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Lecture Comments (12)

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


  • 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.


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

Transcription: Parabolas

Welcome to

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 = ax2, 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 = x2 + 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 lesson0580

on completing the square as part of this lecture series, but we will review it again now): first, I am going0586

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 b2/4.0608

In this case, b is 6, so this is going to give me 62/4, 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 up0697

getting the y variable terms on the right; but for now, the x variable terms on the right); complete the square0701

by adding the b2/4 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 x2, so it is in this form: y = ax2.0767

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), = x2.0833

Now, let's say I change this slightly: let's say I have another function, g(x) = x2 + 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 = x2 + 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 this.0962

So, this one is y = x2, 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 parabola1013

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 = ay2, 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 functions1208

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 could try1216

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 = y2.1278

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 = -y2.1321

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 y2 + 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 = is my graph of x = y2; over here is x = -y2.1379

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 = y2 over here; right here, this is actually x = y2 + 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 = 3y2 - 12y + 10.1513

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.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 = 3y2 - 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 y2 - 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 b2/4.1589

b is actually 4; so this is going to be 42/4, 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 -2y2 + 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 y2.2021

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 b2/4.2040

b is 4, so I am going to be adding 42/4; 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 back2470

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; thanks for visiting!2483