00:01 So let's not wait any longer and let's go straight into an example. 00:04 Here we have a question where you are given a curve C. 00:08 And curve c has the equation 2x squared - 6x + 4. 00:12 So it's going to be a quadratic, a curve that we've been dealing with. 00:17 And it's asking us to find the gradient of the tangent to see at point x equals to 3. 00:22 Now before we start, let's just see if we can model this. 00:26 So we can kind of figure out what's actually happening. 00:29 So we're not quite answering the question yet. 00:32 We're just going to see if we can think about, what this question is actually asking. 00:36 So I have here, y = 2x2 - 6x + 4. 00:42 Just like we've mentioned already is a quadratic. 00:46 We can solve this. 00:48 So if we say y equals to 0. 00:50 We can try to factorize it. 00:51 And see if we can find the two factors. 00:54 So those points would be the point of intersection on the x axis. 00:57 If I try that here x and x. 01:00 And if we try our two factors, 2 and 2. 01:03 And if you double check this that gives me 2x squared minus 4, minus 2 to give me minus 6x. 01:12 And then I've got minus 2 times minus 2 to give me plus 4. 01:15 So I can factorize it. 01:17 It's fairly nice equation. 01:19 From here you can say you got two solutions. 01:21 So you can say that either 2x - 2 = 0. 01:24 Or x - 2 = 0. 01:27 When you solve this, you get x equals to 1 you get x = 2 here. 01:33 So remember I haven't quite started answering the question yet. 01:36 I'm just trying to imagine what this actually looks like. 01:39 And for mathematicians it's really important for us to be able to visualize to have graphical and algebraic interpretations of the questions we're looking at. 01:48 So here is what I think this looks like. 01:54 So I've got x = 1 and x = 2. 01:57 So those are my two solutions from there. 02:00 I know it's a quadratic so it has a shape like so. 02:05 And now coming back to the question. 02:08 It says find the gradient of the tangent to see at point x to 3. 02:12 So let's just say 3 is somewhere around here. 02:14 So we are looking for the gradient at that point. 02:18 And as a result, we are also looking for the tangent here. 02:25 So this here is our tangent line. 02:30 So we're actually looking for the equation of this line which you know takes the form, y = mx + c. 02:36 Or the alternate form that I mentioned earlier. 02:39 And we could see if we could work with both. 02:41 So now that we understand the problem, let's try and answer the question. 02:45 The question says find the equation of the tangent. 02:49 So we're looking for the equation of this blue line here on this side. 02:53 Right, we've already mentioned that the equation of a tangent takes this form y = mx + c. 02:59 Or we're using y - y1 = m(x - x1). 03:07 Both methods again are valid. 03:09 The first thing that we need to do is to actually calculate the gradient. 03:14 Whether you're using this equation or whether you're using this equation we need to calculate the gradient of the tangent. 03:20 And here's what we start to use all the knowledge that we have just discussed. 03:23 The gradient of the tangent, this tangent, I can calculate by finding the gradient of the curve at this point x = 3. 03:32 And it will be exactly the same gradient of the tangent as it is of the curve at x = 3. 03:38 And we know how to do that. 03:40 So to calculate the gradient of a curve, I want to take my curve. 03:44 I'm going to dy by dx it. 03:47 So that's how I calculate my gradient. 03:49 That gives me 4x - 6. 03:52 This is the general gradient. 03:55 So everywhere around this curve I can use that gradient. 03:58 However, if I need the gradient at the point that the question asks, so we want to know dy by dx at x = 3. 04:07 You simply substitute 3 into here. 04:12 So we will do 4 times 3 minus 6 which gives you 12 minus 6 to give you a positive gradient of 6. 04:20 Which is good news because from our picture it looks like this graph has a positive gradient anyway. 04:26 So it's a nice little way of checking. 04:28 And that you can see it graphically. 04:31 You expect a positive gradient. 04:32 And so you've got a positive gradient. 04:34 Now we can put into our equation. 04:38 So you could put it into y = mx + c or we can substitute this into y - y1 = m(x - x1). 04:49 Both of them are fine. 04:52 But you might find this a little bit easier to use. 04:55 Because you don't have to do it into two parts. 04:58 So you don't have to find your m first and then find your c at the end. 05:03 Okay, let's just talk about what we have and what we need. 05:08 So we found the gradient. 05:09 We know what m is. 05:10 We know an x point here. 05:12 So we know that, that point here is 3. 05:14 However, we don't know this yet. 05:16 So we're missing a y value. 05:19 Now how could you find a y value. 05:22 You go back to your original equation here. 05:24 Which says that y equals to something. 05:27 So all we have to do here is replace our x with 3. 05:31 So using your original function we got y = 2x squared minus 6x plus 4. 05:37 We want to know what this is at x = 3. 05:40 So we're going to substitute that in, 3 squared minus 6 times 3 plus 4. 05:47 That gives you two times 9 to give you 18 minus 6 times 3 which also gives you 18 + 4 to give you an answer of 4. 05:57 Therefore, your y value, y equals to 4, when x equals to 3. 06:06 Now we have everything. 06:08 So we've also calculated at this y value. 06:11 You can put it all into the equation. 06:13 So we have the y minus 4 equals to the gradient which was 6(x minus 3). 06:20 The equation has been found. 06:23 But we can tidy up and make it look more like y = mx + c. 06:28 So we make it look more into that from there. 06:30 We've got y minus 4 equals to 6x minus 18. 06:35 Take the 4 over to the other side, you get y equals to 6x minus 18 + 4 to give you minus 14. 06:43 And this equation now follows the form, y = mx + c. 06:50 So the plus c is going to be minus 14. 06:52 So it meets this y axis here at -14. 06:57 So it gives you everything you need to know about that equation.
The lecture Gradient of a Point on a Curve by Batool Akmal is from the course Calculus Methods: Differentiation.
What is the factorization of 2x² + 6x + 4?
What is the gradient of the curve y = (2x + 2) (x + 2) at x = 3?
What is the equation of the tangent on the curve y = (2x + 2) (x + 2) at x = 3?
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