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Lecture Comments (3)

2 answers

Last reply by: Emily Engle
Fri Oct 11, 2013 7:17 AM

Post by Emily Engle on October 9, 2013

How did you figure out the original graph in ex. 2 ? (x^2=y)

Graphs of Quadratics

  • The graph of a quadratic equation forms a “u” shape known as a parabola.
  • A parabola is symmetric along the axis of symmetry. Where this axis meets the parabola is a point known as the vertex. You can also think of the vertex as the maximum, or minimum point of the parabola.
  • By writing a quadratic in the form y = a(x – h)2 + k, we can quickly determine the following information
    • The location of the vertex. It is located at the point (h, k)
    • The direction the parabola is facing. It is facing up if a is positive and down if a is negative
    • How wide the parabola is when compared to x2. It is narrower if |a| > 1 and wider if |a| < 1
  • If the quadratic is written in the form y = ax2 + bx + c, we can quickly determind the following information
    • The location of the vertex. It is located at the point (-b/(2a), f(-b/(2a) )
    • The direction the parabola is facing. It is facing up if a is positive and down if a is negative
    • How wide the parabola is when compared to x2. It is narrower if |a| > 1 and wider if |a| < 1
    • To location of the y-intercept. It is located at the point (0, c)

Graphs of Quadratics

Find the maximum/minimum point of the function y = - x2 + 4x + 5
  • Find the vertex using x = − [b/2a]x = − ( [4/(2( − 1))] ) = 2y = − ( 2 )2 + 4( 2 ) + 5 = 9
  • Determine which direction the parabola opens
    a < 0 opens downwards
  • Consider vertex and direction to determine maximum/minimum
(2,9) is a maximum
Find the maximum/minimum point of the function y = x2 − 2x − 6
  • Find the vertex using x = − [b/2a]x = − ( [( − 2)/2(1)] ) = 1y = ( 1 )2 − 2( 1 ) − 6 = − 7
  • Determine which direction the parabola opens
    a > 0 opens upwards
  • Consider vertex and direction to determine maximum/minimum
(1, - 7) is a minimum
Find the maximum/minimum point of the function y = [(x2)/4] + x
  • Find the vertex using x = − [b/2a]x = − ( [1/(2([1/4]))] ) = − 2y = [(( − 2 )2)/4] + ( − 2 ) = − 1
  • Determine which direction the parabola opens
    a > 0 opens upwards
  • Consider vertex and direction to determine maximum/minimum
( - 2, - 1) is a minimum
Find the maximum/minimum point of the function y = − 5x2 + 10x + 4
  • Find the vertex using x = − [b/2a]x = − ( [10/(2( − 5))] ) = 1y = − 5( 1 )2 + 10( 1 ) + 4 = 9
  • Determine which direction the parabola opens
    a < 0opens downwards
  • Consider vertex and direction to determine maximum/minimum
(1,9) is a maximum
Find the maximum/minimum point of the function y = [( − x2)/20] + 6
  • Find the vertex using x = − [b/2a]x = − ( [0/(2( − [1/20]))] ) = 0y = [( − ( 0 )2)/20] + 6 = 6
  • Determine which direction the parabola opens
    a < 0opens downwards
  • Consider vertex and direction to determine maximum/minimum
(0,6) is a maximum
Graph y = x2 + 4x + 4
  • Find the vertex using x = − [b/2a]x = − ( [4/2(1)] ) = − 2y = ( − 2 )2 + 4( − 2 ) + 4 = 0
  • Make a table of points
    x
    line
    − 2
    − 5
    − 4
    0
    1
    y
    line
    0
    9
    4
    4
    9
Graph y = − x2 − 6x
  • Find the vertex using x = − [b/2a]x = − ( [( − 6)/(2( − 1))] ) = − 3y = − ( − 3 )2 − 6( − 3 ) = 9
  • Make a table of points
    x
    line
    − 5
    − 4
    − 3
    − 2
    − 1
    y
    line
    5
    8
    9
    8
    5
Graph y = x2 − 10x + 18
  • Find the vertex using x = − [b/2a]x = − ( [( − 10)/2(1)] ) = 5y = ( 5 )2 − 10( 5 ) + 18 = − 7
  • Make a table of points
    x
    line
    3
    4
    5
    6
    7
    y
    line
    − 3
    − 6
    − 7
    − 6
    − 3
Solve y = x2 + 4x utilizing axis of symmetry
  • Find the vertex and line of symmetry using x = − [b/2a]x = − ( [4/2(1)] ) = − 2y = ( − 2 )2 + 4( − 2 ) = − 4
  • Make a table of points
    x
    line
    − 2
    − 1
    0
    1
    2
    y
    line
    − 4
    − 3
    0
    5
    12
Graph to find x - intercepts

