WEBVTT mathematics/math-analysis/selhorst-jones
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Welcome to Educator.com.
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Today, we are going to talk about parabolas.
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And in some earlier lectures in this series on quadratic equations, we talked about parabolas and did some graphing.
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But now, we are going to go on and give a specific definition to parabolas, and learn about some other features of parabolas.
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Although you have seen parabolas previously, when we graphed, we didn't form a specific definition of them.
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So, the definition of a parabola is that it is the set of points in the plane whose distance from a given point,
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called the focus, is equal to its distance from a given line, called the directrix.
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Let's talk about that before we go on to talk about the axis of symmetry.
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So, if you had a parabola (let's say right here; and we will do an upward-facing parabola), you would have some point,
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which is known as the focus, and a line (I'm going to put that right about here) called the directrix.
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By definition, every point on this parabola is equidistant from the focus and the directrix.
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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.
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And this is just a very rough sketch; but these distances actually would be equal; they are theoretically equal.
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Looking right here at the vertex, these distances would be equal; so that would be, say, y.
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If I took some other point, say here, and I measured here to here, these two distances would be equal.
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So, a couple things to note: the focus is inside the parabola; the directrix is outside.
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And this is because the focus and the directrix are on the opposite sides of the vertex.
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So, you could have a parabola facing downward, and then it would have a focus here and a directrix up here.
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We are also going to talk, today, about parabolas that face to the left and right--horizontal parabolas.
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But right now, we are going to stick with just (for this discussion) focusing on vertical ones,
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the definition being that every point in the parabola equidistant between the focus and the directrix.
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The axis of symmetry of the parabola passes through the focus; and it is perpendicular to the directrix.
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In this case, the y-axis is the axis of symmetry; it is right here.
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And you see that it passes through the focus, and it forms a right angle; it is perpendicular to the directrix.
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Again, we talked about some of these concepts in earlier lectures.
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But to review, **vertex**: the vertex of a parabola is the point at which the axis of symmetry intersects the parabola.
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And it is a maximum or minimum point on the parabola, if the axis of symmetry is vertical.
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If the axis of symmetry is horizontal (say we have a parabola like this, then the axis of symmetry would be horizontal),
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we still have a vertex, but it is not a maximum or minimum.
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And again, we are going to focus a little more on vertical parabolas right now, and then we will talk about horizontal parabolas.
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So, if I have a downward-facing parabola, the vertex is here; the axis of symmetry is right here.
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And this vertex is the maximum; this is as large as y gets--it is the largest value that the function attains.
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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.
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In this case, the vertex is a minimum; this is the smallest value that the function will attain.
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The standard form of a parabola with vertex at (h,k) is y = a(x - h)² + k.
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And this is for vertical parabolas; there is a slightly different form when we are talking about horizontal parabolas.
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And you might recall this form of the equation that we covered earlier on, under the lecture on quadratic equations.
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And we called this the vertex form of the equation; now we are going to refer to it as standard form.
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And it is a very useful form, because it tells you a lot about the parabola.
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The axis of symmetry is x = h: so I know a few things just from looking at this.
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I know the vertex, because it is (h,k); I know the axis of symmetry--it is at x = h;
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and if I look at a, I will know if the parabola is upward- or downward-facing.
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If a is greater than 0, the parabola will open upward; and k gives you the minimum.
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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.
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Let's look at an example: y = 2(x - 1)² + 4.
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So, this is in standard form: this means that I have h = 1, k = 4, and a = 2.
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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.
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And since a is greater than 0, this opens upward.
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So, I can sketch this out: I have a vertex at (1,4), right here, and it opens upward.
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And the axis of symmetry is going to be right here at x = 1.
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Here is my vertex at (1,4); and this vertex is a minimum, because this opens upward.
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The minimum value is k, which is 4.
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If I were to take a similar situation, but say y = -2(x - 1)² + 4,
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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.
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What I would end up with would be a parabola here, again, with the vertex at (1,4).
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But it would open downward, and therefore, this would be a maximum.
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Also, if the absolute value of a is greater than 1, you end up with a relatively narrow parabola.
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If the absolute value of a is less than 1, you end up with a relatively wide parabola.
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So, this form is very useful, because just by having the equation in this form, we can at least sketch the graph.
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Let's talk a little bit more about graphing parabolas.
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You can use symmetry and translations to graph a parabola: and by translations, we mean a shift.
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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.
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And then, if you altered what h is, it is going to shift the graph horizontally by that number of units.
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If you alter what k is, it is going to translate or shift that graph upward and downward by a certain number of units.
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In order to graph a parabola, you often need to put it in standard form.
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Let's start out by just talking about putting an equation or a parabola in standard form.
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And then we will go on and look at some graphs, and how different values of h and k can affect the graph.
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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,
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and I want it in this standard form, y = a(x - h)² + k.
