WEBVTT mathematics/pre-calculus/selhorst-jones
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Hi--welcome back to Educator.com.
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Today we are going to talk about the properties of quadratic functions.
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In the previous lesson, we worked on finding the roots of quadratics.
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To that end, we learned how to complete the square, and we derived the quadratic formula, which tells us the roots of any quadratic polynomial.
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In this lesson, we will continue working with quadratic polynomials.
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We will explore another important feature of quadratics, the vertex, along with its connection to the shape of the graph.
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In addition, we will also see a new alternative form to write quadratics in that will give us a lot more information
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than that general form, ax² + bx + c, that we have been used to using so far; let's go!
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Let's take a look at the shape of a quadratic function's graph; we call this shape a **parabola**.
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The parabola has this nice curved shape on either side; and we are used to making these--we have been making them
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since probably pretty much right after we learned how to graph in algebra.
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We have been learning how to make parabolas--parabolic shapes--so we are pretty used to this.
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And you see it a lot in nature: if you throw a ball up in the air, it makes a parabolic arc.
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In fact, that is where the roots of the word *parabola* come from--they have to do with throwing, back either in Latin or Greek--I am not quite sure.
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Anyway, we see this kind of shape in every single quadratic function that we graph: we get this parabolic shape, a parabola.
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No matter the specific quadratic, this shape is always there in some form.
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It may have been moved; it may have been stretched; it might have been flipped.
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But it is still a parabola--it still has this cup shape that is symmetric.
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The graph of any quadratic will give us the symmetric, cup-shaped figure.
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As we just mentioned, a parabola is symmetric: the two sides of a parabola are always mirror images.
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If we look at the left side, and we look at the right side, they are doing the same thing all the way to this bottom part.
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And if we go up, we keep seeing the same thing.
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We don't have arrows continuing it up; but it is going to be true forever; as long as we are going up, it is also going to be true on this thing.
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Formally, we can draw a line called the **axis of symmetry** that the parabola is symmetric around.
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So, if we go to the right at some height, and we go to the left at the same height, we will notice that the distance here is equal to the distance here.
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If we do that at a different height (like, say, here), these two are going to be equal, as well.
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So, that is what we have for symmetry: there is this axis that we can draw down the middle, this imaginary line
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that we can put down the middle, that is going to make it so that each side is exactly the same as the other--
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that each side is a mirror reflection--that they are symmetric around that axis of symmetry.
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We call the point where the axis of symmetry crosses the graph the vertex.
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So, where that imaginary line intersects our parabola, we call that the **vertex**.
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Informally, we can think of this as the point where the parabola turns.
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It is going down; it is going down; it is going down; and then, at some point, it changes back, and it starts to go up.
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That point in changing, where it switches the direction it is going, whether up or down,
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or if it is coming from the bottom, going up, to going down--that is the place of turning.
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It is changing its direction; and we call that the vertex.
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Formally, it is the midpoint for the symmetry breakdown; but we can also think of it as where it changes directions, the turning--
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sort of a corner, as much as a nice, smooth curve can have a corner.
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Just as a graph of every quadratic has a parabola, we can now see that every quadratic has an axis of symmetry and a vertex.
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If we go back through all of our previous ones, our original function, f(x)...that has a vertex down here;
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and then, it has an axis of symmetry through the middle.
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Our g(x) right here...that one has its vertex down at the bottom, and an axis of symmetry, as well.
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Our h(x) in green has its vertex right there; and while it is not quite even on the halves,
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because of just the nature of the graphing window we are looking at, it still has that same axis of symmetry.
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If we could see what happened farther out to the left, we would see that it is, indeed, symmetric.
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And finally, I don't actually have the color purple, so I will use yellow, awkwardly, for this purple color right here.
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It, too, has the same vertex, and it has this axis of symmetry.
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So, whenever we draw a parabola, we are going to have this vertex, and we are going to have this axis of symmetry.
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Now, how can we find where they are located--where does this vertex occur; where does this axis of symmetry occur?
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We know where the vertex and axis of symmetry are for the fundamental parabola, our normal square function, x².
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That one is going to be easy; the vertex is at (0,0); we just saw that; and the axis of symmetry will be a vertical line, x = 0.
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A really quick tangent: why is it called x = 0--why does that give us a vertical line?
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Well, remember: one way to think of a graph is all of the solutions that are possible--all of the things that would be true.
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Well, if we have x = 0 here, all that that means is that every point on this graph where the x component is 0.
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So, that is going to be everything on this vertical thing.
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So, we are going to have everything on the vertical axis end up being x = 0.
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We might have (0,-5) or (0,5) or (0,1) or (0,0); but they all end up being the case, that x = 0 for each one of these.
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So, that is why we end up seeing that a vertical line is defined as x = something,
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because we are fixing the x, but we are letting the y component, the vertical component, have free rein.
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All right, that is the end of my tangent; what about those other different parabolas,
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the ones that aren't our normal square function that we are used to, where it is pretty easy to figure out where it is going to be?
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Well, we noticed earlier, when we were just looking at all of those parabolas side-by-side on the same graph--
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we noticed earlier that all parabolas are similar to each other.
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It is just a matter of being shifted, stretched, or flipped, compared to f(x) = x², that fundamental square function.
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Now, shifted/stretched/flipped--that sounds exactly like transformations.
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Remember when we learned about transformations, when we were first talking about functions?
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So, we go back to our lesson on transformations, and we refresh ourselves.
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Let's pretend that we just went back, and we grabbed the transformations for how vertical shifts, horizontal shifts, stretches, and vertical flips work.
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And if any of this stuff is really confusing in this lesson, go back and watch that there.
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And it will give you a refresher, and you will think, "Oh, that makes sense, how these things are all coming into being."
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It all gets fully explained in that lesson; so if you didn't watch it before, it might be a good thing to check out now.