x = − 4,0
Solve y = − x2 + 16x − 55 utilizing axis of symmetry
  • Find the vertex and line of symmetry using x = − [b/2a]x = − ( [16/(2( − 1))] ) = 8y = − ( 8 )2 + 16( 8 ) − 55 = 9
  • Make a table of points
    x
    line
    4
    5
    6
    7
    8
    y
    line
    − 7
    0
    5
    8
    9
Graph to find x - intercepts

x = 5,11

*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

Graphs of Quadratics

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
  • Objectives 0:06
  • Graphs of Quadratics 0:39
    • Axis of Symmetry
    • Vertex
    • Transformations
    • Graphing in Quadratic Standard Form
  • Example 1 5:06
  • Example 2 6:02
  • Example 3 9:07
  • Graphs of Quadratics Cont. 11:26
    • Completing the Square
    • Vertex Shortcut
  • Example 4 13:49
  • Example 5 17:25
  • Example 6 20:07
  • Example 7 23:43

Transcription: Graphs of Quadratics

Welcome back to www.educator.com.0000

In this lesson we are going to take a look at the graphs of quadratics.0002

We will first start off by looking at the basic shape of a quadratic and how it forms the shape known as the parabola.0009

We will pick up some new vocabulary such as the axis of symmetry and the vertex.0015

We will look at a few quadratics that are not written in standard form. 0021

This will lead to a new technique that we have to use in order to apply out their graph.0025

We have seen this technique once before and this is known as completing the square, 0031

but we will also use what is known as the vertex formula to do the same thing.0034

When you are looking at the shape of a quadratic it forms a parabola.0042

One way you can see this is it take the nice basic shape of x2 and plot out lots of different points on a graph.0048

Maybe choose some values like – 2, - 1, 0, 1 and 2, just to see what you get.0061

All of these guys are being squared.0067

-2 would give you 4, 12 would be 1, 02 would be 1 and 12 will be 1.0070

22 will be 4, and it will keep going in both directions.0077

Let us plot those out. 0082

I’m not sure if you can see that it forms a nice parabola. 0088

All other quadratics will have this parabola shaped to them.0097

Here are some features that you want to know about these parabolas.0103

One, they are only symmetric to align known as the axis of symmetry and runs right down the middle of that parabola.0107

Here are two examples that I have the axis of symmetry marked out.0116

Whatever that axis of symmetry intersects with the actual graph that point is known as the vertex0127

and you can see that on my little diagrams right here and right here.0134

Let us call those the vertex.0140

The reason why we want to know about the vertex is it will help us create a nice shortcut method for graphing these parabolas. 0146

That way we would not necessarily have to go back to a large table of values every single time we want to graph them.0153

Think of this kind of in the context of when we are graphing out lines, when we are applying a whole bunch of points for those lines,0159

we simply had to find a y intercept and a slope.0165

With parabolas we do not have a notion of slope, but knowing where the vertex is can often help us plot out the rest of it. 0169

We can use techniques known as transformations in order to figure out where the graph is.0181

This assumes that your quadratic is written in the following form of something like y= a × x - h2 + k.0187

I did not put in the function notation here but we could also imagine that being a y, either way will work out just fine. 0196

If your quadratic is written in this very special form, there are lots of different things that it will tell you.0203

