WEBVTT mathematics/math-analysis/selhorst-jones 00:00:00.000 --> 00:00:01.900 Welcome to Educator.com. 00:00:01.900 --> 00:00:04.200 Today, we are going to talk about parabolas. 00:00:04.200 --> 00:00:10.800 And in some earlier lectures in this series on quadratic equations, we talked about parabolas and did some graphing. 00:00:10.800 --> 00:00:21.500 But now, we are going to go on and give a specific definition to parabolas, and learn about some other features of parabolas. 00:00:21.500 --> 00:00:27.000 Although you have seen parabolas previously, when we graphed, we didn't form a specific definition of them. 00:00:27.000 --> 00:00:33.300 So, the definition of a parabola is that it is the set of points in the plane whose distance from a given point, 00:00:33.300 --> 00:00:41.000 called the focus, is equal to its distance from a given line, called the directrix. 00:00:41.000 --> 00:00:45.200 Let's talk about that before we go on to talk about the axis of symmetry. 00:00:45.200 --> 00:00:59.100 So, if you had a parabola (let's say right here; and we will do an upward-facing parabola), you would have some point, 00:00:59.100 --> 00:01:20.000 which is known as the focus, and a line (I'm going to put that right about here) called the directrix. 00:01:20.000 --> 00:01:25.100 By definition, every point on this parabola is equidistant from the focus and the directrix. 00:01:25.100 --> 00:01:35.500 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. 00:01:35.500 --> 00:01:48.500 And this is just a very rough sketch; but these distances actually would be equal; they are theoretically equal. 00:01:48.500 --> 00:01:57.600 Looking right here at the vertex, these distances would be equal; so that would be, say, y. 00:01:57.600 --> 00:02:08.100 If I took some other point, say here, and I measured here to here, these two distances would be equal. 00:02:08.100 --> 00:02:30.100 So, a couple things to note: the focus is inside the parabola; the directrix is outside. 00:02:30.100 --> 00:02:35.400 And this is because the focus and the directrix are on the opposite sides of the vertex. 00:02:35.400 --> 00:02:45.000 So, you could have a parabola facing downward, and then it would have a focus here and a directrix up here. 00:02:45.000 --> 00:02:53.400 We are also going to talk, today, about parabolas that face to the left and right--horizontal parabolas. 00:02:53.400 --> 00:02:59.700 But right now, we are going to stick with just (for this discussion) focusing on vertical ones, 00:02:59.700 --> 00:03:08.800 the definition being that every point in the parabola equidistant between the focus and the directrix. 00:03:08.800 --> 00:03:15.100 The axis of symmetry of the parabola passes through the focus; and it is perpendicular to the directrix. 00:03:15.100 --> 00:03:23.700 In this case, the y-axis is the axis of symmetry; it is right here. 00:03:23.700 --> 00:03:33.800 And you see that it passes through the focus, and it forms a right angle; it is perpendicular to the directrix. 00:03:33.800 --> 00:03:36.500 Again, we talked about some of these concepts in earlier lectures. 00:03:36.500 --> 00:03:44.100 But to review, vertex: the vertex of a parabola is the point at which the axis of symmetry intersects the parabola. 00:03:44.100 --> 00:03:50.000 And it is a maximum or minimum point on the parabola, if the axis of symmetry is vertical. 00:03:50.000 --> 00:04:02.400 If the axis of symmetry is horizontal (say we have a parabola like this, then the axis of symmetry would be horizontal), 00:04:02.400 --> 00:04:07.400 we still have a vertex, but it is not a maximum or minimum. 00:04:07.400 --> 00:04:13.700 And again, we are going to focus a little more on vertical parabolas right now, and then we will talk about horizontal parabolas. 00:04:13.700 --> 00:04:25.000 So, if I have a downward-facing parabola, the vertex is here; the axis of symmetry is right here. 00:04:25.000 --> 00:04:33.800 And this vertex is the maximum; this is as large as y gets--it is the largest value that the function attains. 00:04:33.800 --> 00:04:45.800 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. 00:04:45.800 --> 00:04:59.600 In this case, the vertex is a minimum; this is the smallest value that the function will attain. 