4. Polynomial and Rational Functions

Polynomial Functions and Their Graphs

# Attributes of Polynomial Functions

Pearson

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Hi, my name is Rebecca Muller. During this session, we're going to look at higher degree polynomial functions. Now in the classes that I teach at the university, I always begin this section by asking my students whether or not they know what a polynomial is. 100% of the time, the students will raise their hand, yes they know what that is. And then I immediately say, OK, how about if someone defines that for me? And invariably, what will happen is someone will say, it's going to be an expression that has more than one term. I think that what they're doing is playing off of the prefix poly-. Well, that could be true, but it doesn't have to be true. And so I want to take some time and make sure that we're speaking the same language. Let's look at the following list of expressions. And what I'd like you to do is to pause this recording and see if you can identify which one of these would be polynomial and which ones will not be polynomial. So again, please take your time. Look at each one individually. Pause the video and try this. Now, let's see how you did. What I've done is I've highlighted the ones that are actually polynomials. So that's going to be numbers 1, 4, 7, and 10. And now before we go on to why these are not going to be polynomials for the other ones, and why these are polynomials, let's look at a definition of what a polynomial function is. A polynomial function of x with degree n can be written in the form f of x equals a sub n x to the n-th power plus a sub n minus 1 x to the n minus 1 power plus a sub n minus 2 times x to the n minus second power plus, and then maybe something in between, all the way down to a sub 2 x squared plus a sub 1 x plus a sub 0. Where n is a non-negative integer and the a sub i for i equal to 0, 1, 2, 3, all the way to n are real numbers and a sub n does not equal 0. Now that's a pretty technical definition. Let's look at it again, and I'm going to try to interpret it for you. So this is what we had on the screen earlier. Another way to say that is that we have a sum of terms-- and when I say some, it could be an addition of a negative, so it might be a sum or a difference. And each term is going to have a format where it's a number, and here they're defined by a sub n, a sub n minus 1-- these are just real numbers-- multiplied times a value x raised to a power. And the key thing here is that the power that it's raised to has to be a real number. Now, we say it's degree n, and therefore what I'm saying is that a sub n can't be 0, which means that my first term is going to end up having the x to the n-th power. And another way to think about that is that the greatest power that we see on the variable is going to determine the degree of that polynomial. Now, let's go back to the list that we had earlier, and now use this definition. So on number one, we said that this was going to be a polynomial function. Notice that we have 1/2, which is a number, multiplied times x to the first power, plus a constant 4. So 1/2 x plus 4 is polynomial. In number two, why isn't this one going to be polynomial? Well, it really has to do with the fact that that x is in the denominator. And notice that if you rewrite that first term, it would be x to the negative 1 power and we know that negative 1 is not going to be a non-negative integer. So that's going to be eliminating number two from the list of polynomials. In number three, it's kind of similar to what we had in number two. What we notice that we have a variable in the denominator. And that's going to mess up that whole whole-number power that we're looking at with polynomials. In number four, this is a polynomial function. The number pi is just a real number multiplied times x to the fourth power, which is a non-negative integer. Plus the middle term is 7x. That certainly fits the format. And then minus 1, which is our constant. Number five is not a polynomial. Reason why is that first term, that square root of x. The square root of x is the same as x to the 1/2 power, and 1/2 is not going to fit the format. It's not going to be a whole number or non-negative integer. And so therefore we have to eliminate that. Number six, well, that's a trigonometric function, so that's not going to be polynomial. In number seven, that is polynomial. Now, we have a fraction in number seven, just like we had in number two. But notice that the difference is that in number two, which is not polynomial, we have a variable in the denominator. In number seven, we could rewrite that as 1/4 x squared-- so that fits-- minus x cubed. So number seven is polynomial. Number eight is going to end up having a base which is a number, 3, raised to a power x. You'll end up looking at that, later on, when you talk about exponential functions. Number nine, well, that doesn't look like it at all. That's a logarithmic function, so that's not polynomial. And number 10, which is polynomial, has the number square root 6 multiplied times x squared, and that fits our format. So the here our polynomial functions in this list. Now, what we're going to do is concentrate on the ones that we decided were polynomial. And let's identify for those what the degree is and what the leading coefficient would be. So we have our first one, which was 1/2 x plus 4. And that was polynomial, we said. Now, notice that the degree is going to be the greatest exponent power that we see, which is going to be x to the first. So the degree is going to equal 1. And what we do is-- when it has a degree 1-- we call that linear. And our leading coefficient, which I'm just going to abbreviate with an l and a c, leading coefficient, is going to equal the value that is the coefficient of the term that gave us the degree. Which in this case is the term 1/2 x. So we have 1/2. Let's look at our next one. We had pi x to the fourth plus 7x minus 1. Here, the greatest power that we see occurs on the first term, where we have x to the fourth power. So this is a degree 4 polynomial. And a degree 4, we end up calling quartic, which may or may not be important to you at this point in time. But why not have some words