The (a) value out front will tell you whether it is been reflected over the x axis or if it had any type of vertical stretches.0210

If (a) is negative then your parabola will be facing down.0220

If it is positive your parabola will be facing up. 0226

It is how we can also determine how wide or narrow to make that parabola. 0231

If the absolute value of (a) is say greater than 1, then it will be a narrow parabola.0237

If the absolute value of a is less than 1, then you know it will be fairly wide.0248

It gives you a little bit about its shape.0259

The h and k are probably two of the most important things you can get rid of here and they help you determine where that vertex is.0262

h and k together would be the point of that vertex. 0273

What h is helping you figure out is how far left or right the parabola ended up moving from its original spot said 0, 0.0277

The k helps you figure out how far up or down that parabola had to move from its original spot.0290

We will look at the h and k more as finding a vertex and then use the value of h to determine how wide or narrow it should be.0298

Let us see if we can identify some of these key things with the following problem. 0310

First of all where is this axis of symmetry?0314

We will imagine a line straight down the middle of it, if we are to fold it up along this line it should match up with itself, axis of symmetry. 0318

Now that we have that we can see where it intersects the graph.0335

We will say that this is the vertex.0341

We will go ahead and say what that location is.0345

It is that -4, 2.0347

Notice how this parabola is facing up, in the actual equation of it the (a) value should be positive.0350

Let us see if we can take a problem written in the special form and put it onto our graph.0364

I’m going to write down the general form, just so we can reference it.0372

It is y= ax - h2 + k.0376

The first thing you want to know is where is that vertex?0385

It is coming from looking at the h in the k.0388

Be very careful notice how in the formula here, it has a list of that as -h.0393

What we want to think of is if we had to rewrite this as x -, 0401

what would that value need to be to make it match up with that 4 we have in the original.0407

It would have had to have been -4 and that will help us figure out the vertex a little bit better. 0413

This tells us our vertex is at -4, 5.0421

Let us make that our first point on the graph. 0432

-1, 2, 3, 4 and 1, 2, 3, 4, 5 right there 0435

We can also figure out a little bit more information about it. 0442

If we look at the (a) value out front, it is negative.0446

We know that our parabola should be facing down.0455

Do you know if it should be wide? Or should be narrow?0460

What else can we figure out about this thing?0463

If we look at the absolute value of -2, we get 2 and that is greater than 1.0466

It is going to be more narrow than the standard parabola of y = x2.0473

Let us put this in comparison. 0481

Suppose I were to graph a parabola facing down that was related to x2, that would be these points right here.0484

I want to write this light because that is not the parabola we want.0500

Our parabola is going to be narrower than this one. 0503

In fact the values will be twice as much.0507

2, 3, 4, 5, 6, 7, 8.0514

Ours will look more like this, narrower than the normal x2.0521

A bit of a rough sketch we get an idea what is going on here.0530

If you want some more accurate things to put in here what you could do is plot out a few points or even just to add a few additional ones.0534

Let us try some more.0545

Let us see if we can work backwards by starting with the graph of a quadratic and see if we can make its equation.0549

Like before, one of the first important points we want to know is where is the location of that vertex?0557

That one is located at 4, 1.0563

Now that we have that I know what at least my h and k values are.0570

Y = ax -42 + 1.0577

It looks like the only thing I do not know right now is the value of a, and because the problems facing up, this should be a positive value.0586

It looks like it might be a little wider than our normal parabola of x2 so I'm thinking maybe it is a fraction of some sort.0600

To be able to figure this out, I’m going to take another point on the parabola that I know it is on there and substitute this value into my equation.0607

Then I will be able to solve that a directly.0615

Y is 2, x is 2.0618

You can see the only unknown we have in here is that a value.0627

2 - 4 would be -22.0633

Subtracting one from both sides 1 and -22 would give us 4.0640

We almost have that a value let us divide both sides by 4.0653

Now we can write down the entire equation. 0665

This graph comes from ¼ (x - 42 + 1).0668

That vertex plays a key part in getting a connection between the graph and the equation. 0679