00:04:59.600 --> 00:05:07.800 The standard form of a parabola with vertex at (h,k) is y = a(x - h)² + k. 00:05:07.800 --> 00:05:14.300 And this is for vertical parabolas; there is a slightly different form when we are talking about horizontal parabolas. 00:05:14.300 --> 00:05:19.500 And you might recall this form of the equation that we covered earlier on, under the lecture on quadratic equations. 00:05:19.500 --> 00:05:24.200 And we called this the vertex form of the equation; now we are going to refer to it as standard form. 00:05:24.200 --> 00:05:29.700 And it is a very useful form, because it tells you a lot about the parabola. 00:05:29.700 --> 00:05:34.700 The axis of symmetry is x = h: so I know a few things just from looking at this. 00:05:34.700 --> 00:05:40.700 I know the vertex, because it is (h,k); I know the axis of symmetry--it is at x = h; 00:05:40.700 --> 00:05:48.300 and if I look at a, I will know if the parabola is upward- or downward-facing. 00:05:48.300 --> 00:05:57.000 If a is greater than 0, the parabola will open upward; and k gives you the minimum. 00:05:57.000 --> 00:06:06.900 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. 00:06:06.900 --> 00:06:14.600 Let's look at an example: y = 2(x - 1)² + 4. 00:06:14.600 --> 00:06:24.100 So, this is in standard form: this means that I have h = 1, k = 4, and a = 2. 00:06:24.100 --> 00:06:44.100 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. 00:06:44.100 --> 00:06:51.300 And since a is greater than 0, this opens upward. 00:06:51.300 --> 00:07:01.400 So, I can sketch this out: I have a vertex at (1,4), right here, and it opens upward. 00:07:01.400 --> 00:07:08.100 And the axis of symmetry is going to be right here at x = 1. 00:07:08.100 --> 00:07:13.600 Here is my vertex at (1,4); and this vertex is a minimum, because this opens upward. 00:07:13.600 --> 00:07:21.200 The minimum value is k, which is 4. 00:07:21.200 --> 00:07:33.200 If I were to take a similar situation, but say y = -2(x - 1)² + 4, 00:07:33.200 --> 00:07:45.100 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. 00:07:45.100 --> 00:07:53.500 What I would end up with would be a parabola here, again, with the vertex at (1,4). 00:07:53.500 --> 00:08:01.100 But it would open downward, and therefore, this would be a maximum. 00:08:01.100 --> 00:08:10.600 Also, if the absolute value of a is greater than 1, you end up with a relatively narrow parabola. 00:08:10.600 --> 00:08:20.500 If the absolute value of a is less than 1, you end up with a relatively wide parabola. 00:08:20.500 --> 00:08:28.600 So, this form is very useful, because just by having the equation in this form, we can at least sketch the graph. 00:08:28.600 --> 00:08:31.600 Let's talk a little bit more about graphing parabolas. 00:08:31.600 --> 00:08:39.400 You can use symmetry and translations to graph a parabola: and by translations, we mean a shift. 00:08:39.400 --> 00:08:50.600 Looking at the standard form: what this really is: if you took a graph of y = ax², this is letting h equal 0 and k equal 0. 00:08:50.600 --> 00:08:58.000 And then, if you altered what h is, it is going to shift the graph horizontally by that number of units. 00:08:58.000 --> 00:09:06.700 If you alter what k is, it is going to translate or shift that graph upward and downward by a certain number of units. 00:09:06.700 --> 00:09:11.800 In order to graph a parabola, you often need to put it in standard form. 00:09:11.800 --> 00:09:15.900 Let's start out by just talking about putting an equation or a parabola in standard form. 00:09:15.900 --> 00:09:22.100 And then we will go on and look at some graphs, and how different values of h and k can affect the graph. 00:09:22.100 --> 00:09:32.600 So, in order to put the equation into standard form...let's say you are given an equation such as this, y = x² + 6x - 8, 00:09:32.600 --> 00:09:40.400 and I want it in this standard form, y = a(x - h)² + k. 00:09:40.400 --> 00:09:46.100 The first thing to do (and this is, again, review from an earlier lesson--you can go back and look at the lesson 00:09:46.100 --> 00:09:52.500 on completing the square as part of this lecture series, but we will review it again now): first, I am going 00:09:52.500 --> 00:09:56.400 to isolate the x variable terms on the right side of the equation. 