Graphing Rational Functions - Video Tutorials & Practice Problems

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Graphing Rational Functions Using Transformations

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Hey, everyone, we now know all of the components of the graph of a rational function. But taking those components and putting them all together to form an actual graph can be the hardest part. So here we're going to take the most basic of rational functions one over X and simply apply rules of transformations to it in order to graph a more complicated rational function. So here I'm gonna walk you through using transformations to graph this function without doing a single calculation. We're just gonna take our function one over X and pick it up and move it around a little bit to get graph of our new function, which will end up looking at something like this. So let's go ahead and walk through how to get there. Now, with our function G of X here we see that it looks really similar to one over X. We even have this one over X. Here, we just have these extra numbers added in. Now these numbers represent transformations to our function. And when working with rational functions, we're going to focus on two types of transformations in particular reflections and shifts because they are the most commonly occurring transformations when working with rational functions. Now remember whenever we have a negative outside of our function that tells us we're going to have a reflection over the X axis. Whereas having a negative inside of our function is a reflection over the Y axis. Now, with H and K over here X minus hh represents a horizontal shift side to side by some number H units. And then K represents a vertical shift up and down by some number K units. So let's go ahead and apply these transformations to our function here. Now, our function one over X are graph here, we see that we have asymptotes at our X and Y axis. And we also have these points on our graph at 11 and negative one negative one that are going to serve as sort of reference or test points for us as we graph our new function. So we're going to focus on transforming our asymptotes and these points in order to get the graph of G of X here. So let's go ahead and start with step one which is to find the vertical Asymptote and plot it at X equals H. Now looking at our function here one over X minus three plus one of X minus three. So H here is three. I'm going to have a vertical Asymptote at X equals three. Now we can go ahead and plot that on our graph using a dashed line because it is an Asymptote. And now we can go ahead and move on to step two, which is to plot our horizontal Asymptote at Y equals K. Now, looking back to our function here, I have this plus one on the end. So I know that K is going to be one, I'm going to plot my horizontal Asymptote at Y equals one. Again using a dash line. Now compared to our original asymptotes at our axis, we see that these got shifted H units over and K units up and we see this transformation happening here. So let's keep transforming our graph. Now, moving on to step number three, we want to go ahead and determine if there is a reflection. Now remember our reflection is if we have these negatives that got put into our function and looking at my function, I don't have any negatives that got added in there. So I do not have a reflection. Now, if I did, I would take those test points 11 and negative one, negative one and simply reflect them over X or Y accordingly. But I don't need to do that here. So let's go ahead and move on to step three B and go ahead and shift our test points by H comma K. Now we know that H comma K is just three and one because we already plotted those at our awesome totes. So I'm going to take my test points, 11 and negative one negative one and shift them accordingly. So my first point here, I'm gonna go ahead and shift that three units over to the right and then one unit up landing me at the 0.42 then my other test point which is right here at negative one, negative one. I'm gonna shift that again, three units to the right and one unit up landing me at 20. So I've now shifted both my asymptotes and those reference points. So I can go ahead and move on to my final step here, which is going to be to sketch my curves approaching the asymptotes. So looking at this first point here, I'm gonna approach that Asymptote going up and then approach my horizontal Asymptote this way, then my other test point, same thing approaching my asymptotes on either side. And now I have the final graph of my function. Now these are my curves here. I'm going to highlight them just so you can see them a little bit bit better. And you'll notice that this looks the almost the exact same as our graph of one over X. It's just picked up and moved over just like I said. Now, we want to look at one other thing here, our domain and our range. Now our domain and our range are going to be split by our asymptotes. And we're going to write them in set notation using this U symbol which just represents a union between two sets. Now, our domain is actually always going to go from negative infinity until I reach my Asymptote at H and then continue on from H to infinity. So looking at the graph I have here, I'm going to go from negative infinity until I reach that Asymptote at three. And then I have a break at that Asymptote because my domain is not defined there and I pick back up and go from three to infinity. Then for my range, the same exact thing is going to happen, we're going to go from negative infinity to K and then from K to infinity because K is where my Asymptote is. So looking at my function here, I know I go from negative infinity until I reach that Asymptote where my domain is not defined at one and then continuing on from one to infinity. You now you'll notice that these are enclosed in parentheses rather than square brackets because we know that these numbers are not included in our domain or range. So now that we know how to graph using transformations, let's get some more practice.