to go along with it? And our leading coefficient, that's going to equal the coefficient of the x to the fourth term, and that's going to equal pi. Our next example was x squared divided by 4 minus x cubed. Well, notice that the greatest exponent occurs on the second term. That's the x cubed. So our degree is going to equal 3, and we're going to call that a cubic polynomial. And the coefficient of that term is going to equal the negative 1. Notice, you have to pick up the sign that accompanies the coefficient. And finally, we had x squared times the square root of 6. We can see that the greatest exponent power that occurs is going to be a 2. So we have degree 2. We call that quadratic. And you may have experience with that already. And our leading coefficient is going to equal the square root of 6, which is the value that is multiplied times x squared. So now let's look at a list of subtopics that we're going to cover during this video. We're going to look at identifying polynomial functions in their degree, and we just finished that one. We're going to look next at power functions. Then we'll look at something called end behavior for polynomial functions, symmetry in graphs, relative extrema, behavior at the x-intercepts for polynomial functions. And finally, we'll finish up by sketching a polynomial function itself. It's time for a quick quiz. Which of the following polynomials is of the greatest degree with the greatest leading coefficient? Is it A. 7x minus x to the fourth plus 5x cubed. B. 8x squared minus x plus 5x to the fourth or C. 9x cubed minus 2x squared plus 5x. Choose A, B, or C. You're correct, the polynomial with the greatest degree and with the greatest leading coefficient is B. 8x squared minus x plus 5x to the fourth. Sorry the correct answer is B. The polynomial of the greatest degree with the greatest leading coefficient is 8x squared minus x plus 5x to the fourth. To compare the polynomials it's best to write them in descending powers of x. So in part A, we note that the greatest power that we see on the variable x occurs in this middle term, so I'm gonna rewrite that beginning with negative x to the fourth. Then descending powers of x I'll pick up the 5x cube term next those will be plus 5x cubed and then finally right the third term which is plus 7x. In part B, we again note that we have the greatest power of x occurring as its number four. So I'm going to begin with nailed that last term which is 5x to the fourth we have no x cubed term but there is an x squared term that I will write next, which is going to be plus 8x squared and then finally minus x is my third term. In part C, the greatest power that I see on my variable occurs in the first term. So this is going to be 9x cubed and then going descending and descending powers will have the minus 2x squared plus the 5x. It turns out the part C was already written in descending powers of x. Now we're looking for the polynomial is of the greatest degree. Well the greatest power that we see occurs into the polynomials that's going to be to the fourth power. So I can note that we have a negative x to the forth in the x first term in part A and a 5x to the fourth in the re-written first term of part B. But now we're told that we want to also have the greatest leading coefficient as part of our answer and so we're going to choose B because the coefficient of 5 is greater than the coefficient of negative 1. Now, the next thing we want to do is look at a definition for what a power function would be. A power function has the form f of x equals a x to the n-th power, where a is a real number. If the value of n is a non-negative integer, then we'll have polynomial functions. So we're going to concentrate on the power functions that happen to be polynomial now. Let's look at n equal to 0 in this list. That means that the power function where you substitute-- the exponent's now zero-- will have x to the 0 power, which is 1. That gives us f of x equals the value a. And you should recognize that as graphing as a horizontal line. If the n value is equal to 1, then substituting that for the exponent gives us f of x equals a times x. Now, when our a value is positive, we end up with the picture that's depicted here, which is going to be a straight line that, notice, is increasing from left to right. If a is negative, we end up with a straight line that is decreasing from left to right. Both of these are going to be linear functions. And again, this should be a topic that you've already covered. For n equal to 2, we're looking at f of x equals a x squared. Again, this is going to be a topic that's familiar to you. When our a value is positive, you're going to have this parabola that's going to open upward. When the a value is negative, we end up with a parabola that opens downward. And in both cases, we're looking at quadratic functions. Now, we're going to look at some of the higher degree polynomials. So now, let's look at a couple of functions that you haven't seen before. These are our higher degree polynomial functions. We're going to start with our n value equal to 3. That's going to give us f of x equals a times x to the third power. If our a value is positive, then we end up with this characteristic shape that ends up in the first and the third quadrants. If our value is negative, then we end up with that same shape, only now we're in the second and fourth quadrants. And you can notice that what happened is we ended up having a reflection there. These are going to be our basic cubic functions. For our n value equal to 4, we end up with f of x equals a times x to the fourth power. Again let's look at what happens with a being positive. Now, this looks like a parabola, but it turns out it's not quite a parabola. It turns out that the part down near the bottom is kind of more compressed. So if you kind of think about-- you know, it normally did like this, and now it's kind of shrunk down a little bit-- that portion that is close to the origin is a lot closer when you're having small values of x. And of course, when you're raising to the fourth power, you're doing it a lot more rapidly than you would if you were squaring. If our a value is negative, we end up with the reflection. And both of these are going to be quartic