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So, we go back; we grab that information; it is always useful to think in terms of "I learned this before; that would be useful now."
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Then, you just go back and look it up in a book; you find it on the Internet; you watch a lesson like this at Educator.com;
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you re-learn what you need for what you are doing right now.
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Reviewing our work in transformations, we get that vertical shift is f(x) + k,
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so we have that f(x) is just x², because that is our basic, fundamental square function.
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And we add k, and that just shifts it up and down by k.
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Horizontal shift is f(x - h); so that is going to be x - h plugging into that square...so it will be (x - h), and then that whole thing gets squared.
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And that is going to shift it by h; so positive h will end up going to the right; -h will end up going to the left.
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Stretch: if we want to stretch it, we multiply by a, a times f(x).
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So, as a gets larger and larger, it will be more stretched vertically, because it is taking it up.
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And as a gets smaller and smaller (closer and closer to 0), it is going to squish it down, because it is taking it and making any given value output smaller.
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And then finally, vertical flip: vertical flip is just based on the sign,
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so it is going to be a negative in front that will cause us to go from being up to being down,
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because everything will get flipped to its negative output.
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Now, one quick note: we are now using a slightly different form to shift horizontally.
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Previously, we were using f(x + k); but now we are using f(x - h); f(x - h) is what we are using right now, (x - h)².
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And that is going to help us with parabolas; and we will see why in just a moment.
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It is going to just make it a little bit easier for us to write out a formula for the vertex.
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Now, for x + k, positive k shifted to the left; but now, when we are using x - h, positive h will shift to the right.
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Now, why is that? Remember: if for x + k, when we plugged in the -1, it caused us to shift to the right;
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then if we plug in a positive h, that is just going to be the same thing as a negative k.
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A positive h is equivalent to a negative k; so a positive h goes right; a negative k goes right.
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A negative h goes left; a positive k goes left.
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For now, we can just switch our thinking entirely to this (x - h)² form, because that is what we will be working with right now.
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But it is useful to see what the connection is between when we first introduced the idea of transformations,
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in our previous lesson, to what we are working with now.
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All right, we can combine all of this together into a new form for quadratic polynomials: a(x - h)² + k.
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This shows all of the transformations a parabola can have.
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Our vertical shift is the + k over here; the horizontal shift is the - h portion inside of the squared.
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The stretch or squeeze on it is the a right there; and then, the vertical flip is also shown by the a, just based on the sign of the a.
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If the a is positive, then we are pointed up; if the a is negative, then we are pointed down.
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So, all of the information about a parabola can be put into just these three values: k, h, and a.
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And this also makes sense, because, in our normal form, ax² + bx + c,
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all of the information about a parabola can be put in 1, 2, 3 (a, b, c).
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All of the information you need to make up a quadratic can be put into a, b, and c.
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So, clearly, it just takes three pieces of information to say exactly what your parabola is.
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So, it shouldn't be too surprising that there are 1, 2, 3 pieces of information if we swap things around into a different order of looking at it.
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We can convert any quadratic we have into this form, this a(x - h)² + k form.
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For example, all of the various parabolas we saw earlier...f(x) is the one in red; g(x) is the one in blue; h(x) is the one in green;
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and j(x) is the one in purple that I confusingly I am using yellow to highlight (sorry about that).
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So, if we look at this, 1 times (x - 0)² + 0...this means a shift horizontal of 0, a shift vertical of 0, and it is a non-stretched, normal-looking thing.
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And that is exactly what we have right here with our red parabola; it seems pretty reasonable.
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The blue one: x - 2 means that we have shifted to the right by 2, and we have shifted down by 3.
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And that is exactly what has happened here; we are at 2 in terms of horizontal location, and 3 in terms of height.
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And then, the 4 here means that we have stretched up, which seems to make sense,
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because it seems to be pulled up more than when we compare it to the green one.
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Similarly, we have similar things with green; it has been moved; the -1/5 causes it to flip and sort of stretch out.
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It has been squished out, so it is not quite as long.
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And then, once again, our confusing yellow is purple: -11 has been flipped down, and it is even more stretched out than the blue one is.
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It is even more stretched out, because it has a larger value (11 versus 4).
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The negative just causes it to flip.
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We see how this stuff connects; how do we convert from our general form, ax² + bx + c, into this new alternate form?
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We are used to getting things in the form ax² + bx + c; we are using to working with things in the form ax² + bx + c.
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So, we might want to be able to convert from ax² + bx + c into this new form that seems to give us all of this information.
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It is very easy to graph with this new form, because we just set a vertex, and then we know how much to squish it or stretch it.
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Now, how do we do this?--we do it through completing the square.
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So, if you don't quite remember how we did completing the square--if you didn't watch the previous lesson,
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and you are confused by what is about to happen--just go and check out the previous lesson.
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It will get explained pretty clearly as you work through that one.
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So, assuming you understand completing the square, let's see it get used here.
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We have ax² + bx + c right here; and so, what we do is take the c,
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and we just sort of move it off to the side, so we don't have to work with it right now.
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And then, we put parentheses around this, and we pull the a out of those parentheses.
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So, since we divide this part by a--since we divide the ax² by a to pull it out right here--
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then we also have to divide it from the b, as well; so we get b/a.
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We have a(x² + b/a(x)) + c.
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So, if you multiply that out, you will see that, yes, you have the exact same thing that you just started with.
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The next step: remember, when you are completing the square, you want to take this thing right here;
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and then you want to divide that by 2, square it, and then plug it right back in.
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So, if we do work through this, we get b/a, over 2, would become b/2a; and we would square that, and we would get b²/4a².
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So, that is what we are plugging in right here: b²/4a².
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But since we are putting something in somewhere, we have to keep the scales balanced.
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If we put one thing into some place, we can't just have nothing else counteract that.