2

example

Example 1

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3m

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In this problem, we're asked to graph the given function as a transformation of one over X squared. And here I have the function G of X is equal to negative one over X minus two squared. So let's go ahead and graph this starting with step one which is to find our vertical Asymptote and plot it at X equals H. Now looking at my function, knowing that this is a transformation of one over X squared. I want to go ahead and identify H which in this case is just two. So I'm gonna plot my vertical Asymptote at X equals two using a dash line. Now, for step number two, we want to plot our horizontal Asymptote at Y equals K. In this case, I actually don't have a value for K. It's just zero because it didn't get transformed. So Y is simply zero for my horizontal Asymptote and it's actually at the same exact place as the horizontal Asymptote of my original function. So that didn't really get shipped in. Now, we want to move on to step three and identify if there is a reflection happening, which we can tell whether there is a negative outside or inside of our function here, my negative is right here and it makes my entire function negative. So I do have a reflection and that reflection is over the X axis. So I'm gonna take these two test points or reference points that I have here and I'm gonna go ahead and reflect them over the X axis accordingly. So looking at my points starting with 11, remember we can think of a reflection as folding it in half at the X axis and kind of stamping that point on the other side. So here 11 will end up right here and then negative 11 will end up here reflected over that X axis. Now, we want to go ahead and shift these points by H comma K and then H here is two K is simply zero. So we're just gonna shift these two points or two units over to the right. So starting with my point at one negative one shifting that two over, it's gonna end up right over here. And then my other point is going to get shifted to over and it's actually going to end up exactly where my other point was right here. So I have my two new points they that have been both reflected and shifted and we can go ahead and move on to our final step and simply sketch our curves approaching our asymptotes. So let's go ahead and do that here approaching my asym tote on either side, I know that the shape is the exact same as my original one over X squared, but we've just been reflected and shifted. So here is my new function, we fully craft it. Let's go ahead and identify our domain and our range. Now, our domain is going to go from negative infinity until we reach our Asymptote where it's going to stop and then continue on the other side of my Asymptote. So from negative infinity to my Asymptote here is negative infinity to two and then from two to infinity. Now as for my range, because we're only on one side of our X axis here, we only go from negative infinity up until we reach that Asymptote, which happens to be at the X axis. If we are up here, our range would be different. But since we're down here, it goes from negative infinity until zero where that Asymptote is OK. Now, now that we've graphed with this transformation and identified our domain in our range. Thanks for watching and I'll see you in the next video.

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Problem

Problem

Graph the rational function using transformations. ï»¿$f\left(x\right)=-\frac{1}{x}+3$ï»¿

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Problem

Problem

Graph the rational function using transformations. ï»¿$f\left(x\right)=\frac{1}{\left(x+3\right)^2}-2$ï»¿

