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About this Lesson
 Type: Video Tutorial
 Length: 17:53
 Media: Video/mp4
 Use: Watch Online & Download
 Access Period: Unrestricted
 Download: MP4 (iPod compatible)
 Size: 194 MB
 Posted: 11/18/2008
This lesson is part of the following series:
Calculus (279 lessons, $198.00)
Calculus Review (48 lessons, $95.04)
Calculus: Final Exam Test Prep and Review (45 lessons, $64.35)
Calculus: Intro to Derivatives (10 lessons, $17.82)
Calculus: Using the Derivative (4 lessons, $8.91)
In this lesson, you will cover how to find the equation of a line tangent to a curve, equations of tangent lines, how to find the derivative, and how to find the point where the tangent line is horizontal. To find the equation of the tangent line, you'll start by taking the derivative of the curve and then evaluate that derivative at the point of tangency. You'll then substitute the coordinates of the point of tangency as well as the calculated slope into the pointslope form of the line: yy1 = m*(xx1) to get the equation of the tangent line. To find where the tangent line is horizontal, you'll set the derivative of the function equal to zero and solve for x.
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 at http://www.thinkwell.com/student/product/calculus. The full course covers limits, derivatives, implicit differentiation, integration or antidifferentiation, L'Hôpital's Rule, functions and their inverses, improper integrals, integral calculus, differential calculus, sequences, series, differential equations, parametric equations, polar coordinates, vector calculus and a variety of other AP Calculus, College Calculus and Calculus 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
 Thinkwell
 2174 lessons
 Joined:
11/13/2008
Founded in 1997, Thinkwell has succeeded in creating "nextgeneration" 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 technologybased 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 starstudded cast of outstanding university professors: Edward Burger (PreAlgebra through...
More..Recent Reviews
 Calculus: The Equation of a Tangent Line
 10/01/2009

WHY CAN'T YOU BE MY COLLEGE TEACHER!!!! Awesome JOB
 Great Explanation to a difficult problem!
 01/20/2009

He explains how to find the equation of the line tangent to f(x) at the point x=l. With this in mind he goes over several problems solving the equation of a tangent line.
He also shows you how to find the slope and derivative of the function. This is a difficult lesson but he does a good job breaking down each section.
An Introduction to Derivatives
Using the Derivative
The Equation of a Tangent Line Page [1 of 1]
Okay. Well, this question is going to be not as scary as the crab question. I’m going to start off, basically, by just giving you this function. Let’s consider the function f of x equals five x squared minus x plus three. Okay? And the question is—well, can we find the equation of the line that’s tangent to the function f of x at the point x equals one.
So, you have to visualize now this curve, and I want to find the tangent line. In fact, let me get my line—oh, wait—did I—where’s the line? What did we do with the line? I need—I need—I need the line. I need the line. Who has it? It’s—well! Look! Look, ladies and gentlemen, it’s calcy—it’s Lucky, the calculus dolphin. Well, thank you! Oh, isn’t this sweet. Look at this. He is a very nice—this little dolphin thing is very nice and—ow! Okay. Get down. (Clears throat.)
Well, anyway. I’ve got the line. So, imagine now if you will this curve and I want to find the tangent line—the equation, in fact, of the tangent line—at x equals one. So how do you do that? Mmm, Lucky was an idiot. He should have eaten the line.
Okay. Well, what do you need to do to find the equation of a line? Two pieces of information are required. Remember, we talked about lines a while back. What do we need to know? Well, we need to know a point on the line, and we need to know the slope of the line. Well, how do we find the slope of the line? Well, what do we know about the line? We know that it’s the line that’s tangent to the function. So, to find slope of a tangent line, what do we do? Well, I remind you, the derivative of the function actually represents the slope of the tangent line.
So, look at that! Surprise, surprise. It turns out that this problem involves finding the derivative. Are you shocked? Are you shaken? Are you excited? I am. At least I’m some of those things, but I’m not going to tell you which ones. Now, the question is  we need to first find the derivative of the function. How do we find the derivative of the function? Once we find the derivative of the function, that will give us the slope, but then we’re not quite finished yet, because we’ll still have to actually find the equation of the line.
Here’s a really important point, by the way. When you’re doing these problems that have a lot of parts, you might get so excited to find one answer, you just stop. Remember the question here is find the equation of the line. So at the end of the day, we have to come back and the answer has to be not seven, four, but an equation for the line. We’ll come back to that. In fact, I’ll put that over there. There it is.
And—oh, ladies and gentlemen, it’s Lucky with a marker. Oh, and—Lucky, are you trying to make up with us now? Oh, look at this. (Stammering.) You can just guess that it’s very late in the evening tonight. Thank you. All right, that was Lucky, the calculus dolphin, ladies and gentlemen. Give him a hand. That’s great. That’s great. Okay. Okay. I think that Lucky can retire now to the seas of calculus—or maybe not. Okay.