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The first thing to do (and this is, again, review from an earlier lesson--you can go back and look at the lesson
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on completing the square as part of this lecture series, but we will review it again now): first, I am going
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to isolate the x variable terms on the right side of the equation.
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I am going to add 8 to both sides: now I am going to complete the square.
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I am going to focus on this, and I need to add a term to it to make this a perfect square trinomial.
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The term I am going to add is going to be b²/4.
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In this case, b is 6, so this is going to give me 6²/4, which is 36/4, which is equal to 9.
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So, that is what I need to add in here: y + 8, plus I need to add 9 to both sides.
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It is easy to forget to add it to the other side, because you get so focused on completing the square.
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But if you don't, the equation will no longer be balanced.
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So, I am going to add 9 to both sides.
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And I want this to end up in this form; so I am going to rewrite this.
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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.
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And I look at what I have, and it is almost in this form, but not quite.
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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.
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And this is in this form: a happens to be equal to 1 in this case.
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And so, if you are given an equation that is not in standard form, and you want to get it in standard form,
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isolate the x variable values on the right (although if we are working with horizontal parabolas,
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it is going to be the other way around, as we will see in a minute--we are actually going to end up
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getting the y variable terms on the right; but for now, the x variable terms on the right); complete the square
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by adding the b²/4 term to both sides of the equation; and then simplify;
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shift things around as needed to get it in this form.
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Remember, also, that if you have a leading coefficient that is something other than 1,
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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.
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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.
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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.
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And translation is knowing how h and k, and changes in h and k, affect the graph, in order to graph.
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All right, so let's just start out with something in this form--a very basic equation for a parabola.
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Let's let f(x) equal x², so it is in this form: y = ax².
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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.
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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.
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And you can also very easily find some points to graph this.
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right now, I am just going to sketch it out, and not worry about exact points, just so you get the idea.
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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);
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it is upward-opening; and it is going to look something like this.
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So, this is my graph here of y, or f(x), = x².
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Now, let's say I change this slightly: let's say I have another function, g(x) = x² + 2.
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So, looking at this form, h is still 0; but now I have k = 2.
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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.
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I have a similar graph, but it is going to be with the vertex right here at (0,2).
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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.
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This is shifted upward; it still opens upward, because a is positive.
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So, now I am just going to have a similar idea, but shifted upward by 2.
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So here, I have y = x² + 2.
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If this had been a -2, then it would have been shifted down by 2, and I would have had a graph right here.
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So, let's see what happens when I change h.
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Let's get a third function: we will call it h(x) = (x - 1)².
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OK, now what I have here is h = 1; k, if I look here, is 0.
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Therefore, the vertex (this is the vertex right here) equals (1,0), and the axis of symmetry is going to be at x = 1.
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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.
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So, this one is y = x², and this is y = (x - 1)².
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Important take-home points: a change in h will shift the graph horizontally, to the right or left.
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A change in k will shift the basic graph either up or down, by k number of units.
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Using symmetry: if I were to graph these out exactly, I would need to find points.
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And I don't need to find all of the points: for example, if I had a parabola that was a downward-facing parabola
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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.
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All right, this concept is another one adding on to our knowledge of parabolas from prior lessons.
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And it is defining a segment called the latus rectum.
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The **latus rectum** is the segment passing through the focus and perpendicular to the axis of symmetry.
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Let's see what that means--let's visualize that.
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Let's say I have a parabola like this, and let's say the focus is here.
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So, this passes through the focus, and is perpendicular to the axis of symmetry.
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This is the focus, and here we have the axis of symmetry.
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That means the latus rectum is going to pass through here, and it is going to be perpendicular to the axis of symmetry.
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So, that is this line; this is the latus rectum.
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The equation for its length is the absolute value of 1/a; and if you have the equation of the parabola in standard form,
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then this a is the same a as you will see in that formula.
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So, this is something you might occasionally need to use.
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For example, if I were given an equation of a parabola y = 2(x - 3)² + 5,
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and I was asked to find the length of the latus rectum of this parabola, then I would just say,
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"OK, a equals 2; therefore, the length equals the absolute value of 1/2."
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Horizontal parabolas: I mentioned that you can also have parabolas that open to the right or left, not just up and down,
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although up to this point in the course, we have just talked about vertical parabolas, or parabolas that open upward or downward.
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For parabolas whose axis of symmetry is horizontal, we end up with equations in this form: y = a(x - k)² + h.
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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.
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In the vertical formula, the h was in here, and the k was out here.
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So, be careful when you are working with this formula to notice that the positions of h and k are reversed.
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And there are translations of x = ay², and then again, h and k shift this graph around horizontally and vertically.
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So, it would look something like this, for example: the axis of symmetry would be right here;
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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.
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These do not represent functions; and you can see that they don't represent functions
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by trying to pass a vertical line through them: they fail the vertical line test.