A

B

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D

5

concept

How to Graph Rational Functions

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8m

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Hey, everyone, we now know all of the components of the graph of a rational function. And we're not always going to be able to graph rational functions using basic transformations sometimes like with this function here, there's gonna be a little bit more going on. So we're gonna take all of these components and put them together. Now, in order to fully graph a rational function from scratch, now, it might feel like a lot and that's OK. I'm going to walk you through it step by step. And if you follow these steps, you'll be able to graph any rational function. So let's go ahead and not waste any time and get into graphing. So looking at my function here, I have two X minus three over X minus one. Now, starting with step number one, we want to factor and find the domain. Now, looking at my function, it can't be factored any more than it already is. So we can just go ahead and skip straight to finding our domain by setting our denominator equal to zero. Now my denominator is X minus one. So adding one to both sides here, I end up with X is equal to one. Now, since this is my domain, I know that this is what my X value cannot be because it would make my denominator zero. So my domain is simply X such that X cannot be equal to one. Now we finish step one. Let's move on to step two and go ahead and find any holes that may exist in our graph by setting our common factors equal to zero. Now, looking at my function, I again, don't have any common factors here. So I don't have to worry about the step and we can move on to step two B and go ahead and put our function in lowest terms. Now taking another look at our function, I can't cancel anything out because it already didn't have common factors. So I can go ahead and move on to step three and find my X intercepts. Now to find my X intercepts, I'm going to set my numerator equal to zero, which is the equivalent of simply solving the equation F of X equals zero. So my numerator here two X minus three, if I add three to both sides, it cancels leaving me with two X is equal to three and then to isolate X, I wanna divide both sides by two. Now that leaves me with X is equal to three halves or if you want that as a decimal, this is also just 1.5. So I can go ahead and plot that on my graph at X equals 1.5. But we also want to determine the behavior of our graph at this X intercept. The same way we did with polynomials by looking at the multiplicity. Now, looking back at my original function two X minus three over X 12, X minus three is a factor that only occurs once, which tells us that our multiplicity is one. Now because our multiplicity is one, this is an odd number. So that tells me that my graph is fully going to cross the X axis at that point which I can denote here. And then I can fix that later whenever I fully sketch my graph. OK, let's move on to our Y intercept in step four and compute F of zero by plugging zero into our function, plugging zero into my function. Here I get two times zero minus three which gives me negative three over zero minus one, negative one. Now, this simplifies to just positive three. So my Y intercept is three and I can go ahead and plot that on my graph as well right on that Y axis. Now that we have our intercepts, let's go ahead and move on to finding our asymptotes. So my first Asymptote that I want to find is my vertical Asymptote. And here I'm just gonna set my denominator equal to zero now because we didn't have any common factors that canceled when we put our function in lowest terms. This is be the same thing as finding our domain and we just have X minus one equals zero. So I can go ahead and add one to both sides. And I'm simply left with X is equal to one because that cancels and that's all I'm left with. So let's go ahead and plot our vertical Asymptote on our graph at X equals one. Now, I'm using a dash line because this is an Asymptote. Let's go ahead and move on to step six and find our horizontal Asymptote. So here to find our horizontal Asymptote, we're looking at the degree of both the numerator and the denominator. And looking at my function here, two X minus three over X minus one, I have a degree of one in that numerator. I also have a degree of one in the denominator. So these degrees are equal to each other now because they're equal to each other that tells me I need to divide my leading coefficients and my leading coefficient in my numerator is simply two. And then in my denominator, I have this one here. So taking those and dividing them 2/1, gives me a horizontal Asymptote at Y equals two. So let's go ahead and plot that Asymptote on our graph. Uh Y equals to again using a dash line because it is an Asymptote. OK? Now that we have all of this information, we still don't know exactly what's happening all around these knows known components. So we're gonna have to do something similar or the exact same that we did with polynomials and divide our graph up into intervals. Now, our intervals with our rational function are going to be determined by our vertical asymptotes and our X intercepts. So we're still going to take a look at our graph from left to right. And along the way, we're going to look for those known components in order to break our graph into intervals and find the behavior within each of those intervals. So let's go ahead and go back up to our graph here in order to determine those intervals. So starting on that left side, going until I reach my first known component, either a vertical Asymptote or an X intercept. The first thing I reach is this vertical Asymptote at X equals one. So from negative infinity to that first known point at one is my first interval. Then my next interval starting from that one going until I reach the next known component I have this X intercept almost right next to it. So my next interval is simply from 1 to 3 halves, a really small interval, but an interval nonetheless, now finally, from this three halves to infinity is my final interval here because there are no other known components. Now that we have these intervals, we need to choose a value for X in each of them in order to plot a point to determine the behavior. So in this first interval, negative infinity to one, remember you want to choose something that's in your graph so that you're actually able to plot it. So the first point I'm going to choose is X equals negative two. Then from 1 to 3 halves, we don't have a ton of options here because this is such a tiny, tiny interval. So I'm going to choose five fourths, which is the same thing as 1.25. Now, for my final interval from three halves to infinity, I'm gonna go ahead and choose positive three here. And now I want to go ahead and plug each of these values of X into my original function in order to get an ordered pair, I'm going to go ahead and give you these values. But if you'd like to pause here and plug these values for X into your function to double check with my answers now would be a great time to do that. So starting with negative two, if I plug that value into my function, I'm going to get seven thirds or as a decimal. If that makes it easier for you to eventually graph, it is 2.3 repeating. Now for five fourths, if I plug that into my function, I'm going to get an even negative two. And then finally, if I plug in three to my function F of three will end up giving me three halves, which is just 1.5 as a decimal. Now that I have these points, I can go ahead and plot them on my graph to give me the complete picture of my function. So starting with a negative 22.3 let's go up to our graph and plot that negative 22.3 right about here. Then my next point five fourths negative two. This is in that really tiny interval. So five fourths negative two will go right here. Then for my last 0.31 0.5 I can go ahead and plot that at 31.5 right on my graph there. OK. Now that we have all of these points, we can move on to our final step which is to go ahead and just connect everything and then have them approach the asymptotes. So let's go up to our graph and complete the graph of this rational function. So connecting everything on this side and approaching my asymptotes and then from the other side as well. And then my other curve up to that Asymptote and then finally connecting these points approaching that horizontal Asymptote, OK. Your curves aren't always going to be perfect. Mine almost never are. And that's OK. We just want to have a nice accurate sketch of the graph of our rational function which we have by all of these points that we found. OK. Now that we know how to fully graph rational functions. Let's get into some more practice.

