In the last lesson, we looked at the behavior of functions, including their maxima and their minima, and where they're increasing or decreasing. In this lesson, we're going to focus on polynomial functions in particular. We're going to look at their graphs as a whole, and then, also zoom in on certain regions.
Now we've learned a lot about some basic functions, like this one, y equals x squared, who's graph is this lovely purple parabola. You also learned about some other polynomial functions. For example, we have y equals x cubed also, which produced this really different and really interesting graph. So clearly the large scale behavior of different functions Is really different on what their equation looks like. And it also changes from region to region of the domain. I wonder what would happen if we increased the degree of polynomial function even more. Like, let's say we have y to the x to the fourth. Well that's totally different again, that's really interesting. Maybe I'll just get rid of these two that we already know a lot about and just focus on this one. Let's do something else, y equals x to the fourth plus. x. Well that's cool. That's totally different from anything we've ever seen before, together. Of course we can add and change terms. If we made this x squared, that changes the behavior again. What if I made this x cubed instead? Everything makes these little, minute adjustments in the middle of the graph here. That's pretty cool. Now I wonder, I just changed one exponent. What if I change this first one? The exponent of the leading term. Wow, now this graph is completely different. Once again changing the degree of the polynomial, seems to really change the overall behavior. What if I make this sign negative? Well that's really interesting, oh my gosh. Clearly changing different terms in polynomials, especially the leading term, the term at the highest power has major affects on the graph. I think this merits some more exploration. In this lesson we're going to investigate the overall appearance of different polynomial functions in terms of how they look when we're zoomed way, way out, and when we're zoomed in on the middle of the graph. In particular we're going to see where these functions intersect the x axis. And how this relates to factors in routes of polynominals. I'm really exicted about the content of this lesson.
We've seen that if we have a constant function like y equals 1 it looks like this just a straight horizontal line. Now if we increase the degree of this by one, straighten out this is a zeroth degree polynomial, if we up it by one we get something like y equals x which takes that horizontal line and just rotates it. Now it has a nonzero slope. Now if we increase the power by one more we get a parabola. A quadratic function. And we know if we increase it by one more once again, we get this really cool cubic function. Now, you just saw your first higher order polynomial, in the last quiz. And I think that, we should focus on one of these in particular for right now. Let's look at the graph of y equals x to the 5th minus x. Let's just look at the overall appearance of this function. And then build up some information about polynomial functions in general based on that. We saw a big picture of this graph earlier showing pretty big range on the y axis in a pretty big part of the domain on the x. But now I've zoomed into the middle where some of the most interesting stuff is happening. Remember that the ends of the graph, this way and this way, are just sort of continuing on for a long time in the same direction. So it seems like all the changes are happening in here. Because this is the only area where things are changing, we can split the domain into sections based on what's happening in here. Now I see one change happening right around here. And another change happening right around here. So with these lines I have effectively divided the domain into three chunks. Three different intervals. Now let's label these intervals interval one, interval two, and interval three. It also seems like there are two special points. These two right here. I'll color them in this bright blue color and label them points A and B. These are of course the points where these lines separating the parts the domain from one another intersect the graph of our function. So let's talk about what the graph looks like within each of these intervals and why these two points are special. Now first of all, can you tell me in which of these intervals the graph is increasing? And which of them the graph is decreasing. And second which of these two points is a relative maximum and which one is a relative minimum?
The graph is increasing in intervals one and three. And the graph is decreasing only in between them in interval two. The graph reaches a relative maximum at point A and reaches a relative minimum at point B.
Now let's think about what another graph that we've already seen before looked like in general. Here's y equals x cubed. A very basic cubic function. Now on the left hand side over here, is the graph of this function coming down from up high or is it coming up from down low? And on the right is it rising up high or is it falling down low?
On the left hand side the function is increasing so that means it's coming up from being way down low. And on the right hand side it's increasing as well so it's rising up high.
In the previous question we had this polynomial, y equals x cubed, which has an odd degree. It starts really low on the left hand side and then it increases and increases until it's ending up high on the right side. Now what if instead of a polynomial was an odd degree like this one? We have one with an even degree, like this one, y equals x to the 4th. Tell me about this graph on the left hand side over here. Is it coming from up high or is it coming from down low?
It's coming from up high. It starts way high on the left side of the graph, goes down as we follow it to the right, and then goes back up. We'll talk about why this happens in the next video.
Lets analysis for a second why this graph has this particular behavior with the ends pointing in the same direction on either side up really, really high. To do that I like you to form this x, y table for this function y equals x to the familiar each of these pairs of numbers one is just the negative version of the previous one I think you'll see why in just a second. I'd actually recommend doing most of these calculations by hand so you can see why y is taking on the value that it is.
