For more information, please see full course syllabus of Math Analysis

For more information, please see full course syllabus of Math Analysis

## Discussion

## Study Guides

## Download Lecture Slides

## Table of Contents

## Transcription

## Related Books

### Properties of Quadratic Functions

- The graph of a quadratic function gives a
*parabola*: a symmetric, cup-shaped figure. - Since the parabola is symmetric, we can draw a line that the parabola is symmetric around. This is called the
*axis of symmetry*. - The point where the axis of symmetry crosses the graph is the
*vertex*. Informally, we can think of this point as where the parabola "turns"-the place where the graph changes directions. - While parabolas can be very different from one another, they are still fundamentally similar. Any parabola can be turned into another parabola through transformations (see the lesson on
*Transformations of Functions*). - With this realization in mind, we can convert any quadratic into the form

This form shows all the transformations that have been applied to the fundamental square function of xa ·(x−h) ^{2}+ k.^{2}. - We can convert into this format by completing the square (see the previous lesson,
*Completing the Square and the Quadratic Formula*, for more information). - In this new form, it's easy to find the vertex. From transformations, we see the the vertex has been shifted horizontally by h and vertically by k. So, in this form, our vertex is at (h, k).
- If our quadratic is in the standard form of f(x) = ax
^{2}+bx+c, the vertex is at⎛

⎝

− b 2a

, f ⎛

⎝

− b 2a

⎞

⎠

⎞

⎠

. - By knowing the vertex of a parabola, we can find the minimum or maximum that the quadratic attains.
- Once we know the location of the vertex, it's extremely easy to find the axis of symmetry. The axis of symmetry runs through the vertex, so if the vertex is at (h,k), then the axis of symmetry is the vertical line x=h.

### Properties of Quadratic Functions

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro
- Introduction
- Parabolas
- Axis of Symmetry and Vertex
- Drawing an Axis of Symmetry
- Placing the Vertex
- Looking at the Axis of Symmetry and Vertex for other Parabolas
- Transformations
- An Alternate Form to Quadratics
- Switching Forms by Completing the Square
- Vertex of a Parabola
- Minimum/Maximum at Vertex
- Axis of Symmetry
- Incredibly Minor Note on Grammar
- Example 1
- Example 2
- Example 3
- Example 4

- Intro 0:00
- Introduction 0:05
- Parabolas 0:35
- Examples of Different Parabolas
- Axis of Symmetry and Vertex 1:28
- Drawing an Axis of Symmetry
- Placing the Vertex
- Looking at the Axis of Symmetry and Vertex for other Parabolas
- Transformations 4:18
- Reviewing Transformation Rules
- Note the Different Horizontal Shift Form
- An Alternate Form to Quadratics 8:54
- The Constants: k, h, a
- Transformations Formed
- Analyzing Different Parabolas
- Switching Forms by Completing the Square 11:43
- Vertex of a Parabola 16:30
- Vertex at (h, k)
- Vertex in Terms of a, b, and c Coefficients
- Minimum/Maximum at Vertex 18:19
- When a is Positive
- When a is Negative
- Axis of Symmetry 19:54
- Incredibly Minor Note on Grammar 20:52
- Example 1 21:48
- Example 2 26:35
- Example 3 28:55
- Example 4 31:40

### Math Analysis Online

### Transcription: Properties of Quadratic Functions

*Hi--welcome back to Educator.com.*0000

*Today we are going to talk about the properties of quadratic functions.*0002

*In the previous lesson, we worked on finding the roots of quadratics.*0005

*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.*0008

*In this lesson, we will continue working with quadratic polynomials.*0014

*We will explore another important feature of quadratics, the vertex, along with its connection to the shape of the graph.*0017

*In addition, we will also see a new alternative form to write quadratics in that will give us a lot more information*0023

*than that general form, ax ^{2} + bx + c, that we have been used to using so far; let's go!*0028

*Let's take a look at the shape of a quadratic function's graph; we call this shape a parabola.*0036

*The parabola has this nice curved shape on either side; and we are used to making these--we have been making them*0040

*since probably pretty much right after we learned how to graph in algebra.*0046

*We have been learning how to make parabolas--parabolic shapes--so we are pretty used to this.*0050

*And you see it a lot in nature: if you throw a ball up in the air, it makes a parabolic arc.*0054

*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.*0058

*Anyway, we see this kind of shape in every single quadratic function that we graph: we get this parabolic shape, a parabola.*0065