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Remember: with a function, the vertical line test tells us that a vertical line drawn...you could try
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any possible area of the curve, and the vertical line will only cross the curve once.
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If the vertical line crosses the curve more than once, it is not a function.
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So, this fails the vertical line test.
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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.
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I am working on graphing some horizontal parabolas.
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When you look at the equation in standard form, y = a...and remember, the k and h are in opposite positions;
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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.
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If a is negative, then the parabola is going to open to the left.
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So, let's look at a very simple horizontal parabola, x = y².
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OK, the vertex is at (h,k); and I can see that h and k are both 0, so the vertex equals (0,0).
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The axis of symmetry is at y = k, so that is going to be at y = 0.
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And the a here is 1: a = 1, so this opens to the right.
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So, you are going to have a parabola that looks something like this.
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You could have another parabola, x = -y².
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Here we would have the same vertex and the same axis of symmetry; here the x-axis is actually the axis of symmetry.
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And I look at a now, and a equals -1, so this parabola is going to open to the left.
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So, I am going to end up with a parabola like this.
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Now again, change in h or change in k is going to shift this parabola a bit.
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Let's change h and see what happens: let's let x equal y² + 2.
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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.
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If I look at this, x = y²...here is my graph of x = y²; over here is x = -y².
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Now, I am going to have h = 2, so that is going to shift horizontally by 2.
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(2,0) will be the vertex; and it is going to open to the right.
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So, this is x = y² over here; right here, this is actually x = y² + 2 now.
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And k, as discussed before, shifts the graph of a parabola vertically.
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The same idea here: if I were to change k, then I would shift this graph up or down by k units.
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So, with horizontal parabolas, you need to be familiar with this equation.
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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.
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The vertex is at (h,k), and the axis of symmetry is y = k.
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And you also need to keep in mind that these do not represent functions.
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In the beginning of today's lesson, we talked about the focus and directrix.
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And here are formulas to allow you to find those if you need to.
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If you have a vertical parabola, the coordinates of the focus are h for the x-coordinate, and k plus 1/4a.
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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.
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And this would be for a vertical parabola; for a horizontal parabola, the focus is found at the coordinates h + 1/4a;
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and then the y-coordinate is k, so the focus is a point, and this gives the coordinates of that point.
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The directrix is a line, and the equation for this line for a horizontal parabola is x = h - 1/4a.
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And you might need to occasionally use these when we are working problems.
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And we will see that in one of the examples, actually, shortly.
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Starting out with Example 1: Write in standard form and identify the key features: x = 3y² - 12y + 10.
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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.
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So, the standard form of this equation is going to be x = a(y - k)² + h.
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Remember, h and k are going to be in opposite positions.
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In order to get this equation in standard form, we need to complete the square.
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This time, since I am working with a horizontal parabola, I am going to isolate all of the y variable terms on the right.
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And I am going to do that by subtracting 10 from both sides to get x - 10 = 3y² - 12y.
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This leading coefficient is not 1, so I have to factor it out.
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And then, I have to be really careful when I am adding to both sides of the equation, because this is factored out.
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So, factor out a 3 to get y² - 4y.
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I need to complete the square: that means I need to add something over here.
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And the term that I need to add is going to be b²/4.
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b is actually 4; so this is going to be 4²/4, equals 16/4, equals 4.
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Here is where I need to be careful: on the right, I am adding 4 inside these parentheses, which is pretty straightforward.
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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.
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So, if I were just to add 4, this equation would not be balanced,
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because in reality, what I am doing over here is adding 3 times 4.
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So, on the right, I am going to add 12; and I got that from 3 times 4.
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Simplifying the left: 12 - 10 is 2; on the right, inside here, I now have a perfect square.
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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.
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I am almost done; I just need to move this constant over to the right to have it in this form.
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x = 3 times (y - 2)², minus 2.
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So, now that I have this in standard form, I can identify key features.
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Key features: 1: this is a horizontal parabola, as you can see from looking at this equation.
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2: The vertex is at (h,k); h is 2, and k is also 2.
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Actually, being careful with the signs, h is actually -2, because remember, standard form has a plus here.
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I don't have a plus here; I could rewrite this so that I do, and that would give me + -2.
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And it is good practice, actually, to write it exactly in this form, although this is correct--you could leave it like this.
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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,
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because here I need a negative to be in standard form; if I ended up with a plus here,
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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).
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And because a equals 3, that means that a is greater than 0; a is positive, so the parabola opens to the right.
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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.
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We can also say that the axis of symmetry is at y = k, and therefore the axis of symmetry is at y = 2.
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OK, in Example 2, we are asked to graph.
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And you will notice that this is the same equation that we worked with in Example 1.
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We already figured out standard form: and standard form is x = 3(y - 2)² - 2.
