### Parabolas

- Understand the geometric significance of the sign of the coefficient of the squared term in the equation of a parabola.
- Use the axis of symmetry to help you graph a parabola.
- Know the standard formula for a parabola.
- Review how to complete the square.
- If the coefficient of the squared term is not 1, then before completing the square, you must first factor this coefficient out of both the squared term and the linear term.

### Parabolas

x = y

^{2}− 4y + 1

- Write in standard form by completing the square, isolate the y on the right.
- x − 1 = y
^{2}− 4y - Add [(b
^{2})/4] on both sides - x − 1 + [(b
^{2})/4] = y^{2}− 4y + [(b^{2})/4] - x − 1 + [(( − 4)
^{2})/4] = y^{2}− 4y + [(( − 4)^{2})/4] - x − 1 + 4 = (y − 2)
^{2} - x + 3 = (y − 2)
^{2} - x = (y − 2)
^{2}− 3 - Identify the key features:

b) Vertex = (h,k) = ( − 3,2)

c) a = 1,a > 0; Parabola opens to the right

d) Axis of symmetry = k = 2 = y = 2

^{2}− 4y + 1

- Notice this is the same problem as question 1 Notice this is the same problem as question # 1
- x = (y − 2)
^{2}− 3 - Sketch the graph with the following information
- a) Horizontal Parabola
- b) Vertex = (h,k) = ( − 3,2)
- c) a = 1,a > 0 Parabola opens to the right
- d) Axis of symmetry = k = 2 = y = 2

x = − 7y

^{2}− 14y − 11

- Write in standard form by completing the square, isolate the y on the right.
- x + 11 = − 7y
^{2}− 14y - Factor out a − 7 in order to complete the square
- x + 11 = − 7(y
^{2}+ 2y) - Add [(b
^{2})/4] on both sides. When adding to the left side, multiply by − 7 - x + 11 + − 7( [(b
^{2})/4] ) = − 7(y^{2}+ 2y + [(b^{2})/4]) - x + 11 + − 7( [(2
^{2})/4] ) = − 7(y^{2}+ 2y + [(2^{2})/4]) - x + 11 − 7 = − 7(y + 1)
^{2} - x + 4 = − 7(y + 1)
^{2} - x = − 7(y + 1)
^{2}− 4 - Identify the key features:

b) Vertex = (h,k) = ( − 4, − 1)

c) a = − 7,a < 0 Parabola opens to the left

d) Axis of symmetry = k = − 4 = y = − 4

^{2}− 14y − 111

- Notice this is the same problem as question 3 Notice this is the same problem as question # 3
- x = − 7(y + 1)
^{2}− 4 - Sketch the graph with the following information
- a) Horizontal Parabola
- b) Vertex = (h,k) = ( − 4, − 1)
- c) a = − 7,a < 0 Parabola opens to the left
- d) Axis of symmetry = k = − 4 = y = − 4
- Since a is a whole number, your parabola will be thinner. Use couple of points to get the right shape.

x = 3y

^{2}− 24y + 43

- Write in standard form by completing the square, isolate the y on the right.
- x − 43 = 3y
^{2}− 24y - Factor out a 3 in order to complete the square
- x − 43 = 3(y
^{2}− 8y) - Add [(b
^{2})/4] on both sides. When adding to the left side, multiply by 3 - x − 43 + 3( [(b
^{2})/4] ) = 3(y^{2}− 8y + [(b^{2})/4]) - x − 43 + 3( [( − 8
^{2})/4] ) = 3(y^{2}− 8y + [( − 8^{2})/4]) - x − 43 + 48 = 3(y − 4)
^{2} - x + 5 = 3(y − 4)
^{2} - x = 3(y − 4)
^{2}− 5 - Identify the key features:

b) Vertex = (h,k) = ( − 5,4)

c) a = 3,a > 0 Parabola opens to the right

d) Axis of symmetry = k = 4 = y = 4

^{2}− 24y + 43

- Notice this is the same problem as question 5 Notice this is the same problem as question # 5
- x = 3(y − 4)
^{2}− 5 - Sketch the graph with the following information
- a) Horizontal Parabola
- b) Vertex = (h,k) = ( − 5,4)
- c) a = 3,a > 0 Parabola opens to the right
- d) Axis of symmetry = k = 4 = y = 4
- Since a is a whole number, your parabola will be somewhat thinner. Use couple of points to get the right shape.