functions. So if we're looking at polynomial functions, what do you think would happen if we had a degree 5 polynomial, or a degree 6? Would the degree 5 look more like the x squared function or the x cubed function? Well, you're probably guessing is going to look more like the x cubed function because of the fact that it's an odd power. And, in fact, you'd be correct. So when we're looking at our f of x equals a x to the n-th power, and our n value is odd, then we can look at what the ends are going to do. Think about the linear. The ends went in opposite directions. For the cubic, ends went in opposite directions. And that's going to be true, in general. So when n is odd, we're going to have the section over the right going up, the section over the left going down as long as our a value is positive. Or vice versa. If we have our section-- our a value negative, then we're going to have this kind of look to it. And so what you can see is that when we're looking to the far right of our graph, if our a value is positive-- that's going to be our leading coefficient-- then we're going to have that portion of the graph going up. If are a value is negative, if our leading coefficient is negative, then that portion to the right is going to end up going down. Now, what happens if our n value is even? Well, this is going to be similar to the x squared function that you saw, or the quartics, the x to the fourth that we looked at earlier. And we end up having the two ends going the same direction. If we have our a value as being positive, then both ends go up. If our value is negative, then both ends go down. So using the power that we're looking at, that is the degree, and the value of the leading coefficient, we know what the end behavior of the polynomial function will look like. So now, we're going to look at some other things dealing with graphs. And we're going to look, first of all, at symmetry. So let's just think about a function that looks something like this, that resembles a cubic. And I notice that if this were a cubic function, think about the fact you'd go over a certain amount. Let's just say if I went over two units, I would end up, up here, at the point 2 comma 8. So again, I'm just going to go ahead and call this y equals x cubed and plot that point. Well, what happens if I go over two units to the left? It turns out that I end up down here at a point negative 2 comma negative 8. And so what we notice is that for a value in the first quadrant, I can think about it as-- let's just kind of draw that in a little bit-- that what you've done is you've kind of flipped it. Think about that and that, compared to this and this, and play like I've drawn it really well. And what you notice is that the point that was here has been flipped over and then down. And so what we end up with is something that's called symmetry to the origin. So we have a graph that is symmetric to the origin. Now, if I want to generalize and just call this point x, f of x, then what happens over here is I end up with negative x. And of course that would be f of negative x as the y value. But what we also notice is that this value when I plug in negative x is the negative of what I got when I plugged in x. All right, that's a long way to say when I plugged in a 2, I got back an 8. Over here, when I plugged in a negative 2, I got back not 8, but I got back negative 8 as my result. It's symmetric to the origin, which means it's symmetric to a point. Now, this is actually rotational symmetry. But if I want to generalize this, what I see is that when I plug in negative x, what I get back is the opposite of what I got when I plugged in x. And when this is the case, we call this an odd function. And I think that's because of the relationship to these power functions, where they're raised to an odd power. There's a second type of symmetry that we want to discuss now, and it's called even symmetry. And for the even symmetry, we're going to look at some of our degrees that are even for our polynomial functions. And, in particular, let's just consider what happens if we're looking at y equals x squared. And, again, let's just look at particular points in order to see what's going on. Now, the y equals x squared is probably more familiar to you because you've already dealt with quadratic functions at this point. And you know that there's an axis of symmetry that goes through a parabola. But what's important here is that the axis of symmetry in this picture is actually the y-axis. So notice that if I substitute in the value of 2, for instance, I'll come up with the point that is at 2 comma 4. But if I substitute in the value of negative 2, we end up at the point negative 2 comma 4. Now, in this picture, notice that it's going to be true that if I look across the y-axis, that I'm going to end up always having-- at all of these points, you can see that they're the same height. And what that means is that, for instance, if I go to x comma f of x here, over here, that is negative x comma f of negative x, of course. But even more importantly, what we notice is that the f of negative x-- that is, the y value that we come up with when we plug in negative x-- is the same y value as we got when we plugged in x. So we're going to have negative x comma f of x. And I can write that down in the following manner. If I take negative x and substitute it into my function, I'm going to end up with the same y value as when I substitute it in x. When that is true, we end up with a graph that is symmetric to the y-axis. And that is going to be called our even symmetry. Now, let's see what we can do with what we just found out about odd symmetry and even symmetry. If we have a picture of the graph, it's pretty easy to look at it and just say, well, did you look at one side and flip it over the y-axis to get the other side? Or, for instance, did you take something in the first quadrant, flip it over to the second, and then flip it down to get to the third quadrant? But what if we don't have a picture of the graphs that we're looking at to start with? What if we're going to end up trying to get there? And that's what we're going to look at now. So I'm going to start off with a function f of x equals x to the fourth minus x cubed. And what I'm interested