 Welcome back to our lecture series Math 1050, College Algebra for Students at Southern Utah University. As usual, I'm your professor today, Dr. Andrew Missildine. At the start of lecture 19, we are really at the heart of this chapter about quadratic functions. In the previous two sections, 3.1 and 3.2, we've been reviewing techniques and skills necessary for working with quadratic equations, involving things like factoring, completing the square, the quadratic formula, discriminants, and complex numbers in general. We're now ready to put those into place here. We start studying quadratic functions. Functions of the form, I should mention, that f of x equals a x squared plus b x plus c. We're interested in functions like this, but our primary goal, or at least our first goal, is going to be graphing these things. Now, one thing that's really nice about quadratic equations is that every quadratic equation is a transformation of the very basic graph, y equals x squared. The y equals x squared here, this is just our standard parabola, for which case if we were to graph it, we might get something like the following. Let me slide this to the side real quick. If we were to graph some lines, let's say here's our x-axis and this is our y-axis. Graph in a quadratic equation, y equals x squared would look something like the following y equals x squared. Well, you have the point zero, zero, zero squared, gives you a point right here, one squared is one, two squared is four, so we get one, two, three, four, like so. I think I'm gonna slide this up a little bit. And then if we do the next one, three squared, we're gonna get nine. So that gives us one, two, three, four, five, six, seven, eight, nine. I have to go one more. If I draw this to scale again, remember one, two, three, four, five, six, seven, eight, nine. So we get this point right here. And this is gonna be a nice continuous curve, something like the following, like so. So this gives us kind of like half of the picture. The other half you can actually get from reflection. I've kind of missed my point there. Let's try that again. All right, not perfect, but you get the idea. This is the graph of a standard parabola y equals x squared. Now every other parabola that you're gonna deal with is going to be a translation, or yeah, some type of transformation of this graph. And what I mean is the following. If you start off with the equation y equals x squared, what we could do is we could perform some type of like vertical stretch or compression. So if we wanna do like a vertical reflection or a vertical scaling of some kind, stretch it or compress it, this is gonna be some factor A that we use. And so what we've seen is you'll just replace, you'll just put an A in front of your function, A equals x squared. If A is negative, it means you reflect across the x-axis. If A is bigger than one, it means you do some type of stretching. And of course it's actually values what I'm talking about. Or you can do some type of compression. I wanna mention that if you wanna do, you know, are you doing a vertical or horizontal? If you wanna do a horizontal stretch or reflection, you're gonna get some like x over B squared, which notice because this is a x squared here, x over B squared is the same thing as x squared over B squared. And so you end up with just one over B squared, x squared. And remember, this is the so-called Play-Doh effect in action right here, that when you try to horizontally stretch this function, so by dividing by B in the horizontal zone, that actually does a horizontal stretch by a factor of B. You're actually dividing outside the function by a factor of B squared. So this tells you that if you horizontally stretch the graph, that actually vertically compresses the graph. And vice versa, if you horizontally compress the graph, this will do a vertical stretch to the graph. I also wanna mention that if you take y equals negative x square, if you reflect across the y-axis, this just gives you x squared again. This function is an even function. It's symmetric with respect to the y-axis. That reflection is invisible. So because of that, when we work with a quadratic, well, when we work with a parabola, we don't worry about horizontal stretching or reflections whatsoever, because every horizontal reflection is ignored because of symmetry and every vertical, sorry, every horizontal stretch or compress can actually be written as a vertical compress or stretch because of the Play-Doh effect. So it suffices to only have to worry about vertical reflection and scaling when it comes to these quadratic equations. So the next thing to do is we could do some type of vertical, we could do a vertical shift, vertical shift by a factor of say, like k. We could also do a horizontal shift, maybe by a factor of h. And so this would give us y equals a x squared plus k if you do a vertical shift. And then you get y equals a times x minus h squared plus k. And so this right here gives you the most general, generally transformed parabola there is. And so this right here, I wanna mention to you is what we refer to as the vertex form, the vertex form of the quadratic, y equals a times x minus h squared plus k. In the