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And for clarity, we can actually write this as I did at the end, which is 3(y - 2)² + -2,
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so that we truly have it in standard form, with the plus here to make it easy to see what is going on.
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To graph this, I want to know the vertex: the vertex is (h,k): h here is -2; k is 2.
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The axis of symmetry is going to be at y = k; k is 2, so it is going to be at y = 2.
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I know that this opens to the right, so I have a general sense of this graph.
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But I can also just find a few points.
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And we are used to working with a situation where x is the input and y is the output.
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It is the opposite here, so we need to be really careful.
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I also want to note that, since the vertex is here at (-2,2), and this opens to the right,
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for this graph, we are not going to have values of x that are smaller than -2.
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So, if I end up with something where an x is smaller than -2, then it is going to be off the graph.
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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.
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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.
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And you can see, as I mentioned, that this is not a function; it failed the vertical line test (as horizontal parabolas do).
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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.
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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.
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So, that is off this graph; but it gives us an idea of the shape.
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So, I know that my axis of symmetry is going to be here; and I have a point at (1,1);
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I have another point at (1,3); and then I have a point way out here at (10,0).
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I know that this is going to be a fairly narrow graph, because a equals 3.
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This is the graph of the horizontal parabola described by this equation; and here it is, written in standard form.
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So, it opens to the right; it is fairly narrow, because a equals 3.
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It has a vertex at (-2,2), and it has an axis of symmetry at y = 2.
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Example 3: we are asked to graph; this is also going to be a horizontal parabola.
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We are going to start out by putting it in the standard form, x = (y - k)² + h.
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We need to complete the square; start out by isolating the y variable terms on the right.
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So, I am going to add 6 to both sides to get -2y² + 8y.
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Since the leading coefficient is not 1, I need to factor it out; so I am going to factor this -2 to get y².
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Factoring a -2 from here would give me a -4.
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And I need to add something to this to complete the square.
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What I need to add is b²/4.
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b is 4, so I am going to be adding 4²/4; that is 16, divided by 4; that is 4.
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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.
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So, we subtract 8 from that side; to this side, since I am adding inside the parentheses, I am just adding 4.
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But then, 4 times -2--that is how I got the -8 on the left.
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This gives me x - 2 = -2; and I want it in this form, so I am going to rewrite this as (y - 2)².
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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.
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The vertex is going to be at (h,k); so h is here; k is here; the vertex is at (2,2).
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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.
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Now, to finish out graphing this, I am going to find a few points.
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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.
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So, I know it is going to look something like this; but I will find a couple of points.
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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.
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The graph is just going to go this way.
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So, I can't use values that end up giving me an x that is greater than 2.
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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.
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And 3: 3 minus 2 is 1, squared is 1; 1 times -2 is -2, plus 2 is 0.
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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.
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And this is going to give me a parabola shaped like this, opening to the left with a vertex at (2,2).
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The axis of symmetry would be right through here; and I have a couple of points, just to make it a bit more precise.
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So, the first step in graphing a parabola is always to get it into this form by completing the square.
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And then, using the features you can see from here, sketch it out, and finding a few points, make the graph more accurate.
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Find the equation of the parabola with a vertex of (2,3) and focus at (2,7); draw the graph.
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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.
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Well, I am given that the vertex is at (2,3); so I know that the vertex is right here; that is the vertex.
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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.
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So, the vertex is (2,3); the focus is (2,7).
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Remember, in the beginning of this lesson, I mentioned that the focus is always inside the parabola.
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Since the focus is inside the parabola, I already know that this has to open upward.
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So, I know something about the shape of the graph.
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Let's find the equation: now, I know that this is a vertical parabola, because the focus is inside the parabola.
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That told me that this has to open upward, so I know I am dealing with a vertical parabola.
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And that helps me to find the equation, because the standard form is going to be y = a(x - h)² + k.
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I am given the vertex, so I am given h and k: I know that h = 2 and k = 3.
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In order to write this equation, I need a, h, and k; all I am missing is a.
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I am given the piece of information, though, that the focus is (2,7).
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And that is going to allow me to find a.
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You will recall that I mentioned the formulas for focus and directrix.
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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.
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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.
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So, I know k; so I can solve for a.
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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,
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or just take the reciprocal of each side (essentially, that is what you are doing) to get a = 1/16.
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Now, I have h and k given; I was able to figure out a, based on the definition of focus.
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So, I end up with the equation y = 1/16(x - 2)² + 3.
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So, this is the equation.
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And as you know, once we have the equation, the graphing is pretty easy.
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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.
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I am going to have a parabola that opens upward, with a vertex of (2,3), and fairly wide in shape.
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That was a pretty challenging problem, because you had to go back
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and think about how you could use a formula to find the focus; and knowing the focus allowed you to find a.
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That concludes this lesson on parabolas at Educator.com; thanks for visiting!