x = 2y

^{2}+ 8y + 6

- Write in standard form by completing the square, isolate the y on the right.
- x − 6 = 2y
^{2}+ 8y - Factor out a 2 in order to complete the square
- x − 6 = 2(y
^{2}+ 4y) - Add [(b
^{2})/4] on both sides. When adding to the left side, multiply by 2 - x − 6 + 2( [(b
^{2})/4] ) = 2(y^{2}+ 4y + [(b^{2})/4]) - x − 6 + 2( [(4
^{2})/4] ) = 2(y^{2}+ 4y + [(4^{2})/4]) - x − 6 + 8 = 2(y + 2)
^{2} - x + 2 = 2(y + 2)
^{2} - x = 2(y + 2)
^{2}− 2 - Identify the key features:

b) Vertex = (h,k) = ( − 2, − 2)

c) a = 3,a > 0 Parabola opens to the right

d) Axis of symmetry = k = − 2 = y = − 2

^{2}+ 8y + 6

- Notice this is the same problem as question 7 Notice this is the same problem as question # 7
- x = 2(y + 2)
^{2}− 2 - Sketch the graph with the following information
- a) Horizontal Parabola
- b) Vertex = (h,k) = ( − 2, − 2)
- c)a = 3,a > 0 Parabola opens to the right
- d) Axis of symmetry = k = − 2 = y = − 2
- Since a is a whole number, your parabola will be somewhat thinner. Use couple of points to get the right shape.

- Standard form of a parabola is y = a(x − h)
^{2}+ k where vertex = (h,k) - Focus is located at (h,k + [1/4a])
- Find a given the focus ( − 7,4)
- k + [1/4a] = 4
- 5 + [1/4a] = 4
- [1/4a] = − 1
- [1/a] = − 4
- a = − [1/4]
- Write the quation given a and vertex
- y = − [1/4](x + 7)
^{2}+ 5

- Standard form of a parabola is y = a(x − h)
^{2}+ k where vertex = (h,k) - Focus is located at (h,k + [1/4a])
- Find a given the focus ( − 2,2)
- k + [1/4a] = 2
- 1 + [1/4a] = 2
- [1/4a] = 1
- [1/a] = 4
- a = [1/4]
- Write the quation given a and vertex
- y = [1/4](x + 2)
^{2}+ 1

*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

### Parabolas

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
- What is a Parabola? 0:20
- Definition of a Parabola
- Focus
- Directrix
- Axis of Symmetry
- Vertex 3:33
- Minimum or Maximum
- Standard Form 4:59
- Horizontal Parabolas
- Vertex Form
- Upward or Downward
- Example: Standard Form
- Graphing Parabolas 8:31
- Shifting
- Example: Completing the Square
- Symmetry and Translation
- Example: Graph Parabola
- Latus Rectum 17:13
- Length
- Example: Latus Rectum
- Horizontal Parabolas 18:57
- Not Functions
- Example: Horizontal Parabola
- Focus and Directrix 24:11
- Horizontal
- Example 1: Parabola Standard Form 25:12
- Example 2: Graph Parabola 30:00
- Example 3: Graph Parabola 33:13
- Example 4: Parabola Equation 37:28

### Algebra 2

### Transcription: Parabolas

*Welcome to Educator.com.*0000

*Today, we are going to talk about parabolas.*0002

*And in some earlier lectures in this series on quadratic equations, we talked about parabolas and did some graphing.*0004

*But now, we are going to go on and give a specific definition to parabolas, and learn about some other features of parabolas.*0011

*Although you have seen parabolas previously, when we graphed, we didn't form a specific definition of them.*0021

*So, the definition of a parabola is that it is the set of points in the plane whose distance from a given point,*0027

*called the focus, is equal to its distance from a given line, called the directrix.*0033

*Let's talk about that before we go on to talk about the axis of symmetry.*0041

*So, if you had a parabola (let's say right here; and we will do an upward-facing parabola), you would have some point,*0045

*which is known as the focus, and a line (I'm going to put that right about here) called the directrix.*0059

*By definition, every point on this parabola is equidistant from the focus and the directrix.*0080

*So, if I took a point right here, and I measured the distance from the focus, it would be equal to the distance from the directrix.*0085

*And this is just a very rough sketch; but these distances actually would be equal; they are theoretically equal.*0095

*Looking right here at the vertex, these distances would be equal; so that would be, say, y.*0108

*If I took some other point, say here, and I measured here to here, these two distances would be equal.*0117

*So, a couple things to note: the focus is inside the parabola; the directrix is outside.*0128

*And this is because the focus and the directrix are on the opposite sides of the vertex.*0150

*So, you could have a parabola facing downward, and then it would have a focus here and a directrix up here.*0155

*We are also going to talk, today, about parabolas that face to the left and right--horizontal parabolas.*0165