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For example, if I had 5, and I wanted to put in a 3 (just, for some reason, I felt like putting in a 3):
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5 is totally different than 5 + 3; so you have to put in something else to counteract that.
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What will counteract + 3? Minus 3 will counteract that.
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Now, notice, though: we have this a standing in front; so if we put b²/4a² right here,
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this a is going to multiply it, so it is effectively like we put in more than that.
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Going back to 5 + 3, if we had a 2 standing out front, and we wanted to have this be just the same thing as 2 times 5--
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we wanted this to be the same as this--then 2(5 + 3)...well, this may be the same thing as a 6 inside of it.
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So, we would have to put a -6 on the outside; 2 times 3; we are having this thing on the outside that has to be dealt with on what we just added in.
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Otherwise, we are not going to have equal scales.
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We have this thing on the outside, this extra weight modifying things, this a on the outside of our quantity.
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And so, if we don't deal with that, when we figure out how much to put on the outside part (-6 or -3),
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it is going to end up not being equal on the two sides anymore.
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So, if that is the case, then we have -b²/4a, because if we think about it,
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we have the yellow a (just because yellow is a little bit hard to read...) times this thing right here, b²/4a².
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So, that ends up being the a here and the a²...the a² cancels to just a, and that cancels out a right there.
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So, we have b²/4a; and so, that right there is what we are going to need to subtract by.
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We will subtract by that, and we will end up having it still be equal.
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And if you work that out, if you multiply that a out into that entire quantity and check it out,
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you will see that we haven't changed it at all from ax² + bx + c; it is still equivalent.
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So, at this point, we can now collapse this thing right here into a squared form: (x + b/2a)².
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Let's check and see that that actually makes sense still: (x + b/2a)².
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So, we get x², x(b/2a) + (b/2a)x...so that becomes b/a(x), because they get added together twice; + b²/4a².
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So sure enough, that checks out; that is the same thing.
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-b²/4a + c...we just want to put that over a common denominator.
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So, if we have c, that is going to be the same as 4ac, all over 4a; so now they are over a common denominator.
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So, this + c here becomes the 4ac right there; so we have (-b² + 4ac)/a.
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And now, we are back in that original alternate form...sorry, not in that original alternate form...
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We have gone from our original, normal form of ax² + bx + c into our new alternate form.
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We will see how it parallels in just a moment.
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In our new form, it is really easy to find the vertex; we have a(x - h)² + k;
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so, from our information about transformations, we see that the vertex has been shifted horizontally by h:
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a horizontal shift of h, and then vertically by k.
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So, in this form, our vertex is at (h,k); it is as simple as that.
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What about ax² + bx + c, if we start them that way?
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Well, we just figured out that ax² + bx + c is just the same thing as this right here.
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Now, that is kind of a big thing; but we can see how this parallels right here.
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So, we have -h here; we have +b/2a here; so with h, it must be the case that we have -b/2a,
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because otherwise we wouldn't be able to deal with that + sign right there.
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Then, we also have the +(-b² + 4ac), all over 4a.
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And that is the exact same thing as our k.
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Since they are both plus signs here, they end up saying the same value.
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We have -b² + 4ac, over 4a; so that gives us our vertex.
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Now, I think it is pretty difficult to remember -b² + 4ac, all over 4a, since it is also different than our quadratic formula.
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But it is a lot like it, so we might get the two confused.
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So, what I would recommend is: just remember that the horizontal location of the vertex occurs at -b/2a.
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And then, if you want to figure out what the vertical location is, just plug it into your function.
00:17:55.400 --> 00:18:02.300
You have to have your function to know what your polynomial is; so you have -b/2a; that will tell you what the horizontal location of the vertex is.
00:18:02.300 --> 00:18:08.600
And then, you just plug that right into your function, and that will, after you work it through, give out what the vertical location is.
00:18:08.600 --> 00:18:13.000
It seems pretty easy to me--and much easier, I think, than trying to memorize this complicated formula.
00:18:13.000 --> 00:18:19.300
So, I would just remember that vertices occur horizontally at -b/2a.
00:18:19.300 --> 00:18:24.900
Great; by knowing the vertex of a parabola, we also know the minimum or maximum that the quadratic attains.
00:18:24.900 --> 00:18:32.400
So, if a is positive, it cups up; it goes up; so if it cups up, then we are going to have a minimum at the bottom.
00:18:32.400 --> 00:18:44.700
So, if a is positive, it cups up; then we will have a minimum at x = -b/2a, because that is where the vertex is.
00:18:44.700 --> 00:18:52.500
And clearly, that is going to be the most extreme point; the vertex is going to be the extreme point of our parabola, whether it is high or low, depending.
00:18:52.500 --> 00:18:59.200
Then, if we have that a is negative, then that means we are cupping down.
00:18:59.200 --> 00:19:04.600
It points down; so if we are cupping down, then it must be the high point that is going to be the vertex.
00:19:04.600 --> 00:19:11.200
So, it is a maximum if we have an a that is negative, because we are cupping down; so it will be x = -b/2a.
00:19:11.200 --> 00:19:17.000
Now, you might have a little difficulty remembering that a is positive means minimum; that seems a little bit counterintuitive, probably.
00:19:17.000 --> 00:19:23.200
a is negative means maximum...what I would recommend is just to think, "a is positive; that means it is going to have to cup up."
00:19:23.200 --> 00:19:26.000
Oh, the vertex has to come at the bottom.
00:19:26.000 --> 00:19:30.900
If a is negative, it means it has to cup down; oh, the vertex is going to come at the maximum.
00:19:30.900 --> 00:19:34.600
Remember it in terms of that, just honestly making a picture in front of your face,
00:19:34.600 --> 00:19:39.900
being able to just use your hands and gesticulate in the air in front of yourself; and see what you are making in the air.