in it is for this particular function, which is a polynomial function but is not a power function anymore because we have the extra term in here. We can see that it's a quartic. But does it have any kind of symmetry? Does that even symmetry for the f of x equal x to the fourth power follow when we end up having that minus x cubed as another term here? To use that, we're going to notice that in both cases, we were interested in what happened when we substituted in the value of negative x. So if I can take the value of negative x, substitute it into this function, and see what the result is, then that's going to tell us whether or not we have symmetry of the two types that we were looking at earlier. So here, let's go ahead and accomplish this. We're going to have negative x raised to the fourth power, minus negative x raised to the third power. And now we're going to simplify. Now we know that if we have negative x multiplied times itself 4 times, we're going to have the coefficient negative 1 raised to the fourth power, which is going to give us a positive 1. And that x to the fourth. So we end up with just simply x to the fourth. When we end up with a negative x cubed-- that's negative x times negative x times negative x-- that's going to end up giving us negative x cubed. And so notice that when we subtract a negative, we end up with x to the fourth plus x cubed. Now, What do we do with this result? Well, in order to have odd symmetry, we had to have f of negative x equal to negative f of x. In order to have even symmetry, we had to have f of negative x equals just f of x. So again, this was our odd symmetry. And this was our even symmetry. So we obviously don't have the same thing we started with when we plugged in negative x. We don't have x to the fourth minus x cubed. So we can say, OK, it can't be this even symmetry because it does not end up equaling itself. Is it odd symmetry? Do I end up with exactly the opposite of what I started with? Well, if we think about negative f of x, we would have negative f of x. That would have to be the negative of x to the fourth minus x cubed. We would distribute the negative sign here. We would end up with a negative x to the fourth, right? And then a positive x cubed, or a plus x cubed. Well, notice down here, we do not have the negative x to the fourth. So it turns out, that we do not the odd symmetry either. In this case, we would say that this is neither even nor odd. Now, let's look at a second example. But this example, I'm going to look at g of x equals x times x minus 1 times x plus 1. And again, the way we're going to look at the symmetry is to substitute in the negative x to see what happens. So we substitute negative x wherever we saw x in the expression that we had above. So we're going to have negative x. This is going to be multiplied times negative x minus 1. And that's going to multiplied times negative x plus 1. We need to simplify this somewhat, so our g of negative x equals-- and what I'm going to do is factor out a negative 1 from each of our second factors that we have here. So this is going to be negative x. We're going to factor out a negative 1 from the first set of parentheses, leaving us with x plus 1. And then I'm going to factor out a negative 1 from the second set of parentheses, leaving us with x minus 1. Continuing down, we get g of negative x equals. Now, the negative 1 times the negative 1 is going to be simply positive 1. So we'll have our negative x times our x plus 1 times our x minus 1. Let's compare that to what we started with. Well, it's certainly not exactly the same in that we have a negative sign in front of it. But you can see that it is exactly the opposite. That is, that we-- if we put a negative sign in front of our g of x, we would just simply put a negative sign in front of that x. So because we have that g of negative x equals the negative of g of x, we can say that we are going to end up with odd symmetry, which means that this function would be symmetric to the origin. Now, we're going to look at a couple more graphing attributes. I'm going to just sketch in a couple of pictures here. Here's one. Let's do another one, have it do something like this. And what I notice on here is that I can look at this. It looks like the top of a hill-- that point. This point looks kind of like the bottom of a valley. Over here I can identify these two-- they're kind of tops the hill, and there's the bottom of the valley. And, you know, in mathematical terms, we normally don't say top of the hill, bottom of the valley. So we have names for these. In the first one, notice that this point is going to give us a y value that is a maximum relative to the points that are close to it. And, appropriately, we call this a relative maximum. And, you may be thinking, what do we call this term? Well, it's going to end up being called a relative minimum. Because in relation to the points near that, relative to that point, it is going to give us a y value that is a minimum. And so on the second graph, we notice that we have these two values. These would both be relative max, maxima. And this point would have-- be a relative min. Now, if it's a polynomial function, we note that we have a relative max and a relative min, and that's called two relative extrema. And on the second picture, I have 1, 2, 3 relative extrema. Well, the relative-- number of relative extrema has something to do with the degree of-- that a polynomial could possibly have. If we had n relative extrema, then the degree of the polynomial has to be at least n plus 1. It's going to be greater than or equal to n plus 1. Now, what that means is, for instance, here I notice that I have two relative extrema. The degree of the polynomial that would fit that graph would be a 3 or maybe a 5 or maybe a 7. And of, course, it's odd because it goes in-- the ends go in different directions, so I know that I'm going to end up with an odd degree. The smallest degree that it can have is going to be 3. In our second picture, we have three relative extrema. We notice that the ends are both heading down. And we can say that if this is polynomial, then the degree for 3 relative extrema would have to be at least 4, and maybe it's 6 and maybe it's 8. But we can tell something