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About this Lesson
 Type: Video Tutorial
 Length: 8:25
 Media: Video/mp4
 Use: Watch Online & Download
 Access Period: Unrestricted
 Download: MP4 (iPod compatible)
 Size: 90 MB
 Posted: 06/26/2009
This lesson is part of the following series:
College Algebra: Full Course (258 lessons, $198.00)
College Algebra: Relations and Functions (57 lessons, $74.25)
College Algebra: Using the Quadratic Formula (4 lessons, $5.94)
Taught by Professor Edward Burger, this lesson was selected from a broader, comprehensive course, College Algebra. This course and others are available from Thinkwell, Inc. The full course can be found athttp://www.thinkwell.com/student/product/collegealgebra. The full course covers equations and inequalities, relations and functions, polynomial and rational functions, exponential and logarithmic functions, systems of equations, conic sections and a variety of other AP algebra, advanced algebra and Algebra II topics.
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 UTAustin 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, padic 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
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Now, you may remember the discriminate of a quadratic. Let me remind you what that is really fast. If you have a quadratic, let's say, f(x) = ax^2 + bx + c, then the discriminate, well, that's the thing that would be under the square root if you use the quadratic equation to solve that equals zero. And remember the quadratic equation says if you set that equal to zero and solve, then the solution is x = b the square root of b^2  4ac all divided by 2a. That's how you find the two roots if that equals zero. So that's actually finding you the x intercept for this parabola, because you set that equal to zero and solve.
But remember, when will there be two real roots? Well, think about it. There will be two real roots when you have ± the square root and that square root is, in fact, positive. So when the thing inside the square rootremember, that's b^2  4acwhen that's positive then we know we're going to have two real roots. If that equals zero, then I'm going to have, in the formula, b ± 0. Well, that means there's only one real root because I have ± 0. So if this number were to be 0, then I know I only have one real root, only one x intercept. And finally, if I see that this number's actually negative I can't take the square root of a negative number, so there must be no real roots, which would mean that the parabola would never cross the xaxis.
Now, you may remember this is called the discriminate, sometimes known as D or some other way. So just knowing the discriminate, in fact, just computing b^2  4ac, you can actually get and sketch a pretty accurate picture of what the graph of the function would look like, because all you have to do is say, okay, first of all, is it going to cross in two places, which would be where the discriminate is positive, because I'd have two real solutions. Is it going to cross just in one place, which means the parabola comes down, just raises, and then comes up. That's when the discriminate equals zero, because there's only one root. And finally, maybe it doesn't cross at all. This is when the discriminate is, in fact, negative.
So just knowing the discriminate and looking at the sign of a, seeing if it's a happy face parabola or a sad face parabola, you can actually make a rough, but a pretty reasonable sketch of what the function looks like, just looking at the discriminate.
And to really drive this home, folks, it's time for that favorite latenight game show you've been waiting for all your lives. You can't wait; every time it comes on in the middle of the week you get excited. Yes, folks, it's time for "Match Game," with your host, Ed Burger. Now, how do we play match game? Let me remind you. We show you the candidates, a > 0, and then I tell you exactly what the discriminate is, and of course, here I'm taking a look at the function f(x) = ax^2 + bx + c. So here's the discriminate and there's the a coefficient. And these are candidates for this evening's show. We have a is positive and the discriminate is positive. We have a is negative and the discriminate is negative. We have a is positive and the discriminate is zeroquite unusual. We have that a is negative and the discriminate is zero. And we have that a is negative but the discriminate is positive, and finally, ladies and gentlemen, for your viewing pleasure we have that a is positive and the discriminate is negative.
Your task, if you choose to accept it, which we hope you do, because we have some wonderful parting gifts for you in either case, is for each graph that I show you, figure out which particular scenario we're in. Let's begin.
Our first graphyour job is to figure out which of these, if any, corresponds to a graph, a parabola, that looks roughly like this. Try it now.
Okay, well let's see. I see that this is a sad face parabola, so that tells us, and the viewing audience at home, that the a coefficient, the a here, must be a negative number. So this is not in the candidacy, so we're only looking at this, this, or this. Now, what else do we know? Well, I see that it crosses the xaxis at two points. That means there are two real roots, and if there are two real roots; that means that the discriminate must be positive. So I want a to be negative and I want the discriminate to be positive. Yes, ladies and gentlemen, it's the penultimate answer right herea is negative and the discriminate is positive. It has two real rootsthe discriminate is positivebut a is negative, it's a sad face parabola. So roughly speaking, that's sort of how the graph of that would go. Neat! Congratulations to those of you who got it correctly. Let's try graph number 2. See if you can find the appropriate match.
Well, I see a happy face parabola so that means that a must be positive. The coefficient must be positive, so that takes care of this and this possibilities. But I see that the parabola comes down and just grazes this xaxis at one point and comes back up, so there's only one real root. The only way there could be one real root is if the discriminate were to be zero. So therefore the discriminate is zero, so a must be positive, so therefore I see we must have a positive, it's going up like this, happy face, but the discriminate is zero, it just grazes the xaxis. Neat, now we're moving. Let's take a look at graph number 3. See if you can make a match.
Well, I see that this is a happy face parabola, so a is positive, and I see that I have two real roots. It crosses the xaxis twice, so I must have the discriminate is positive. So I must have a is positive, discriminate is positive, that is choice number one, right here. We have a is positive, it's a happy face, the discriminate is positive, that means there will be two roots that crosses the xaxis in two different places. Neato! Okay, graph number... I think we're up to 4, but I'm not quite sure. Graph number whatever. See if you can find a match.
Okay, well, I see that a is negative because this thing is going down, it's a sad face parabola, so that means actually there's only one possibility then. Oh no, it's two possibilities. It could be either one of these two things. So let's see. Well, this just grazes the xaxis, so it only has one real root. The only way that can happen is if the thing under the square root were to be zero, so I don't have two roots, so that means the discriminate must be zero. So a is negative means it's going like thissad faceand discriminate zero, just grazes the xaxis. Congratulations to all. And now, our penultimate one. The viewers at home have a 50/50 chance of getting this right. It's sort of a great thing about this game when you get down to the end. Can you figure out which of these two things is the appropriate answer? Good luck.
Let's see. I see that a must be negative because this is a sad face parabola, and notice it does not cross the axis at all. That means there won't be any real solutions where this thing is going to actually equal zero. That must mean that you have complex solutions, which means the square root thing underneath must be negative. So I must have a negative square root, but it's a sad face parabola, so a must be negative, so there is the answer.
And then, we will not even do this one, but let me show you the next one, the final one, is this one. Ha ha! Fooled you. You thought I was going to have a different placard. But no, no, no. In fact, this corresponds perfectly to this and you can see why. This is a happy face parabola, so a is positive, and yet it still does not cross the axis, so this number, the discriminate, must be negative. Congratulations, and whether you won a lot or won a little, remember doing math, you're always a winner. Try some of these and see how much money you can make.
Relations and Functions
Quadratic Functions  Basics
Using Discriminants to Graph Parabolas Page [1 of 2]
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