00:09:56.400 --> 00:10:02.300 I am going to add 8 to both sides: now I am going to complete the square. 00:10:02.300 --> 00:10:08.300 I am going to focus on this, and I need to add a term to it to make this a perfect square trinomial. 00:10:08.300 --> 00:10:13.200 The term I am going to add is going to be b²/4. 00:10:13.200 --> 00:10:26.800 In this case, b is 6, so this is going to give me 6²/4, which is 36/4, which is equal to 9. 00:10:26.800 --> 00:10:40.400 So, that is what I need to add in here: y + 8, plus I need to add 9 to both sides. 00:10:40.400 --> 00:10:45.300 It is easy to forget to add it to the other side, because you get so focused on completing the square. 00:10:45.300 --> 00:10:48.200 But if you don't, the equation will no longer be balanced. 00:10:48.200 --> 00:10:52.900 So, I am going to add 9 to both sides. 00:10:52.900 --> 00:10:57.300 And I want this to end up in this form; so I am going to rewrite this. 00:10:57.300 --> 00:11:07.800 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. 00:11:07.800 --> 00:11:11.600 And I look at what I have, and it is almost in this form, but not quite. 00:11:11.600 --> 00:11:19.300 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. 00:11:19.300 --> 00:11:23.300 And this is in this form: a happens to be equal to 1 in this case. 00:11:23.300 --> 00:11:30.500 And so, if you are given an equation that is not in standard form, and you want to get it in standard form, 00:11:30.500 --> 00:11:37.300 isolate the x variable values on the right (although if we are working with horizontal parabolas, 00:11:37.300 --> 00:11:41.100 it is going to be the other way around, as we will see in a minute--we are actually going to end up 00:11:41.100 --> 00:11:47.300 getting the y variable terms on the right; but for now, the x variable terms on the right); complete the square 00:11:47.300 --> 00:11:55.900 by adding the b²/4 term to both sides of the equation; and then simplify; 00:11:55.900 --> 00:11:58.600 shift things around as needed to get it in this form. 00:11:58.600 --> 00:12:04.100 Remember, also, that if you have a leading coefficient that is something other than 1, 00:12:04.100 --> 00:12:14.500 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. 00:12:14.500 --> 00:12:21.600 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. 00:12:21.600 --> 00:12:29.500 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. 00:12:29.500 --> 00:12:40.000 And translation is knowing how h and k, and changes in h and k, affect the graph, in order to graph. 00:12:40.000 --> 00:12:47.100 All right, so let's just start out with something in this form--a very basic equation for a parabola. 00:12:47.100 --> 00:12:55.100 Let's let f(x) equal x², so it is in this form: y = ax². 00:12:55.100 --> 00:13:04.600 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. 00:13:04.600 --> 00:13:16.300 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. 00:13:16.300 --> 00:13:19.700 And you can also very easily find some points to graph this. 00:13:19.700 --> 00:13:25.600 right now, I am just going to sketch it out, and not worry about exact points, just so you get the idea. 00:13:25.600 --> 00:13:39.100 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); 00:13:39.100 --> 00:13:52.700 it is upward-opening; and it is going to look something like this. 00:13:52.700 --> 00:13:58.600 So, this is my graph here of y, or f(x), = x². 00:13:58.600 --> 00:14:07.400 Now, let's say I change this slightly: let's say I have another function, g(x) = x² + 2. 00:14:07.400 --> 00:14:14.700 So, looking at this form, h is still 0; but now I have k = 2. 00:14:14.700 --> 00:14:27.200 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. 00:14:27.200 --> 00:14:40.100 I have a similar graph, but it is going to be with the vertex right here at (0,2). 00:14:40.100 --> 00:14:48.600 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. 00:14:48.600 --> 00:14:52.600 This is shifted upward; it still opens upward, because a is positive. 00:14:52.600 --> 00:15:01.400 So, now I am just going to have a similar idea, but shifted upward by 2. 00:15:01.400 --> 00:15:07.400 So here, I have y = x² + 2. 