Well, okay, here’s the pen and I—I wish I had a cap for that—okay, let’s get moving, though. We are—I know you’re waiting to see the solution to this. We have to find the derivative of this, first. Let me remind you that the formula for the derivative is given by—again—this limit. We’re hopefully getting a little bit more comfortable with that. Now we’ve got to actually try to nail this thing down. This is going to be a longy, folks, not a hardy, but it is going to be a longy. So, let’s try to work this out here together.
This equals the limit. Notice, by the way, that I’m going to really tighten this thing up, because I’ve got so much to write in here. So, it’s the limit as delta x approaches zero. And now, what is f of x plus delta x? Well, remember the recipe. Wherever I see an x, I’m going to erase it and replace it by this entire thing—x plus delta x. So, I see it there. I put an x plus delta x. I see it there. I put an x plus delta x. I don’t see it here, so I don’t put anything in there, I just report the three.
So, what I see is five times the quantity x plus delta x, all squared minus x plus delta x—notice I’m putting that in parentheses so I subtract all of it—plus three. That little piece right here—that little piece right here is equal to that piece right there. I’ll give you a second to look at that and verify that really is exactly what’s going on if you plug into that function.
Now, I’ve got to subtract off from that f of x, which equals five x squared minus x plus three. But remember, I’ve got to subtract off all of that. In fact, I’m going to use Lucky’s pen and put a lucky parentheses around that so everything gets subtracted off—everything. And then I divide that whole thing by delta x. Looks pretty long? You ain’t seen nothing yet. Because I’ve got to now expand that out.
So what does this equal? This equals the limit as delta x approaches zero. And now, I’m going to expand all that out, and what do I see? I’m going to see an x squared. I’m going to see a two x delta x, and I’m going to see a delta x squared. And each and every one of those terms has to be multiplied by a five. So, I’m going to write this out with the five there. That’s five x squared—there’s the first term. That inside and the outside gives me a two x delta x, but I’ve got to multiply that by five. So, that gives me a two times five, which is a plus ten x delta x. And that’s just going to be a plus five delta x squared. All that stuff, right there, is just that little piece if you square it out and multiply through by five.
Now, I’ve got to subtract this off. So, I’m going to subtract off each of those terms. I’m going to distribute—I’m going to distribute that. And so, what I see is minus x and then minus delta x. And then I have a plus three. Look at that thing. And now, I’m going to distribute that minus sign to every single person there. So, I’m going to see a minus five x squared—a minus and a minus is a plus x minus three. And that’s all divided by delta x. Looks really, really, really, really long and indeed it is. Look how long that is, but look what happens. There’s all sorts of great stuff.
First of all, the five x squared –there’s a minus five x squared. I love this stuff. They add to give zero. I see a minus x, and then I see a plus x. They add to give zero. I see a plus three and a minus three. They add to give zero. What’s left? This term, which has a delta x in it, this term, which has a delta x in it, this term, which has a delta x in it. I can factor out all those delta x’s, and what do I see? Well, what I see is the limit as delta x approaches zero of delta x times ten x plus five delta x minus one.
So, all I did there was take out the delta x here. See it’s gone. Take out one of the delta x’s here. I’m still left with one. And that delta x, I’m left with a minus one. Okay? And that’s all divided by delta x. And now again, I’ve isolated the zero over zero term. It’s a common theme. I can cancel it away, as long as you promise at this point that delta x won’t be zero, and at this point, I tell you it’s not, because I’m only taking the limit. And so then, what do I see? Well, I see this equals the same thing as this new limit problem—this different limit problem. The limit as delta x approaches zero of ten x plus five x minus one—I’m sorry—five delta x minus one. That sloppy thing there is a delta x.
Don’t get too attached to in, folks, because the delta x is going to zero, now. That whole sloppy term that I messed up here, that now is approaching zero. So therefore, this equals—I take the limit ten x minus one. Now, you know, Amy asked last time—in case you remember—she said, “Well, that’s sort of funny. You promised me that delta x wasn’t going to equal zero, and then you let delta x be zero here. Why is that?”
And the answer is because delta x can’t be zero here. But what we’re seeing by this equal sign is that this limit is the same as that limit. Now, I admit that these two functions aren’t the same, because that is defined when delta x equals zero. There’s no denominator here. So, basically, if you wanted to look at the graph of this, it would look this, but there would be hole when delta x equals zero. But the limits are the same. They’re heading to the same thing. But here, since I can plug in delta x to be zero and I get a nice answer that, in fact, that’s the limit here, which equals this limit here. That was a great question she asked.
Okay. Well you might be so excited after all this long—but not very hard, you have to admit—algebra—but it was long algebra. We actually found the derivative of this function. So, the moral of the story is that we found the derivative and we see it equals ten x minus one.
But, that is not what the question asks. The question asked us—in fact, you might have forgotten already; let me remind you. You’ve got to find the equation of that line tangent at this point. And we did all this work just to find the slope of the tangent line. Well, now I’m ready to tell you what the slope of the tangent line is at that point, because remember that the derivative of the function at a point represents the slope of the tangent line.
So, I put the tangent—the slope formula over there. And you can see it right there—ten x minus one. And if I plug in the point we’re interested in—and remember, the point we’re interested in is x equals one—then I see that the slope equals f prime at one. And what is f prime at one? Well, f prime at one is going to be—now, I plug into the derivative machine, and the derivative machine is going to push out a slope.