*No matter the specific quadratic, this shape is always there in some form.*0073

*It may have been moved; it may have been stretched; it might have been flipped.*0077

*But it is still a parabola--it still has this cup shape that is symmetric.*0080

*The graph of any quadratic will give us the symmetric, cup-shaped figure.*0084

*As we just mentioned, a parabola is symmetric: the two sides of a parabola are always mirror images.*0089

*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.*0094

*And if we go up, we keep seeing the same thing.*0101

*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.*0104

*Formally, we can draw a line called the axis of symmetry that the parabola is symmetric around.*0111

*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.*0117

*If we do that at a different height (like, say, here), these two are going to be equal, as well.*0127

*So, that is what we have for symmetry: there is this axis that we can draw down the middle, this imaginary line*0134

*that we can put down the middle, that is going to make it so that each side is exactly the same as the other--*0139

*that each side is a mirror reflection--that they are symmetric around that axis of symmetry.*0144

*We call the point where the axis of symmetry crosses the graph the vertex.*0150

*So, where that imaginary line intersects our parabola, we call that the vertex.*0153

*Informally, we can think of this as the point where the parabola turns.*0158

*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.*0162

*That point in changing, where it switches the direction it is going, whether up or down,*0166

*or if it is coming from the bottom, going up, to going down--that is the place of turning.*0170

*It is changing its direction; and we call that the vertex.*0175

*Formally, it is the midpoint for the symmetry breakdown; but we can also think of it as where it changes directions, the turning--*0178

*sort of a corner, as much as a nice, smooth curve can have a corner.*0185

*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.*0191

*If we go back through all of our previous ones, our original function, f(x)...that has a vertex down here;*0198

*and then, it has an axis of symmetry through the middle.*0205

*Our g(x) right here...that one has its vertex down at the bottom, and an axis of symmetry, as well.*0208

*Our h(x) in green has its vertex right there; and while it is not quite even on the halves,*0218

*because of just the nature of the graphing window we are looking at, it still has that same axis of symmetry.*0225

*If we could see what happened farther out to the left, we would see that it is, indeed, symmetric.*0231

*And finally, I don't actually have the color purple, so I will use yellow, awkwardly, for this purple color right here.*0235

*It, too, has the same vertex, and it has this axis of symmetry.*0244

*So, whenever we draw a parabola, we are going to have this vertex, and we are going to have this axis of symmetry.*0247

*Now, how can we find where they are located--where does this vertex occur; where does this axis of symmetry occur?*0252

*We know where the vertex and axis of symmetry are for the fundamental parabola, our normal square function, x ^{2}.*0260

*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.*0265

*A really quick tangent: why is it called x = 0--why does that give us a vertical line?*0272

*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.*0279

*Well, if we have x = 0 here, all that that means is that every point on this graph where the x component is 0.*0285

*So, that is going to be everything on this vertical thing.*0293

*So, we are going to have everything on the vertical axis end up being x = 0.*0297

*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.*0303

*So, that is why we end up seeing that a vertical line is defined as x = something,*0318

*because we are fixing the x, but we are letting the y component, the vertical component, have free rein.*0324

*All right, that is the end of my tangent; what about those other different parabolas,*0329

*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?*0333

*Well, we noticed earlier, when we were just looking at all of those parabolas side-by-side on the same graph--*0339

*we noticed earlier that all parabolas are similar to each other.*0344

*It is just a matter of being shifted, stretched, or flipped, compared to f(x) = x ^{2}, that fundamental square function.*0348

*Now, shifted/stretched/flipped--that sounds exactly like transformations.*0357

*Remember when we learned about transformations, when we were first talking about functions?*0361

*So, we go back to our lesson on transformations, and we refresh ourselves.*0365

*Let's pretend that we just went back, and we grabbed the transformations for how vertical shifts, horizontal shifts, stretches, and vertical flips work.*0369

*And if any of this stuff is really confusing in this lesson, go back and watch that there.*0375

*And it will give you a refresher, and you will think, "Oh, that makes sense, how these things are all coming into being."*0378

*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.*0381

*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."*0386

*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;*0392

*you re-learn what you need for what you are doing right now.*0397

*Reviewing our work in transformations, we get that vertical shift is f(x) + k,*0400

*so we have that f(x) is just x ^{2}, because that is our basic, fundamental square function.*0406

*And we add k, and that just shifts it up and down by k.*0412

*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.*0416

*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.*0429

*Stretch: if we want to stretch it, we multiply by a, a times f(x).*0434

*So, as a gets larger and larger, it will be more stretched vertically, because it is taking it up.*0439