00:15:07.400 --> 00:15:12.600 If this had been a -2, then it would have been shifted down by 2, and I would have had a graph right here. 00:15:12.600 --> 00:15:17.000 So, let's see what happens when I change h. 00:15:17.000 --> 00:15:34.700 Let's get a third function: we will call it h(x) = (x - 1)². 00:15:34.700 --> 00:15:46.700 OK, now what I have here is h = 1; k, if I look here, is 0. 00:15:46.700 --> 00:16:01.900 Therefore, the vertex (this is the vertex right here) equals (1,0), and the axis of symmetry is going to be at x = 1. 00:16:01.900 --> 00:16:19.200 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. 00:16:19.200 --> 00:16:31.400 So, this one is y = x², and this is y = (x - 1)². 00:16:31.400 --> 00:16:41.300 Important take-home points: a change in h will shift the graph horizontally, to the right or left. 00:16:41.300 --> 00:16:49.500 A change in k will shift the basic graph either up or down, by k number of units. 00:16:49.500 --> 00:16:53.400 Using symmetry: if I were to graph these out exactly, I would need to find points. 00:16:53.400 --> 00:17:01.300 And I don't need to find all of the points: for example, if I had a parabola that was a downward-facing parabola 00:17:01.300 --> 00:17:14.200 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. 00:17:14.200 --> 00:17:20.000 All right, this concept is another one adding on to our knowledge of parabolas from prior lessons. 00:17:20.000 --> 00:17:23.700 And it is defining a segment called the latus rectum. 00:17:23.700 --> 00:17:30.800 The latus rectum is the segment passing through the focus and perpendicular to the axis of symmetry. 00:17:30.800 --> 00:17:33.700 Let's see what that means--let's visualize that. 00:17:33.700 --> 00:17:45.000 Let's say I have a parabola like this, and let's say the focus is here. 00:17:45.000 --> 00:17:52.500 So, this passes through the focus, and is perpendicular to the axis of symmetry. 00:17:52.500 --> 00:18:02.800 This is the focus, and here we have the axis of symmetry. 00:18:02.800 --> 00:18:10.000 That means the latus rectum is going to pass through here, and it is going to be perpendicular to the axis of symmetry. 00:18:10.000 --> 00:18:15.500 So, that is this line; this is the latus rectum. 00:18:15.500 --> 00:18:28.100 The equation for its length is the absolute value of 1/a; and if you have the equation of the parabola in standard form, 00:18:28.100 --> 00:18:32.400 then this a is the same a as you will see in that formula. 00:18:32.400 --> 00:18:35.100 So, this is something you might occasionally need to use. 00:18:35.100 --> 00:18:42.400 For example, if I were given an equation of a parabola y = 2(x - 3)² + 5, 00:18:42.400 --> 00:18:46.800 and I was asked to find the length of the latus rectum of this parabola, then I would just say, 00:18:46.800 --> 00:18:57.700 "OK, a equals 2; therefore, the length equals the absolute value of 1/2." 00:18:57.700 --> 00:19:03.100 Horizontal parabolas: I mentioned that you can also have parabolas that open to the right or left, not just up and down, 00:19:03.100 --> 00:19:09.700 although up to this point in the course, we have just talked about vertical parabolas, or parabolas that open upward or downward. 00:19:09.700 --> 00:19:20.100 For parabolas whose axis of symmetry is horizontal, we end up with equations in this form: y = a(x - k)² + h. 00:19:20.100 --> 00:19:27.700 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. 00:19:27.700 --> 00:19:31.100 In the vertical formula, the h was in here, and the k was out here. 00:19:31.100 --> 00:19:37.600 So, be careful when you are working with this formula to notice that the positions of h and k are reversed. 00:19:37.600 --> 00:19:49.600 And there are translations of x = ay², and then again, h and k shift this graph around horizontally and vertically. 00:19:49.600 --> 00:19:56.700 So, it would look something like this, for example: the axis of symmetry would be right here; 00:19:56.700 --> 00:20:08.400 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. 00:20:08.400 --> 00:20:12.100 These do not represent functions; and you can see that they don't represent functions 00:20:12.100 --> 00:20:16.400 by trying to pass a vertical line through them: they fail the vertical line test. 00:20:16.400 --> 00:20:23.900 Remember: with a function, the vertical line test tells us that a vertical line drawn...you could try 00:20:23.900 --> 00:20:29.800 any possible area of the curve, and the vertical line will only cross the curve once. 00:20:29.800 --> 00:20:33.500 If the vertical line crosses the curve more than once, it is not a function. 