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Int Algebra: Find Vertex by Completing the Square

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About this Lesson

  • Type: Video Tutorial
  • Length: 6:07
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 65 MB
  • Posted: 12/02/2008

This lesson is part of the following series:

College Algebra: Full Course (258 lessons, $198.00)
Trigonometry: Full Course (152 lessons, $148.50)
College Algebra: Relations and Functions (57 lessons, $74.25)
Trigonometry: Algebra Prerequisites (60 lessons, $69.30)
Intermediate Algebra Review (25 lessons, $49.50)
College Algebra: Quadratic Functions: The Vertex (4 lessons, $6.93)
College Algebra: Solving by Completing the Square (3 lessons, $4.95)
Int Algebra: Solving by Completing the Square (3 lessons, $4.95)

In this lesson, you will learn how to find the vertex of a parabola given the formula for the parabola. To do this, you will complete the square. By completing the square of the parabola equation, you will be able to get the equation into a standard form that can be more easily evaluated. A parabola is a conic section in which the locus of points constructing it are equidistant from the focus and the directrix. Once we've identified the vertex of a parabola, we can get a good sense for how the parabola is positioned on the Cartesian coordinate plane.

Taught by Professor Edward Burger, this lesson was selected from a broader, comprehensive course, Intermediate Algebra. This course and others are available from Thinkwell, Inc. The full course can be found at http://www.thinkwell.com/student/product/intermediatealgebra. The full course covers real numbers, equations and inequalities, exponents and polynomials, rational expressions, roots and radicals, relations and functions, the straight line, systems of equations, quadratic equations and quadratic inequalities, conic sections, inverse and exponential and logarithmic functions, and a variety of other AP algebra and advanced algebra.

Edward Burger, Professor of Mathematics at Williams College, earned his Ph.D. at the University of Texas at Austin, having graduated summa cum laude with distinction in mathematics from Connecticut College.

He has also taught at UT-Austin and the University of Colorado at Boulder, and he served as a fellow at the University of Waterloo in Canada and at Macquarie University in Australia. Prof. Burger has won many awards, including the 2001 Haimo Award for Distinguished Teaching of Mathematics, the 2004 Chauvenet Prize, and the 2006 Lester R. Ford Award, all from the Mathematical Association of America. In 2006, Reader's Digest named him in the "100 Best of America".

Prof. Burger is the author of over 50 articles, videos, and books, including the trade book, Coincidences, Chaos, and All That Math Jazz: Making Light of Weighty Ideas and of the textbook The Heart of Mathematics: An Invitation to Effective Thinking. He also speaks frequently to professional and public audiences, referees professional journals, and publishes articles in leading math journals, including The Journal of Number Theory and American Mathematical Monthly. His areas of specialty include number theory, Diophantine approximation, p-adic analysis, the geometry of numbers, and the theory of continued fractions.

Prof. Burger's unique sense of humor and his teaching expertise combine to make him the ideal presenter of Thinkwell's entertaining and informative video lectures.

About this Author

Thinkwell
Thinkwell
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Founded in 1997, Thinkwell has succeeded in creating "next-generation" textbooks that help students learn and teachers teach. Capitalizing on the power of new technology, Thinkwell products prepare students more effectively for their coursework than any printed textbook can. Thinkwell has assembled a group of talented industry professionals who have shaped the company into the leading provider of technology-based textbooks. For more information about Thinkwell, please visit www.thinkwell.com or visit Thinkwell's Video Lesson Store at http://thinkwell.mindbites.com/.

Thinkwell lessons feature a star-studded cast of outstanding university professors: Edward Burger (Pre-Algebra through...

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Recent Reviews

Nopic_gry
Helpful
02/24/2012
~ Priscilla3

This is very helpful, but would be even better if some of the problems had had an a- coefficient, such as f(x)=3x^2-9x=14

Nopic_tan
Great quality video!
12/31/2008
~ brittanie

This video series is really easy to follow and the video quality is amazing. The colors stand out and make it really interesting to watch. I'm a big fan of this higher video production quality versus the at-you-home style backgrounds and lighting.

Nopic_gry
Helpful
02/24/2012
~ Priscilla3

This is very helpful, but would be even better if some of the problems had had an a- coefficient, such as f(x)=3x^2-9x=14

Nopic_tan
Great quality video!
12/31/2008
~ brittanie

This video series is really easy to follow and the video quality is amazing. The colors stand out and make it really interesting to watch. I'm a big fan of this higher video production quality versus the at-you-home style backgrounds and lighting.