*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.*0444

*And then finally, vertical flip: vertical flip is just based on the sign,*0453

*so it is going to be a negative in front that will cause us to go from being up to being down,*0457

*because everything will get flipped to its negative output.*0462

*Now, one quick note: we are now using a slightly different form to shift horizontally.*0465

*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}.*0470

*And that is going to help us with parabolas; and we will see why in just a moment.*0483

*It is going to just make it a little bit easier for us to write out a formula for the vertex.*0486

*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.*0490

*Now, why is that? Remember: if for x + k, when we plugged in the -1, it caused us to shift to the right;*0501

*then if we plug in a positive h, that is just going to be the same thing as a negative k.*0507

*A positive h is equivalent to a negative k; so a positive h goes right; a negative k goes right.*0512

*A negative h goes left; a positive k goes left.*0518

*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.*0521

*But it is useful to see what the connection is between when we first introduced the idea of transformations,*0527

*in our previous lesson, to what we are working with now.*0531

*All right, we can combine all of this together into a new form for quadratic polynomials: a(x - h) ^{2} + k.*0534

*This shows all of the transformations a parabola can have.*0543

*Our vertical shift is the + k over here; the horizontal shift is the - h portion inside of the squared.*0546

*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.*0556

*If the a is positive, then we are pointed up; if the a is negative, then we are pointed down.*0565

*So, all of the information about a parabola can be put into just these three values: k, h, and a.*0569

*And this also makes sense, because, in our normal form, ax ^{2} + bx + c,*0574

*all of the information about a parabola can be put in 1, 2, 3 (a, b, c).*0579

*All of the information you need to make up a quadratic can be put into a, b, and c.*0585

*So, clearly, it just takes three pieces of information to say exactly what your parabola is.*0589

*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.*0594

*We can convert any quadratic we have into this form, this a(x - h) ^{2} + k form.*0601

*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;*0610

*and j(x) is the one in purple that I confusingly I am using yellow to highlight (sorry about that).*0620

*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.*0626

*And that is exactly what we have right here with our red parabola; it seems pretty reasonable.*0645

*The blue one: x - 2 means that we have shifted to the right by 2, and we have shifted down by 3.*0650

*And that is exactly what has happened here; we are at 2 in terms of horizontal location, and 3 in terms of height.*0659

*And then, the 4 here means that we have stretched up, which seems to make sense,*0668

*because it seems to be pulled up more than when we compare it to the green one.*0672

*Similarly, we have similar things with green; it has been moved; the -1/5 causes it to flip and sort of stretch out.*0676

*It has been squished out, so it is not quite as long.*0683

*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.*0688

*It is even more stretched out, because it has a larger value (11 versus 4).*0696

*The negative just causes it to flip.*0700

*We see how this stuff connects; how do we convert from our general form, ax ^{2} + bx + c, into this new alternate form?*0702

*We are used to getting things in the form ax ^{2} + bx + c; we are using to working with things in the form ax^{2} + bx + c.*0710

*So, we might want to be able to convert from ax ^{2} + bx + c into this new form that seems to give us all of this information.*0717

*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.*0722

*Now, how do we do this?--we do it through completing the square.*0729

*So, if you don't quite remember how we did completing the square--if you didn't watch the previous lesson,*0733

*and you are confused by what is about to happen--just go and check out the previous lesson.*0736

*It will get explained pretty clearly as you work through that one.*0739

*So, assuming you understand completing the square, let's see it get used here.*0742

*We have ax ^{2} + bx + c right here; and so, what we do is take the c,*0745

*and we just sort of move it off to the side, so we don't have to work with it right now.*0750

*And then, we put parentheses around this, and we pull the a out of those parentheses.*0753

*So, since we divide this part by a--since we divide the ax ^{2} by a to pull it out right here--*0758

*then we also have to divide it from the b, as well; so we get b/a.*0764

*We have a(x ^{2} + b/a(x)) + c.*0768

*So, if you multiply that out, you will see that, yes, you have the exact same thing that you just started with.*0773

*The next step: remember, when you are completing the square, you want to take this thing right here;*0777

*and then you want to divide that by 2, square it, and then plug it right back in.*0783

*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 b ^{2}/4a^{2}.*0789

*So, that is what we are plugging in right here: b ^{2}/4a^{2}.*0799

*But since we are putting something in somewhere, we have to keep the scales balanced.*0804

*If we put one thing into some place, we can't just have nothing else counteract that.*0808