*But right now, we are going to stick with just (for this discussion) focusing on vertical ones,*0173

*the definition being that every point in the parabola equidistant between the focus and the directrix.*0180

*The axis of symmetry of the parabola passes through the focus; and it is perpendicular to the directrix.*0189

*In this case, the y-axis is the axis of symmetry; it is right here.*0195

*And you see that it passes through the focus, and it forms a right angle; it is perpendicular to the directrix.*0204

*Again, we talked about some of these concepts in earlier lectures.*0214

*But to review, vertex: the vertex of a parabola is the point at which the axis of symmetry intersects the parabola.*0216

*And it is a maximum or minimum point on the parabola, if the axis of symmetry is vertical.*0224

*If the axis of symmetry is horizontal (say we have a parabola like this, then the axis of symmetry would be horizontal),*0230

*we still have a vertex, but it is not a maximum or minimum.*0242

*And again, we are going to focus a little more on vertical parabolas right now, and then we will talk about horizontal parabolas.*0247

*So, if I have a downward-facing parabola, the vertex is here; the axis of symmetry is right here.*0254

*And this vertex is the maximum; this is as large as y gets--it is the largest value that the function attains.*0265

*If I am looking at a vertex that is upward-facing, then the axis of symmetry...we will put it right here; and the vertex is here.*0274

*In this case, the vertex is a minimum; this is the smallest value that the function will attain.*0286

*The standard form of a parabola with vertex at (h,k) is y = a(x - h) ^{2} + k.*0299

*And this is for vertical parabolas; there is a slightly different form when we are talking about horizontal parabolas.*0308

*And you might recall this form of the equation that we covered earlier on, under the lecture on quadratic equations.*0314

*And we called this the vertex form of the equation; now we are going to refer to it as standard form.*0319

*And it is a very useful form, because it tells you a lot about the parabola.*0324

*The axis of symmetry is x = h: so I know a few things just from looking at this.*0330

*I know the vertex, because it is (h,k); I know the axis of symmetry--it is at x = h;*0335

*and if I look at a, I will know if the parabola is upward- or downward-facing.*0341

*If a is greater than 0, the parabola will open upward; and k gives you the minimum.*0348

*If a is a negative value--if it is less than 0--the parabola opens downward, and k is the maximum value of the function.*0357

*Let's look at an example: y = 2(x - 1) ^{2} + 4.*0367

*So, this is in standard form: this means that I have h = 1, k = 4, and a = 2.*0374

*So, I know that my vertex is going to be at (1,4); the axis of symmetry is going to be at x = h, so at x = 1.*0384

*And since a is greater than 0, this opens upward.*0404

*So, I can sketch this out: I have a vertex at (1,4), right here, and it opens upward.*0411

*And the axis of symmetry is going to be right here at x = 1.*0421

*Here is my vertex at (1,4); and this vertex is a minimum, because this opens upward.*0428

*The minimum value is k, which is 4.*0433

*If I were to take a similar situation, but say y = -2(x - 1) ^{2} + 4,*0441

*I would have, again, an h equal to 1 and a k equal to 4, but this time a would be -2, so this would open downward.*0453

*What I would end up with would be a parabola here, again, with the vertex at (1,4).*0465

*But it would open downward, and therefore, this would be a maximum.*0473

*Also, if the absolute value of a is greater than 1, you end up with a relatively narrow parabola.*0481

*If the absolute value of a is less than 1, you end up with a relatively wide parabola.*0490

*So, this form is very useful, because just by having the equation in this form, we can at least sketch the graph.*0500

*Let's talk a little bit more about graphing parabolas.*0508

*You can use symmetry and translations to graph a parabola: and by translations, we mean a shift.*0511

*Looking at the standard form: what this really is: if you took a graph of y = ax ^{2}, this is letting h equal 0 and k equal 0.*0519

*And then, if you altered what h is, it is going to shift the graph horizontally by that number of units.*0530

*If you alter what k is, it is going to translate or shift that graph upward and downward by a certain number of units.*0538

*In order to graph a parabola, you often need to put it in standard form.*0547

*Let's start out by just talking about putting an equation or a parabola in standard form.*0552

*And then we will go on and look at some graphs, and how different values of h and k can affect the graph.*0556

*So, in order to put the equation into standard form...let's say you are given an equation such as this, y = x ^{2} + 6x - 8,*0562

*and I want it in this standard form, y = a(x - h) ^{2} + k.*0572

*The first thing to do (and this is, again, review from an earlier lesson--you can go back and look at the lesson*0580