6

example

Example 1

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10m

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Hey, everyone, let's work through this example problem together here we want to graph the rational function F of X is equal to two X squared over X squared minus one. So let's go ahead and get started with step one, which is to factor and find our domain. So here when we factor our numerator isn't going to factor anymore. It's simply just two X squared. But my denominator does factor into X plus one times X minus one. So from here, I can go ahead and find my domain by setting my denom nominator equal to zero. So I take my denominator X plus one equals zero and X minus one equals zero. Now subtracting one from both sides here gives me X is equal to negative one. And then here adding one to both sides leaves you with X is equal to positive one. So here my domain restriction is X such that X cannot be equal to negative one or positive one. Now it doesn't matter the order you write that in as long as your restriction as those two numbers. So let's move on to step number two and find the holes of our function So we want to set any common factors equal to zero, but we actually don't have any common factors here. So we don't have to worry about that. And then, since we also don't have any common factors that tells us that we're already in lowest terms, so let's go ahead and move on to step three and find our X intercepts and their behavior. So we're going to set our numerator equal to zero here, which is two X squared. So setting that equal to zero, if I divide both sides by two, canceling that out, I get X squared is equal to zero. And then if I take the square root of zero, I will simply end up with zero. Now, you might have already noticed that before you solved. And if that's the case, don't worry about solving and just go straight to X is equal to zero. Now, we want to check the multiplicity of this and looking at our original function since this is two X squared, this comes from a factor that is squared, it occurs twice, it has a multiplicity of two. So that is an even number which tells us we're just going to touch the X axis at this point and not fully cross it. So let's go ahead and graph our X intercept here. So it's right at our origin at zero. And we're not sure if it's going to touch on the top of the X axis or at the bottom of the X axis. So we'll wait to satisfy that. Let's move on to finding our Y intercepts by computing F of zero, plugging zero into our function. So I get two times zero squared over zero squared minus one. Now, since this gives me zero on the top of my fraction, I am simply just going to end up with zero, which you might have already noticed since our X intercept is at that origin, that serves as both our X intercept and our Y intercept here. So we can go ahead and move, move on to finding asymptotes here. So let's start with our vertical asymptotes by setting our denominator equal to zero. Now, since we didn't cancel any factors, this is going to end up being the exact same as our domain. So just again, setting that equal to zero, X minus one equals zero and X plus one equals zero, adding one to both sides gives me X is equal to one and then subtracting one from both sides gives me X is equal to negative one. So you don't have to redo that calculation if you already recognize that that's going to be the same as your domain restriction. But here just to show you there's my calculation and my two vertical asymptotes are at X equals one and X equals negative one. So let's go ahead and plot that on our graph. So X equals one using a dash line and then X equals negative one using another dash line. So those are my two vertical asymptotes. Let's go ahead and find our horizontal Asymptote. So our horizontal Asymptote, we're looking at the degree of our numerator and our denominator here. So let's just take another look at our function. So our function is two X squared over X squared minus one. Now, these both have a degree of two. So my degree of my numerator is equal to the degree of my denominator. Now, because of that, that tells me I need to divide my leading coefficients which looking back up at our function, I have two X squared in the numerator. So my leading coefficient is two and then I simply have X squared minus one and in the denominator. So it's just one. Now, this leaves me with a horizontal Asymptote at Y equals two. So let's go ahead and plot that on our graph again using a dash line. OK. Now that we have all of our asymptotes, we're gonna go ahead and break our graph up into intervals and then plot a point in each of them. So remember when we're looking at our intervals, we want to go from something that we know. So we're gonna start at our, at our vertical Asymptote. That's the first thing we know. So our very first interval is going to be from zero until I reach that first known piece of information here. My vertical Asymptote. So it's gonna go from negative infinity to negative one. Now from negative one to the next piece of information I know is an X intercept. So negative one until I reach that X intercept at zero is my second interval. So negative 1 to 0 and then zero to my next known piece of information, another vertical Asymptote zero to positive one. Now these are really small intervals, but we still