*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):*0813

*5 is totally different than 5 + 3; so you have to put in something else to counteract that.*0820

*What will counteract + 3? Minus 3 will counteract that.*0825

*Now, notice, though: we have this a standing in front; so if we put b ^{2}/4a^{2} right here,*0829

*this a is going to multiply it, so it is effectively like we put in more than that.*0836

*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--*0841

*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.*0849

*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.*0857

*Otherwise, we are not going to have equal scales.*0868

*We have this thing on the outside, this extra weight modifying things, this a on the outside of our quantity.*0871

*And so, if we don't deal with that, when we figure out how much to put on the outside part (-6 or -3),*0877

*it is going to end up not being equal on the two sides anymore.*0886

*So, if that is the case, then we have -b ^{2}/4a, because if we think about it,*0889

*we have the yellow a (just because yellow is a little bit hard to read...) times this thing right here, b ^{2}/4a^{2}.*0897

*So, that ends up being the a here and the a ^{2}...the a^{2} cancels to just a, and that cancels out a right there.*0911

*So, we have b ^{2}/4a; and so, that right there is what we are going to need to subtract by.*0916

*We will subtract by that, and we will end up having it still be equal.*0922

*And if you work that out, if you multiply that a out into that entire quantity and check it out,*0925

*you will see that we haven't changed it at all from ax ^{2} + bx + c; it is still equivalent.*0930

*So, at this point, we can now collapse this thing right here into a squared form: (x + b/2a) ^{2}.*0934

*Let's check and see that that actually makes sense still: (x + b/2a) ^{2}.*0942

*So, we get x ^{2}, x(b/2a) + (b/2a)x...so that becomes b/a(x), because they get added together twice; + b^{2}/4a^{2}.*0946

*So sure enough, that checks out; that is the same thing.*0958

*-b ^{2}/4a + c...we just want to put that over a common denominator.*0960

*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.*0965

*So, this + c here becomes the 4ac right there; so we have (-b ^{2} + 4ac)/a.*0972

*And now, we are back in that original alternate form...sorry, not in that original alternate form...*0977

*We have gone from our original, normal form of ax ^{2} + bx + c into our new alternate form.*0983

*We will see how it parallels in just a moment.*0988

*In our new form, it is really easy to find the vertex; we have a(x - h) ^{2} + k;*0991

*so, from our information about transformations, we see that the vertex has been shifted horizontally by h:*0996

*a horizontal shift of h, and then vertically by k.*1002

*So, in this form, our vertex is at (h,k); it is as simple as that.*1007

*What about ax ^{2} + bx + c, if we start them that way?*1013

*Well, we just figured out that ax ^{2} + bx + c is just the same thing as this right here.*1016

*Now, that is kind of a big thing; but we can see how this parallels right here.*1021

*So, we have -h here; we have +b/2a here; so with h, it must be the case that we have -b/2a,*1025

*because otherwise we wouldn't be able to deal with that + sign right there.*1035

*Then, we also have the +(-b ^{2} + 4ac), all over 4a.*1038

*And that is the exact same thing as our k.*1042

*Since they are both plus signs here, they end up saying the same value.*1044

*We have -b ^{2} + 4ac, over 4a; so that gives us our vertex.*1048

*Now, I think it is pretty difficult to remember -b ^{2} + 4ac, all over 4a, since it is also different than our quadratic formula.*1052

*But it is a lot like it, so we might get the two confused.*1060

*So, what I would recommend is: just remember that the horizontal location of the vertex occurs at -b/2a.*1062

*And then, if you want to figure out what the vertical location is, just plug it into your function.*1070

*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.*1075

*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.*1082

*It seems pretty easy to me--and much easier, I think, than trying to memorize this complicated formula.*1088

*So, I would just remember that vertices occur horizontally at -b/2a.*1093

*Great; by knowing the vertex of a parabola, we also know the minimum or maximum that the quadratic attains.*1099

*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.*1105

*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.*1112

*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.*1125

*Then, if we have that a is negative, then that means we are cupping down.*1132

*It points down; so if we are cupping down, then it must be the high point that is going to be the vertex.*1139

*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.*1144

*Now, you might have a little difficulty remembering that a is positive means minimum; that seems a little bit counterintuitive, probably.*1151

*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."*1157

*Oh, the vertex has to come at the bottom.*1163

*If a is negative, it means it has to cup down; oh, the vertex is going to come at the maximum.*1166

*Remember it in terms of that, just honestly making a picture in front of your face,*1171