00:20:33.500 --> 00:20:40.700 So, this fails the vertical line test. 00:20:40.700 --> 00:20:50.100 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. 00:20:50.100 --> 00:20:53.100 I am working on graphing some horizontal parabolas. 00:20:53.100 --> 00:21:03.100 When you look at the equation in standard form, y = a...and remember, the k and h are in opposite positions; 00:21:03.100 --> 00:21:13.600 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. 00:21:13.600 --> 00:21:17.700 If a is negative, then the parabola is going to open to the left. 00:21:17.700 --> 00:21:24.600 So, let's look at a very simple horizontal parabola, x = y². 00:21:24.600 --> 00:21:33.800 OK, the vertex is at (h,k); and I can see that h and k are both 0, so the vertex equals (0,0). 00:21:33.800 --> 00:21:38.100 The axis of symmetry is at y = k, so that is going to be at y = 0. 00:21:38.100 --> 00:21:48.500 And the a here is 1: a = 1, so this opens to the right. 00:21:48.500 --> 00:22:01.100 So, you are going to have a parabola that looks something like this. 00:22:01.100 --> 00:22:05.600 You could have another parabola, x = -y². 00:22:05.600 --> 00:22:13.300 Here we would have the same vertex and the same axis of symmetry; here the x-axis is actually the axis of symmetry. 00:22:13.300 --> 00:22:23.000 And I look at a now, and a equals -1, so this parabola is going to open to the left. 00:22:23.000 --> 00:22:30.000 So, I am going to end up with a parabola like this. 00:22:30.000 --> 00:22:38.200 Now again, change in h or change in k is going to shift this parabola a bit. 00:22:38.200 --> 00:22:46.400 Let's change h and see what happens: let's let x equal y² + 2. 00:22:46.400 --> 00:22:59.400 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. 00:22:59.400 --> 00:23:06.200 If I look at this, x = y²...here is my graph of x = y²; over here is x = -y². 00:23:06.200 --> 00:23:17.100 Now, I am going to have h = 2, so that is going to shift horizontally by 2. 00:23:17.100 --> 00:23:22.500 (2,0) will be the vertex; and it is going to open to the right. 00:23:22.500 --> 00:23:36.200 So, this is x = y² over here; right here, this is actually x = y² + 2 now. 00:23:36.200 --> 00:23:42.800 And k, as discussed before, shifts the graph of a parabola vertically. 00:23:42.800 --> 00:23:50.600 The same idea here: if I were to change k, then I would shift this graph up or down by k units. 00:23:50.600 --> 00:23:55.200 So, with horizontal parabolas, you need to be familiar with this equation. 00:23:55.200 --> 00:24:01.200 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. 00:24:01.200 --> 00:24:05.900 The vertex is at (h,k), and the axis of symmetry is y = k. 00:24:05.900 --> 00:24:11.700 And you also need to keep in mind that these do not represent functions. 00:24:11.700 --> 00:24:15.600 In the beginning of today's lesson, we talked about the focus and directrix. 00:24:15.600 --> 00:24:20.300 And here are formulas to allow you to find those if you need to. 00:24:20.300 --> 00:24:33.200 If you have a vertical parabola, the coordinates of the focus are h for the x-coordinate, and k plus 1/4a. 00:24:33.200 --> 00:24:45.600 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. 00:24:45.600 --> 00:24:53.000 And this would be for a vertical parabola; for a horizontal parabola, the focus is found at the coordinates h + 1/4a; 00:24:53.000 --> 00:24:57.800 and then the y-coordinate is k, so the focus is a point, and this gives the coordinates of that point. 00:24:57.800 --> 00:25:04.900 The directrix is a line, and the equation for this line for a horizontal parabola is x = h - 1/4a. 00:25:04.900 --> 00:25:08.400 And you might need to occasionally use these when we are working problems. 00:25:08.400 --> 00:25:13.400 And we will see that in one of the examples, actually, shortly. 00:25:13.400 --> 00:25:23.700 Starting out with Example 1: Write in standard form and identify the key features: x = 3y² - 12y + 10. 