Derivatives spit out a slope. And what does that equal? It equals ten times one minus one. And that equals nine. So, that’s the slope. The slope of the tangent line—the way it’s pitched—let me bring back Lucky’s line. It’s a slope of nine. So it looks like that. Very steep.
But we don’t know, of course, where that line is located. I need to anchor it. So, how could I actually find the equation of the line? Well, there’s something that we haven’t used yet. There’s something we haven’t used yet. Find the equation of the line tangent to the function at the point. So, in fact, it’s tangent. Let me try to draw you a little sketch here, so you can get a sense of this.
Some sort of curve—actually some sort of parabola—and let’s see—looks something like this probably. I’m just making this up. Well, let’s see if this looks good or not. Roughly speaking, it’s some sort of parabola. We’re looking at the point one, and we’ve just discovered that the tangent line has slope nine. And you can see the curve that—there, the way I’ve drawn it—that slope does look pretty steep. It could be nine. And it’s nine, we see.
But how do you find the equation of that? Well, we need a point on the line. We need a point on this tangent line. But what does it mean to be a tangent line? It means that this line grazes the curve at that point—at that point, x equals one. So that means that that’s a point that both the line and the curve share in common. The point one—and then, what’s the value of the function, by the way? Look at the function there. Well, I’ll bring it back up for you to see. You can see it right over here.
That’s the function. When x equals one, what’s the height? I plug in here and produce f of one. That would tell me what this value is right there. And what is that value? Well, that will be five times one minus one plus three. So, that’s five minus one, which is four, plus three would be seven. So, this number is seven. So, at one, f of one is seven. But that’s a point where this line is touching the curve. So, they share that point in common.
So, therefore, one comma seven must also be on the line. Let me say that again, the idea is that this point of tangency is a point that is shared by both the line and the curve. So, since I can find out what the curve is at that point by just plugging back into the original function—I won’t plug back into the derivative, by the way, because the derivative is the machine that produces slopes. The function tells me how high I am. So, I plug in one for x, and I see that I get seven. But, since that’s the point of tangency—since that’s the point where the line and curve touch, then in fact that must be a point also on the line. So, therefore, what we see is that one comma seven is a point on the line
We’ve got the slope. We’ve got a point on the line. We now have the two pieces of information required to give me the equation of the line. There’s two ways of doing it. There’s the slope/intercept form or there’s the point/slope form. My favorite—point/slope form. Why? Point/slope form says if you have a point, you can just insert it into this formula where the point is x one, y one and there’s the slope.
So the answer, without doing any work at all is y minus the y value, we know equals the slope times x minus the x value. That’s the answer. And maybe now, you’re going to get a sense of why I love point/slope form. No work is involved. You don’t define the yintercept of anything. That’s the answer. So, that answers the question “Find the equation of the line tangent to the function at this point.” Well, there’s the equation of the line that’s tangent to the point.
Now, this problem actually has a slight followup problem I’ll run for you really fast. At what x is the tangent line horizontal? Look at this for a second. What does it mean for the tangent line to be horizontal? What would the slope have to equal? Well, the slope—here, it’s not horizontal; the slope is nine. What would have to happen for the slope to be horizontal—uh, for the line to be horizontal? For the line to be horizontal, the slope would actually have to be zero.
So, how would I find out where the slope is zero? Well, what would I do? I would go back to the derivative, which is over there—ten x minus one. And I would ask, “When is the slope zero?” I would ask, “When is ten x minus one equal to zero?” So, let me actually write that down here for you. So the question is, “At what x value is the tangent line horizontal?” And what I’m saying here, if we draw a picture of this thing—so, we’ve got this thing here—the tangent line being horizontal, that means that its slope equals zero.
Well, where’s the slope machine? The slope machine is given by the derivative. And we know the derivative. We just found that. The derivative is equal to ten x minus one. So, when is slope equal to zero? Should I plug in zero for x? Should I plug in zero for x? No. Because that will then spit out a slope. I don’t want to spit out a slope; I want to spit an x value where the slope equals zero. So I should take the slope machine, find out or ask when does that equal to zero.
So, I set this equal to zero and solve for x. And if I do that, what do I see? I see x equals—I bring the minus one over, it becomes a plus one divide by ten. I see one tenth. So when x equals onetenth, the slope is zero. So, in fact one tenth must be right here. So, when is the slope—when is the tangent line horizontal? Well, when the slope is zero. And I solved and found out that happens exactly when x equals one tenth. And if you think about it, since it’s a parabola, that’s sort of the place where everything bends—goes from being going this way to going that way, right at the point where this is zero.
Okay. Great job. Take a look at this one and we’ll be right back.
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WHY CAN'T YOU BE MY COLLEGE TEACHER!!!! Awesome JOB
He explains how to find the equation of the line tangent to f(x) at the point x=l. With this in mind he goes over several problems solving the equation of a tangent line.
He also shows you how to find the slope and derivative of the function. This is a difficult lesson but he does a good job breaking down each section.