Graphing Polynomial Functions - Video Tutorials & Practice Problems
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1
concept
Identifying Intervals of Unknown Behavior
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Hey, everyone, we now know how to find a bunch of different individual elements of the graph of a polynomial function like the end behavior. Whether a graph is rising or falling on either end the X intercept where a graph is either crossing or touching our X axis, our Y intercept where a graph crosses our Y axis and how many turning points our graph might have where it changes direction from increasing to decreasing or vice versa. Now, even with all of this great information, we still are missing something, we're not quite sure what's happening in between all of these points. So how do we fill in that missing information in order to give us the complete graph of a polynomial function? Well, here, I'm going to show you how we're going to do that by simply finding and plotting points in these areas that we are not quite sure what's happening yet. So let's go ahead and get started. So because we do have all of these points that we know we're simply going to break our graph down into intervals of unknown behavior. So these places that I don't know what's happening and simply find and plot a point in each of these intervals. So let's take a look at our graph down here. Now, I'm going to go in this graph from left to right, looking at each of my known points and going in between there to create intervals. So looking at my graph starting on this left side, going into the right, I'm not really sure what's happening here until I get to this known point of negative two. So all the way from negative infinity until I get to this point, negative two, I'm not quite sure what's happening. I don't know how steep this end behavior is going to be even if I already know it's going down. So that represents my first interval of unknown information from negative infinity until negative two. Let's keep going. So from negative two, where is my next known point? Well, my next known point is going to be my Y intercept at X equals zero. So from negative two until I reach that point at zero, which will actually always be a known point of information. I don't know what's happening. I don't know if this is going to be a lower turning point or it's gonna go much steeper. So that represents my next interval. Now from zero, where's my next known point? Well, it's right here at X equals negative three. So from 0 to 3 again, I'm not quite sure what's happening here. So this represents my next interval of unknown behavior. And then finally, from three and beyond again, I'm not quite sure how steep or not so steep, this end behavior is going to be. So from three to infinity represents my last interval. So these four intervals represent all of the places that I don't know the behavior of my graph. I'm not quite sure how steep it's going to be, whether it's going really high or really low. So in each of these intervals, we want to find one X value for which we calculate F in order to give us some ordered pair that we can plot on our graph. So inside of my first interval, negative infinity to negative two, I wanna choose a point that I can calculate F of X four and then plot on my graph. So there is some strategy to picking this point because I don't wanna pick something like negative 100 because I'm not going to be able to plot that on my graph and it's not really going to help me out here. So I want to choose something that is actually going to help me on my graph. So because I have this point negative two, I might want to know what's happening at negative three. So I'm going to choose X equals negative three and then I would simply calculate F of negative three, giving me an ordered pair to plot on my graph, giving me some more information in that interval. So moving on to our next interval from negative 2 to 0 in this interval. It might be a little more obvious some number that we should choose for X. I'm just going to go directly in between at X equals negative one. And then I would want to compute f of negative one in order to give me an ordered pair to plot on my graph. Then into my next interval from 0 to 3, remember, you can choose anything in this interval as long as you're going to be able to graph it. So I'm just going to choose at X equals two. And then I would simply calculate F of two and plot that point on my graph, then my final interval from three to infinity. Remember, we don't want to choose anything crazy. That's not actually going to help us here. So I'm simply going to choose X equals four. And then I would want to calculate F of four giving me my final point of unknown behavior. OK? Now that we have all of these points and we've plotted them on our graph. This is where we would actually want to connect all of these points to give us a clear picture of what's happening here. So with all of that information, I now am left with the complete graph of a polynomial function. Now, if you want to plot some more points here, that's totally fine. You can always choose some more values for X to find your F of X in order to give you an even more complete picture of the graph of your polynomial function. But now that we know everything that we need to, to graph any polynomial function. Let's go ahead and get to graphing.
2
Problem
Problem
Based on the known points plotted on the graph, determine what intervals the graph should be broken into.