want to know that information. Then lastly my final interval is going to be from one to infinity because I don't know anything else here. So from one to infinity, OK. From here, we want to go ahead and choose a number in each interval to plot a 0.4. So we want to choose our value of X within each interval. So for my first interval from negative infinity to negative one, I'm gonna go ahead and choose negative three and then from negative 1 to 0, this is a rather small interval. So I'm just going to choose a fraction right in the middle at negative one half. And I'm going to do the same from 0 to 1 and just choose positive one half, then from one to infinity, I'm gonna choose positive two. And this will give me a good picture of what's going on in my graph. OK. From here, we want to go ahead and actually plug these values into our original function and find F of X in order to get ordered pairs. So let's go ahead and start with F of negative three and plug negative three into our original function. So F of negative three, looking back at my original function, I have two times negative three squared over negative three squared minus one. Going ahead and multiplying that out. I know that negative three squared is just gonna be nine. So this is two times nine over nine minus one which gives me 18/8. Now simplifying that fraction, these are both divisible by two. So this ends up giving me 9/4. Now, if I were to plug that into a calculator, if it's easier for you to see this as a decimal, this is going to end up being 2.25 which will help us plot it on our graph. So here at negative three, I get 2.25 as F of X. Now you want to go ahead and plug in negative one half. So I have F of negative one half and plug one half into my function. So two times negative one half squared over negative one half squared minus one. Now, whenever we're squaring a negative, it just goes away. So this ends up being two times 1/4 squaring that and then 1/4 minus one. Now this is equal to 2/4 on the top there and then 1/4 minus one is going to give me negative 3/4. Now fraction over a fraction, not my favorite thing to deal with. But if we just flip these, since they have this exact same denominator or denominators are actually going to end up canceling out. And this is simply going to give me negative two thirds. Now, negative two thirds as a fraction or as a decimal, if that helps you out here is just going to be right around negative 0.667 because it's just 66 repeating. Now for if I go ahead and put this in my table, I have negative 0.66 and then positive one half. If I plug this in, since we're squaring these values of X, it's actually going to end up being the exact same thing. So this is also going to end up being negative 0.667. So putting that in my table here, I know that this is a really time consuming part, but we're almost done here. One final value F of two. OK? So with F of two, plugging that into our function, I get two times two squared over two squared minus one. Now this gives me two times four on the top and then four minus one in the bottom. Now this leaves me with two times four, which is eight and then four minus one gives me 38 divided by three. If I put this into my calculator, it will end up being right about 2.667. If that helps you graph it. So let's go ahead and take all of these ordered pairs and plot them on our graph. So we have quite a lot going on here. But let's start with this first point negative 32.25. So going back up to our graph here, negative three and then 2.25 it's gonna be right about here. Then my next point is negative one half, negative 0.66. So negative one half, negative 0.66 right about here. The same thing on the other side because that's positive one half and then the same point. And then finally, F of two is 2.66. So 22.66 right about here. OK. So now we have finally plotted all of those points. We've made it through all of those calculations. Now we have one final step here which is to connect everything and draw everything approaching our asymptotes. So let's go back up to our graph and complete this final step to form our graph. So looking at this side, I'm going to go ahead and have this approach, my horizontal Asymptote from there and then approach my vertical Asymptote on the other side. Now, looking at this middle section, we remember that our multiplicity of that X intercept told us that we're simply going to touch the X axis at that point. So we're not going to cross it. We're just going to touch it and then approach our asymptotes on either side. Now, finally, our last point, we want to approach our vertical Asymptote here and our horizontal Asymptote here. Now you'll notice that the shape of all of our rational functions is going to be really similar. We just need to make sure that we're always approaching our asymptotes and paying attention to where our additional points are. Remember that if you need to calculate extra points because you want a better idea, that's totally fine. Just pick some more values for X, plug them into your function and you're good to go now that we finish graphing this. Thanks for watching and I'll see you in the next one.

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Problem

Problem

Graph the rational function. ï»¿$f\left(x\right)=\frac{x+3}{x^2+5x+6}$ï»¿

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