vertex form, we can identify what were the transformations one did to the standard parabola right here. Now, if you were to foil out the x minus h squared, distribute the h and add k and such, if you kind of expanded it, you could transform it into something like this, y equals a x squared plus b x plus c, the so-called standard form we saw before, some people would call it the quadratic form or the general form, all of that, whatever. One could translate back and forth between it, right? Given the vertex form, you can just translate this into the quadratic form just by expanding it, multiplying, expanding it. But then you can reverse this process. You can go from the standard form to the vertex form by completing the square. We'll see an example of that in just a moment. And so it's useful to go back and forth. Now, if I had a graph of quadratic function, I think this vertex form is a lot easier to keep track of for the following reason. Let's come back to our picture over here. So this picture, you see in yellow, this is the standard quadratic, right? That is just the basic one, y equals x squared. Now, if we were to do, if we had something like y equals a x minus h squared plus k, what I want you to do is think of the transformations. The original graph has the point zero, zero, and it has the point one, one. If you were to translate these things, you know, you shifted things by h and up by k, you might get something like one, two, three, and then like one, two, three, four. Let's say that the origin's gonna move to the point right here, h comma k. This is what's referred to as the vertex of the quadratic. And you can find the vertex just by knowing h and k. Now, this other point one, one, I want you to notice that the original parabola, it had this point one, one, there's this one to one ratio between the vertex and a second point on the graph. This has been affected by a factor of a. And so imagine we've stretched the graph by like a factor of two. Then this point is gonna work and become h comma, you know, let's see, we're gonna have h comma a plus k, like so. And so you can find this other point here by this coefficient, the leading coefficient a, it's gonna tell you the right and run to go from the vertex to another point. Now, every parabola has an axis of symmetry that's gonna run vertically and it goes through the vertex of the graph. Once you've graphed the right-hand side, you can use the left-hand side symmetry to graph it and you get something like the following. So we can graph a quadratic function very easily if it's in vertex form. The vertex form, of course, coming from the name, gives you the vertex. Now, if you wanna switch from vertex form to standard form, like I said, that's pretty easy to multiply the thing out. If you wanna go from the standard form, the quadratic form to the vertex form, then that's a little bit more challenging, but you can do this by completing the square. So to complete the square here, we're gonna do what we did before. So we're gonna take f of x equals, well, when we did this for solving quadratic equations, we would move the 23 to the other side of the equation. But since we have a function, we kinda wanna leave the f of x all by itself. So in terms of moving it, we don't have to move to the other side, but we wanna separate the constant from the variables. And so we can actually just do that with parenthesis here. Two x squared minus 16x, we're gonna leave a spot for our guest of honor. And then we just have a plus 23 right here, okay? When your leading coefficient is not a one, we have to factor out the two, but we have to factor from both coefficients there. So we end up with two times x squared minus eight x, leave a space plus 23. And so now we have to identify who our guest of honor is gonna be. So we look at this middle coefficient, we take half of it, negative eight times one half is negative four, then we have this, cause this is our b value, then we have to square b and we get positive 16. So we're gonna add a 16 right here, and we have to then subtract a 16. The idea is if we add and subtract on the same side of the equation, things will be balanced. But wait a second, this two distributes on to all three terms, we didn't add a 16, we actually added a two times 16. So we have to subtract two times 16 right here. Don't forget to multiply by the coefficient that's in front of your quadratic right there. We then simplify this thing. So what we get here is you're gonna have two times x minus four quantity squared, and then we get 23 minus 32, which should give us a negative nine. So our quadratic will look like f of x equals two times x minus four squared minus nine. So we can identify the transformations directly from this graph right here. So notice that we've vertically stretched the graph by a factor of two. It's been horizontally shifted by a factor of four, and it's been shifted down by a factor of nine. This tells us that the vertex of this parabola is gonna be four comma negative nine. Did I say this was shifted? Which direction did I say that was? I don't remember. The graph got shifted to the right by a factor of four, and it got shifted down by a factor of nine there. So if