaller and smaller  values for x, those y-values are actually getting larger and larger. One divided by something that's  really, really small gives us a very large output.

[00:01:54]The graph is approaching the y-axis both on  the right and the left side of x equals zero.

[00:02:04]What we actually have here are known as  asymptotes. As x goes to positive infinity on the right side of the graph, the y-values are  approaching zero. On the left side of the graph, the y-values are also approaching  zero. We can describe this behavior by saying y equals zero is a  horizontal asymptote. Again, that means the y-values are approaching  zero at the ends of the graph.

[00:02:31]In the middle of the graph, we have another  asymptote. What's happening here is as we approach x equals zero on the right side, the y-values are  going to infinity. As we approach x equals zero, the y-values are getting really, really big.  As we approach x equals zero on the left side, the y-values are now going to negative infinity.  That's because x equals zero is a vertical asymptote. The function is approaching that line  x equals zero but never actually reaching it.

[00:03:07]Many rational functions are going to have  asymptotes. Just to summarize, we say that we have a vertical asymptote if, as our x-values approach  that number, our y-values go either to infinity or negative infinity. We have a horizontal asymptote  if, as our x-values go to positive or negative infinity, our y-values go to that value. These  are more formal definitions of asymptotes.

[00:03:37]Let's take a look at another example and see if we can graph and identify our asymptotes.  Let's try to sketch the graph of f of x equals one over (x plus one), plus three,  then give the equations of all asymptotes.

[00:03:52]I can use my idea of graphing transformations  to sketch this graph. I can start with my parent function, y equals one over x.  That's what we just graphed in the previous slide. y equals one over  x goes through the point one one, has a horizontal asymptote of zero, and  a vertical asymptote of x equals zero.

[00:04:17]This plus three that is tacked on outside  of the fraction is an output transformation that's going to shift the graph up three  units. The plus one that's inside the fraction is an input transformation  that's going to impact our x-values.

[00:04:34]We do the opposite of what it says, so  we're actually going to subtract one from all of our x's. That's going to take our  graph and shift it to the left one unit.

[00:04:43]So I'm going to take this parent  function, which I have drawn in blue, and I am going to move it up three units  and to the left one unit. I like doing that by drawing in the asymptotes. That  helps me kind of center my graph.

[00:05:00]Notice my vertical asymptote moved left one  unit. So there's the line x equals negative one. My horizontal asymptote went up one, two,  three units. It's now at y equals three. Then I took a couple of the points on my parent  function and I moved them up one, two, three—here we go, let's try that again—one,  two, three, and then to the left one unit.

[00:05:31]In black is a sketch of this function: f of  x equals one over (x plus one), plus three. I can see where the asymptotes are. Those asymptotes  were moved through my shifts. I have a horizontal asymptote of y equals three. The graph is  flattening out at the ends and approaching the line y equals three. I have a vertical  asymptote of x equals negative one. That is where my function is undefined, and those y-values are  going to positive infinity and negative infinity.

[00:06:06]All right. Let's see if we can  summarize the important features of our rational functions without  actually having to create a graph.

[00:06:15]We're going to discuss finding domain, vertical  asymptotes, horizontal asymptotes, and intercepts.

[00:06:23]If you want to find the domain of a rational  function, this is actually something we've talked about before. It hasn't changed. The  idea is: division by zero is undefined. We can't divide by zero. So if we want to  find the domain of a rational function, we're going to take our denominator  and say that that cannot equal zero.

[00:06:44]In my first example, the denominator is x minus four. x minus four cannot equal zero.

[00:06:52]I add four to both sides.  I get x cannot equal four.

[00:06:56]So the domain is every single number except for  four. Four makes that denominator equal to zero.

[00:07:03]In interval notation, I would write that as  negative infinity to four, union four to infinity, making sure to use those parentheses with the  four because four is actually not in that domain.

[00:07:16]Same idea works for the next function.  I'm going to take my denominator, x squared minus 2x minus three. That cannot equal zero.

[00:07:24]Denominator can't equal zero. This is something that we can solve by factoring.

[00:07:32]Product of negative three, sum of negative  two would be negative three and positive one.

[00:07:39]So this factors as x minus three, x plus one. That means x cannot equal positive three, and x cannot equal negative one. So in interval notation, I've got negative infinity to negative one, union  negative one to three, union three to infinity.

[00:07:59]Again using those parentheses because we are not  including the points at negative one and three.

[00:08:07]Okay, so if we want to find the  domain of a rational function, we always look at the denominator. Doesn't matter what's in the numerator. The denominator can't be  zero because we can't divide by zero.

[00:08:20]All right, well what about  asymptotes? How do we find asymptotes?

[00:08:23]Let's start with our vertical asymptotes. So vertical asymptotes are going to occur at zeros of the denominator, as long as that  zero doesn't come from a common factor.

