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Lecture Comments (2)

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

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
    a ·(x−h)2 + k.
    This form shows all the transformations that have been applied to the fundamental square function of x2.
  • 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) = ax2+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


f(x) = −2x2 + 16x − 25
Put the above function in the form f(x) = a(x−h)2 + k, then identify what a, h, and k are.
  • We put the original function into this special form by completing the square (which was taught in the previous lesson). First, we need to pull out whatever coefficient is on the x2 from all the variables:
    f(x) = −2 (x2 −8x) − 25
  • Next we need to complete the square by adding in ([(−8)/2])2 = 16. However, because we can't really add this to either side of the equation, we need to introduce it by effectively changing nothing. We add 0 = 16−16:
    f(x) = −2 (x2 −8x+16−16) − 25
  • Use distribution to pull the −16 out and then complete the square:
    f(x) = −2 (x2 −8x+16)+32 − 25     ⇒     −2 (x − 4)2 + 7
  • We now have f(x) = −2 (x − 4)2 + 7. We're matching it to the form of f(x) = a(x−h)2 + k, so we must keep that in mind while identifying a, h, and k.
f(x) = −2 (x − 4)2 + 7        a = −2, h = 4, k = 7

g(t) = 9t2 + 42t +38
Put the above function in the form g(t) = a(t−h)2 + k, then identify what a, h, and k are.
  • We put the original function into this special form by completing the square (which was taught in the previous lesson). First, we need to pull out whatever coefficient is on the t2 from all the variables:
    g(t) = 9
    t2 + 14

    3
    t
    2

     
    + 38
  • Next we need to complete the square by adding in ([1/2] ·[14/3])2 = [49/9]. However, because we can't really add this to either side of the equation, we need to introduce it by effectively changing nothing. We add 0 = [49/9] − [49/9]:
    g(t) = 9
    t2 + 14

    3
    t+ 49

    9
    49

    9

    2

     
    + 38
  • Use distribution to pull the − [49/9] out and then complete the square:
    g(t) = 9
    t2 + 14

    3
    t+ 49

    9

    2

     
    −49 + 38     ⇒     9
    t + 7

    3

    2

     
    −11
  • We now have g(t) = 9(t + [7/3])2 −11. We're matching it to the form of g(t) = a(t−h)2 + k, so we must keep that in mind while identifying a, h, and k.
g(t) = 9(t + [7/3])2 −11        a = 9, h = − [7/3], k = −11
Give a function for the parabola graphed below.
  • Any parabola can be put in the form f(x) = a(x−h)2 + k, where (h,k) is the location of the vertex and a is a constant number representing how "squeezed" the parabola is and whether or not it is flipped. [Remember, the vertex is where the parabola "turns".]
  • Looking at the graph of the parabola, we see that the vertex occurs at (3, −8), so we have h=3 and k = −8.
  • We can plug these into the function we're building to get
    f(x) = a(x−3)2 − 8.
    Furthermore, we know that (0,  10) is another point on the parabola, so that must come out of the function: f(0) = 10.
  • We can use this to create an equation and solve for a:
    10 = a(0−3)2 − 8
  • Once we find out that a=2, we plug that into the function we're building and we have finished creating a function that describes the parabola.
f(x) = 2(x−3)2 − 8
Give a function for the parabola graphed below.
  • Any parabola can be put in the form f(x) = a(x−h)2 + k, where (h,k) is the location of the vertex and a is a constant number representing how "squeezed" the parabola is and whether or not it is flipped. [Remember, the vertex is where the parabola "turns".]
  • Looking at the graph of the parabola, we see that the vertex occurs at (−4, 4), so we have h=−4 and k = 4.
  • We can plug these into the function we're building to get
    f(x) = a(x+4)2 +4.
    Furthermore, we know that (1,  −6) is another point on the parabola, so that must come out of the function: f(1) = −6.
  • We can use this to create an equation and solve for a:
    −6 = a(1+4)2 +4
  • Once we find out that a=−[2/5], we plug that into the function we're building and we have finished creating a function that describes the parabola.
f(x) = −[2/5](x+4)2+4
Consider a parabola that has a vertex at (1, −4) and that passes through the point (5,  12). What are the x-intercepts of this parabola?
  • Remember, x-intercept is just another word for root/zero. That means we are looking to find the roots of this parabola. But to do that, we first need a function that describes this parabola.
  • We can create a function for the parabola from the vertex and a point on the parabola (the same as we did in previous problems). Any parabola f(x) = a(x−h)2 + k, so we just need to find a, h, and k.
  • The vertex is at (1, −4), so we have h=1 and k=−4. Plugging those in, we have f(x) = a(x−1)2 −4. We can figure out a by plugging in the point (5, 12) then solving for a:
    12 = a(5−1)2 −4
  • We find that a=1, so we have f(x) = (x−1)2 −4. Now we want to find the roots, so we set f(x) = 0. Normally, we'd find roots by factoring or using the quadratic formula. However, things are even easier this time because of the format it's in:
    0 = (x−1)2 −4     ⇒     4 = (x−1)2     ⇒     ±2 = x−1     ⇒     1 ±2 = x
x = 1 ±2 or, equivalently, x=−1 and x = 3
Consider a parabola that has a vertex at (−8, 45) and that passes through the point (−12,  −35). What are the x-intercepts of this parabola?
  • Remember, x-intercept is just another word for root/zero. That means we are looking to find the roots of this parabola. But to do that, we first need a function that describes this parabola.
  • We can create a function for the parabola from the vertex and a point on the parabola. Any parabola f(x) = a(x−h)2 + k, so we just need to find a, h, and k.
  • The vertex is at (−8, 45), so we have h=−8 and k=45. Plugging those in, we have f(x) = a(x+8)2 +45. We can figure out a by plugging in the point (−12, −35) then solving for a:
    −35 = a(−12+8)2 +45
  • We find that a=−5, so we have f(x) = −5(x+8)2 +45. Now we want to find the roots, so we set f(x) = 0. Normally, we'd find roots by factoring or using the quadratic formula. However, things are even easier this time because of the format it's in:
    0 = −5(x+8)2 +45     ⇒     9 = (x+8)2     ⇒     ±3 = x+8     ⇒     −8 ±3 = x
x = −8 ±3 or, equivalently, x=−11 and x = −5

