Algebra 2 Analyzing the Graphs of Quadratic Functions

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In today's lesson, we will be working more on quadratic equations and functions.0002

And we are going to start out by analyzing the graphs of quadratic functions.0007

Recall: in previous lessons, we discussed quadratic functions in standard form.0012

Today, I am going to introduce a new form, which is the vertex form of a quadratic function.0019

And this is a very useful form of the function, and it is given as y = a, times (x - h) squared, plus k.0024

h and k are the vertex of the parabola; and x = h is the line that is the axis of symmetry.0033

And as you will recall, the vertex is the maximum point for a downward-opening parabola;0041

and it is the minimum point for an upward-opening parabola.0049

So, by having a function in this form, we already have information about what the graph is going to look like.0053

So, let's start out with just a very simple function, y = x².0059

OK, so looking at it in this form, we have just an a that is 1, and h and k are 0.0068

So, the vertex is going to be right here; so h equals 0 and k equals 0, so the vertex is (0,0).0076

Now, drawing some points, x and y: when x is -1, y is 1; when x is 1, y is 1; when x is 2, y is 4; and when x is -2, y is also 4.0087

So, it is a pretty simple graph.0106

Now, by starting out with this graph, we can look at what happens--the significance of h and k on the shape of the graph.0114

So, we are starting out with just this graph y = x².0122

Now, let's look at a situation where y = x² + 2.0131

All right, so now, I still have an h that equals 0; but here, k equals 2; therefore, the vertex is at (0,2).0142

So, my vertex is right here; now let's look at some points, x and y.0155

When x is -1, then x² will be 1, plus 2--that gives me 3.0166

When x is 1, again, y is 3; when x is 2, y is 6; and when x is -2, that is -2 squared is 4, plus 2 is 6.0175

So, look at what happens with this graph: when x is 1, y is 3; when x is -1, y is 3; when x is 2, y is going to be up right about here at 6, and right here.0188

OK, so as you can see, the graph here is shifted starting right up 2; so, the vertex is going to be right here;0209

and when x is 1, y is 3; when x is 2...there we go.0224

So, the vertex is right here; OK, so what this is going to give me is a graph that is shifted up by 2.0231

So, you see what k does to the graph: it is the same basic graph, but it is shifted up by 2.0244

If k were to be -2, then the graph would be shifted down by 2.0253

Now, h also has an effect on the graph; so let's look at what h could do to the graph.0257

Let's take the example y = (x - 1)².0267

Here, I have h = 1, k = 0; so the vertex is at (1,0); now let's find some points.0278

The vertex is over here at (1,0): some points x and y for this function: when x is -1, then that is going to give me -1 and -1 is -2, squared is 4.0293

When x is 0, 0 minus 1 is -1; squared is going to give me 1.0314

I already have the vertex at (1,0), so let's pick 2: 2 - 1 is 1, squared is 1.0320

One more point: 3 - 1 is 2, squared is 4; that is going to give me (-1,4), (0,1), (1,0), which is the vertex, (2,1), and then finally (3,4).0328

So, I have this graph right here; this graph is y = (x - 1)².0352

You can see, what happened is: the graph here is shifted to the right by 1.0368

h shifts to the right or left; if this had been -1, the graph would have shifted over to the left.0378

So again, k is going to shift the graph either up or down; h is going to shift to the left or right.0386

And we will talk about a in a minute.0394

So again, vertex form is very useful, because it gives you a lot of information.0396

It gives you the vertex, which is (h,k), and it also gives you the axis of symmetry.0400

And recall that the axis of symmetry is the line at x = h, which bisects this parabola into two symmetrical halves.0407

For example, if I had a parabola here, a downward-facing one, and the vertex is right here, the axis of symmetry would be right here0417

at the line x = h, where the vertex is some point (h,k).0428

Now, let's talk more about the coefficient a; what the coefficient a tells you is how steep or flat the parabola is.0434

For example, if I had one parabola, and let's say it had a = 1, and it looked like this, for example:0443

and then I had a very similar parabola, but this time with a = 2; well, as it says here,0455

if the absolute value of a is greater than 1, the parabola is narrower.0463

I have a = 1; and if I have a = 2, it is going to be a narrower parabola, meaning that, for every change in x, y changes more dramatically.0479