we were to graph this thing real quick, we might look at something like the following. Now let's graph a line right there. Whoops, that's not horizontal. Let's graph this right here, and then something like this maybe. And just for simplicity, I'm gonna do this. The x-axis will be like one, two, three, four. But the y-axis I'm gonna do three, six, nine. Just change the scale a little bit. And so when I'm graphing a quadratic polynomial, I wanna first graph the vertex. And the vertex, remember, we saw was four comma negative nine. So we come over to four, we go down by nine, one, two, three. So our vertex right here is gonna be four comma negative nine. Then using, so our vertex, remember, h comma k, we found out was four comma negative nine. We also wanna use the stretch there. It was stretched by a factor of two. So what this tells us is we're gonna go up by two spaces and over by one to find another point on the graph. This would be the point. And I guess I changed the scale here, so one has to be a little bit more careful. Cause when I use this stretch, I can only go up, I have to go up two spaces and over by one. Since this is a scale of three, I have to be a little bit more careful. Basically what we're gonna do is we're gonna add, let me, let me, I'm gonna move this right here down below. So we get four comma negative nine. And so we wanna go up by two spaces and over by one space. So we add one to the x-coordinate, which gives us five. We then add two to the y-coordinate, which is negative seven. So we get something like this. And then we could try to graph the parabola from there. Now those points are kind of close to the vertex. So it might be difficult to go from there, but when you're graphing this like on a computer application, it's pretty easy. You just kind of click the vertex and then you click another point that's on the graph, which again, you can use this two to find it. Personally, if it was me, I would be looking for x-intercepts, but that's a conversation we're gonna have in a different video. Let's just stick with the graph we have for right here. Let's take a look at another example of this thing. Let's graph the function f of x equals two x square plus eight x plus five. If we wanna graph this, I would recommend using the vertex form. So if we try to complete the square here, you're gonna have a two x squared plus eight x plus something, right? The guess of honor, we get a plus five, factor out the two. We get two times x squared plus four x plus five. I'm then going to add, so taking my middle number right here, half of that is two, that's your b values, so b squared is equal to four. We're gonna add four, but then we're gonna subtract eight. Where do the eight come from? Four times two, four times two. And this, thus we get y equals two times x minus x plus two squared, and then you get five minus eight, which is negative three. So this tells us that the vertex of the parabola is gonna be negative two comma negative three. You're always gonna pick the opposite of the sign you see because this is in the horizontal zone. So you get negative two, negative three, so we go over by two to the left, and then one, two, three down. That gives you the vertex of this thing. Our vertical stretch was also by a factor of two, so going from the vertex, which the vertex here is gonna be negative two, negative three, like I said, you can go up by two, one, two over one. That finds you another point on the graph, this point right here, which would be negative one comma negative one. And then by symmetry, we also get this point over here, which would be negative three comma negative one. And using those three points, we can find the whole graph right here. And so unlike the last one, I do wanna mention that you, I wanted to use one that was actually graphed by a computer so you could see how things are drawn to scale. And because you have to be careful when you change the scale that a quadratic function is not aligned. So in the last one, I kind of was getting in trouble for a second because I wanted to use this slope of two. I want to go up two over one. That doesn't work on that previous graph because the scale, the Y scale was different from the X scale. You have to have a one to one ratio in that regard in order to use it correctly. You can't just go up two and then over one like you came with a line. And that's because of the line, the rate of change is the same no matter where you are. For a parabola though, the arbitrary change gets steeper and steeper and steeper the farther away you get from the vertex. And so if you're going to use this slope right here to find another point on the graph, you can only go one step to the right or to the left on the X axis. Go beyond that, it's going to mess up. And so I have to go one step to the right and then two steps up to get this point right here. Like I said, this is going to be a lot easier if we just focus on X intercepts which is what we'll do in a forthcoming video. But this shows us how we can, we can graph a quadratic equation, use the vertex form, we can see the transformations. And in particular, we can find this very important point which we call the vertex of the parabola.