*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.*1174

*That makes it very easy to be able to remember this stuff.*1180

*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!"*1182

*You are remembering primary concepts; and working from there, it is much easier to work things out.*1188

*All right, also, if we know the vertex, it is pretty easy to find the axis of symmetry.*1193

*The axis of symmetry runs through the vertex; since it has to go through the vertex,*1199

*and the vertex is at a horizontal location of h, then it is just going to end up being a vertical line, x = h.*1204

*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.*1210

*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.*1222

*We don't really care about the -1.5, since we are looking for a vertical line that is just x = h.*1232

*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.*1237

*And sure enough, that splits it right down the middle--a nice axis of symmetry.*1248

*And here is an incredibly minor note on grammar--this is really minor, but a lot of people are confused by this grammar point.*1253

*So, I just want you to have this clear, so you don't accidentally say the wrong thing, and it ends up being embarrassing.*1259

*You might as well know what it is: the singular form, if you want to just talk about one, is "vertex";*1264

*the singular form, if you want to just talk about one, is "axis."*1270

*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.*1273

*Vertexes is kind of hard to say, so that is why it transforms into vertices.*1284

*Axis becomes axes; so axises, once again, sounds a little bit awkward; so we make axis become axes.*1289

*So, it is not just one vertex, but many vertices; the earth has one axis, but the plane has two axes.*1297

*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.*1304

*All right, we are ready for our first example.*1309

*If we have this polynomial, f(x) = -3x ^{2} - 24x - 55,*1311

*and we want to put this in the form a(x - h) ^{2} + k, then we will identify what a, h, and k are.*1316

*So, how do we end up doing this? We have to complete the square.*1323

*So, we have to complete the square; and we are going to complete the square on our polynomial.*1326

*-3x ^{2} - 24x - 55; all right, so once again, if you don't quite remember how to complete the square,*1333

*you can also get a chance to see more completing the square in the previous lesson.*1341

*But you also might be able to just pick it up right here.*1344

*So, the first thing we have to do is separate out that -55, so we can see things a little more easily.*1346

*We aren't actually going to do anything to it; we are just going to shift it, literally shift it to the side,*1352

*just so we can see things and keep our head clear of the -55.*1356

*It is still part of the expression; we are just moving it over right now.*1359

*Now, we want to pull out the -3 to clear things out.*1362

*So, if we are pulling out -3 on the left, we have to pull out -3...*1365

*If we are pulling -3 out of -3x ^{2}, we also have to pull it out of -24x.*1369

*If we are going to do that--if we are pulling out -3 out front, dividing into -3x ^{2}: -3x^{2},*1373

*divided by -3, becomes just positive x ^{2}; great; and then -24x/-3 becomes + 8x; minus 55 still.*1380

*Now, we might be a little bit unsure of this, doing the distribution property in reverse.*1390

*It might be a little bit worrisome; we are not used to doing that yet.*1395

*So, we might want to check this: -3 times (x ^{2} + 8x); -3x^{2} - 24x; great; that checks out, just like what we originally had.*1398

*So, -3(x ^{2} + 8x) - 55 is the exact same thing as what we started with right here.*1408

*So, checking like this--you will probably find that you can just do it in your head.*1415

*You could do -3 times that quantity in your head, and think, "Yes, this makes sense."*1418

*But if it is an exam, if you have something like that, you definitely want to check under those situations.*1421

*Make sure that if it is something really important, you are really checking and thinking about your work,*1425

*because it is really easy to make mistakes, especially if it is new to you.*1429

*All right, -3(x ^{2} + 8x)--how do we get this to collapse into a square?*1432

*How do we get this to actually pull into a square?*1438

*Well, if you remember, we were talking about completing the squares; we want to take this number,*1440

*and we want to add in 8 divided by 2, and then squared: (8/2) ^{2} is 4^{2}, which is equal to 16.*1445

*So, we want to add 16 inside; so we have -3(x ^{2} + 8x + 16); so we are adding 16 in.*1455

*But now, remember, if we are putting something into the expression, we have to make sure that we keep those scales balanced.*1466

*If you put 16 in, we have to take away however much we just put in.*1472

*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.*1477

*So, what did we put, in total, into the expression?*1484

*We put -3 times 16, in total, into this expression.*1487

*-3 times 16...3 times 10 is 30; 3 times 6 is 18; so it is -48 in total.*1491

*So, if we put -48 into the expression, we need to take -48 out of the expression.*1498

*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.*1505

*We have that positive 16 inside of the quantity; but because of that -3 out front,*1515