[00:08:39]This will be easiest for me  to explain with some examples, but I do have kind of a summary  here typed up on the screen.

[00:08:47]If we want to find vertical asymptotes, we're  going to factor everything as much as possible.

[00:08:53]This is a good time to note  any restrictions in the domain, the idea that the denominator can't equal zero.

[00:08:59]We're going to reduce the expression  by canceling common factors.

[00:09:04]After we reduce, then zeros of the denominator  are places where our vertical asymptotes occur.

[00:09:12]If we have restrictions in the  domain where asymptotes do not occur, those are actually going to be holes in  the graph, or removable discontinuities.

[00:09:20]All right, so again, I think this will be easier  for me to explain by showing you an example.

[00:09:25]So if I want to find vertical asymptotes,  the first thing I'm going to do is see if I can factor my  numerator and/or my denominator.

[00:09:35]In my first example, I have the  function f of x equals 4x over x plus 3.

[00:09:42]I can't factor this at all, so all I'm  going to do is take my denominator, x plus 3, set it equal to zero and solve. x plus 3 equals zero when x equals negative three.

[00:09:56]So negative three is going  to be a vertical asymptote.

[00:10:00]If we plug negative three into this function,  we're going to get zero in the denominator.

[00:10:05]That means the function is going to be  going to infinity or negative infinity.

[00:10:10]Let's compare that to my second example. In my second example, we have the function f of x equals 2x minus 6 over x squared minus 9. My first step is to factor as much as possible.

[00:10:24]In this case, I can factor the  numerator by pulling out a two.

[00:10:28]We've got a common factor of two in the numerator. So if I pull that out, I'm left with 2 times x minus 3. And the denominator is a difference of perfect squares: x squared minus 9. That can be factored as x plus 3, x minus 3.

[00:10:48]Now notice that we have a  common factor of x minus 3.

[00:10:53]If you have a common factor—something that  exists in both the top and the bottom—that is going to actually be a place where you have a  hole in the graph or a removable discontinuity.

[00:11:06]To find my asymptote, I only look  at factors of the denominator that are not part of common factors. So the only thing I'm going to look at here is the x plus 3. I set it equal to zero.

[00:11:20]x plus 3 equals zero when x equals negative three. So x equals negative three is a vertical asymptote.

[00:11:28]So vertical asymptotes occur at zeros  of the denominator as long as that zero comes from something that's not a common factor. That x minus 3, because that is a common factor, that actually means that we're going to  have a hole in the graph at x equals three.

[00:11:48]And holes are sometimes referred  to as removable discontinuities.

[00:11:55]All right. What about our horizontal asymptotes?

[00:12:00]The kind of horizontal asymptote we have is  going to depend on the degree of the numerator and the degree of the denominator,  degree being that largest power of x.

[00:12:10]So the first situation would be the degree of  the numerator is smaller than the degree of the denominator. An example would be x, which has  a degree of one, over x squared plus one, which has a degree of two. Anytime the degree of  the numerator is smaller than the degree of the denominator, the line y equals zero  is going to be a horizontal asymptote.

[00:12:38]To explain why, the end behavior is always  determined by our leading terms, our largest powers of x. So the end behavior of this function  is going to be determined by x over x squared, and x over x squared simplifies down to one over x. If  you plug in a really, really big number for x, one divided by a huge number is going to go to zero,  and y equals zero is our horizontal asymptote.

[00:13:13]Anytime the degree of the numerator is  smaller than the degree of the denominator, that denominator is growing  much faster than the numerator, and the function is going to approach  y equals zero when x is very large.

[00:13:27]The second situation we might have is when the  degree of the numerator is equal to the degree of the denominator. An example would be f of x  equals two x plus one over five x plus three. In this case, the degree of the numerator is one,  and the degree of the denominator is also one, so they have the same degree. To  find our horizontal asymptote, again we can think about the leading term, the  largest power of x in the top and the bottom.

[00:14:01]When x gets really, really big, this function  is going to approach two x divided by five x, which simplifies down to two fifths. So y  equals two fifths is my horizontal asymptote.

[00:14:17]Our horizontal asymptote is always going to be  the ratio of those leading coefficients — so leading coefficient two over five. Y equals  two fifths is my horizontal asymptote.

[00:14:32]Finally, the third thing we might see is the  degree of the numerator could be bigger than the degree of the denominator. For example, x  squared over x plus four. In this situation, there's actually going to be no horizontal  asymptote. If you take your leading term in the top — in this example it was  x squared — divided by the leading term in the bottom — in this case x — that  simplifies down to x. If I plug in a really, really big number, that's actually going to  go to infinity. It's not going to approach a finite number. So this is a situation  where we have no horizontal asymptote.