00:25:23.700 --> 00:25:36.600 We have x equal to all of this; so this tells me, since I have x set equal to this y² term, that I am looking at a horizontal parabola. 00:25:36.600 --> 00:25:46.800 So, the standard form of this equation is going to be x = a(y - k)² + h. 00:25:46.800 --> 00:25:50.600 Remember, h and k are going to be in opposite positions. 00:25:50.600 --> 00:25:54.100 In order to get this equation in standard form, we need to complete the square. 00:25:54.100 --> 00:26:01.300 This time, since I am working with a horizontal parabola, I am going to isolate all of the y variable terms on the right. 00:26:01.300 --> 00:26:09.200 And I am going to do that by subtracting 10 from both sides to get x - 10 = 3y² - 12y. 00:26:09.200 --> 00:26:12.900 This leading coefficient is not 1, so I have to factor it out. 00:26:12.900 --> 00:26:20.100 And then, I have to be really careful when I am adding to both sides of the equation, because this is factored out. 00:26:20.100 --> 00:26:25.100 So, factor out a 3 to get y² - 4y. 00:26:25.100 --> 00:26:29.300 I need to complete the square: that means I need to add something over here. 00:26:29.300 --> 00:26:34.600 And the term that I need to add is going to be b²/4. 00:26:34.600 --> 00:26:43.700 b is actually 4; so this is going to be 4²/4, equals 16/4, equals 4. 00:26:43.700 --> 00:26:53.100 Here is where I need to be careful: on the right, I am adding 4 inside these parentheses, which is pretty straightforward. 00:26:53.100 --> 00:27:08.100 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. 00:27:08.100 --> 00:27:11.600 So, if I were just to add 4, this equation would not be balanced, 00:27:11.600 --> 00:27:15.600 because in reality, what I am doing over here is adding 3 times 4. 00:27:15.600 --> 00:27:20.600 So, on the right, I am going to add 12; and I got that from 3 times 4. 00:27:20.600 --> 00:27:29.100 Simplifying the left: 12 - 10 is 2; on the right, inside here, I now have a perfect square. 00:27:29.100 --> 00:27:40.700 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. 00:27:40.700 --> 00:27:45.900 I am almost done; I just need to move this constant over to the right to have it in this form. 00:27:45.900 --> 00:27:51.500 x = 3 times (y - 2)², minus 2. 00:27:51.500 --> 00:27:56.800 So, now that I have this in standard form, I can identify key features. 00:27:56.800 --> 00:28:06.300 Key features: 1: this is a horizontal parabola, as you can see from looking at this equation. 00:28:06.300 --> 00:28:22.800 2: The vertex is at (h,k); h is 2, and k is also 2. 00:28:22.800 --> 00:28:30.300 Actually, being careful with the signs, h is actually -2, because remember, standard form has a plus here. 00:28:30.300 --> 00:28:38.400 I don't have a plus here; I could rewrite this so that I do, and that would give me + -2. 00:28:38.400 --> 00:28:44.700 And it is good practice, actually, to write it exactly in this form, although this is correct--you could leave it like this. 00:28:44.700 --> 00:28:56.100 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, 00:28:56.100 --> 00:29:01.200 because here I need a negative to be in standard form; if I ended up with a plus here, 00:29:01.200 --> 00:29:08.300 then I would have needed to rewrite that, as well, which would have been equal to minus -2. 00:29:08.300 --> 00:29:14.700 Standard form, just like this, looking here, gives me a vertex at (-2,2). 00:29:14.700 --> 00:29:33.200 And because a equals 3, that means that a is greater than 0; a is positive, so the parabola opens to the right. 00:29:33.200 --> 00:29:43.800 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. 00:29:43.800 --> 00:29:57.500 We can also say that the axis of symmetry is at y = k, and therefore the axis of symmetry is at y = 2. 00:29:57.500 --> 00:30:01.900 OK, in Example 2, we are asked to graph. 00:30:01.900 --> 00:30:06.300 And you will notice that this is the same equation that we worked with in Example 1. 00:30:06.300 --> 00:30:16.400 We already figured out standard form: and standard form is x = 3(y - 2)² - 2. 00:30:16.400 --> 00:30:26.100 And for clarity, we can actually write this as I did at the end, which is 3(y - 2)² + -2, 00:30:26.100 --> 00:30:31.600 so that we truly have it in standard form, with the plus here to make it easy to see what is going on. 00:30:31.600 --> 00:30:40.600 To graph this, I want to know the vertex: the vertex is (h,k): h here is -2; k is 2. 