*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.*1521

*So, it is still equivalent to what we started with: + 48 - 55...at this point, we just finish things up and collapse the stuff.*1529

*x ^{2} + 8x + 16 becomes (x + 4)^{2}; we know that because it is going to end up being 8/2, which is 4.*1535

*Let's check it really quickly: (x + 4) ^{2}: x^{2}; x(4) + 4(x) is 8x; 4 times 4 is 16; great, it checks out.*1543

*And then, plus 48 minus 55; 48 - 55 becomes + -7; great.*1552

*At this point, we are ready to identify: we have a = -3, because that one is in front of our multiplication.*1559

*Then, h is equal to -4, because it is the one right here; but notice it is -h, and what we have is + 4.*1567

*So, since we have + 4, it has to be h = -4; otherwise we are breaking from that form.*1579

*And then finally, k is equal to -7.*1585

*Great; and we have everything we need to be able to put it in that form.*1591

*All right, the next example: Give a function for the parabola graph below.*1594

*We have this great new form; and look at that--it looks to me very like we have the vertex right here.*1599

*It is pretty clearly the lowest point on that parabola; so it must be the vertex,*1605

*if it is the absolute lowest point on a parabola that is pointing up.*1609

*So, f(x) = a(x - h) ^{2} + k.*1612

*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).*1621

*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.*1631

*All right, so we are close, but we still don't know what a is.*1645

*But look over here: we have this other point over here with additional information.*1648

*We know that, when we plug in -1, we get out 4; f(-1) = 4.*1654

*So now, if f(-1) = 4, then we can say that 4 = a(-1 - 2) ^{2} + 1.*1661

*4 = a(-3) ^{2} + 1; 4 = a(9) + 1; we subtract the 1 from both sides; we get 3 = a(9).*1676

*We divide out the 9, and we get 1/3 (3/9 becomes 1/3) equals a.*1689

*So finally, our function is f(x) = 1/3(x - 2) ^{2} + 1; great.*1695

*And if the problem had asked us to put it in that general form, ax ^{2} + bx + c, that we are used to,*1708

*at this point you could also just expand 1/3(x - 2) ^{2} + 1; you would be able to expand it*1713

*and work through it, and you would also be able to get into that form, ax ^{2} + bx + c.*1719

*Remember: you can just switch from one to the other.*1722

*To switch from this new form into our old, general form, you just expand.*1724

*If you want to go from the old general form, you complete the square, like we just did in the previous example.*1730

*All right, the third example: f(x) = 6x ^{2} - 18x + 5; does f have a global minimum or a global maximum?*1735

*And then, which one and at what point?*1742

*So, if f(x) = 6x ^{2}, then notice a: ax^{2} + bx + c...they end up being the same in f(x) = a(x - h)^{2} + k.*1746

*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.*1756

*What it has is a global minimum on it.*1774

*Now, where is it going to happen? The vertex horizontally is x = -b/2a.*1779

*What is our b? Our b is -18 (remember, because it is ax ^{2} + bx...so if it is -18, then b is -18);*1793

*so we have negative...what is b? -18, over 2 times...what was our a? 6.*1802

*We simplify this out; so we get positive 18 (when those negatives cancel), over 12, equals 3/2.*1808

*So, our x location is at 3/2; and now, what is our y going to be at?*1814

*Well, we can figure that out by f of...plug in 3/2...equals 6(3/2) ^{2} - 18(3/2) + 5.*1821

*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.*1835

*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;*1845

*and we have 3(9) up top; 27/2 minus...what is 27 + 5? That becomes 22.*1858

*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.*1866

*So, that is our y location; so that means that the point where we have our minimum is point (3/2,-17/2).*1886

*And there is the point of our global minimum.*1896

*Great; all right, the final one: this one is a big word problem, but it is a really great problem.*1900

*A Norman window has the shape of a semicircle on top of a rectangle.*1905

*If the perimeter of the window is 6 meters in total, what height (h) and width (w) will give a window with maximum area?*1910

*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.*1917

*This seems to make sense: we have a semicircle--that is a semicircle right here.*1922

*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.*1928

*It is on top of a rectangle, and we have our rectangle right here; so that seems to make sense, right?*1935

*We have a semicircle on top of a rectangle box; OK, that idea makes sense.*1940

*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.*1945

*A nice, hard-to-see yellow...we go outside: perimeter, perimeter, perimeter, perimeter, perimeter, perimeter, perimeter, perimeter, perimeter...*1952