[00:15:13]All right, let's see if we can put  this together and do some examples.

[00:15:17]Find all horizontal asymptotes. So my first  example: four x minus three over two minus x.

[00:15:27]This is the second situation, where the degree  of the top and the degree of the bottom is the same. So if I look at those leading terms,  largest power of x, four x over negative x simplifies down to negative four. That means y  equals negative four is my horizontal asymptote.

[00:15:50]Second example, we have the function f of x equals  five x plus two over x squared plus one. In this case, the degree of the top is smaller than the  degree of the bottom. If I look at those leading terms, five x over x squared, that simplifies to  five over x. If we plug in a really big number for x, that is going to go to zero, and y equals zero  is my asymptote. Anytime the degree of the top is smaller than the degree of the bottom, it's always  going to have an asymptote of y equals zero.

[00:16:29]Finally, my last example: f of x equals x cubed plus three x over x squared minus x  minus two. This is a situation where the degree of the top is bigger than the degree of  the bottom. Third-degree polynomial on the top, second degree on the bottom, and so that means  we are going to have no horizontal asymptote.

[00:16:53]Okay. So vertical asymptotes are zeros of the  denominator as long as they don't come from a common factor. Horizontal asymptotes are  related to the degree of the numerator and the denominator, and horizontal asymptotes really  are describing our end behavior of our function.

[00:17:11]Now, in that third case where we  don't have a horizontal asymptote, sometimes we have a slanted asymptote. So I  have graphed on my computer the function f of x equals two x squared minus five x plus three  over x minus two. If I take a look at the ends of the graph, there's no horizontal asymptote.  They're not approaching a horizontal line. I know that's going to happen because the degree of  the top is bigger than the degree of the bottom.

[00:17:46]Instead, I see a slanted asymptote. The  ends of the graph are approaching a line, but it's actually not a horizontal line. If we  want to figure out what this slanted asymptote is, we can use division. You can use synthetic division or long division. I'll go  ahead and use synthetic division.

[00:18:08]So my numerator was two x squared minus five x  plus three, so I jot those numbers down. If I'm dividing by x minus two, I write down a positive  two. Bring down the two. Two times two is four.

[00:18:29]Add, I get negative one. Negative one  times two is negative two. Add, I get one.

[00:18:41]So what this quotient, what this division is  telling me, is that I can rewrite this function as two x minus one plus my remainder, which was  one over x minus two. So this rational function that we started with is equivalent or equal  to the line two x minus one plus one over x minus two. The slant asymptote is always  going to be your quotient. So in this case, the two x minus one — my slant asymptote is  the line y equals two x minus one. That's because the remainder, as x gets really, really  big, that remainder is going to go to zero, and we're just going to be left with  that line y equals two x minus one.

[00:19:32]So if you want to find a slant asymptote, you  can use synthetic division or long division.

[00:19:36]You ignore the remainder, and your answer is  just the quotient. That is your asymptote.

[00:19:46]All right, let's see if we can put all of  this information together and sketch a graph.

[00:19:52]So we've got the rational function f of x equals x minus one over x plus two. We're going to find  all of the asymptotes, then we'll find intercepts, and then finally we'll put all of that  information together and draw our picture.

[00:20:11]All right. To find my vertical asymptote,  I noticed that I can't do any factoring, so I'm just going to take my denominator,  x plus two, and set it equal to zero. That gives me x equals negative two. So I have a  vertical asymptote at x equals negative two.

[00:20:32]For my horizontal asymptote, this is case  two. The degree of the numerator and the degree of the denominator is equal.  So if I take those leading terms, those largest powers of x, x over x is one.  So y equals one is my horizontal asymptote.

[00:20:53]All right. To find the intercepts,  we're going to plug in zero.

[00:20:58]To find my x-intercept, I'm going to plug  in zero for y. If I plug in zero for f of x, I actually get zero equals x minus one over x plus  two. The only way a fraction can equal zero is if the numerator equals zero. So if you want to find  an x-intercept, x-intercepts are going to occur at zeros of the numerator. Vertical asymptotes  occur at zeros of the denominator. X-intercepts occur at zeros of the numerator. So  I set my numerator equal to zero, solve for x. I get x equals one.  So I have an x-intercept at one.

[00:21:41]To find my y-intercept, I just plug in zero  for x, and I get negative one over two. So my y-intercept is the point zero comma one-half.  My x-intercept is the point one comma zero.

[00:21:56]Finally, let's put this all  together and draw a graph.

[00:22:00]To graph a rational function, I usually like  drawing in my asymptotes first. So I've got a vertical asymptote x equals negative  two, and my horizontal asymptote is y equals one. Those asymptotes are  going to help me guide my graph.