about the degree of the polynomial according to how many relative extrema we see pictured in the graph. Now that we have some tools at our disposal, dealing with polynomial functions, we're going to now try to get a graph of this function that is given here. We have f of x equals x to the fourth minus 25x squared. First of all, we can identify that is a degree 4. We notice that the leading coefficient is 1, which means that the a value is positive. So we know that the end behavior, both ends are going to end up going up. The next thing we might consider is maybe it has some symmetry to the graph. Let's discuss that now by using our definition for symmetry, which means we're going to substitute in negative x and find out whether or not we get back the same or the opposite or neither one. Let's go ahead that evaluate this. This is negative x raised to the fourth power minus 25 times negative x squared. When we raised to the fourth power, we end up with x to the fourth for the first term. We subtract 25. And negative x squared is going to give us x squared. So we can see that we end up with the same result when we plug-in negative x as we had when we plugged in positive x. This means that we're going to have even symmetry. And, of course, that means that we're going to have our graph symmetric to the y-axis. Now, the next thing that we're going look at will be to find the intercepts of that graph. And that means that-- remember that the y-intercept is found by evaluating f of 0. So when we substitute in 0 to our function, we end up with 0 to the fourth power, which is 0, minus 25 times 0 squared, which is also 0. So again, 0 minus 0, which is 0. To find our x-intercepts, we're going to need to take our function, x to the fourth minus 25x squared, and set it equal to 0. Now, this is going to require us to factor. And we notice that we can factor out the x squared from both terms. We can then factor the difference of two squares, so we're going to have x squared times x minus 5 times x plus 5 equals zero. And so we end up with setting each factor equal to zero to give us our x-intercepts. So x squared equals 0 gives us back the value of x equal to zero which corresponds, of course, to origin. And that makes sense here. And for our x minus 5 equal to 0, we end up with x equals 5. And for x plus 5 equals 0, we get x equals negative 5. So we end up having x-intercepts at 0, negative 5, and positive 5. So now, I've just transferred the information that we just found out onto another screen. And what we'd like to do is look at how we can put this together into a graph. So we have the intercepts. Here's negative 5. There's zero. Here is positive 5. And, you know, there's a couple of things we can do here. One is we don't want to spend time really plotting a whole lot of points. If we do that, then we're kind of missing the point of getting this whole idea behind what's going on with the graph and just being able to sketch a quick graph in order to see what's important. There are a couple of things we can't do, and that is because we haven't taken calculus yet. We're not going to be able to figure out our relative maxima and relative minima values. But recall that we had this factored a few minutes ago. We had x squared times x minus 5 times x plus 5. And that's our function f of x. Well, because it's a fourth degree polynomial, we know that the ends are going to go in the same direction and we'd already determined that the leading coefficient was positive, so the two ends were going to go up. We're now going to look at each intercept and think about the factor from which that intercept comes. In other words, when I look at the negative 5, I notice that it came from the factor x plus 5. Now, x plus 5 is a linear factor. It turns out that as I get close to the x-axis, the graph is going to resemble the look of that factor in that it's going to be linear or another degree. It's not going to end up being increasing, OK, because we're not going to be substituting in values. We're not going to find out what the value in front of it might be, like the a value there. But we are going to notice that if this is linear, then when I get to the negative 5, I want to have that same shape. I'm going to start on the far left, here. I know that, again, the ends go in the same direction. And I'm going to say, well, that end has to be going up. I'm going to come down and I may get close to my x-intercept. And because it came from a linear factor, I'm going to cross through the axis so that this section of it looks like a straight line, to some extent. When I come back, I know I need to connect dots. I need to come back up and end up at the origin. Well, where did that intercept come from? It came from the factor that we had from x squared. Now, x squared is a quadratic factor. If we think about the graph of a quadratic, we know that it's a parabola shape. So what occurs is, I'm going to dip down and come back up. And when I'm drawing polynomials, I always want to draw them as being what's called smooth and continuous. Now, smooth means I'm going to not have any sharp points that are going on. I'm going to do like a really sharp V shape. It's going to be this nice little smooth rounding out. And I'm going to come back up here, and as I get close to the origin, I don't want to pass through the axis. I want it to kind of look, in this general area, like that parabolic shape that a quadratic would have. So I'm going to come here. I going to end up hitting that point and bouncing off of the axis. And again, if you just look at that little bitty section, do you see that parabolic shape that I was talking about? Where it's going to be like at the top of a vertex? I now need to go ahead and get to this other x-intercept, which is at 5. I need to do it in a manner that is symmetric to the y-axis. And notice that if I do that, I'll just come down the same amount. And then come back up. I'm going to, because of the symmetry, I know I'm going to end up doing this. But again, let's just relate it to the factor from which that intercept came. We ended up getting that x-intercept at 5 from the factor x minus 5. That's a linear factor. So when I look at what's going on at the-- with the graph-- at the x-axis, it's going to have a