So, it goes up more steeply--each section of the parabola.0493

Now, if you have an absolute value of a that is less than 1 (that is a fraction), the parabola is going to be wider.0497

So, a = 1/2 might look something like this; for every change in x, the change in y is less dramatic.0507

OK, so the first thing that the absolute value of a tells us is how narrow or wide the parabola is.0517

A larger number is going to be narrower; a smaller number, especially a fraction, is going to have a much flatter, wider parabola.0524

The second thing that a tells us is which direction the parabola opens.0531

All of these parabolas that I showed you opened upward, and that means that a is positive.0535

If a is greater than 0, the parabola opens upward; if a is less than 0, the parabola opens downward.0541

And here, I have drawn a pretty wide parabola, so it would have a fairly small value of a; and it would have a negative value.0553

The absolute value of a would be small, and the value of a itself would actually be negative.0561

Here, all of these have an absolute value of a that is...these two are larger; this one is smaller; but all of them have a value of a that is positive.0569

So they open upward; so if a is positive, the parabola opens upward; if a is negative, it opens downward.0585

If the absolute value is larger, you have a narrower parabola; if the absolute value of a is smaller, you have a wider, flatter parabola.0592

Between what we just learned about h and k and the vertex,0601

and what we learned here about a, you can get a lot of information0606

just by looking at the equation about the shape of the graph of the parabola0609

before you have even found points beyond the vertex.0615

OK, now since this is an important form of the equation, you need to know how to write it in vertex form.0620

And you may actually be given the quadratic equation in standard form and be asked to write it in vertex form.0627

And the way you go about that, if you are given a quadratic equation in standard form, is to complete the square.0635

And that will allow you to write the equation in vertex form.0643

Now, it says here that, if a does not equal 1, then you need to factor out a before you go on to complete the square.0647

So, we will deal with that in a minute; let's start out with a simpler case where y equals x² + 6x - 8.0655

So, I have standard form; and I am going to go ahead and write it in vertex form.0679

Let's just recall vertex form: y - a(x - h)² + k.0687

So, I am starting out with standard form; and when I complete the square, the first thing I need to do is isolate the variable terms on the right.0699

Here, I have my x variable terms, but I also have a constant here.0715

So, what I am going to do is add 8 to both sides to get this.0720

Now, I have my x variable terms isolated on the right.0724

Isolate the x; I should specify x variable terms on the right; the y stays on the left.0729

OK, next complete the square; and recall that, in order to complete the square, you need to add b²/4 to both sides.0738

So, here let's figure out what this would give me...y + 8, b is 6, so that is going to be plus 6²/4 = x² 6x + 6²/4.0763

Doing some simplification: 6² is 36, divided by 4 is simply 9.0779

OK, I have completed the square on the right; I am going to do some simplification here, because this is actually y + 15.0787

And this is not quite in vertex form, because if I look up here, vertex form would have y isolated on the left.0794

So, I am going to actually change this, and I am going to subtract 15 from both sides.0800

And this is going to give me, let's see, -8.0808

Actually, one more step: before we do that, I have completed the square; so I want to go ahead0819

and write this out closer to vertex form before I even go on.0827

And this is x² + 6x + 9, and that is a perfect square, so I am going to write it in this form.0831

Then, I am going to go ahead and move my 15 over.0838

OK, so I have written this in vertex form; and I accomplished that by first isolating the x variable terms on the right,0842

then adding b²/4 to both sides; I did that, and I ended up with this right here.0853

And I see, on the right, I have a perfect square; so I can rewrite that as (x + 3)².0861

And then, finally, this is vertex form.0867

And although vertex form has a negative here, it is fine to write it like this.0869

If I wanted to, I could have written it as x - -3, but this is actually acceptable, as well.0874

Now, it gets a little bit more complicated if the a term is not equal to 1; so let's go ahead and look at that situation.0881

If I had something like y = 2x² - 8x + 5,0895

I am first going to isolate the x variable terms on the right (that is going to give me y - 5 = 2x² - 8x),0904

and then, as it says here, I need to factor a out, since a is not 1, before I complete the square.0913