[00:22:18]Then I'm going to plot my intercepts. I  have a y-intercept at zero comma one-half and an x-intercept at one comma zero. So I can  connect those points, and I want to make sure that on the right side of the graph, my curve  is approaching my horizontal asymptote. Then, on the left side, I'm going to draw the graph so  it's approaching that vertical asymptote. I know that the curve goes down towards the vertical  asymptote and not up because there's no other intercepts. It can't cross the axis again,  so it has to go down towards that asymptote.

[00:23:03]There's going to be another half of the graph,  and to help me figure out where the rest of the function is, I am going to try to plot  another point to the left of this asymptote.

[00:23:16]So I have a vertical asymptote at negative  two. I decided to choose negative three. I'm just going to choose any number that's  smaller than negative two. I'm going to plug that into my function just to guide my  graph, give me another point to plot. So I took negative three, and I plugged that into  the function. I got negative three minus one, that's negative four, over negative three plus  two, that's negative one. That simplifies to four.

[00:23:43]So if f of negative three is four, that  means my graph has to contain the point negative three comma four. I'm going to draw  this so that as I go towards x equals infinity, the graph goes down towards my horizontal  asymptote. Then I'm going to have it go up towards my vertical asymptote. And there's  a rough sketch of this rational function.

[00:24:14]All right, let's take a look at  another example. Let's see if we can find all asymptotes, find  all holes, find all intercepts.

[00:24:23]I have the function f of x equals x squared  minus three x minus ten over x squared plus x minus two. I'm going to start by factoring as  much as I can. The numerator can be factored because the product of negative ten  and the sum of negative three would be negative five and positive two. So I can  factor the numerator as x minus five, x plus two.

[00:24:52]I can also factor the denominator.  Product of negative two, sum of one would be two times negative  one, or x plus two, x minus one.

[00:25:04]Now I notice I do have a common  factor of x plus two. That is a hole in the graph. That's not going to  give me an asymptote or an intercept.

[00:25:15]I look at x minus one. That is the only factor  in the denominator that can't be canceled, that isn't part of a common factor.  I set that x minus one equal to zero. I get x equals one. So I have  a vertical asymptote at x equals one.

[00:25:35]To find my horizontal asymptote, I look  at my leading terms. The largest power in the numerator is x squared. The  largest power of x in the denominator is also x squared. x squared over x squared is  one, so y equals one is my horizontal asymptote.

[00:25:56]Since I have my function in factored form, I  can find my hole or my removable discontinuity by looking at that common factor. That  common factor x plus two equals zero when x equals negative two. So that tells me  there's a hole at x equals negative two.

[00:26:15]To find my intercepts, my x-intercepts are  going to occur at zeros of the numerator, as long as we're not talking about  the hole. The common factor is not going to give you an  asymptote or an intercept.

[00:26:32]To find my x-intercept, I just  look at the x minus five. x minus five equals zero when x equals five, so  the point five, zero is my x-intercept.

[00:26:45]To find my y-intercept, I plug  in zero for x. In the numerator I get zero minus zero minus ten, and in the  denominator I get zero plus zero minus two.

[00:27:01]Negative ten over negative two is five.  So I have a y-intercept at zero, five.

[00:27:08]Okay, so common factors are holes. Zeros of the denominator are vertical asymptotes.  Zeros of the numerator are x-intercepts.

[00:27:21]For my last example, I want to work backwards.  Now we're given a picture or a graph of a rational function. Let's see if we can come  up with an equation that matches this graph.

[00:27:32]I'm going to start by identifying  my asymptotes. We have a vertical asymptote at x equals negative one. I have a  horizontal asymptote of y equals positive one.

[00:27:47]Oh, and then one more thing—let's identify our intercept as well. It looks like we  have an x-intercept at two. Two, zero.

[00:27:59]Let's see if we can use this information  to create our graph—or our equation, I should say. If we have a vertical asymptote at  x equals negative one, x plus one has to be in my denominator. Vertical asymptotes  are zeros of the denominator, so I have to have a factor of  x plus one in my denominator.

[00:28:25]My x-intercept is at two. X-intercepts are zeros of the numerator, so that means x  minus two has to be in my numerator.

[00:28:38]All right, so is this it, or should there be any  other constants in this problem? To determine that, I am going to look at my horizontal  asymptote. We need to have a horizontal asymptote of one, and I'm going to check:  does this have a horizontal asymptote of one?

[00:28:56]Oh yeah, because if I look at my leading  terms—x and x—x over x becomes one. So this does have a horizontal asymptote of one. If I  wanted to have a horizontal asymptote of two, I could just put a GCF or a factor of two in  front of the x minus two. But we don't have to do that. x over x gives us one, so this  function matches the graph that is given.

[00:29:24]All right, that was my last example. Please let  me know if anything in this video is unclear.