00:19:39.900 --> 00:19:42.400
That makes it very easy to be able to remember this stuff.
00:19:42.400 --> 00:19:48.400
Just trying to memorize it as cold, dry facts is not as easy as just being able to think, "Oh, I see it--that makes sense!"
00:19:48.400 --> 00:19:53.100
You are remembering primary concepts; and working from there, it is much easier to work things out.
00:19:53.100 --> 00:19:58.900
All right, also, if we know the vertex, it is pretty easy to find the axis of symmetry.
00:19:58.900 --> 00:20:04.000
The axis of symmetry runs through the vertex; since it has to go through the vertex,
00:20:04.000 --> 00:20:10.500
and the vertex is at a horizontal location of h, then it is just going to end up being a vertical line, x = h.
00:20:10.500 --> 00:20:21.800
So, for example, if we had f(x) = (x - 1)² - 1.5, then we would know that that would give us a vertex of x - h + k.
00:20:21.800 --> 00:20:32.000
So, it would give us a vertex of...-1 becomes h; it is just 1; and then k...since it is + k, it means that k must be -1.5.
00:20:32.000 --> 00:20:36.900
We don't really care about the -1.5, since we are looking for a vertical line that is just x = h.
00:20:36.900 --> 00:20:48.600
So, we take that x = h, and we make x = 1; so we get an axis of symmetry that runs right through the parabola at a horizontal location of 1.
00:20:48.600 --> 00:20:53.100
And sure enough, that splits it right down the middle--a nice axis of symmetry.
00:20:53.100 --> 00:20:58.700
And here is an incredibly minor note on grammar--this is really minor, but a lot of people are confused by this grammar point.
00:20:58.700 --> 00:21:03.800
So, I just want you to have this clear, so you don't accidentally say the wrong thing, and it ends up being embarrassing.
00:21:03.800 --> 00:21:09.800
You might as well know what it is: the singular form, if you want to just talk about one, is "vertex";
00:21:09.800 --> 00:21:13.100
the singular form, if you want to just talk about one, is "axis."
00:21:13.100 --> 00:21:24.200
On the other hand, if you want to talk about multiple of a vertex, that is going to be "vertices"; multiple vertex...multiple vertices is what it becomes.
00:21:24.200 --> 00:21:28.700
Vertexes is kind of hard to say, so that is why it transforms into vertices.
00:21:28.700 --> 00:21:37.100
Axis becomes axes; so axises, once again, sounds a little bit awkward; so we make axis become axes.
00:21:37.100 --> 00:21:43.700
So, it is not just one vertex, but many vertices; the earth has one axis, but the plane has two axes.
00:21:43.700 --> 00:21:49.300
So, it is a really, really minor thing; but you might as well know it, because you occasionally have to say this stuff out loud.
00:21:49.300 --> 00:21:51.000
All right, we are ready for our first example.
00:21:51.000 --> 00:21:56.600
If we have this polynomial, f(x) = -3x² - 24x - 55,
00:21:56.600 --> 00:22:03.500
and we want to put this in the form a(x - h)² + k, then we will identify what a, h, and k are.
00:22:03.500 --> 00:22:06.600
So, how do we end up doing this? We have to complete the square.
00:22:06.600 --> 00:22:13.100
So, we have to complete the square; and we are going to complete the square on our polynomial.
00:22:13.100 --> 00:22:20.900
-3x² - 24x - 55; all right, so once again, if you don't quite remember how to complete the square,
00:22:20.900 --> 00:22:24.300
you can also get a chance to see more completing the square in the previous lesson.
00:22:24.300 --> 00:22:26.400
But you also might be able to just pick it up right here.
00:22:26.400 --> 00:22:31.700
So, the first thing we have to do is separate out that -55, so we can see things a little more easily.
00:22:31.700 --> 00:22:36.300
We aren't actually going to do anything to it; we are just going to shift it, literally shift it to the side,
00:22:36.300 --> 00:22:39.600
just so we can see things and keep our head clear of the -55.
00:22:39.600 --> 00:22:42.000
It is still part of the expression; we are just moving it over right now.
00:22:42.000 --> 00:22:45.400
Now, we want to pull out the -3 to clear things out.
00:22:45.400 --> 00:22:48.700
So, if we are pulling out -3 on the left, we have to pull out -3...
00:22:48.700 --> 00:22:53.000
If we are pulling -3 out of -3x², we also have to pull it out of -24x.
00:22:53.000 --> 00:22:59.900
If we are going to do that--if we are pulling out -3 out front, dividing into -3x²: -3x²,
00:22:59.900 --> 00:23:10.400
divided by -3, becomes just positive x²; great; and then -24x/-3 becomes + 8x; minus 55 still.
00:23:10.400 --> 00:23:15.000
Now, we might be a little bit unsure of this, doing the distribution property in reverse.
00:23:15.000 --> 00:23:17.800
It might be a little bit worrisome; we are not used to doing that yet.
00:23:17.800 --> 00:23:28.200
So, we might want to check this: -3 times (x² + 8x); -3x² - 24x; great; that checks out, just like what we originally had.
00:23:28.200 --> 00:23:35.200
So, -3(x² + 8x) - 55 is the exact same thing as what we started with right here.
00:23:35.200 --> 00:23:37.800
So, checking like this--you will probably find that you can just do it in your head.
00:23:37.800 --> 00:23:41.200
You could do -3 times that quantity in your head, and think, "Yes, this makes sense."
00:23:41.200 --> 00:23:45.400
But if it is an exam, if you have something like that, you definitely want to check under those situations.
00:23:45.400 --> 00:23:49.400
Make sure that if it is something really important, you are really checking and thinking about your work,
00:23:49.400 --> 00:23:52.500
because it is really easy to make mistakes, especially if it is new to you.