Quadratic Equations and Inequalities
Solving by Completing the Square
Finding the Vertex by Completing the Square Page [1 of 1]
Okay, so the standard form of a quadratic or of a parabola is of the form f(x) = a(x - h)2 + k. So if you can ever get a
parabola or quadratic in that form, you can just read off where the vertex is, it’s at (h, k). Well, how can you do that?
Well, one way is to plug into the formula, which is the way I like, but another way that some people might ask you to
do, is to actually do this by completing the square. Let me illustrate how you would actually do this. Suppose I give
you the following function: f(x) = x2 + 4x + 1. Sort of hard to read off the h and the k that make up the vertex, that
bottom point right there, of the parabola. So how can you find it? You can find it by writing this in this form by
completing the square. You have sort of part of a square; you want to add the other part to make it a perfect square.
So what would I do? Well, I would behave just like we used to behave when we completed the square. I would put a
4x, write that down, and I’m going to now give a big piece of space here, put a plus, and I’m going to add some
number. But I can’t change the value of the function, so if I add a number, I have to immediately subtract that number.
So if you’re going to add in a number like 17, you have to immediately subtract 17, otherwise, you’re going to change
the function. But if I add 17, subtract 17, they cancel, I don’t change the function. Now, what should I add? Well, the
complete the square method is always the same. I take half of this coefficient, which is 2, and square it, and I get a 4.
So I put a 4 in there and I immediately take it out. If I put it in, take it out, nothing happens. But now look what goes
on. This piece here can be factored perfectly as (x + 2)2. And here I’m left with just a -3.
So look, voila! I have written this complicated quadratic in standard form. And now I can just read off where the
vertex is. The vertex is at -2--because remember I take the opposite sign here, (-2, -3). So this is standard parabola
that’s centered at (-2, -3), so actually I have a good sense of what that looks like, right? If here’s a standard parabola,
I go -2, so that would be… And then I go -3. So actually, using the completing the square method is a neat way to
write it in standard form and then you can read off what this thing looks like.
Let’s try another example. How about if the quadratic was given to us as x2 - 8x + 5? Well, again, it’s not in the nice
form that I want, so I want to complete the square. So what would I do? Well, I’d write x2 - 8x, and then I’m going to
add something over here--plus the 5--but then I have to immediately subtract it, because otherwise I’m going to
change the value. So whatever I write in here, I’m going to have to subtract it away immediately. What do I write?
Well, I take half of -8, which would be -4, and then I square it, which is 16. So I’m going to put in a 16 and
immediately take it out, 16 - 16 = 0, it’s still the same thing as this. But, the point is writing this way allows me to
factor. This factor is a perfect square. This is going to be (x - 4)2 and this gives me a -11. And now I can just read off
what I’ve got here. If I read off what I’ve got, this is the same function as this, but now I can see the vertex is located
at (4, -11). And why? Because remember, I take the opposite of the sign for--and then, -11. So it’s just a good, oldfashioned
parabola. So I’m going to try to do this for you. But now I’m going to go 4 units over, so I go one, two,
three, four, and then -11--one, two, three, four, five, six, seven, eight, nine, ten, eleven. So the parabola looks just like
that. See? So you can get a really good sense of how the parabola sits by just finding where the vertex sits.
Let’s try one last one. f(x) = -x2 + 4x + 2. Now, I want to complete the square here. I’m a little nervous, to be honest
with you, because I have that negative sign way out in front. I’m sort of nervous about that, so you know what I’m
going to do first? I’m just going to factor it out of everything so I don’t have to worry about it. If I factor it out I’m going
to get an x2 then -4x and then I’m going to have a -2. Let me just hold that out so that I’m all done with my factoring
and then I’ll worry about it later. Now I’m going to complete the squares as normal. So I have an x2 - 4x. I’m going to
add something in--put the -2--but then I’m going to subtract it out immediately, so I’m not going to change the value.
What do I do? I take half of -4, which is -2 and square it, and I get 4, and I’m going to subtract it off immediately. And
now what do I see? And this thing, notice, factors really, really nicely. That’s just (x - 2)2. And then I have a -2, -4 is -
6, and I can distribute that negative sign now to both of these terms, and I see -(x - 2)2 + 6. And so what do I have? I
see that, in fact, this is going to be a sad face parabola, and the vertex is at (2, 6). So what I’ve got is a good, oldfashioned
parabola, but now what do I do? I go 2 units over and I go 6 units up. But it’s sad faced, so it’s just like
that. So that’s what the parabola looks like. You get a very accurate sense of what it looks like even graphically, just
by finding the vertex, just by writing it in standard form.

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