For more information, please see full course syllabus of Pre Calculus
For more information, please see full course syllabus of Pre Calculus
Properties of Quadratic Functions
 The graph of a quadratic function gives a parabola: a symmetric, cupshaped 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 x^{2}.a ·(x−h)^{2} + k.  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

 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 x^{2} from all the variables:
f(x) = −2 (x^{2} −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 (x^{2} −8x+16−16) − 25  Use distribution to pull the −16 out and then complete the square:
f(x) = −2 (x^{2} −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.

 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 t^{2} from all the variables:
g(t) = 9 ⎛
⎝t^{2} + 14 3t ⎞
⎠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 ⎛
⎝t^{2} + 14 3t+ 49 9− 49 9⎞
⎠2
+ 38  Use distribution to pull the − [49/9] out and then complete the square:
g(t) = 9 ⎛
⎝t^{2} + 14 3t+ 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.
 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
Furthermore, we know that (0, 10) is another point on the parabola, so that must come out of the function: f(0) = 10.f(x) = a(x−3)^{2} − 8.  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.
 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
Furthermore, we know that (1, −6) is another point on the parabola, so that must come out of the function: f(1) = −6.f(x) = a(x+4)^{2} +4.  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.
 Remember, xintercept 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
 Remember, xintercept 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

 For a parabola in the form f(x) = ax^{2} + bx+c, the xlocation of the vertex occurs at x = [(−b)/2a].
 Plugging in, we have that the vertex's xlocation is x=4. To find the ylocation, we can now plug in x=4:
Thus, the vertex occurs at (4, −3).f(4) = 4^{2} − 8·4 + 13 = −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.]

 For a parabola in the form f(x) = ax^{2} + bx+c, the xlocation of the vertex occurs at x = [(−b)/2a].
 Plugging in, we have that the vertex's xlocation is x=[(−(−14))/(2(−3))] = − [7/3]. To find the ylocation, we can now plug in x=−[7/3]:
Thus, the vertex occurs at ( −[7/3], [19/3] ).f ⎛
⎝− 7 3⎞
⎠= −3 ⎛
⎝− 7 3⎞
⎠2
−14 ⎛
⎝− 7 3⎞
⎠−10 = 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.]
 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:
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 usP = x·y
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).P = x(94−x)  Expanding the above, we have
The maximum of this parabola (and thus the maximum product) will occur at the parabola's vertex. We can find the xlocation with x = [(−b)/2a]. Plugging in, we have that the maximum occurs atP = −x^{2} + 94x. 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.
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
Plugging in, we can find the profit for making and selling n computers:P(n) = S(n) − C(n).
Expand and simplify the above to obtainP(n) = n(600−0.5n) − [175n + 5000] P(n) = −0.5n^{2} + 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
Thus, Ada makes her maximum profit when she sells n=425 computers.n = −b 2a= −425 2(−0.5= 425  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
*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 screencaptured 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
Precalculus with Limits Online Course
Transcription: Properties of Quadratic Functions
Hiwelcome 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, 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 thesewe 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 parabolasparabolic shapesso 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 fromthey have to do with throwing, back either in Latin or GreekI 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 parabolait still has this cup shape that is symmetric.0080
The graph of any quadratic will give us the symmetric, cupshaped 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 other0139
that each side is a mirror reflectionthat 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 downthat 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 turning0178
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 locatedwhere 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 = 0why does that give us a vertical line?0272
Well, remember: one way to think of a graph is all of the solutions that are possibleall 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 sidebyside on the same graph0339
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/flippedthat 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 relearn 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 nonstretched, normallooking 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 squareif you didn't watch the previous lesson,0733
and you are confused by what is about to happenjust 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 asince we divide the ax^{2} by a to pull it out right here0758
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 50841
we wanted this to be the same as thisthen 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 meand 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 itthat 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 middlea nice axis of symmetry.1248
And here is an incredibly minor note on grammarthis 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 thatif 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 thisyou 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 thatit 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 it1713
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 semicirclethat 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, hardtosee 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 = areawe 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 halfcircle,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 perimeterwhat 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 somethingmaybe 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 downwe will make it as2386
(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 herethis whole thingis 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 thisit is cupping down.2487
At this point, we have figured outrememberthe 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 realizedI 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 minimum2713
the horizontal location for the vertexis 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 latergoodbye!2732
1 answer
Last reply by: Professor SelhorstJones
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