00:23:52.500 --> 00:23:57.800
All right, -3(x² + 8x)--how do we get this to collapse into a square?
00:23:57.800 --> 00:24:00.000
How do we get this to actually pull into a square?
00:24:00.000 --> 00:24:04.800
Well, if you remember, we were talking about completing the squares; we want to take this number,
00:24:04.800 --> 00:24:14.800
and we want to add in 8 divided by 2, and then squared: (8/2)² is 4², which is equal to 16.
00:24:14.800 --> 00:24:26.300
So, we want to add 16 inside; so we have -3(x² + 8x + 16); so we are adding 16 in.
00:24:26.300 --> 00:24:32.300
But now, remember, if we are putting something into the expression, we have to make sure that we keep those scales balanced.
00:24:32.300 --> 00:24:36.700
If you put 16 in, we have to take away however much we just put in.
00:24:36.700 --> 00:24:44.400
Now, we didn't just put 16 in; there is also this -3 up front, so we put 16 into the quantity, but that gets multiplied by -3, as well.
00:24:44.400 --> 00:24:47.000
So, what did we put, in total, into the expression?
00:24:47.000 --> 00:24:51.500
We put -3 times 16, in total, into this expression.
00:24:51.500 --> 00:24:58.100
-3 times 16...3 times 10 is 30; 3 times 6 is 18; so it is -48 in total.
00:24:58.100 --> 00:25:04.800
So, if we put -48 into the expression, we need to take -48 out of the expression.
00:25:04.800 --> 00:25:15.400
What is the opposite to -48? Positive 48; so if we put -48 in, and we put in positive 48 at the same time, it is going to be as if we had done nothing.
00:25:15.400 --> 00:25:21.000
We have that positive 16 inside of the quantity; but because of that -3 out front,
00:25:21.000 --> 00:25:28.800
it is as if we had put in a -48; so we put in a positive 48 outside of the parentheses, and it is as if we had had no effect at all.
00:25:28.800 --> 00:25:35.400
So, it is still equivalent to what we started with: + 48 - 55...at this point, we just finish things up and collapse the stuff.
00:25:35.400 --> 00:25:43.600
x² + 8x + 16 becomes (x + 4)²; we know that because it is going to end up being 8/2, which is 4.
00:25:43.600 --> 00:25:52.000
Let's check it really quickly: (x + 4)²: x²; x(4) + 4(x) is 8x; 4 times 4 is 16; great, it checks out.
00:25:52.000 --> 00:25:59.300
And then, plus 48 minus 55; 48 - 55 becomes + -7; great.
00:25:59.300 --> 00:26:07.300
At this point, we are ready to identify: we have a = -3, because that one is in front of our multiplication.
00:26:07.300 --> 00:26:19.400
Then, h is equal to -4, because it is the one right here; but notice it is -h, and what we have is + 4.
00:26:19.400 --> 00:26:25.600
So, since we have + 4, it has to be h = -4; otherwise we are breaking from that form.
00:26:25.600 --> 00:26:31.000
And then finally, k is equal to -7.
00:26:31.000 --> 00:26:34.200
Great; and we have everything we need to be able to put it in that form.
00:26:34.200 --> 00:26:38.700
All right, the next example: Give a function for the parabola graph below.
00:26:38.700 --> 00:26:44.900
We have this great new form; and look at that--it looks to me very like we have the vertex right here.
00:26:44.900 --> 00:26:49.100
It is pretty clearly the lowest point on that parabola; so it must be the vertex,
00:26:49.100 --> 00:26:52.500
if it is the absolute lowest point on a parabola that is pointing up.
00:26:52.500 --> 00:27:01.400
So, f(x) = a(x - h)² + k.
00:27:01.400 --> 00:27:11.500
Great; so what is our (h,k)? Well, (h,k) is going to come out of this, so (h,k) is what our vertex is, which is (2,1).
00:27:11.500 --> 00:27:25.400
So, h = 2; k = 1; we plug that information in; and we are going to get that f(x) is now equal to a(x - 2)² + 1.
00:27:25.400 --> 00:27:28.500
All right, so we are close, but we still don't know what a is.
00:27:28.500 --> 00:27:33.900
But look over here: we have this other point over here with additional information.
00:27:33.900 --> 00:27:41.500
We know that, when we plug in -1, we get out 4; f(-1) = 4.
00:27:41.500 --> 00:27:56.400
So now, if f(-1) = 4, then we can say that 4 = a(-1 - 2)² + 1.
00:27:56.400 --> 00:28:09.400
4 = a(-3)² + 1; 4 = a(9) + 1; we subtract the 1 from both sides; we get 3 = a(9).
00:28:09.400 --> 00:28:15.100
We divide out the 9, and we get 1/3 (3/9 becomes 1/3) equals a.
00:28:15.100 --> 00:28:28.200
So finally, our function is f(x) = 1/3(x - 2)² + 1; great.
00:28:28.200 --> 00:28:33.300
And if the problem had asked us to put it in that general form, ax² + bx + c, that we are used to,
00:28:33.300 --> 00:28:38.900
at this point you could also just expand 1/3(x - 2)² + 1; you would be able to expand it
00:28:38.900 --> 00:28:42.300
and work through it, and you would also be able to get into that form, ax² + bx + c.
00:28:42.300 --> 00:28:44.100
Remember: you can just switch from one to the other.
00:28:44.100 --> 00:28:50.200
To switch from this new form into our old, general form, you just expand.
00:28:50.200 --> 00:28:55.500
If you want to go from the old general form, you complete the square, like we just did in the previous example.
00:28:55.500 --> 00:29:02.600
All right, the third example: f(x) = 6x² - 18x + 5; does f have a global minimum or a global maximum?
00:29:02.600 --> 00:29:06.100
And then, which one and at what point?
