### 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 = ax
^{2}+ 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

^{2}+ 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}− 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

^{2})/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}+ 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

^{2})/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

^{2}+ 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

^{2}− 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

^{2}− 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

^{2}+ 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

x = − 4,0

^{2}+ 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

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

### Algebra 1 Online Course

### 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 x ^{2} 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, 1^{2} would be 1, 0^{2} would be 1 and 1^{2} will be 1.*0070

*2 ^{2} 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 vertex*0127

*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 - h ^{2} + 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 - h ^{2} + 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 = x ^{2}.*0473

*Let us put this in comparison.*0481

*Suppose I were to graph a parabola facing down that was related to x ^{2}, 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 x ^{2}.*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 -4 ^{2} + 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 x ^{2} 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 -2 ^{2}.*0633

*Subtracting one from both sides 1 and -2 ^{2} 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 - 4 ^{2} + 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 - h ^{2} + 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 x ^{2} + 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) = 4x ^{2} - 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 x ^{2} 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 – ½ ^{2}.*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) = - x ^{2} + 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, -1 ^{2} + (2 × 1) + 5.*1108

*1 ^{2} 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 x ^{2}.*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 x ^{2} + 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 9/2 ^{2}) + (3 × 9/2) and then we will subtract 6.*1272

*9/2 ^{2} 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) – 6*1311

*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 × ½ ^{2} + 2× ½ + 4*1485

*A little bit of simplifying but not too bad -2 × ¼ ^{2} + 1 + 4*1493

*-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

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)