Hey, everyone, you now know absolutely everything that you need in order to fully graph a polynomial function. So here we're going to put all of that together in order to graph a polynomial function from scratch. Now, it might feel like a lot. And that's OK. I'm gonna walk you through it step by step. And if you follow these steps, you'll be able to graph any polynomial function correctly every single time. So let's not waste any time here and get straight into graphing. So the polynomial function that I we have here is two X cubed minus six X squared plus six X minus two. Now let's start at step one and find the end behavior of our polynomial function by looking at our leading coefficient A sub nx to the end, which in this case is two X cubed. So first looking at our leading coefficient A sub N, this is a positive two. And because this is positive, that tells me that my graph is going to rise on that right side, which I, I can go ahead and sketch here. Then looking at the degree of my polynomial, it is three, which is an odd number, which tells me that the behavior on the end is going to be the opposite. So sketching that on my graph, if my right side was rising the opposite on the left, it is going to fall. Now, let's go ahead and move on to step number two and find our X intercepts and their behavior. So here we want to go ahead and solve F of X is equal to zero. Now, you might have to factor this on your own sometimes. But we already have this pre factor here. So let's go ahead and solve four X. So I'm going to take this factor X minus one and simply set it equal to zero. Now, if I add one to both sides, it will cancel, leaving me with X is equal to one and that is my single X intercept. X equals one. Now, what is the multiplicity of this X intercept or remember it's just the number of times that our factor occurs. So here that is three, which is an odd number. So that tells me that it is going to cross the X axis at that point. So plotting that on my graph at X equals one, I know that it's going to fully cross the axis at that point. Let's move on to step number three and find our Y intercept. So our Y intercept, we're going to go ahead and compute F of zero by plugging zero into our original function. Right here. This leaves me with two times negative one cubed, which is really just two times negative one, which is simply negative two. So this is my Y intercept that I can go ahead and plot on my graph. So my Y intercept is at negative two. OK. So we've already plotted a bunch of stuff. We have a bunch of known elements. Let's go ahead and move on to step number four, which is going to be to determine our intervals where we're not quite sure what's happening yet and then plot a point in each of them. So let's determine our intervals by going through our graph from left to right. So looking at that left side and going until I reach my first known point, my first known point is actually my Y intercept. So from negative infinity to zero is going to represent my first interval of unknown behavior. Then going to my next known point here from 0 to 1 which it's really close together. But that's my next interval, 0 to 1. And then finally going from one and beyond because I don't have any other known points there from one to infinity is that f final interval here. So now let's find a point in each of these intervals in order to get a better picture of what's going on in our graph. So in this interval, negative infinity to zero, remember I want to choose something that's on my graph. So here, I'm going to go ahead and choose X is equal to negative one to get that point. Then from 0 to 1, this is a rather small interval. So it is OK to choose a fraction here, I'm gonna go ahead and choose one half in that interval then from one to infinity. Remember again, I don't want to choose anything too crazy. That's not actually going to help me. So I'm simply going to choose the point X equals three. Now you can go ahead and pause here and plug each of these into your function in order to get F of X and then come back and check that you've got the same answer as me. So here for X equals negative one, if I plug negative one in, I'm going to end up with negative 16. So the first point that I can plot here is negative one negative 16, then one half, if I plug that into my function, I will end up getting negative 1/4 which I can also go ahead and plot on my graph. Now, I know that that doesn't look like it helps the ton here, but it was an unknown interval. And then lastly, I have this X equals three, which if I plug in, I end up getting positive 16, which is the last point that I'm gonna plot here at three positive 16. OK? Now we have a ton of information about our graph and it looks like we can go ahead and move on to step number six and simply connect all of our points with a smooth and continuous curve because it's a polynomial function. So starting with the point that I have up here, I'm gonna go ahead and connect with a smooth curve, all of my points. Now, you'll notice that I didn't go through those original end behavior lines and that's totally OK. I still am matching the end behavior. I just found out some more information so I can sketch it more accurately. So we're completely done. We have finally fully graphed our polynomial function. Let's perform one final check here using our turning points. So with our points, remember, the maximum number we can have is our degree minus one here. My degree is three. So three minus one gives me a maximum number of turning points of two. Now, looking at my graph, does this have more than two turning points? No, it doesn't even have one turning points. So my last check is good and I am completely done. Now, you have fully graphed your polynomial function. Let's get some more practice.