Now, unfortunately, a lot of quadratic equations and functions are not written in that nice standard form. 0688

In fact, we might end up with something that is written more like this.0696

When it is written like that, how are we supposed to figure out where that vertex is or which direction it is facing?0702

There are a couple of things you could do. 0708

One is you directly take that and end up rewriting it into that nice standard form the a x - h2 + k + 4.0711

The way you accomplish that is you will go through the process of completing the square. 0721

Once you have it in your nice standard form then you can go ahead and identify your vertex, which way it is going, is it going be wider.0727

You will know everything you need from there.0733

If you recall, completing the square can be quite a clunky process with lots of steps involved.0738

Fortunately, just like the quadratic formula, there is a way around. 0745

Maybe not having to use completing a square and still plot out quadratics that look like this, like a x2 + bx + c.0751

You can package up a lot of these steps of completing the square and get what is known as your vertex formula.0760

It is this one right here. 0768

It does look a little bit odd, but here is how you can interpret it. 0770

This first value has given you a small formula so you can figure out the x coordinate of your vertex.0775

The second part of this, it has that –b/2a and what is that trying to tell you is that you should take the value you found for your x 0784

and you should put it back into your quadratic.0792

You should evaluate it at that value. 0795

One other thing to note, the vertex can sometimes be thought of as the minimum or the maximum of your quadratic.0801

Here it is a minimum point and here it is a maximum point.0812

We will use this vertex formula to help us out if our equation is written in this form.0818

Let us use completing the square to see how you could potentially use that as well to figure out what your vertex is.0832

In this one I have f(x) = 4x2 - 4x + 21 and in the method of completing the square first I need to organize things.0839

I will get my variables to one side.0848

I have f(x) and I’m going to subtract 21 from both sides.0851

That looks pretty good.0860

We need to make sure that the coefficient in front of x2 is 1.0862

Now it is 4, let us divide everything by 4, looking pretty good. 0866

We need to go ahead and find the new number that we need to add to both sides to factor nicely.0886

We need to check our term in front of x in order to do that.0893

-1 ÷ 2 and then squared.0897

That would equal to ¼.0902

We need to add ¼ to both sides so that this will go ahead and factor.0904

On the right side there, it will factor nicely.0923

In fact this factors into x – ½2.0928

You can see that we are starting to build that similar form on the right side.0933

To see all of our other terms we need to go ahead and clean this up a bit and get that f(x) all by itself again. 0942

I’m going to go ahead and combine the 21 and the 1 since they do have the same denominator down here. 0952

That will be f(x) -20 ÷ 4.0960

Let us multiply both sides by 4.0972

Let us go ahead and add 20.0983

That was quite a bit of work, but at least now we can see what is going on here.0993

These two values here, we can get rid of our vertex.0999

I'm at ½ , 20.1006

For my (a) value out here we can see that it should be facing up and the absolute value of 4 is larger than 1, it is going to be a narrow parabola.1013

Notice how we could have also figured out lots of information about our a value from the original equation out here.1026

This value here is the same a value as this one here.1034

You could get rid of that information at the very beginning.1040

Let us see if we can go ahead and graph another quadratic.1047

This one is f(x) = - x2 + 2x +5.1052

If we are going to graph this one, there are few things that we can do.1057

One, we could go through the process of completing the square, so we can figure out where its vertex is.1062

Or since it is in this form we might as well just use the vertex formula and get it directly.1066

That formula is –b ÷ 2a and once we find it we will put it back into the original problem.1076

Negative the value b is 2 ÷ 2 × a.1087

I have -2 ÷ -2 = 1.1094

This tells me that the x value of my vertex is 1.1099

We will borrow that and evaluate it back into the function.1105

What happens when we evaluate this at 1, -12 + (2 × 1) + 5.1108

12 is 1 that is -1 + 2 + 5 we have 6.1119

We have the location of the vertex.1130

It looks like it is at 1, 6.1131

Let us see what else we can say about this.1140

What we need to know about a value to see which direction it is facing and looks like it is negative.1142