So here are all the y values for these corresponding x values. And we can see that each of these pairs of numbers, where one is just the negative version of the previous one, all have matching y values, despite their opposite signs in x. Let's think about why this is. Well, one way we can rewrite y equals x to the 4th is y equals x squared squared. We can do the same thing for something like y equals x to the 6th. That can be written as y equals x to the 3rd squared. And we know that whenever we're squaring something it's going to involve multiplying that thing by itself. So if what's inside here, like x to the 3rd. In the case of y plus x to the 6th. Is negative, squaring it is going to make it positive. That means that any sine g/ differences in x. Are going to be cancelled out in y. So that helps tell us why, even when we move way over to the left. Way to the negative side of the graph. On the x axis. The general behavior of the graph of a polynomial with an even degree. Is going to match the behavior on the right side, because of that squaring action here. Now, if we had a more complex polynomial here instead of just y equals x to the 4th. The y values to these numbers might not line up exactly. They might not be exactly equal, because of the terms following this. But by the time we move way far out in x. Either positively or negatively. This overall pattern would re-emerge and we'll see the same general behavior on either side of the graph.
Now let's go back to something a little bit more familiar, quadratic functions. By looking at the coefficient over here, of our x squared term, either 1 or negative 1, can you tell me which of these graphs matches each of these equations?
This graph over here on the left is the graph of y equals negative x squared. As we learned a few lessons ago, when a problem has a negative leading coefficient, it points downward. Both of its ends are reaching toward negative infinity. This other graph over here is representing y equals x squared, with a positive leading coefficient. Note that these graphs are just flipped versions of one-another. You get from this one to this one by just taking this graph and flipping it across the x axis.
Now we've already seen the graph of y equals x cubed before it just looked like this. But notice the graph over here on the right side of the screen. Its exaclty the same as this graph just flipped upside down across the x axis. What function do you think this graph represents?
This graph represents y equal to negative x to the third. Just like in the previous quiz with the parabola, changing the sign in the leading coefficient simply flips the graph upside down. Note that for y equals x cubed, like we said before, the graph is starting way down low on the left and reaching up way high to the right. Since y equals negative x cubed is going to have all the opposite function values that y equals x cubed has, it starts really high on the left and reaches way down low on the right.
Though when we zoom in and look at graphs, we're really only seeing part of the entire representation of the function. We're just zooming in on this little tiny portion of the domain. If we zoom out even just a bit though, we can see that all these changes that are happening in the middle of the graph. Are really just tiny details the overall behavior of the function is determined by what's happening at either end of the graph on the right side and on the left side. We've already learned a lot about the general shape of polynomial functions so let's see if you can apply what you've learned. To work out how each of these graphs will look. For each function I've listed here please tell me what the behavior on the left side of the graph is going to look like. In other words, what direction the graph is coming from? And then also tell me what's happening on the right side of the graph. What direction is the graph pointing in as we move over to the right hand side?
You've already seen what the graph of x cubed looks like. So if we made the coefficient at the front of the term 3 instead of 1 but keep it positive the overall behavior will still be the same. It's still going to start low on the left and reach up way high on the right. Since 5 x to the 4th times x squared has an even degree. We know that the left and right side are going to point in the same direction. Since the leading coefficient is positive it's going to start way up high on the left hand side. And then also rise up high on the right hand side. For this function with a negative 2 x to the 5th, since the degree is odd the ends are going to point in opposite directions. And since this leading coefficient is negative, it's going to start up high on the left, and reach way down low on the right.
We've just reviewed a bit of new material on the shape of a function and how that's affected by its degree, and also how changing the sign of the coefficient flips the graph over the x axis. Now, I've drawn four different graphs here, and labeled them a, b, c, and d. They represent 4 types of functions which I've listed right here. One of them shows a polynomial with an even degree and a positive leading coefficient. One shows a polynomial with an even degree and a negative leading coefficient. Another shows a polynomial with an odd degree and positive leading coefficient. And the last one shows a polynomial with odd degree and a negative leading coefficient. I'd like you to tell me which one of these graphs matches each of these categories. Please put the letter of each graph next to the description over here that it matches.
The first two graphs here, which are the ones with even degree, are these two graphs down here. The one with the positive leading coefficient is graph D, the one whose ends are both pointing upward. It's coming from up high on the left and rising up again on the right. B is the graph that has an even degree, but a negative leading coefficient. It's coming up from negative infinity on the left side and dropping back down there again on the right. The top two graphs both have odd degree. Graph A has a negative leading coefficient, since it's starting way up high and then dropping down low on the right. And graph C has a positive leading coefficient, since it's starting low and then rising high.