shape that appears to be linear as you get closer and closer to that intercept. So one thing you'll notice that I don't have on this graph is I don't have any y values. I mean I have, of course, the origin. But I didn't put any units on my y-axis. At this point, what we're more interested in is coming up with the graph where are you going to be able to kind of see what's going on with the x-intercepts. And you have the graph being positive in the correct areas. That is, the y value's positive here and here. And you have it being negative in the correct regions. And so, again, putting all of this together, we're looking at finding intercepts, determining what's going to be going on at those x-intercepts, and then putting together a picture depending on whether we have symmetry or not in that manner. We just used the degree of each factor to help us determine what was going on with our graphs. Let's formalize that process now. If a polynomial function has a factor x minus c that appears exactly m times, then c is a zero of multiplicity m. Now what that means is that our value c is going to be an x-intercept on our graph. If we have x minus c raised to the power m-- so in other words, you could see it as x minus c squared or cubed or whatever. Then we would say it was a zero of multiplicity, whatever that power is. If the value of m is odd, the graph of the function will pass through the x-axis at c. So in our previous example, we saw that when we had the linear, it looked like a linear. It passed through the axis. If it turns out-- if it's a cubic, it also will pass through the axis. If the m is even, like when we had the quadratic factor x squared, we had it bouncing off of the x-axis at our x-intercept. So that's going to be true. When the m value is even, the graph of the function will bounce off of the x-axis at c. Now, depending on your instructor, you may have them saying that the graph will touch the axis. When I'm talking about it bouncing off, they'll just say it just touches the axis. And the other one, they'll say that the graph will cross the axis. So it's just different terminology, but it means the same thing. So let's look, now, at a graph. And we're going to kind of take it from the opposite direction. This time, I'm going to give you a depiction of a graph. And we're going to, from this graph, try to come up with a possible equation for our f of x. What could the function possibly be? So we're going to note that we're going to have our x-intercepts. And we have an x-intercept here at negative 2. And we have another x-intercept at positive 5. So from those, we can come up with some factors that are going to correspond to those intercepts. In particular, we can say, well, if I have f of x equals-- for the intercept that occurred at negative 2-- what I'm going to say is I have the factor x plus 2. Remember that we just said for an intercept or a zero of c, we had the factor x minus c. So notice that we have to look at the opposite sign in our factor. That kind of makes sense because if I take that factor, x plus 2, and set it equal to zero, I need to come up with the negative 2 that I had as the intercept. We noticed that for that factor, this graph is bouncing off of the axis. So I'm going to need to give that power an even value. I'm going to take the smallest even value I can come up with, now. So that's going to be squared. Then I look at my other x-intercept. Now that's occurring over here at the value of 5. That means I'm going to have a factor that corresponds to that. That's going to be x minus 5. That's a c value. This is x minus c. And because the graph is crossing through the axis there, or passing through the axis, we're going to give it an odd power. Which means if I take the smallest odd power that's possible, I'm just going to leave it at the first power. Now, this is a possible equation that would fit that format. Notice that I don't have any values on my y-axis. So, again, we're not really worried about how high or how low it's going, right now. We're just worried about-- well actually, I am kind of worried about how low it goes here, but I'm not worried about this relative maxima that I see over here. I'm just going to try to figure out a possible equation that is going to mimic what we see going on with the x-intercepts. The other thing to note is, you know, you've got to be a little careful. Sometimes, we could write this in a different format and it could still be equivalent. So for instance, I could have written f of x equals, and I could have that x plus 2 squared. And instead of writing x minus 5, I could have written it as 5 minus x, as long as I factor out a negative. So in some textbooks, you're going to look at trying to match up an equation with a graph. And they may do things like this to you to see if you're paying attention. So, of course, 5 minus x multiplied times negative 1 is the same as x minus 5. Now if we look back at our function, f of x equals x plus 2 squared times x minus 5, we notice that we're only concentrating on the values for the x-intercept. Notice that on the y-axis, there are no units given. And this is not necessarily drawn to scale. Since I had no units, I wanted to just take the value of a equal to 1. But what if I now know some more information? What if, for instance, knowing that the y-intercept is equal to the value of 40, it allows me to now figure out what the leading coefficient might be if I multiplied this out. So in other words, I'm going to think about our f of x equals some value a times x plus 2 squared times x minus 5. And again, if I'm given that I have a y-intercept of 40, then I know that the point 0 comma 40 would end up on this graph. Let's now use that in order to find the value of a. So we're going to be able to do that because notice that the zero value is going to replace x in our equation, and 40 will replace f of x. So we'll have 40 equals our value a, which is what we're going to be solving for, times 0 plus 2 quantity squared times 0 minus 5. Simplifying, we have 40 equals a times 2 squared, which is going to be 4, times 0 minus 5, which is negative 5. So equals a, multiplied times negative 20. And then dividing both sides of this by negative 20, we end up with negative 2 equals the value of