00:29:06.100 --> 00:29:16.400
So, if f(x) = 6x², then notice a: ax² + bx + c...they end up being the same in f(x) = a(x - h)² + k.
00:29:16.400 --> 00:29:34.400
So, a = 6; if a equals 6, then a is positive; if a is positive, we have that it cups up; if it cups up, then it is going to have a minimum.
00:29:34.400 --> 00:29:39.100
What it has is a global minimum on it.
00:29:39.100 --> 00:29:52.900
Now, where is it going to happen? The vertex horizontally is x = -b/2a.
00:29:52.900 --> 00:30:02.400
What is our b? Our b is -18 (remember, because it is ax² + bx...so if it is -18, then b is -18);
00:30:02.400 --> 00:30:08.400
so we have negative...what is b? -18, over 2 times...what was our a? 6.
00:30:08.400 --> 00:30:14.600
We simplify this out; so we get positive 18 (when those negatives cancel), over 12, equals 3/2.
00:30:14.600 --> 00:30:21.400
So, our x location is at 3/2; and now, what is our y going to be at?
00:30:21.400 --> 00:30:34.800
Well, we can figure that out by f of...plug in 3/2...equals 6(3/2)² - 18(3/2) + 5.
00:30:34.800 --> 00:30:45.300
We work through that: 6(9/4) (we square both the top and the bottom) - 18(3/2)...well, we can knock that out, and this becomes a 9.
00:30:45.300 --> 00:30:57.900
So, we have 9(3), or -27 + 5 equals...6(9/4)...well, this is 2(2); 6 is 3(2); so we knock out the 2's;
00:30:57.900 --> 00:31:06.300
and we have 3(9) up top; 27/2 minus...what is 27 + 5? That becomes 22.
00:31:06.300 --> 00:31:26.000
So, we can put -22 over a common denominator with 27/2; we get 27/2 - 44/2; (27 - 44)/2; 27 - 44...we get -17/2.
00:31:26.000 --> 00:31:36.100
So, that is our y location; so that means that the point where we have our minimum is point (3/2,-17/2).
00:31:36.100 --> 00:31:39.900
And there is the point of our global minimum.
00:31:39.900 --> 00:31:45.200
Great; all right, the final one: this one is a big word problem, but it is a really great problem.
00:31:45.200 --> 00:31:49.700
A Norman window has the shape of a semicircle on top of a rectangle.
00:31:49.700 --> 00:31:57.200
If the perimeter of the window is 6 meters in total, what height (h) and width (w) will give a window with maximum area?
00:31:57.200 --> 00:32:02.500
At this point, how do we do this? The first thing we want to do is that we want to understand what is going on.
00:32:02.500 --> 00:32:08.400
This seems to make sense: we have a semicircle--that is a semicircle right here.
00:32:08.400 --> 00:32:15.600
What is a semicircle? A semicircle is just half of a circle; and yes, that looks like half of a circle, on top of a rectangle.
00:32:15.600 --> 00:32:19.900
It is on top of a rectangle, and we have our rectangle right here; so that seems to make sense, right?
00:32:19.900 --> 00:32:24.900
We have a semicircle on top of a rectangle box; OK, that idea makes sense.
00:32:24.900 --> 00:32:31.700
The perimeter of the window is 6 meters; now, you might have forgotten what perimeter is; but perimeter is just all of the outside edge put together.
00:32:31.700 --> 00:32:39.500
A nice, hard-to-see yellow...we go outside: perimeter, perimeter, perimeter, perimeter, perimeter, perimeter, perimeter, perimeter, perimeter...
00:32:39.500 --> 00:32:44.400
It is going to be all of the outside of that box, but it is not going to go across now;
00:32:44.400 --> 00:32:47.900
it is going to also have to be the outside perimeter of the top part of the window.
00:32:47.900 --> 00:32:51.500
We have that dashed line there to indicate that that is where we split into the semicircle.
00:32:51.500 --> 00:32:55.500
But there is no actual material there; the perimeter is just the very outside part.
00:32:55.500 --> 00:32:59.700
The perimeter will be the top part of that semicircle, and then three sides of our box.
00:32:59.700 --> 00:33:04.000
The fourth side doesn't actually exist; it is just connected into the semicircle.
00:33:04.000 --> 00:33:10.900
All right, that makes sense; we know that it is 6 meters, and then we want to ask what height and width will give a window with maximum area.
00:33:10.900 --> 00:33:18.500
So, if we are looking for area, we are going to need to use area; so A = area--we will define that idea.
00:33:18.500 --> 00:33:28.600
If we also know information about the perimeter, then we will say P = perimeter.
00:33:28.600 --> 00:33:36.600
OK, let's work these things out: area is equal to the area of this part, plus the area of our box;
00:33:36.600 --> 00:33:39.400
so it is the area of our semicircle, plus the area of our box.
00:33:39.400 --> 00:33:45.800
So, area of our semicircle...well, what is the area of our circle?
00:33:45.800 --> 00:33:54.500
Area for a circle is equal to πr²; so a semicircle: if it is a half-circle,
00:33:54.500 --> 00:34:01.100
that is going to be area = 1/2πr², because it is half of a circle.
00:34:01.100 --> 00:34:06.700
So, we have 1/2πr²; well, we don't have an r yet, so we had better introduce an r.
00:34:06.700 --> 00:34:13.400
So, we will say that here is the middle; r is from the middle of our circle, out like that.
00:34:13.400 --> 00:34:22.600
Area equals πr² divided by 2 (because, remember, it is a semicircle, so it is half of it), plus...what is the area of our box?
00:34:22.600 --> 00:34:26.100
That is height times width.
00:34:26.100 --> 00:34:30.800
Now, we don't really want to have to have r; the fewer variables, the better, if we are going to try to solve this.
00:34:30.800 --> 00:34:32.900
So, how does r connect to the rest of it?