*on completing the square as part of this lecture series, but we will review it again now): first, I am going*0586

*to isolate the x variable terms on the right side of the equation.*0592

*I am going to add 8 to both sides: now I am going to complete the square.*0596

*I am going to focus on this, and I need to add a term to it to make this a perfect square trinomial.*0602

*The term I am going to add is going to be b ^{2}/4.*0608

*In this case, b is 6, so this is going to give me 6 ^{2}/4, which is 36/4, which is equal to 9.*0613

*So, that is what I need to add in here: y + 8, plus I need to add 9 to both sides.*0627

*It is easy to forget to add it to the other side, because you get so focused on completing the square.*0640

*But if you don't, the equation will no longer be balanced.*0645

*So, I am going to add 9 to both sides.*0648

*And I want this to end up in this form; so I am going to rewrite this.*0653

*First I will add these two together to simplify to get y + 17 =...well, this is a perfect square trinomial, so I just take (x + 3) ^{2}.*0657

*And I look at what I have, and it is almost in this form, but not quite.*0668

*I want to isolate y on the left, so I am going to subtract 17 from both sides to get y = (x + 3) ^{2} - 17.*0671

*And this is in this form: a happens to be equal to 1 in this case.*0679

*And so, if you are given an equation that is not in standard form, and you want to get it in standard form,*0683

*isolate the x variable values on the right (although if we are working with horizontal parabolas,*0690

*it is going to be the other way around, as we will see in a minute--we are actually going to end up*0697

*getting the y variable terms on the right; but for now, the x variable terms on the right); complete the square*0701

*by adding the b ^{2}/4 term to both sides of the equation; and then simplify;*0707

*shift things around as needed to get it in this form.*0716

*Remember, also, that if you have a leading coefficient that is something other than 1,*0718

*when you get to this step after isolating the x variable terms, you are going to need to factor out that term before completing the square.*0724

*All right, assuming that you have gotten your equation in standard form, and you are ready to graph the parabola, you are going to use symmetry.*0734

*The two halves of the parabola are symmetrical; if you graph half the points, you can use reflection across the axis of symmetry to graph the other points.*0741

*And translation is knowing how h and k, and changes in h and k, affect the graph, in order to graph.*0749

*All right, so let's just start out with something in this form--a very basic equation for a parabola.*0760

*Let's let f(x) equal x ^{2}, so it is in this form: y = ax^{2}.*0767

*And so, here, what is happening is: if you think about what we have, we have a = 1, and then h is 0 and k is 0.*0775

*What this tells me is that the vertex is going to be at (0,0), and the axis of symmetry is going to be at x = 0.*0784

*And you can also very easily find some points to graph this.*0796

*right now, I am just going to sketch it out, and not worry about exact points, just so you get the idea.*0800

*So, since a = 1, this is going to open upward; this is going to be upward-opening, so the vertex is here at (0,0);*0805

*it is upward-opening; and it is going to look something like this.*0819

*So, this is my graph here of y, or f(x), = x ^{2}.*0833

*Now, let's say I change this slightly: let's say I have another function, g(x) = x ^{2} + 2.*0838

*So, looking at this form, h is still 0; but now I have k = 2.*0847

*And according to this, this is going to shift the graph up 2 units; so k is going to translate this graph up 2 units.*0855

*I have a similar graph, but it is going to be with the vertex right here at (0,2).*0867

*And remember that the axis of symmetry is at x = h, so the axis of symmetry is going to still be at x = 0; right here--this is the axis of symmetry.*0880

*This is shifted upward; it still opens upward, because a is positive.*0888

*So, now I am just going to have a similar idea, but shifted upward by 2.*0892

*So here, I have y = x ^{2} + 2.*0901

*If this had been a -2, then it would have been shifted down by 2, and I would have had a graph right here.*0907

*So, let's see what happens when I change h.*0912

*Let's get a third function: we will call it h(x) = (x - 1) ^{2}.*0917

*OK, now what I have here is h = 1; k, if I look here, is 0.*0935

*Therefore, the vertex (this is the vertex right here) equals (1,0), and the axis of symmetry is going to be at x = 1.*0947

*So, this is going to be shifted to the right; so I am going to have a graph something...let me move this out of the way...like this.*0962

*So, this one is y = x ^{2}, and this is y = (x - 1)^{2}.*0979

*Important take-home points: a change in h will shift the graph horizontally, to the right or left.*0991

*A change in k will shift the basic graph either up or down, by k number of units.*1001

*Using symmetry: if I were to graph these out exactly, I would need to find points.*1009

*And I don't need to find all of the points: for example, if I had a parabola that was a downward-facing parabola*1013