*It is going to be all of the outside of that box, but it is not going to go across now;*1959

*it is going to also have to be the outside perimeter of the top part of the window.*1964

*We have that dashed line there to indicate that that is where we split into the semicircle.*1968

*But there is no actual material there; the perimeter is just the very outside part.*1971

*The perimeter will be the top part of that semicircle, and then three sides of our box.*1975

*The fourth side doesn't actually exist; it is just connected into the semicircle.*1980

*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.*1984

*So, if we are looking for area, we are going to need to use area; so A = area--we will define that idea.*1991

*If we also know information about the perimeter, then we will say P = perimeter.*1998

*OK, let's work these things out: area is equal to the area of this part, plus the area of our box;*2008

*so it is the area of our semicircle, plus the area of our box.*2016

*So, area of our semicircle...well, what is the area of our circle?*2019

*Area for a circle is equal to πr ^{2}; so a semicircle: if it is a half-circle,*2026

*that is going to be area = 1/2πr ^{2}, because it is half of a circle.*2034

*So, we have 1/2πr ^{2}; well, we don't have an r yet, so we had better introduce an r.*2041

*So, we will say that here is the middle; r is from the middle of our circle, out like that.*2047

*Area equals πr ^{2} divided by 2 (because, remember, it is a semicircle, so it is half of it), plus...what is the area of our box?*2053

*That is height times width.*2062

*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.*2066

*So, how does r connect to the rest of it?*2071

*Well, r...look, that is connected to our w, because it is just half of that side; so r = w/2.*2073

*We can change this formula into area = π...r ^{2} is (w/2)^{2}...over 2, plus height times width.*2083

*Now, we have managed to get rid of the radius there; so let's simplify that out a bit more.*2093

*Area = π...the w/2 gets us w ^{2}/4, over 2, plus hw.*2097

*We have this whole thing divided by 2; so area equals πw ^{2}, and we are dividing...*2107

*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.*2113

*Great; so we have that area equals πw ^{2} + hw.*2123

*All right, what about perimeter--what can we say about the perimeter?*2129

*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.*2132

*And then, what is this portion here? Right away, we see that we have two h's, plus a w, plus something else.*2139

*Circumference for a circle: if we want the perimeter of a circle, that is circumference.*2148

*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.*2153

*So, perimeter for a semicircle will be 1/2(2πr), which is just going to be πr.*2168

*We have 2h + w + (the amount of perimeter that our semicircle puts in is) πr.*2177

*Now, once again, we don't really want to have extra variables floating around.*2183

*So, we want to get rid of that; so perimeter equals 2h + w + π(w/2).*2186

*Great; at this point, we see that we have h and w; we have area.*2195

*So, if we could turn this into...we have that area is unknown; we don't really know what the maximum area is,*2201

*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.*2205

*So, it is not even something that has a fixed value, necessarily.*2210

*What about perimeter, though? Perimeter is something we do know.*2213

*Remember: perimeter is 6 meters; so we can plug in 6 = 2h + w + π(w/2).*2216

*Now, it seems like we have more w's than we have h's, in both our area and perimeter stuff.*2225

*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.*2232

*We will move everything over, and we have 6 - w - π(w/2) = 2h.*2242

*We divide by 2 on both sides; we get 3 - w/2 - π(w/4) = h.*2251

*And now, let's just pull those things together, so it will be easier to plug in later, into our one over here,*2259

*because we want to swap out the h here for it.*2264

*3 - 2w/4 + πw...now, that part might be a little confusing; but notice that this has minus, and this has minus;*2267

*so when we combine them together, they are just one minus, because they are actually working together before they do their subtraction.*2279

*It equals h; and we can even pull out the w onto the outside; so we get 3 - (2 + π)/4 times w = h.*2284

*Great; now we have an expression about area, and we have an expression about h and w.*2296

*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.*2304

*And we will have area equals...just stuff involving w; and since it is just stuff involving w, we will have just one variable.*2312

*Maybe we can figure things out; maybe it looks like something--maybe it looks like a quadratic,*2319

*since, after all, this was all about how quadratics work.*2323

*So (student logic), it seems pretty likely that it is going to end up looking like a quadratic.*2326

*And we can apply the knowledge that we just learned.*2329

*So, the maximum area is at what h and w?*2332

*We have area = πw ^{2}/8 + wh; and h = 3 - (2 + π)/4 times w.*2335

*OK, great; so at this point, we take h here, and we plug it in over here.*2344

*We have area equals πw ^{2}/8 (that is still the same thing), plus w times h,*2349