a. Well, what that means is I can rewrite my function using the value of a as negative 2. Now, one thing you may want to end up doing is multiplying this out and seeing it in a standard form where you have the first term plus the second terms, so on and so forth. We're going to practice that, now. So I'm going to rewrite my f of x as, instead of the a value, I'm going to put in the negative 2. This is going to be multiplied times x plus 2, quantity squared, times x minus 5. And, again, we're just going to practice some of our techniques and our skills, now. So I'm going to multiply this out. I'm going to start with my squared term here, the x plus 2 quantity squared. Now remember, that that means x plus 2 times x plus 2. You could write it like that, and then use FOIL method on it. Or you could remember that special product. And so I'm going to do that now. I'm going to think of this as f of x equals-- I'm just going to keep the negative 2 for the time being. We're going to have the first term squared, which is going to be x squared, plus twice that product. So the product would be 2x. Multiply it by 2. We get 4x. Plus the last term squared, which is going to be 4. And then that value is going to be multiplied-- or that whole expression, that whole factor will be multiplied times x minus 5. Now how do we do that? Well, we need to use distributive property. And the way we can do it is we can take the first term in this factor and multiply it times each term in the second factor, and then come back and get the second term, and so on and so forth. So let's demonstrate that, f of x will equal, we'll keep that negative 2 till the end. I'm going to take the x squared, and multiply it times x. So x squared times x is x cubed. Then I'm going to take x squared and multiply it times what I'm going to think of as negative 5, since it's a subtraction. So x squared times negative 5 is negative 5x squared. Then I'm going to move to the second term. And that's going to be the 4x. I'm going to multiply it times each term in that second factor here. So I have 4x times x, which is going to be plus 4x squared. And then we're going to have 4x times negative 5. That's negative 20x. And then, finally, I pick up the third term in this factor-- that's a 4-- and multiply it times each term in the last factor. So that's going to be plus 4x minus 20. Now, what we have to do is just multiply everything times negative 2. I'm going to do ahead and simplify as I go, so one more step. Let's go ahead and the negative 2 one more time, and I'll simplify first. We'll have x cubed-- because that's the only time that has x cubed in it. We have minus 5x squared plus 4x squared. That will be a minus x squared. We have a minus 20x and a plus 4x, so that's going to be minus 16x. And then minus 20. And finally, multiplying everything by negative 2, we end up with f of x equals negative 2x cubed plus 2x squared plus 32x. And then we end up with plus 40. Notice that if I substitute in the value of 0, we're going to end up getting 0 plus 0 plus 0 equals 40. And that shows us, again, that we end up with that y-intercept equal to the value of 40 in this expression. So there are a few generalizations we can make when we're dealing with polynomial functions. Let's look at those now. We can say for a polynomial function of degree n, the domain will always be the set of all real numbers. That is, from negative infinity all the way to infinity. If you think about it, we're always either just raising to a power, and we can do that for all real numbers. Or we're multiplying by a number. We can do that for all real numbers. And then we're adding those results together. And again, all real numbers. So the polynomial functions will always have this domain. The number of x-intercepts will always end up being less than or equal to the degree of the polynomial. They'll be less than or equal to the value of n. So now, if we-- again, kind of use your experience. You know that for a linear function we could have one and only one x-intercept for a quadratic. The most number of intercepts we would have is two. Well, it turns out for our cubic, the most number of intercepts for x would be three, and so on and so forth. The maximum number of relative extrema is going to end up being n minus 1. So we're just stating this in another manner. That is, if the degree is 3, the maximum number of relative extrema would be 2. If the degree was 5, the maximum number of relative extrema would be 4, and so on and so forth. And the graph of a polynomial function is smooth and continuous. And I talked about this a little bit earlier. Let me just go ahead and reinforce that. Smooth means there's going to be no sharp points in the graph, so always going to have a nice smooth curve. Continuous means we can draw the entire thing from left to right without picking up a pencil is one way to think about it. So you're always going to be able to connect the entire thing. So now, here's a problem that you should try on your own. And then what we'd like you to do is to try this and then come back and we'll do it together. Here's the problem. Find all real zeros of the function, which means find all the x-intercepts, and sketch the graph. And the function is f of x equals x to the fourth minus 2x cubed minus 8x squared. Now what we want to do here is use your information dealing with the degree of the function and multiplicity of roots and maybe look at symmetry. So all the things that we've discussed so far in this video, try that with this particular function. Stop the video, try it on your own, and then when you're ready to try it-- to look at it together, then come on back and we'll do it as one. So let's see how you were able to do with this problem. So we want to find all real zeros of the function. So the first thing you may decide to do is to go ahead and take, x to the fourth minus 2x minus 8x squared, and set it equal to zero. And now, we're going to factor this completely. So we're going to notice we have a common factor of x squared that can be factored out. That's going to leave us with x squared minus 2x minus 8 equals zero. And then we'll factor this using the product of two linear factors. We can take x squared and write it as x times x. We can look at factors of 8 