Now, we've looked at the overall shape of a graph, talking about what happens at either end of it. So now we're going to focus back in again on the middle, mainly on the x-intercepts. So here's a graph that you are definitely familiar with, the graph of a parabola. You know three different ways to write the equation of a parabola. But the one we're going to focus on today is its factored form. The function we're concerned with is f of x equals the quantity, x minus 5 times the quantity x plus 3. Just to make sure we're on the same page, what are the x-coordinates of the x- intercepts of the graph of this function? Please fill them in here.
There's one x intercept whose x coordinate is negative 3, and there's one whose coordinate is 5. We can see this from the equation as well based on these factors. We know that at the x intercepts the value of this function is 0. So these are the two x values that make either of these factors 0, which then makes the entire expression equal to what would be on the left hand side, 0.
Let's press on with this function we were working with before. Now I've talked before about 0's or roots of functions. We know that these are values of x that when substituted into our function make the function itself f of x equal 0. So, for this function in particular If we know that f of a equals 0, then what numbers can a equal? What values can it take on to make this true? Please write your answers in this box and separate them with commas.
We know that f of a is what we get if we plug a in the slot of x into our function. So that means that the quantity a minus 5 times the quantity a plus 3 equals 0. If we want to find out what a equals, we just need to solve for a in this equation. There are two ways that this expression right here can equal 0. If either this factor equals 0 or if this factor equals 0. And we just solve these two equations for A. As before we A equals 5 and A equals negative 3. So these are the two roots of this function. They're the two values of X that make F of X equal 0.
Let's summarize what we've been talking about for the past few quizzes. We started with a function, f of x equals x minus 5 times x plus 3. We figured out, it's two factors are these two quantities inside parentheses. That told us that the x coordinates of the x intercepts are 5 and negative 3, and these two numbers are the same as the zero's, or the roots of the function. Let's see if we can fill out all this information for a new function. This time we have f of x equals x minus 2 times x plus 7 times x minus 1. Please tell me what the factors of this function are, what the x coordinates of it's x intercepts are and what the 0s are. Notice I've already written an x equals here for you. So, in the zero box, I'd like you to just write, the values that x needs to equal in order for the function to equal zero. In each of these three boxes, please separate your answers by commas.
The factors of this polynomial function are each of these expressions that are multiplied together. X minus 2, x plus 7 and x minus 1. The x coordinates and the x intercepts are the numbers we get if we set each of these factors to 0. does all this information mean for us and mean for the graph of the function? Well we'll figure that out pretty soon. You can start to think about it on your own for now though.
Let's consider a new function. We have f of x equals x minus six, times x plus two, times x minus four. I'd like to discuss how to evaluate this function and a few very special points. And this is going to help us figure out how to factor polynomials better. So if we consider this function, then what do we get for f of four? And what do we get for f of negative two?
For both of these we get 0. This means then that 4 and negative 2 are both zeros or roots of this function.
We've already found two zeros of this function negative 2 and 4. So what is the third zero? Please fill your answer in, in this box.
The answer is 6. That's because the last factor we haven't dealt with is x minus 6. If we set x minus 6 equal to 0 then that means that x equals 6.
Let's say we have some function f of x. And we know that f of 7 is equal to 0. Then, what's one factor of this function that we can know right off the bat?
We know that x minus 7 is going to be a factor. Since if we plug 7 in to this expression it's going to equal 0. Which would make the entire function equal 0.
Now let's say we still have a function called f of x but this time we know that f of a equals 0. Then what factor do we know about?
If that's the case, then we know that x minus a is a factor of our function.
We saw in the last quiz that x minus a is a factor of a function, f of x, if and only if f of a equals 0. So let's take another function as an example to see if this rule that we just figured out really makes sense. If we have f of x equals x squared plus 3 x minus 10. Then what do we get for f of two. Please write that in this box. Now based on your answer here, you may or may not have information that tells you about a factor of f of x. If you think you can figure one of theme out, please write it right here. Remember to find f of two, just replace every x here with a two.
When we plug 2 into the function we get 0. Based on the theorem that we came up with on the last quiz that tells us that one of our factors is going to be x minus 2.