*somewhere, then I could use the axis of symmetry, and I could just find the points over here and reflect across that axis in order to graph.*1021

*All right, this concept is another one adding on to our knowledge of parabolas from prior lessons.*1034

*And it is defining a segment called the latus rectum.*1040

*The latus rectum is the segment passing through the focus and perpendicular to the axis of symmetry.*1044

*Let's see what that means--let's visualize that.*1051

*Let's say I have a parabola like this, and let's say the focus is here.*1054

*So, this passes through the focus, and is perpendicular to the axis of symmetry.*1065

*This is the focus, and here we have the axis of symmetry.*1072

*That means the latus rectum is going to pass through here, and it is going to be perpendicular to the axis of symmetry.*1083

*So, that is this line; this is the latus rectum.*1090

*The equation for its length is the absolute value of 1/a; and if you have the equation of the parabola in standard form,*1095

*then this a is the same a as you will see in that formula.*1108

*So, this is something you might occasionally need to use.*1112

*For example, if I were given an equation of a parabola y = 2(x - 3) ^{2} + 5,*1115

*and I was asked to find the length of the latus rectum of this parabola, then I would just say,*1122

*"OK, a equals 2; therefore, the length equals the absolute value of 1/2."*1127

*Horizontal parabolas: I mentioned that you can also have parabolas that open to the right or left, not just up and down,*1138

*although up to this point in the course, we have just talked about vertical parabolas, or parabolas that open upward or downward.*1143

*For parabolas whose axis of symmetry is horizontal, we end up with equations in this form: y = a(x - k) ^{2} + h.*1150

*So, one thing to note: the positions of the x's and y's are reversed, but so are the h's and k's.*1160

*In the vertical formula, the h was in here, and the k was out here.*1168

*So, be careful when you are working with this formula to notice that the positions of h and k are reversed.*1171

*And there are translations of x = ay ^{2}, and then again, h and k shift this graph around horizontally and vertically.*1177

*So, it would look something like this, for example: the axis of symmetry would be right here;*1189

*and it would be a horizontal axis of symmetry; or maybe I have one that opens to the left, and it has an axis of symmetry right here.*1197

*These do not represent functions; and you can see that they don't represent functions*1208

*by trying to pass a vertical line through them: they fail the vertical line test.*1212

*Remember: with a function, the vertical line test tells us that a vertical line drawn...you could try*1216

*any possible area of the curve, and the vertical line will only cross the curve once.*1224

*If the vertical line crosses the curve more than once, it is not a function.*1230

*So, this fails the vertical line test.*1233

*It is not a function; it is still an equation--you can still make a graph of it; but horizontal parabolas do not represent functions.*1241

*I am working on graphing some horizontal parabolas.*1250

*When you look at the equation in standard form, y = a...and remember, the k and h are in opposite positions;*1253

*they are reversed...looking at a, if a is positive (if a is greater than 0), then the parabola is going to open to the right.*1263

*If a is negative, then the parabola is going to open to the left.*1273

*So, let's look at a very simple horizontal parabola, x = y ^{2}.*1278

*OK, the vertex is at (h,k); and I can see that h and k are both 0, so the vertex equals (0,0).*1284

*The axis of symmetry is at y = k, so that is going to be at y = 0.*1294

*And the a here is 1: a = 1, so this opens to the right.*1298

*So, you are going to have a parabola that looks something like this.*1308

*You could have another parabola, x = -y ^{2}.*1321

*Here we would have the same vertex and the same axis of symmetry; here the x-axis is actually the axis of symmetry.*1325

*And I look at a now, and a equals -1, so this parabola is going to open to the left.*1333

*So, I am going to end up with a parabola like this.*1343

*Now again, change in h or change in k is going to shift this parabola a bit.*1350

*Let's change h and see what happens: let's let x equal y ^{2} + 2.*1358

*Here I have h = 2, k = 0; so (2,0) is the vertex; a = 1, so it is positive, so this still opens to the right.*1366

*If I look at this, x = y ^{2}...here is my graph of x = y^{2}; over here is x = -y^{2}.*1379

*Now, I am going to have h = 2, so that is going to shift horizontally by 2.*1386

*(2,0) will be the vertex; and it is going to open to the right.*1397

*So, this is x = y ^{2} over here; right here, this is actually x = y^{2} + 2 now.*1402

*And k, as discussed before, shifts the graph of a parabola vertically.*1416

*The same idea here: if I were to change k, then I would shift this graph up or down by k units.*1423