So, this is going to give me 2x² - 4x; now, I complete the square by adding b²/4 to both sides.0922

But you have to be careful on the left; let's see what happens.0932

This is going to give me x² - 4x, and then b²/4 is actually going to be (-4)²/4.0936

Now, I can't just add that to the right, because in reality, what I am adding is not just (-4)²/4.0945

But it is actually (-4)²/4, times 2; that is what I need to add to the left.0952

Otherwise, the equation won't be balanced.0958

So, let me go ahead and simplify this (-4)²/4; that equals 16/4, which equals 4.0960

Now, what I am really adding to both sides is actually 4 times 2; 4 times 2 equals 8.0967

This is 2 times 4; so that is what I need to add to this side, as well.0977

So, this is y - 5 + 8; it is going to give me y + 3 = 2 times x² - 4x + 4.0982

Now, I am almost there, because I have a perfect square, which is (x - 2)².0995

So then, I can rewrite this in vertex form.1001

And as you see, it was more complicated, because I started out with a leading coefficient that is other than 1.1008

In that case, I have to first go ahead and isolate my x variables, and then factor that out, so that this becomes 1.1016

Once I have done that, I can complete the square; but I have to be very careful, on the left,1025

that I am not just adding b²/4, but that I am adding b²/4,1029

times what I factored out, to keep the equation balanced.1034

And now, you see, I have vertex form; and here a equals 2; h equals 2; and k equals -3.1038

OK, finally, if you are given the vertex and another point on the graph, you can write the equation of the graph in vertex form.1046

For example, if I am given that the vertex equals (2,11), and I am also told that a point on the graph1055

is at (1,6), then let's look at what I have.1066

Well, I have y, because I have a point on the graph; so this is h; this is k; this is x; this is y.1071

So, I have y; I have x; I have h; and I have k.1078

The only thing that I am missing is a, but since I have everything else, I can solve for a.1083

So, I am going to go ahead and do that: y is 6; 6 equals a times...x is 1, minus h squared, plus k.1088

OK, so this gives me 6 = a, and this is -1 squared plus 11, so 6 equals a times 1, plus 11; 6 = 11 + a; therefore, a equals -5.1101

Now that I have solved for a, I can go ahead and write this in vertex form.1136

I have y = -5(x -...I know that h is 2) + k, which is 11.1141

So again, if I am given vertex form and a point on the graph, then I can write this, substituting in everything I have, and then solving for a.1154

And here, I found that a equals -5; so now, I have h, k, and a, and all I need to write an equation in vertex form is h, k, and a.1165

So, if you are either given the quadratic equation in standard form, or you are given1174

the vertex and a point on the graph, either of those is enough to rewrite the equation in vertex form.1180

OK, practicing this with some examples: here I am given a quadratic equation in standard form, and I am asked to write it in vertex form.1188

Remember that the first step is to isolate the x variable; I am going to do that by subtracting 7 from both sides.1195

OK, so I have y - 7 = x² -4x; next, I am going to complete the square.1209

And remember, you do that by adding b²/4 to both sides.1216

And here, b is -4; so this is going to be (-4)²/4 = x² - 4x + (-4)²/4.1223

Simplifying: this gives me 16/4; simplifying further, I get simply 4.1239

OK, so I have -7 + 4; that is going to give me y - 3 = x² - 4x + 4.1254

And what I have on the right is a perfect square, so I am going to rewrite that as y - 3 = (x - 2)².1261

One more step to getting this in vertex form is to move the constant over to the right.1279

And recall that vertex form is y = a(x - h)² + k.1285

So, here I have a = 1; h is 2; and k equals 3; and the vertex is at (2,3).1292

And I accomplished that by isolating the x variables on the right, completing the square1308

by making sure I added b²/4 to both sides, and then once I had that completed,1314

I could rewrite this in this form, the (x - h)² form, and then moving the constant over to the right.1322

OK, here: find an equation in vertex form for the parabola with vertex at (-4,3), and passing through the point (-2,15).1329