00:30:40.600 --> 00:30:53.800 The axis of symmetry is going to be at y = k; k is 2, so it is going to be at y = 2. 00:30:53.800 --> 00:30:59.200 I know that this opens to the right, so I have a general sense of this graph. 00:30:59.200 --> 00:31:08.200 But I can also just find a few points. 00:31:08.200 --> 00:31:13.300 And we are used to working with a situation where x is the input and y is the output. 00:31:13.300 --> 00:31:15.700 It is the opposite here, so we need to be really careful. 00:31:15.700 --> 00:31:20.900 I also want to note that, since the vertex is here at (-2,2), and this opens to the right, 00:31:20.900 --> 00:31:24.800 for this graph, we are not going to have values of x that are smaller than -2. 00:31:24.800 --> 00:31:31.100 So, if I end up with something where an x is smaller than -2, then it is going to be off the graph. 00:31:31.100 --> 00:31:46.100 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. 00:31:46.100 --> 00:31:55.300 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. 00:31:55.300 --> 00:32:01.100 And you can see, as I mentioned, that this is not a function; it failed the vertical line test (as horizontal parabolas do). 00:32:01.100 --> 00:32:09.000 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. 00:32:09.000 --> 00:32:20.800 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. 00:32:20.800 --> 00:32:25.200 So, that is off this graph; but it gives us an idea of the shape. 00:32:25.200 --> 00:32:33.700 So, I know that my axis of symmetry is going to be here; and I have a point at (1,1); 00:32:33.700 --> 00:32:40.100 I have another point at (1,3); and then I have a point way out here at (10,0). 00:32:40.100 --> 00:32:51.100 I know that this is going to be a fairly narrow graph, because a equals 3. 00:32:51.100 --> 00:32:59.300 This is the graph of the horizontal parabola described by this equation; and here it is, written in standard form. 00:32:59.300 --> 00:33:03.800 So, it opens to the right; it is fairly narrow, because a equals 3. 00:33:03.800 --> 00:33:14.300 It has a vertex at (-2,2), and it has an axis of symmetry at y = 2. 00:33:14.300 --> 00:33:18.800 Example 3: we are asked to graph; this is also going to be a horizontal parabola. 00:33:18.800 --> 00:33:28.800 We are going to start out by putting it in the standard form, x = (y - k)² + h. 00:33:28.800 --> 00:33:34.400 We need to complete the square; start out by isolating the y variable terms on the right. 00:33:34.400 --> 00:33:40.700 So, I am going to add 6 to both sides to get -2y² + 8y. 00:33:40.700 --> 00:33:51.700 Since the leading coefficient is not 1, I need to factor it out; so I am going to factor this -2 to get y². 00:33:51.700 --> 00:33:55.400 Factoring a -2 from here would give me a -4. 00:33:55.400 --> 00:33:59.700 And I need to add something to this to complete the square. 00:33:59.700 --> 00:34:04.200 What I need to add is b²/4. 00:34:04.200 --> 00:34:21.200 b is 4, so I am going to be adding 4²/4; that is 16, divided by 4; that is 4. 00:34:21.200 --> 00:34:33.200 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. 00:34:33.200 --> 00:34:42.400 So, we subtract 8 from that side; to this side, since I am adding inside the parentheses, I am just adding 4. 00:34:42.400 --> 00:34:45.600 But then, 4 times -2--that is how I got the -8 on the left. 00:34:45.600 --> 00:34:56.800 This gives me x - 2 = -2; and I want it in this form, so I am going to rewrite this as (y - 2)². 00:34:56.800 --> 00:35:06.600 The last thing I need to do is add 2 to both sides; and I have it in standard form. 00:35:06.600 --> 00:35:09.800 Now that I have this in standard form, it is much easier to graph. 00:35:09.800 --> 00:35:19.500 The vertex is going to be at (h,k); so h is here; k is here; the vertex is at (2,2). 00:35:19.500 --> 00:35:42.400 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. 00:35:42.400 --> 00:35:45.700 Now, to finish out graphing this, I am going to find a few points. 00:35:45.700 --> 00:35:55.500 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. 00:35:55.500 --> 00:35:58.500 So, I know it is going to look something like this; but I will find a couple of points. 