*So, with horizontal parabolas, you need to be familiar with this equation.*1430

*You need to know that they open to the right if a is greater than 0; they open to the left if a is less than 0.*1435

*The vertex is at (h,k), and the axis of symmetry is y = k.*1441

*And you also need to keep in mind that these do not represent functions.*1446

*In the beginning of today's lesson, we talked about the focus and directrix.*1452

*And here are formulas to allow you to find those if you need to.*1455

*If you have a vertical parabola, the coordinates of the focus are h for the x-coordinate, and k plus 1/4a.*1460

*And the equation for the directrix is y = k - 1/4a; remember that the directrix is a line, so this is giving you the equation for that line.*1473

*And this would be for a vertical parabola; for a horizontal parabola, the focus is found at the coordinates h + 1/4a;*1485

*and then the y-coordinate is k, so the focus is a point, and this gives the coordinates of that point.*1493

*The directrix is a line, and the equation for this line for a horizontal parabola is x = h - 1/4a.*1498

*And you might need to occasionally use these when we are working problems.*1505

*And we will see that in one of the examples, actually, shortly.*1508

*Starting out with Example 1: Write in standard form and identify the key features: x = 3y ^{2} - 12y + 10.*1513

*We have x equal to all of this; so this tells me, since I have x set equal to this y ^{2} term, that I am looking at a horizontal parabola.*1524

*So, the standard form of this equation is going to be x = a(y - k) ^{2} + h.*1536

*Remember, h and k are going to be in opposite positions.*1547

*In order to get this equation in standard form, we need to complete the square.*1550

*This time, since I am working with a horizontal parabola, I am going to isolate all of the y variable terms on the right.*1554

*And I am going to do that by subtracting 10 from both sides to get x - 10 = 3y ^{2} - 12y.*1561

*This leading coefficient is not 1, so I have to factor it out.*1569

*And then, I have to be really careful when I am adding to both sides of the equation, because this is factored out.*1573

*So, factor out a 3 to get y ^{2} - 4y.*1580

*I need to complete the square: that means I need to add something over here.*1585

*And the term that I need to add is going to be b ^{2}/4.*1589

*b is actually 4; so this is going to be 4 ^{2}/4, equals 16/4, equals 4.*1594

*Here is where I need to be careful: on the right, I am adding 4 inside these parentheses, which is pretty straightforward.*1604

*But what I need to do on the left is realize that I am actually going to be adding 3 times 4, which is 12.*1613

*So, if I were just to add 4, this equation would not be balanced,*1628

*because in reality, what I am doing over here is adding 3 times 4.*1631

*So, on the right, I am going to add 12; and I got that from 3 times 4.*1635

*Simplifying the left: 12 - 10 is 2; on the right, inside here, I now have a perfect square.*1640

*And I want this to end up in this form, so I am going to write this as (y - 2) (and it is negative, because I end up with a negative sign in here) squared.*1649

*I am almost done; I just need to move this constant over to the right to have it in this form.*1661

*x = 3 times (y - 2) ^{2}, minus 2.*1666

*So, now that I have this in standard form, I can identify key features.*1671

*Key features: 1: this is a horizontal parabola, as you can see from looking at this equation.*1677

*2: The vertex is at (h,k); h is 2, and k is also 2.*1686

*Actually, being careful with the signs, h is actually -2, because remember, standard form has a plus here.*1703

*I don't have a plus here; I could rewrite this so that I do, and that would give me + -2.*1710

*And it is good practice, actually, to write it exactly in this form, although this is correct--you could leave it like this.*1718

*By writing it in this form...and the same thing if I had ended up with a plus here--then I would need to rewrite that,*1725

*because here I need a negative to be in standard form; if I ended up with a plus here,*1736

*then I would have needed to rewrite that, as well, which would have been equal to minus -2.*1741

*Standard form, just like this, looking here, gives me a vertex at (-2,2).*1748

*And because a equals 3, that means that a is greater than 0; a is positive, so the parabola opens to the right.*1755

*OK, so key features: horizontal parabola; it has a vertex at (-2,2); a = 3, so this tells me that the parabola opens to the right.*1773

*We can also say that the axis of symmetry is at y = k, and therefore the axis of symmetry is at y = 2.*1784

*OK, in Example 2, we are asked to graph.*1797

*And you will notice that this is the same equation that we worked with in Example 1.*1802

*We already figured out standard form: and standard form is x = 3(y - 2) ^{2} - 2.*1806

*And for clarity, we can actually write this as I did at the end, which is 3(y - 2) ^{2} + -2,*1816

*so that we truly have it in standard form, with the plus here to make it easy to see what is going on.*1826