So far, we've learned what happens when we substitute in special values of x that make the function equal to zero. And we've seen that this connects directly to factors. Now we're going to look at how to find special values, like a in this case. Working backwards from what we have written here, if we could look at a function and find its factors, then we would know these special values. So let's figure out how to do that. Now you already know how to factor quadratic expressions and you had a lot of practice with that earlier in the course. So let's see if we can look at another function and use similar methods to this to find one factor and then figure out what that factor needs to be multiplied by to give us the original function. Let's say we have a function that's already in factored form maybe it's f of x equals x minus 6 times x plus hand side out, and then simplify, what is the constant term that we'll get? Now what you want to be looking for here is a shortcut to do this. So that you don't have to actually multiply out everything. Think about which combinations of terms here are going to give us just constants.
The constant term is going to be 48. And we can actually get this really quickly. Let's think about how. We can start to simplify this by multiplying out the first two sets of parentheses. If we do this, then the first expression we have is going to be x squared minus 4 x minus 12. And then this, of course, is still multiplied by x minus 4. Now to further simplify this, we know that we have to multiply every term in this first set of parentheses by each of the terms over here. So let's think carefully about this for a moment. If we multiply anything over here by x squared, it's going to have a factor of x squared in it. So that's not going to give us a constant. The same is true if we multiply anything by negative 4x, since there are no terms over here that would cancel out that x. Now if instead, we multiply something over here by negative 12, we do have the potential to come up with a constant term. If we multiply negative 12 times x, well, that's going to give us negative 12x. That won't work. It has an x in it. So the last combination of terms is what's going to give us the answer we're looking for, a constant. Negative 12 times negative constants in the first 2 expressions that we multiplied together. So the real way we get 48 is by multiplying all of the constants involved in these expressions in the original factored form. Negative 6 times 2 times negative 4 is what gives us 48.
So you've seen that the constant turn that we would get if we multiplied this entire right hand side of the equation out, is the product of the constant terms negative six, two and negative four from each of the original factors that we had. So that means that this number 48 here is pretty intimately related to the factors of this polynomial. That means that we should be able to use some information involving 48 to get to the roots of this polynomial. There's gotta be a pretty close connection there. So let's say we have some fourth degree polynomials to add. Maybe it's this one, f of x equals x to the guesses for how to factor this polynomial. Let me write them down for you. Which of these three choices down here do you think is the correct factorization of this function? Remember to think about what we need in terms of this constant up here.
This middle choice, f of x equals x plus 1 times x minus 2 times x plus 5 times x minus 4, is the only one that could be the factorization of this polynomial. Remember that all the constants we have in our original factors must multiply to equal this constant out here, 40. We can double check that 1 times negative either of these.
So how can we use what we've just learned? Well let's say that we have a polynomial like this one that's pretty tough to factor. Here we have y equals x cubed minus 4 x squared plus x plus 6. And I'll say It again. I think this is really hard to factor. So how can we guess what the zeros might be? Remember that's going to tell us what our factors are. Well, we know that the constant terms in the factors need to multiply to equal 6. So, what then are all the factors of 6? Please separate your answers by commas and don't forget the negatives. They're perfectly valid factors, too, as long as they're integers.
The factors of six are one, negative 1, 2, negative 2, 3, negative 3, 6 and negative 6.
We saw before that x minus a is a factor of f of x if f of a equals 0. So that means we can test the constants that we came up with in the last quiz by substituting them in for x here and seeing which ones of them gives us y equals notation up here instead of using y. Let's make this f of x instead. Great, now it'll be very clear what numbers we're plugging in. Let's start with something from down here, maybe one, since that's first in the list. First tell me what f of 1 equals, using this function, and then tell me whether or not that means that x minus 1 is a factor of this function.
When we plug in 1 to our function we actually get that f of 1 equals 4. So that means that x minus 1 is not a factor of f of x.
Okay, on to a new number. 1 didn't work, so maybe let's count up 1 to 2. What does f of 2 equal? And, does that mean that x minus 2 is a factor of this polynomial or not?
When we plug in two to our function we do in fact get zero. So that means that x minus two is a factor of this polynomial.
So great, we've found a factor. How does this help us though? Eventually I want to get f of x in it's fully factored form, but what do we do with this single factor that we already know to find out the other factors? Well, let's think. If we have a number like 12 and we know that two is a factor of it. Then if we divide 12 by 2 we get 6.As what we need to multiply by two to get twelve. So that means that all of the other factors must multiply to equal six. Then we can also factor six in to two and three. Those are the other two prime factors of 12. So now we have the full prime factorization of 12. It's just equal to 2 times 2 times 3. So looks like if we divide a number by a factor that's known, then we get a simplified number in this case, or expression in this case that is a product of the leftover factors we're still looking for. That means that we shoudl divide this expression by x minus 2. To get one step closer to figuring out our other factors. But how do we do this. Well, you'll have to wait for the next lesson to find out.