f(x) = x2 −8x + 13
Find the vertex of the above parabola. Identify whether the vertex is the maximum or the minimum. Give the parabola's axis of symmetry.
  • For a parabola in the form f(x) = ax2 + bx+c, the x-location of the vertex occurs at x = [(−b)/2a].
  • Plugging in, we have that the vertex's x-location is x=4. To find the y-location, we can now plug in x=4:
    f(4) = 42 − 8·4 + 13   =  −3
    Thus, the vertex occurs at (4, −3).
  • Because the vertex of a parabola occurs where it "turns", that location is always the maximum or the minimum depending on the direction the parabola is pointed. This parabola is pointed up because a is a positive number (a=1), so the vertex must be a minimum. [Make a cup with your hand in the appropriate direction if you don't see this.]
  • The axis of symmetry is a vertical line that runs through the vertex. Thus the equation for the line is x = 4. [Remember, a vertical line is given simply by x=horizontal location of line.]
Vertex: (4,  −3)     Minimum     Axis of symmetry: x=4

f(x) = −3x2 − 14x −10
Find the vertex of the above parabola. Identify whether the vertex is the maximum or the minimum. Give the parabola's axis of symmetry.
  • For a parabola in the form f(x) = ax2 + bx+c, the x-location of the vertex occurs at x = [(−b)/2a].
  • Plugging in, we have that the vertex's x-location is x=[(−(−14))/(2(−3))] = − [7/3]. To find the y-location, we can now plug in x=−[7/3]:
    f
    7

    3

    = −3
    7

    3

    2

     
    −14
    7

    3

    −10   =   19

    3
    Thus, the vertex occurs at ( −[7/3], [19/3] ).
  • Because the vertex of a parabola occurs where it "turns", that location is always the maximum or the minimum depending on the direction the parabola is pointed. This parabola is pointed down because a is a negative number (a=−3), so the vertex must be a maximum. [Make a cup with your hand in the appropriate direction if you don't see this.]
  • The axis of symmetry is a vertical line that runs through the vertex. Thus the equation for the line is x = −[7/3]. [Remember, a vertical line is given simply by x=horizontal location of line.]
Vertex: ( −[7/3], [19/3] )     Maximum     Axis of symmetry: x=−[7/3]
Find two positive numbers whose sum is 94 and whose product is the maximum possible.
  • Begin by understanding what the question is asking for. We need to find a combination of two numbers that add up to 94 and whose product is the largest possible. For example, 1 & 93 add up to 94, and their product is 94. However 2 & 92 also add up to 94, but their product is 184. So we're looking for a number pair that makes the largest possible product.
  • Set up variables to describe things with equations. Let x=the first number  and  y=the second number. Thus we have
    x+y = 94
  • The quantity we care about is the product of x and y. Let's call this P:
    P = x·y
    As it stands, we can't figure out what values would maximize P. However, we can substitute to replace y. We have y = 94−x, which would give us
    P = x(94−x)
    Notice that the above is a parabola (it's easier to see once we expand it), so now we can find what x location maximizes the parabola (where the vertex is).
  • Expanding the above, we have
    P = −x2 + 94x.
    The maximum of this parabola (and thus the maximum product) will occur at the parabola's vertex. We can find the x-location with x = [(−b)/2a]. Plugging in, we have that the maximum occurs at
    x = −94