4
example
Example 1
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9m
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Video transcript
Hey, everyone. Now that we know how to fully graph a polynomial function. Let's work through this example together and graph this polynomial function and then determine both the domain and range. The function I have here is F of X is equals to three X cubed plus 12 X squared plus 12 X. So jumping into step one and determining our end behavior, we want to go ahead and look at our leading term here, which is three X cubed. Now, the very first, first thing I want to look at is my leading coefficient, which in this case is three and it is positive. So that tells me that the right side of my graph is going to rise now because my degree here is three. This is an odd number that tells me that the ends are going to have the opposite behavior. So going ahead and sketching that on my graph here, I know my right side is going to rise and then my left side is going to do the opposite because the ends have the opposite behavior. Remember that this doesn't need to be precise yet. We're just sketching what's eventually going to happen on our graph. So we're done with step one. Let's move on to step two and determine our X intercepts and their behavior. Now, for this one, we need to do a little bit more work and do some calculations. So let's go ahead and set up our equation F of X is equal to zero so that we can solve that. Now, I'm going to take my function here which is three X cubed plus 12 X squared plus 12 X and set it equal to zero. Now, let's come down here and work this problem out. So looking at this function or this equation that I have here, I see that I can go ahead and factor out a greatest common factor out of each of these terms. Now they all have the common factor of three X that I can go ahead and pull out. So taking out that factor I'm left with squared plus four X plus four equals zero. Now, here I don't have to worry about this three X anymore, but I can further factor this and you may recognize that this is a perfect square trinomial. So I can actually factor this into X plus two squared and that's equal to zero. So now this is fully factored and I can go ahead and take each factor and set them equal to zero. So I have three X equals zero and then X plus two equals zero. Now solving for X in each of these, this three isn't going to do anything. I'm simply left with X is equal to zero. And then over here, I can subtract two from both sides canceling that out. Leaving me with X is equal to negative two. So these are my two X intercepts X equals zero and X equals two. Now we need to determine the behavior of the graph at each of these points. So we want to look at the multiplicity. Now, looking at my zero X is equal to zero, I see that it comes from the factor three X which only occurs a once. So this has a multiplicity of one. Now one is an odd number. So that tells me that my graph is fully going to cross the X axis at that point. Now looking at my other factor X is equal to negative two. I see that this comes from the factor X plus two, which is squared. So this happens twice. It has a multiplicity of two. Now, because that multiplicity is even that tells me that my graph is simply going to touch the X axis and bounce right back off at that point. So we have our X intercepts and their behavior. Let's go ahead and put them on our graph. So looking at X equals zero, I know it's going to cross the X axis at that point and then X equals negative too. It is only going to touch and bounce off which might be shaped something like this. Remember that we're just sketching this right now. We'll connect it all with more information later. So we have our X intercepts. Let's move on to look at our Y intercept, which we can find by calculating F of zero, plugging zero into our function. Now, we can go ahead and use our factored version here. We have this three X times X plus two squared or you can always go back to your original function and plug zero in there. I'm going to use the factored version here because I think it's going to be a little bit easier. So plugging zero in I get three times zero times zero plus two squared. And now all of this is getting multiplied by a zero. So that tells me that I'm simply going to be left with zero here. So my Y intercept is just zero, which I don't need to plot on my graph because I already have it at that origin point there from my X intercept. Now that we have those done, let's move on to determine our intervals. So let's take a look at our graph. Now, we want to go in our graph from a left to right, looking for points that we know and identifying those intervals of unknown behavior. So from negative infinity until I reach my first known point, I first know something at X equals a negative two. So my very first interval is from negative infinity until I reach negative two, then going from negative two until I reach my next known point. It is that X equals zero. So my next interval is simply negative 2 to 0. And then finally, from zero, there is no other known point on my graph. So my last interval is from zero all the way to infinity. Now that I have my intervals of unknown behavior, I'm