*To graph this, I want to know the vertex: the vertex is (h,k): h here is -2; k is 2.*1831

*The axis of symmetry is going to be at y = k; k is 2, so it is going to be at y = 2.*1840

*I know that this opens to the right, so I have a general sense of this graph.*1854

*But I can also just find a few points.*1859

*And we are used to working with a situation where x is the input and y is the output.*1868

*It is the opposite here, so we need to be really careful.*1873

*I also want to note that, since the vertex is here at (-2,2), and this opens to the right,*1876

*for this graph, we are not going to have values of x that are smaller than -2.*1881

*So, if I end up with something where an x is smaller than -2, then it is going to be off the graph.*1885

*Let's let y equal 1: if y is 1, 1 - 2 is -1, squared gives me 1; 1 times 3 is 3, minus 2 is 1; so, when y is 1, x is 1.*1891

*Let's let y equal 3: when y is 3, 3 minus 2 is 1, squared is 1; 1 times 3 is 3, minus 2 is 1.*1906

*And you can see, as I mentioned, that this is not a function; it failed the vertical line test (as horizontal parabolas do).*1915

*And you can see that there is an x-value, 1, that is assigned 2 values of y; so it does not meet the definition of a function.*1921

*So, just a couple of points...let's do one more: 0...0 minus 2 is -2; squared is 4; 4 times 3 is 12; 12 minus 2 is 10.*1929

*So, that is off this graph; but it gives us an idea of the shape.*1941

*So, I know that my axis of symmetry is going to be here; and I have a point at (1,1);*1945

*I have another point at (1,3); and then I have a point way out here at (10,0).*1954

*I know that this is going to be a fairly narrow graph, because a equals 3.*1960

*This is the graph of the horizontal parabola described by this equation; and here it is, written in standard form.*1971

*So, it opens to the right; it is fairly narrow, because a equals 3.*1979

*It has a vertex at (-2,2), and it has an axis of symmetry at y = 2.*1984

*Example 3: we are asked to graph; this is also going to be a horizontal parabola.*1994

*We are going to start out by putting it in the standard form, x = (y - k) ^{2} + h.*1999

*We need to complete the square; start out by isolating the y variable terms on the right.*2009

*So, I am going to add 6 to both sides to get -2y ^{2} + 8y.*2014

*Since the leading coefficient is not 1, I need to factor it out; so I am going to factor this -2 to get y ^{2}.*2021

*Factoring a -2 from here would give me a -4.*2032

*And I need to add something to this to complete the square.*2035

*What I need to add is b ^{2}/4.*2040

*b is 4, so I am going to be adding 4 ^{2}/4; that is 16, divided by 4; that is 4.*2044

*So, I am going to be adding 4 to the right; but to the left, I am actually adding -2 times 4, which is -8.*2061

*So, we subtract 8 from that side; to this side, since I am adding inside the parentheses, I am just adding 4.*2073

*But then, 4 times -2--that is how I got the -8 on the left.*2082

*This gives me x - 2 = -2; and I want it in this form, so I am going to rewrite this as (y - 2) ^{2}.*2085

*The last thing I need to do is add 2 to both sides; and I have it in standard form.*2097

*Now that I have this in standard form, it is much easier to graph.*2106

*The vertex is going to be at (h,k); so h is here; k is here; the vertex is at (2,2).*2110

*There is going to be an axis of symmetry at y = k, and so that is going to be at y = 2; my axis of symmetry is going to be at y = 2.*2119

*Now, to finish out graphing this, I am going to find a few points.*2142

*I have the vertex at (2,2); I also know that a is less than 0 (a is negative), so I know this is going to open to the left.*2146

*So, I know it is going to look something like this; but I will find a couple of points.*2155

*And I know that x is (actually, (2,2) is right here)...I know that this opens to the left, and that x is not going to get any larger than that.*2158

*The graph is just going to go this way.*2172

*So, I can't use values that end up giving me an x that is greater than 2.*2174

*Let's try some simple values: I am going to try 1 for y, and looking in standard form, 1 - 2 gives me -1, squared is 1, times -2 is -2, plus 2 is 0.*2181

*And 3: 3 minus 2 is 1, squared is 1; 1 times -2 is -2, plus 2 is 0.*2195

*So, I have a couple of points here: this is at 0...when x is 0, y is 1; when x is 0, y is 3.*2204

*And this is going to give me a parabola shaped like this, opening to the left with a vertex at (2,2).*2214

*The axis of symmetry would be right through here; and I have a couple of points, just to make it a bit more precise.*2226

*So, the first step in graphing a parabola is always to get it into this form by completing the square.*2235