00:34:32.900 --> 00:34:43.500
Well, r...look, that is connected to our w, because it is just half of that side; so r = w/2.
00:34:43.500 --> 00:34:53.100
We can change this formula into area = π...r² is (w/2)²...over 2, plus height times width.
00:34:53.100 --> 00:34:56.700
Now, we have managed to get rid of the radius there; so let's simplify that out a bit more.
00:34:56.700 --> 00:35:07.500
Area = π...the w/2 gets us w²/4, over 2, plus hw.
00:35:07.500 --> 00:35:13.400
We have this whole thing divided by 2; so area equals πw², and we are dividing...
00:35:13.400 --> 00:35:23.500
so with the 4 and the 2, since they are both dividing on π and that stuff, they are going to combine into dividing by 8...plus hw.
00:35:23.500 --> 00:35:28.800
Great; so we have that area equals πw² + hw.
00:35:28.800 --> 00:35:32.200
All right, what about perimeter--what can we say about the perimeter?
00:35:32.200 --> 00:35:39.600
Well, the perimeter is going to be equal to...well, there is an h here; there is a w here; what is here? Just another h.
00:35:39.600 --> 00:35:48.600
And then, what is this portion here? Right away, we see that we have two h's, plus a w, plus something else.
00:35:48.600 --> 00:35:53.000
Circumference for a circle: if we want the perimeter of a circle, that is circumference.
00:35:53.000 --> 00:36:08.500
Circle circumference is equal to 2πr; but if it is a semicircle, then it is half a circle; so it is going to be 2πr/2.
00:36:08.500 --> 00:36:17.200
So, perimeter for a semicircle will be 1/2(2πr), which is just going to be πr.
00:36:17.200 --> 00:36:23.000
We have 2h + w + (the amount of perimeter that our semicircle puts in is) πr.
00:36:23.000 --> 00:36:25.900
Now, once again, we don't really want to have extra variables floating around.
00:36:25.900 --> 00:36:35.300
So, we want to get rid of that; so perimeter equals 2h + w + π(w/2).
00:36:35.300 --> 00:36:40.700
Great; at this point, we see that we have h and w; we have area.
00:36:40.700 --> 00:36:45.100
So, if we could turn this into...we have that area is unknown; we don't really know what the maximum area is,
00:36:45.100 --> 00:36:50.000
or what area we are dealing with; really, it is a function that is going to give area when we plug in h and w.
00:36:50.000 --> 00:36:52.700
So, it is not even something that has a fixed value, necessarily.
00:36:52.700 --> 00:36:55.900
What about perimeter, though? Perimeter is something we do know.
00:36:55.900 --> 00:37:05.600
Remember: perimeter is 6 meters; so we can plug in 6 = 2h + w + π(w/2).
00:37:05.600 --> 00:37:12.000
Now, it seems like we have more w's than we have h's, in both our area and perimeter stuff.
00:37:12.000 --> 00:37:22.100
So, let's try to get rid of area; we will figure out what h is in terms of w, so we can substitute out the h and switch it in for stuff about w.
00:37:22.100 --> 00:37:31.300
We will move everything over, and we have 6 - w - π(w/2) = 2h.
00:37:31.300 --> 00:37:39.200
We divide by 2 on both sides; we get 3 - w/2 - π(w/4) = h.
00:37:39.200 --> 00:37:44.600
And now, let's just pull those things together, so it will be easier to plug in later, into our one over here,
00:37:44.600 --> 00:37:47.500
because we want to swap out the h here for it.
00:37:47.500 --> 00:37:59.000
3 - 2w/4 + πw...now, that part might be a little confusing; but notice that this has minus, and this has minus;
00:37:59.000 --> 00:38:04.000
so when we combine them together, they are just one minus, because they are actually working together before they do their subtraction.
00:38:04.000 --> 00:38:16.300
It equals h; and we can even pull out the w onto the outside; so we get 3 - (2 + π)/4 times w = h.
00:38:16.300 --> 00:38:24.400
Great; now we have an expression about area, and we have an expression about h and w.
00:38:24.400 --> 00:38:32.300
At this point, we can plug in what we know about h, and we can plug it in for the h in the hw in our area.
00:38:32.300 --> 00:38:38.800
And we will have area equals...just stuff involving w; and since it is just stuff involving w, we will have just one variable.
00:38:38.800 --> 00:38:42.900
Maybe we can figure things out; maybe it looks like something--maybe it looks like a quadratic,
00:38:42.900 --> 00:38:45.700
since, after all, this was all about how quadratics work.
00:38:45.700 --> 00:38:49.300
So (student logic), it seems pretty likely that it is going to end up looking like a quadratic.
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And we can apply the knowledge that we just learned.
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So, the maximum area is at what h and w?
00:38:55.300 --> 00:39:04.400
We have area = πw²/8 + wh; and h = 3 - (2 + π)/4 times w.
00:39:04.400 --> 00:39:09.200
OK, great; so at this point, we take h here, and we plug it in over here.
00:39:09.200 --> 00:39:18.200
We have area equals πw²/8 (that is still the same thing), plus w times h,
00:39:18.200 --> 00:39:26.300
so w times...what is h?...3 - (2 + π)/4 times w.
00:39:26.300 --> 00:39:35.200
We expand that out; our area is equal to πw²/8, plus w times [3 - ((2 + π)/4)w].
00:39:35.200 --> 00:39:42.100
So, we get 3w minus...let's keep it as (2 + π)/4, and it just combines with that other w; we have w².
00:39:42.100 --> 00:39:46.600
So now, we have a w² here and a w² here; so let's make them talk to each other.
00:39:46.600 --> 00:39:56.000
We get πw²/8; let's actually pull that down--we will make it as
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(π/8)w² - [(2 + π)/4]w² + 3w.