and, in this case, we pay attention to the fact that the middle term has to have a two in it. So our options are 1 an 8 or 2 and 4. We're definitely going to go with the two and the four. And now, thinking about the signs, we know these have to be opposite signs, and we end up with a negative 2x. So the inner term is 2x. The outer term is 4x. To get a negative 2x as a result, when we add those together, we need a negative 4 here and positive 2 there. So again, just double checking. We'd have x squared plus 2x minus 4x, which gives us a negative 2x. And then, 2 times negative 4 is our negative 8. So this is the factorization. Now, that we have all three factors, we set each factor equal to 0 and solve for x. And I'm going to do this in one step-- x squared equals 0 when x is zero, x plus 2 is equal to 0, when x is equal to negative 2. And x minus 4 is equal to 0 when x is equal to 4. So what we found are the x-intercepts for this graph, which are also called the zeros of the function. We can also look at our y-intercept, but, of course, when we take f of zero and substitute in. We're going to end up with 0 minus 0 minus 0 when we look at the original version. And so we get zero, and that corresponds, of course, to the fact that we have an x value of 0. Let's go out and take a quick sketch of this graph, at this point, to see what we're working with. So again, just a really quick sketch. I'm just going to put my x-intercepts on here. Here's an x-intercept at zero. And then we have an x-intercept at negative 2, and we have another x-intercept at positive 4. This is a polynomial function. It's a degree 4. We know that that means that the ends are going to go in the same direction. The leading coefficient is a positive value, which means that the ends are going to end up both being upward. We're going to have a start over here, and we know it's going to end up somewhere over here. And we're going to try to connect the dots. Now, as we're coming down, we're going to use the multiplicity of each root in order to help us. So we're going to come down toward the value of negative 2 on the x-axis. That's going to come from the factor x plus 2. That's a linear factor that is to the first degree, so the multiplicity is one. Meaning, we come down and we cross through the axis here. Now, one thing that I didn't do yet is to check symmetry. But notice that if it were going to be symmetric to the y-axis, if I had a negative 2 here. I should've had a positive two there. And it's certainly not going to be an odd symmetry because of the fact that the ends are both going in the same direction. So I'm not going-- I know for sure this is not going to have any kind of a symmetry to it. I could certainly figure it out by substituting in the value of negative x into our function to determine that. But just by the diagram that we have so far, we know that can't be the case. The other thing is, remember, we don't know how far down it goes. I'm just going to draw it so that I'm going to come back up toward the next x-intercept. And when I get to the next x-intercept, which is at the origin, I note that that intercept came from the factor that had a-- to the second power, so its multiplicity is two. Meaning that when I get there, I'm going to bounce off of the x-axis and come back down. And I had no symmetry, so I'm not worried about this section of it-- come being, mimicking the next. I know that I'm going to come down, and I'm going to come toward the value of the 4. When I get to the 4, I look at the fact that it comes from the factor x minus 4, the multiplicity is one, which means it's going across through the axis. And that is going to correspond to the fact that it has to come up here and reach that point where it goes up to the right. And so what we've been able to do is to get a quick sketch of this function using the zeros, their multiplicities, and what we know about end behavior. So I hope you did pretty well on this problem. It's time, now, for you to try some higher degree polynomial functions on your own. So go ahead and take time now to get some experience with it. Good luck. It's time for another quick quiz. What are the roots of the function and their behaviors at the x-axis for the function y equals negative 3x squared times the quantity x plus 4 cubed, times quantity x minus 5 to the fourth power. Choose the statement below that is true. A. The graph crosses through the x-axis at negative 3 and negative 4. B. The graph bounces off of the x-axis at x equals 5. C. The graph has three x-intercepts and crosses through the x-axis at each of them. Choose A, B or C. You're correct, the statement that is true is that the graph bounces off of the x-axis at x equals 5. Sorry the correct statement is B the graph bounces off of the x-axis at x equals 5. To find the roots of the function, we look at where the factors of the function are equal to 0, so we end up with three results. In the first factor, we can think of it as negative 3 x squared, that's equal to 0 when x is equal to 0. The second factor, x plus 4 quantity cubed is equal to 0 when x is equal to negative 4. The next factor, x minus 5 raised to the fourth power is going to be equal to 0 when x is equal to the value 5. So there are three x-intercepts that part is true. However, in part A, it says that there is also an x-intercept at negative 3 which is incorrect, so we would have to eliminate A from our choices. In C, it says it has 3x-intercepts and it crosses through the x-axis at each of them. Now this has to do with the multiplicity of the roots. The multiplicity of the root is found by looking at the exponent power that is on the factor that the root came from. So when x is equal to 0, that had a multiplicity of 2. The x-intercept of negative 4 has a multiplicity of 3. The x-intercept at 5 came from a factor that has a multiplicity of 4. When a root has an even multiplicity, it does not cross through the x-axis at that point. Instead, it bounces off of the x-axis. So C is incorrect because it says that the graph will cross through the x-axis at each of those x-intercepts, and that is incorrect. B is the correct answer because the graph does bounce off of the x-axis at the x-intercept of 5.

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