*And then, using the features you can see from here, sketch it out, and finding a few points, make the graph more accurate.*2240

*Find the equation of the parabola with a vertex of (2,3) and focus at (2,7); draw the graph.*2250

*This is a very challenging problem: we are not given an equation--we actually have to find the equation based on some key points that we are given.*2256

*Well, I am given that the vertex is at (2,3); so I know that the vertex is right here; that is the vertex.*2266

*This time, I am also given the focus; the focus is at (2,7), which is going to be up here somewhere...5, 6, 7...about here.*2278

*So, the vertex is (2,3); the focus is (2,7).*2290

*Remember, in the beginning of this lesson, I mentioned that the focus is always inside the parabola.*2296

*Since the focus is inside the parabola, I already know that this has to open upward.*2301

*So, I know something about the shape of the graph.*2306

*Let's find the equation: now, I know that this is a vertical parabola, because the focus is inside the parabola.*2311

*That told me that this has to open upward, so I know I am dealing with a vertical parabola.*2316

*And that helps me to find the equation, because the standard form is going to be y = a(x - h) ^{2} + k.*2320

*I am given the vertex, so I am given h and k: I know that h = 2 and k = 3.*2330

*In order to write this equation, I need a, h, and k; all I am missing is a.*2340

*I am given the piece of information, though, that the focus is (2,7).*2347

*And that is going to allow me to find a.*2350

*You will recall that I mentioned the formulas for focus and directrix.*2353

*And for a vertical parabola, the focus is at h...the x-coordinate is h, which we see here; and the y-coordinate is k + 1/4a.*2359

*And I know the focus is at (2,7): so 2 = h, and 7 = k + 1/4a, according to this definition.*2377

*Well, since I know that k is 3, then I can solve this.*2392

*So, I know k; so I can solve for a.*2401

*Subtract 3 from both sides to get 1/4a; 1/a equals 16; multiply both sides by a, and then divide both sides by 16,*2408

*or just take the reciprocal of each side (essentially, that is what you are doing) to get a = 1/16.*2422

*Now, I have h and k given; I was able to figure out a, based on the definition of focus.*2427

*So, I end up with the equation y = 1/16(x - 2) ^{2} + 3.*2433

*So, this is the equation.*2444

*And as you know, once we have the equation, the graphing is pretty easy.*2446

*I know that this opens upward; and since I know what a is, I know that this is going to be a pretty wide parabola; the a is a small value.*2451

*I am going to have a parabola that opens upward, with a vertex of (2,3), and fairly wide in shape.*2461

*That was a pretty challenging problem, because you had to go back*2470

*and think about how you could use a formula to find the focus; and knowing the focus allowed you to find a.*2473

*That concludes this lesson on parabolas at Educator.com; thanks for visiting!*2483

1 answer

Last reply by: Dr Carleen Eaton

Sun Jun 1, 2014 10:00 PM

Post by Sophie Zhong on April 26, 2014

Does the equation h=-b/2a and k=-(b^2-4ac)/4a also apply in horizontal parabolas? Because when I used it for example 1, I came up with h=-(-12/6)=2, k=-2, which is wrong.

1 answer

Last reply by: Dr Carleen Eaton

Thu May 16, 2013 11:10 PM

Post by Saki Amagai on May 16, 2013

I don't know why but.. I'm having trouble watching this entire conic section due to technical issues. I don't have any problem for the other ones. It's just this section that I get "network failure". I really need to watch this... Can you please check if there's nothing wrong with the server? Thank you.

1 answer

Last reply by: Dr Carleen Eaton

Sun Jan 27, 2013 1:01 PM

Post by Monis Mirza on January 26, 2013

how do you find the maximum and minimum of a parabola using the equation?

i have a test on this on monday and i really need the answer!

0 answers

Post by julius mogyorossy on January 11, 2013

Merc, I think you are correct. I dig it that Educator is being advertised on my blog page. Educator said, learn like you are going to live forever, it seems somebody there knows who I am.

1 answer

Last reply by: Dr Carleen Eaton

Thu Feb 9, 2012 7:43 PM

Post by Edmund Mercado on February 9, 2012

For Horizontal Parabolas at 18:57, should the standard form say x = a(y-k)^2 +h instead of

y = a(x-k)^2 +h with the x and y in opposite positions?

0 answers

Post by norman stradleigh on June 21, 2011

thanks really helped me out

1 answer

Last reply by: Dr Carleen Eaton

Mon Jul 5, 2010 5:15 PM

Post by Timothy miranda on June 21, 2010

thanks that cleared it up for me