    2(−1)
    = 47.
  • If the maximum product will occur when x=47, then we can use that to figure out what y is: y = 94 −x = 94 −47 = 47.
The two numbers are 47 and 47.
Ada builds computers for a living. Building computers has some start-up cost, so the cost of building n computers in a year is C(n) = 175n + 5000.
She is able to sell all the computers that she builds, but she has to set the price based on how many she builds. Since more of a product in a market decreases the demand, she is forced to lower the price per unit based on how many she builds. If she builds n computers in a year, she can sell each computer for 600 − 0.5n dollars.
How many computers should she build in a year to maximize her net profit? What will her net profit be if she builds that many?
  • Begin by understanding the problem. It costs Ada a certain amount of money (C) to build n computers in a year. She sells them, but the more computers she builds, the less she can sell each computer for. We want to figure out the right number of computers to build to maximize her net profit.
  • Net profit is the sales revenue (how much money people pay her) minus the cost (how much money it took to build the computers). We already have a function for cost: C(n) = 175n + 5000. However, we don't yet have a function for sales revenue. Let's use S to denote sales revenue. If she makes 600 − 0.5n per computer she sells, then we can multiply that number by the total number of computers sold (n) to find out the total sales revenue:
    S(n) = n(600 − 0.5n)
  • Let P represent net profit. Since profit is sales minus cost, we can create the function
    P(n) = S(n) − C(n).
    Plugging in, we can find the profit for making and selling n computers:
    P(n) = n(600−0.5n) − [175n + 5000]
    Expand and simplify the above to obtain
    P(n) = −0.5n2 + 425n − 5000
  • We're looking to maximize P. Since P is a parabola, we want to find where the maximum for that parabola occurs (we know it has to be a maximum because the parabola is pointed down). This will occur at
    n = −b

    2a
    = −425

    2(−0.5
    = 425
    Thus, Ada makes her maximum profit when she sells n=425 computers.
  • To find out what the value of that profit is, simply plug n=425 into our profit function P(n):
    P(425) = −0.5(425)2 + 425 ·425 − 5000 = 85312.5
Ada's profit is maximized when she builds and sells 425 computers. Doing so, she will make a net profit of $85 312.50.

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

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 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

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 information0023

than that general form, ax2 + 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 them0040

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 line0134

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, x2.0260

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) = x2, 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 x2, 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, ax2 + 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, ax2 + bx + c, into this new alternate form?0702

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.0710

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.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 ax2 + 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 ax2 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(x2 + 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 b2/4a2.0789

So, that is what we are plugging in right here: b2/4a2.0799

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 b2/4a2 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 -b2/4a, 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, b2/4a2.0897

So, that ends up being the a here and the a2...the a2 cancels to just a, and that cancels out a right there.0911

So, we have b2/4a; 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 ax2 + 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 x2, x(b/2a) + (b/2a)x...so that becomes b/a(x), because they get added together twice; + b2/4a2.0946

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

-b2/4a + 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 (-b2 + 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 ax2 + 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 ax2 + bx + c, if we start them that way?1013

Well, we just figured out that ax2 + 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 +(-b2 + 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 -b2 + 4ac, over 4a; so that gives us our vertex.1048

Now, I think it is pretty difficult to remember -b2 + 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) = -3x2 - 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

-3x2 - 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 -3x2, 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 -3x2: -3x2,1373

divided by -3, becomes just positive x2; 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 (x2 + 8x); -3x2 - 24x; great; that checks out, just like what we originally had.1398

So, -3(x2 + 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(x2 + 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 42, which is equal to 16.1445

So, we want to add 16 inside; so we have -3(x2 + 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

x2 + 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: x2; 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, ax2 + 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 it1713

and work through it, and you would also be able to get into that form, ax2 + 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) = 6x2 - 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) = 6x2, then notice a: ax2 + 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 ax2 + 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 πr2; so a semicircle: if it is a half-circle,2026

that is going to be area = 1/2πr2, because it is half of a circle.2034

So, we have 1/2πr2; 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 πr2 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...r2 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 w2/4, over 2, plus hw.2097

We have this whole thing divided by 2; so area equals πw2, 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 πw2 + 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 = πw2/8 + 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 πw2/8 (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 πw2/8, 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 w2.2375

So now, we have a w2 here and a w2 here; so let's make them talk to each other.2382

We get πw2/8; let's actually pull that down--we will make it as2386

(π/8)w2 - [(2 + π)/4]w2 + 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)]w2 + 3w.2422

We simplify this out: π - 4...so the -4 will come through, and -2π will become -π, over 8, w2 + 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 w2, 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 ax2 + 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