I know that this one is facing down.1151

Let us see what else we have figured out about it.1162

When it is written in this form, this last number will give us the y intercept.1164

I know it will go through 5.1170

Since the absolute value of our (a) term is exactly 1, then it is exactly as wide as our parabola x2. 1175

Let us do that.1183

1, 1 will go out to 1, 2, 3, 4.1184

Now we can sketch out a nice graph of this parabola.1193

That vertex form can help out if you have your quadratic in this particular form.1200

Let us go ahead and sketch a graph of 1/3 x2 + 3x – 6.1209

This looks like another good one we should use our vertex formula on.1214

-b ÷ 2a then we will play it back in, -b ÷ 2a.1217

Negative our value of b is 3 ÷ 2 × -1/3.1229

This will be -3 ÷ -2/3.1238

Negative divided by negative will be a positive, we have 3 ÷ 2/3, which is the same as 3 × 3/2 = 9/2.1244

That will give me the x value of my vertex, something 9/2.1259

Now I have to take that and evaluate that by putting it back into the original function.1267

Take a little bit of work, so (-1/3 9/22) + (3 × 9/2) and then we will subtract 6.1272

9/22 will be 81/4 and I have 3 × 9/2 that will 27/2.1288

I can do a little bit of simplifying here.1305

3 goes into 81 27 times.1307

(-27 ÷ 4) + (27 ÷2) – 61311

That will to do a little bit of work if I'm going to get a common denominator.1319

I have to multiply by this fraction, top and bottom by 2 so 54/4.1325

We need a common denominator down here 24.1335

Slowly combining these together I have 27/4 - 24/4 = ¾ .1340

Quite a bit of work but now I know the location for sure of my vertex.1352

Let us see if we can now plot this out.1362

I need ½ , 2/2. 3/2, 4/2, 5/2, 6/2, 7/2, 8/2, and 9/2.1366

We will go up ¾ 1374

Not all the way up to 1 but just ¾ of that, right about here.1375

It looks like this particular parabola is facing down and because the 1/3 is going to be a fairly wide parabola.1382

How large that will make it?1398

I know it crosses through the y axis at -6.1400

We can put that value on there as well. 1410

I have a good idea on what this looks like.1414

There is our sketch of the quadratic.1420

Let us do one last one just to make sure that we have all of these techniques down.1425

On this one, we want to find the maximum or the minimum value of our function.1429

This is just the same as finding the vertex.1434

We will go ahead and graph it out so you can see what that vertex is.1443

-b ÷ 2a, ½ (-b) ÷ 2a.1447

We will need both of those.1453

Starting off with trying to find the x coordinate.1456

Negative the value b2 ÷ 2 × a.1462

-2 ÷ - 4 will give us ½.1469

I would need to go ahead and put this back into the original.1478

-2 × ½2 + 2× ½ + 41485

A little bit of simplifying but not too bad -2 × ¼2 + 1 + 41493

-2 × ¼ = -1/2, 1 + 4 = 5 and now I can find a common denominator and put these two together.1509

I will get 9/2 for the second half of my vertex.1522

This vertex is located at ½, 9/2.1526

Let us put that on the graph.1533

½ for x and I have 1, 2, 3, 4, 5, 6, 7, 8, 9/2.1535

There is where our vertex is located.1542

The (a) value for this one is negative, so I know that the parabola is facing down. 1545

It looks like it has a y intercept of 4.1553

It goes through that spot.1560

Be just a little bit more narrow than your average parabola because the absolute value of 2 is greater than 1.1563

Now that we sketched that out and from the sketch since it is facing down I can say that this vertex here is a maximum.1571

The maximum value of our function is that 9/2 that comes directly from the vertex which is located at ½ and 9/2.1584

A lot of different ways that you could go about actually sketching the graph of the quadratic and it is actually not so bad.1593

If it is in a nice standard form you can simply get rid of the vertex and its (a) value.1601

But if it is not in that form of then feel free to use the vertex formula.1606

Thank you for watching www.educator.com.1611