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So now, we have π/8, and we can bring this stuff to bear, so it will be minus 2 + π, but it used to be a 4.
00:40:16.400 --> 00:40:22.200
So, it is going to be...to become an 8, we are going to multiply by 2 here, to keep it the same.
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2 times (2 + π) becomes 4 + 2π; so we have [π - (4 + 2π)]w² + 3w.
00:40:32.300 --> 00:40:44.300
We simplify this out: π - 4...so the -4 will come through, and -2π will become -π, over 8, w² + 3w.
00:40:44.300 --> 00:40:54.200
Look, if area equals this, then this right here--this whole thing--is a quadratic.
00:40:54.200 --> 00:40:59.700
If it is a quadratic, then we can use the stuff that we know about where vertices show up.
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Where do vertices show up on the parabola?
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Now, we are looking for the maximum area; so we had better hope that our parabola points down, so it does have a maximum at its vertex.
00:41:09.100 --> 00:41:12.800
Sure enough, we have a negative here and a negative here.
00:41:12.800 --> 00:41:21.600
And since that is on the w², that means we can pull out the negative, and our first a is going to be -(4 + π)/8.
00:41:21.600 --> 00:41:26.700
That whole thing ends up being a negative number; so sure enough, it is going to have a vertex at the top,
00:41:26.700 --> 00:41:31.800
so it will have a maximum area out of this--it is cupping down.
00:41:31.800 --> 00:41:41.400
At this point, we have figured out--remember--the vertex is at our horizontal location (in this case,
00:41:41.400 --> 00:41:48.200
horizontal would be just our w); vertex is going to be at w = -b/2a.
00:41:48.200 --> 00:41:56.700
So, what is that? It is going to be w =...what is our b? That is going to be 3, so we have -3, over...
00:41:56.700 --> 00:42:05.900
what is a? a is this whole thing, (-4 - π)/8.
00:42:05.900 --> 00:42:09.200
This is a little bit confusing; but we have a fraction over a fraction, so if it is like this,
00:42:09.200 --> 00:42:18.600
we can multiply the top and the bottom by 8; 8 on top; 8 on the bottom; we will get -24, and the 8's down here will just cancel out.
00:42:18.600 --> 00:42:25.900
So, we will get -24 - 4 - π we see that there are negatives everywhere, so we can cancel out all of our negatives.
00:42:25.900 --> 00:42:32.600
Multiply the top and the bottom by -1; we get positive 24, over (4 + π).
00:42:32.600 --> 00:42:41.800
The maximum is going to happen at 24/(4 + π); there is our...
00:42:41.800 --> 00:42:48.500
Oh, sorry: one thing that I just realized--I made one tiny mistake: it is 2a; so we have a 2 here.
00:42:48.500 --> 00:42:56.600
So, it is still a 2 up front; so it is not 4 + π.
00:42:56.600 --> 00:42:58.500
It is always important to think about what you are doing.
00:42:58.500 --> 00:43:07.000
24/2(4 + π) cancels to be 12/(4 + π).
00:43:07.000 --> 00:43:13.900
Now, that was clearly a very long, very difficult problem; but we see that it is actually really similar to the previous example that we just did.
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All we are looking for is where the vertex is; the only thing is that it is couched inside of a word problem.
00:43:18.700 --> 00:43:21.600
So, we just have to be carefully thinking: how do we build equations?
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Once we have our equations built, how do we put them together?
00:43:24.000 --> 00:43:28.300
How do we get this to look like something where we can apply what we just learned in this lesson?
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How can I apply this stuff about quadratic properties?
00:43:31.300 --> 00:43:36.100
So, we find out that the maximum occurs when w is equal to 12/(4 + π).
00:43:36.100 --> 00:43:47.500
Now, they asked what h and w; so since we have to figure that out, we know h is equal to 3 - (2 + π)/4 times w.
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So, h is going to be at its maximum, as well, since it is h and w.
00:43:52.100 --> 00:44:07.300
3 - (2 + π)/4 times (12/(4 + π)) (as our w): we notice that 4 can take out the 12, and we will get 3 up top.
00:44:07.300 --> 00:44:15.900
We get that h is equal to 3 - [3(2 + π)] over (4 + π).
00:44:15.900 --> 00:44:19.100
We want to put them over common denominators, so we can get the two pieces talking to each other.
00:44:19.100 --> 00:44:32.200
We have 3(4 + π)/(4 + π); and then it is going to be minus 3(2 + π), so - 6 - 3π.
00:44:32.200 --> 00:44:41.400
We have 12 + 3π, minus 6 - 3π, all over 4 + π.
00:44:41.400 --> 00:44:50.900
So finally, our h is going to be...the -3π's cancel each other out; for 12 - 6, we get 6, over 4 + π.
00:44:50.900 --> 00:44:54.900
And that is our value for what our maximum h will be with our maximum width.
00:44:54.900 --> 00:45:02.200
So, the maximum area will occur when our width is 12/(4 + π), and our h is 6/(4 + π).
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They will both be in units of meters, because meters is what we started with for our perimeter.
00:45:05.900 --> 00:45:10.800
All right, I hope you have a sense of how quadratics work, what their shape is, and this idea of the vertex,
00:45:10.800 --> 00:45:13.300
and that being where maximum and minimum are located.
00:45:13.300 --> 00:45:17.300
Now, remember: you just have to remember that the horizontal location for maximum or minimum--
00:45:17.300 --> 00:45:25.300
the horizontal location for the vertex--is going to occur at -b/2a, when we have it in that standard form of ax² + bx + c.
00:45:25.300 --> 00:45:32.000
As long as you remember -b/2a, you can just plug it in any time that you need to find what the vertical location going along for that vertex is.
00:45:32.000 --> 00:45:34.000
All right, we will see you at Educator.com later--goodbye!