*so w times...what is h?...3 - (2 + π)/4 times w.*2358

*We expand that out; our area is equal to πw ^{2}/8, plus w times [3 - ((2 + π)/4)w].*2366

*So, we get 3w minus...let's keep it as (2 + π)/4, and it just combines with that other w; we have w ^{2}.*2375

*So now, we have a w ^{2} here and a w^{2} here; so let's make them talk to each other.*2382

*We get πw ^{2}/8; let's actually pull that down--we will make it as*2386

*(π/8)w ^{2} - [(2 + π)/4]w^{2} + 3w.*2396

*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.*2407

*So, it is going to be...to become an 8, we are going to multiply by 2 here, to keep it the same.*2416

*2 times (2 + π) becomes 4 + 2π; so we have [π - (4 + 2π)]w ^{2} + 3w.*2422

*We simplify this out: π - 4...so the -4 will come through, and -2π will become -π, over 8, w ^{2} + 3w.*2432

*Look, if area equals this, then this right here--this whole thing--is a quadratic.*2444

*If it is a quadratic, then we can use the stuff that we know about where vertices show up.*2454

*Where do vertices show up on the parabola?*2460

*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.*2462

*Sure enough, we have a negative here and a negative here.*2469

*And since that is on the w ^{2}, that means we can pull out the negative, and our first a is going to be -(4 + π)/8.*2473

*That whole thing ends up being a negative number; so sure enough, it is going to have a vertex at the top,*2481

*so it will have a maximum area out of this--it is cupping down.*2487

*At this point, we have figured out--remember--the vertex is at our horizontal location (in this case,*2492

*horizontal would be just our w); vertex is going to be at w = -b/2a.*2501

*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...*2508

*what is a? a is this whole thing, (-4 - π)/8.*2517

*This is a little bit confusing; but we have a fraction over a fraction, so if it is like this,*2526

*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.*2529

*So, we will get -24 - 4 - π we see that there are negatives everywhere, so we can cancel out all of our negatives.*2538

*Multiply the top and the bottom by -1; we get positive 24, over (4 + π).*2546

*The maximum is going to happen at 24/(4 + π); there is our...*2552

*Oh, sorry: one thing that I just realized--I made one tiny mistake: it is 2a; so we have a 2 here.*2562

*So, it is still a 2 up front; so it is not 4 + π.*2568

*It is always important to think about what you are doing.*2576

*24/2(4 + π) cancels to be 12/(4 + π).*2578

*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.*2587

*All we are looking for is where the vertex is; the only thing is that it is couched inside of a word problem.*2594

*So, we just have to be carefully thinking: how do we build equations?*2599

*Once we have our equations built, how do we put them together?*2601

*How do we get this to look like something where we can apply what we just learned in this lesson?*2604

*How can I apply this stuff about quadratic properties?*2608

*So, we find out that the maximum occurs when w is equal to 12/(4 + π).*2611

*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.*2616

*So, h is going to be at its maximum, as well, since it is h and w.*2627

*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.*2632

*We get that h is equal to 3 - [3(2 + π)] over (4 + π).*2647

*We want to put them over common denominators, so we can get the two pieces talking to each other.*2656

*We have 3(4 + π)/(4 + π); and then it is going to be minus 3(2 + π), so - 6 - 3π.*2659

*We have 12 + 3π, minus 6 - 3π, all over 4 + π.*2672

*So finally, our h is going to be...the -3π's cancel each other out; for 12 - 6, we get 6, over 4 + π.*2681

*And that is our value for what our maximum h will be with our maximum width.*2691

*So, the maximum area will occur when our width is 12/(4 + π), and our h is 6/(4 + π).*2695

*They will both be in units of meters, because meters is what we started with for our perimeter.*2702

*All right, I hope you have a sense of how quadratics work, what their shape is, and this idea of the vertex,*2706

*and that being where maximum and minimum are located.*2711

*Now, remember: you just have to remember that the horizontal location for maximum or minimum--*2713

*the horizontal location for the vertex--is going to occur at -b/2a, when we have it in that standard form of ax ^{2} + bx + c.*2717

*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.*2725

*All right, we will see you at Educator.com later--goodbye!*2732

1 answer

Last reply by: Professor Selhorst-Jones

Wed Mar 25, 2015 11:26 PM

Post by thelma clarke on March 25, 2015

why is this video skipping and going backwards and repeating I have been watching for over an hour and it has not complete because of all the problems with the vidoe