00:35:58.500 --> 00:36:11.900 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. 00:36:11.900 --> 00:36:14.400 The graph is just going to go this way. 00:36:14.400 --> 00:36:20.900 So, I can't use values that end up giving me an x that is greater than 2. 00:36:20.900 --> 00:36:35.600 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. 00:36:35.600 --> 00:36:44.200 And 3: 3 minus 2 is 1, squared is 1; 1 times -2 is -2, plus 2 is 0. 00:36:44.200 --> 00:36:54.400 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. 00:36:54.400 --> 00:37:06.100 And this is going to give me a parabola shaped like this, opening to the left with a vertex at (2,2). 00:37:06.100 --> 00:37:15.100 The axis of symmetry would be right through here; and I have a couple of points, just to make it a bit more precise. 00:37:15.100 --> 00:37:20.600 So, the first step in graphing a parabola is always to get it into this form by completing the square. 00:37:20.600 --> 00:37:30.000 And then, using the features you can see from here, sketch it out, and finding a few points, make the graph more accurate. 00:37:30.000 --> 00:37:36.100 Find the equation of the parabola with a vertex of (2,3) and focus at (2,7); draw the graph. 00:37:36.100 --> 00:37:46.100 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. 00:37:46.100 --> 00:37:58.400 Well, I am given that the vertex is at (2,3); so I know that the vertex is right here; that is the vertex. 00:37:58.400 --> 00:38:09.700 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. 00:38:09.700 --> 00:38:16.400 So, the vertex is (2,3); the focus is (2,7). 00:38:16.400 --> 00:38:21.400 Remember, in the beginning of this lesson, I mentioned that the focus is always inside the parabola. 00:38:21.400 --> 00:38:25.900 Since the focus is inside the parabola, I already know that this has to open upward. 00:38:25.900 --> 00:38:31.000 So, I know something about the shape of the graph. 00:38:31.000 --> 00:38:36.600 Let's find the equation: now, I know that this is a vertical parabola, because the focus is inside the parabola. 00:38:36.600 --> 00:38:40.200 That told me that this has to open upward, so I know I am dealing with a vertical parabola. 00:38:40.200 --> 00:38:50.600 And that helps me to find the equation, because the standard form is going to be y = a(x - h)² + k. 00:38:50.600 --> 00:38:59.700 I am given the vertex, so I am given h and k: I know that h = 2 and k = 3. 00:38:59.700 --> 00:39:06.700 In order to write this equation, I need a, h, and k; all I am missing is a. 00:39:06.700 --> 00:39:09.800 I am given the piece of information, though, that the focus is (2,7). 00:39:09.800 --> 00:39:13.100 And that is going to allow me to find a. 00:39:13.100 --> 00:39:18.700 You will recall that I mentioned the formulas for focus and directrix. 00:39:18.700 --> 00:39:36.700 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. 00:39:36.700 --> 00:39:52.300 And I know the focus is at (2,7): so 2 = h, and 7 = k + 1/4a, according to this definition. 00:39:52.300 --> 00:40:01.500 Well, since I know that k is 3, then I can solve this. 00:40:01.500 --> 00:40:08.100 So, I know k; so I can solve for a. 00:40:08.100 --> 00:40:22.000 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, 00:40:22.000 --> 00:40:26.800 or just take the reciprocal of each side (essentially, that is what you are doing) to get a = 1/16. 00:40:26.800 --> 00:40:33.400 Now, I have h and k given; I was able to figure out a, based on the definition of focus. 00:40:33.400 --> 00:40:44.600 So, I end up with the equation y = 1/16(x - 2)² + 3. 00:40:44.600 --> 00:40:46.100 So, this is the equation. 00:40:46.100 --> 00:40:51.100 And as you know, once we have the equation, the graphing is pretty easy. 00:40:51.100 --> 00:41:00.800 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. 00:41:00.800 --> 00:41:10.400 I am going to have a parabola that opens upward, with a vertex of (2,3), and fairly wide in shape. 00:41:10.400 --> 00:41:12.900 That was a pretty challenging problem, because you had to go back 00:41:12.900 --> 00:41:23.000 and think about how you could use a formula to find the focus; and knowing the focus allowed you to find a. 00:41:23.000 --> 00:41:27.000 That concludes this lesson on parabolas at Educator.com; thanks for visiting!