gonna go ahead and identify a value for X in each of those intervals that I can calculate a point for. So my first interval from negative infinity to negative two, I want to choose a point that I will be able to plot my graph which here, I'm just gonna go ahead and choose a negative three. Then from negative 2 to 0, I'm just gonna go ahead and choose a point right in the middle of that interval at negative one. Then finally, from zero to infinity. I'm just going to see how quickly it increases from there. And I'm gonna go ahead and choose X equals one. OK? Now that we're here, we want to take each of these X values and plug them into our function to calculate F of X to get that ordered pair. So let's come down here and plug all these into our function. Now, I'm first going to calculate F of negative three by plugging negative three in for X to my function. Now again, you can use your factored form or you can use your original equation. I'm again going to use the factored form because I think it will make it a little bit easier. So here I have three times negative three times negative three plus two squared equals this three times negative three. I can go ahead and combine that. That's gonna give me negative nine and then I have negative three plus two, which is going to give me negative one. And that is squared. Now negative nine times negative one squared, negative one squared is just one. So this is just negative nine and that's my first ordered pair negative three, negative nine. Now for my next ordered pair, I'm gonna calculate F of negative one by plugging negative one into my function. So three times negative one times negative one plus two squared. And simplifying that three times negative one is going to give me negative three and then negative one plus two gives me positive one. And that is squared one squared is just a one. So this is just negative three times one, which is negative three. And here is my second ordered pair negative one, negative three. Now lastly I need to go ahead and calculate F of one. So plugging one into my function, I get three times one times one plus two squared. And simplifying that three times one is just 31 plus two is also three, but that three is squared. So this becomes three times nine because three squared is 93 times nine is 27. So we have our final ordered pair here 127. So we're done calculating those points. Finally, let's go ahead and put them up on our graph here. So my first point is negative three, negative nine, which I can go ahead and plot negative three and then negative nine is right around here. And then my second point I have is negative one, negative three which plotting that negative one and negative three. So right about in the middle there and then finally 127. So we see that this increases really quickly on this side. One all the way up to 27. Now filling in those unknown behavior, I can now finally connect everything with a smooth and continuous curve like I know all polynomial functions have. So let's go ahead and connect everything using what we already know. So we know that our end behavior is going to go down to negative infinity on this side. And then we're gonna go through this point and cross and all the way up. Now, remember it's OK. If your curve isn't perfect, it can be really hard to draw these in a continuous line. I'm gonna go ahead and remove all that extra stuff on my graph. So I'm just left with my final graph. Remember the other stuff is just to help us later on so that we can really easily sketch this at the end. So here's my polynomial function. We still have one last thing to check here. And that is our turning points which we want to make sure it does not exceed our maximum number. Now, for our turning points, we want to take our degree which is three here, subtract one, which gives us two and check that we don't have more than two turning points here. So I see that I have a turning point right at negative two. And then somewhere around here is another turning point. I don't have any more here. So it looks like I'm good. I have not exceeded my maximum number of turning points. Now, we've finished everything with our graph, but we're going to take a look at our graph and determine that two more things are domain and our range. Now remember the domain, we actually don't have to determine anything because the domain of all polynomial functions is the same. It is negative infinity to infinity or all real numbers. Now, for our range, we do want to take a look at our graph here because it can vary based on our function. But we know that it continues on to negative infinity on down here on the bottom. And then on the top, we see that this also carries on to positive infinity. So our range is actually going to be the same as our domain here and go from negative infinity to positive infinity. Now, that's not always going to happen with every single polynomial function. It just happens to happen with this function. So we finally have graphed our full polynomial function and identified our domain and our range. Let's get some more practice.
5
Problem
Problem
Graph the polynomial function. Determine the domain and range. f(x)=(3x+2)(x−1)2
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