So, first I am looking at what I have and at what I need.1339

Vertex form: y = a(x - h)² + k.1344

So, to write in vertex form, I need a, h, and k.1348

What do I have? Well, I have the vertex; this is h and k; and I have a point--this is x and y.1354

What I am missing is a; but since I have everything else, I can solve for a.1361

I am going to substitute in here for y; that is 15; I don't know a; I know that x is -2; I know that h is -4; and k is 3.1367

So, solving: a negative and a negative is actually a positive, so this gives me 15 = a...1386

-2 plus 4 is 2, squared; so then, this is 15 =...this is actually...1396

I am down to here; I am going to subtract 3 from both sides to get 4a = 12.1416

Isolate a by dividing both sides by 4 to give me a = 3.1421

So, now that I have a, I can write in vertex form: I have y = 3 times x, minus h (is -4).1428

And I can either write this as x + 4 or x - -4, plus k, which is 3.1437

This is vertex form; and I accomplished that by recognizing that I had everything I needed except for a,1443

substituting in, solving for a, and then going back and writing this in vertex form.1450

And here, a equals 3; h equals -4; and k equals 3 (vertex form).1460

OK, again, I am asked to write the quadratic equation in vertex form.1472

But notice this time: this is with the standard form; a does not equal 1.1478

Recall that I am going to have to do an extra step before completing the square in this situation.1483

So, the first step is to isolate the x variables on the right.1489

Now, the extra step I have to take is to factor out a.1494

Now that I have done that, I can complete the square; and that is accomplished by adding b²/4 to both sides.1504

So, I need to add b²/4 to both sides.1520

And here, we have b² is actually -3, but I need to be careful,1530

because I am not just adding b²/4; I actually have factored out the 4;1548

and if I don't add that back in on the left, my equation won't be balanced.1556

So, this is actually 4 times b²/4.1562

x - 3x + (-3)²/4: now, I already have 4 times b²/4 here, so I needed to balance it by having that on the left side, as well.1566

This is going to give me y - 20 + 4, times 9/4, equals 4 times x, minus 3x, plus 9/4.1585

OK, this is going to give me y - 20; and the 4's cancel here, which is kind of convenient;1600

so that is going to give me...these will cancel; I will just get 9.1606

And -20 + 9 is -11, so this is y - 11 equals 4 times...and this actually is a perfect square, and you can check it out for yourself.1615

This equals (x - 3/2)²; this is a perfect square.1625

Now, I am almost in vertex form; I just have to move this 11 over to the right by adding 11 to both sides.1632

And this is vertex form; so again, vertex form is y = a(x - h)² + k.1640

Here, I have a = 4, h = 3/2, and k = 11; so the vertex is going to be at (3/2,11).1650

This was a little more complicated, because I did have a coefficient here, an a, that did not equal 0.1660

So, right after I isolated my x variable terms, I factored that out.1667

And then, when I complete the square, I need to make sure that I add 4 times b²/4 to the left, since that is what I am adding to the right.1672

And then, I just went on and did that; I got a perfect square on the right.1680

And this is going to end up...this is actually all x²'s, and this is going to end up giving me (x - 3/2)² as a perfect square.1685

And then, I just move the 11 over to the right.1699

OK, now find an equation in vertex form of the parabola with vertex at (2,-3) and passing through the point (4,-19).1702

OK, vertex form: y = a(x - h)² + k.1713

Now, I look at what I am given; and I am given the vertex, so that is h and k, and a point (that is x and y).1724

Since I have all of these, I can find a; I don't know what a is--that is my unknown.1733

Well, I have y = -19; I don't know a; x is 4; h is 2; and k is -3.1739

4 minus 2 gives me 2, squared, minus 3; so that is -19, equals a, times 4, minus 3;1757

-19 equals 4a - 3; so I am going to add 3 to both sides to give me -16 = 4a; therefore, a equals -4.1766

Now that I have found a, I just go back in here and write this as y = -4x, and then I have h as 2, and k is -3.1778

So, here I have vertex form; I know the vertex is (2,-3).1791

And now I know that a equals -4; so this is going to be a parabola that opens downward.1795

OK, this concludes this session of Educator.com.1802

And I will see you again next lesson.1806