0:00:02.050,0:00:07.671 And polynomial functions I would[br]be looking at functions of X 0:00:07.671,0:00:11.248 that represent polynomials of[br]varying degrees, including 0:00:11.248,0:00:13.292 cubics and quadratics and 0:00:13.292,0:00:18.100 Quartics. We will look at some[br]of the properties of these 0:00:18.100,0:00:23.028 curves and then we will go on to[br]look at how to deduce the 0:00:23.028,0:00:27.604 function of the curve if given[br]the roots. So first of all, what 0:00:27.604,0:00:31.476 is a polynomial with a[br]polynomial of degree add is a 0:00:31.476,0:00:33.940 function of the form F of X. 0:00:33.960,0:00:41.928 Equals a an X to the power N[br]plus AN minus One X to the power 0:00:41.928,0:00:48.900 and minus one plus, and it keeps[br]going all the way until we get 0:00:48.900,0:00:55.654 to. A2 X squared[br]plus a 1X plus a 0:00:55.654,0:01:01.711 zero and here the AA[br]represents real numbers and 0:01:01.711,0:01:04.403 often called the coefficients. 0:01:05.240,0:01:11.120 Now, this may seem complicated[br]at first sight, but it's not and 0:01:11.120,0:01:16.510 hopefully a few examples should[br]convince you of this. So, for 0:01:16.510,0:01:22.390 example, let's suppose we had F[br]of X equals 4X cubed, minus 0:01:22.390,0:01:24.350 three X squared +2. 0:01:25.720,0:01:30.021 Now this is a function and a[br]polynomial of degree three. 0:01:30.021,0:01:35.495 Since the highest power of X is[br]3 and is often called a cubic. 0:01:36.650,0:01:43.776 Let's suppose we hard F of X[br]equals X to the Power 7 - 0:01:43.776,0:01:47.339 4 X to the power 5 + 0:01:47.339,0:01:52.200 1. This is a polynomial of[br]degree Seven. Since the highest 0:01:52.200,0:01:56.974 power of X is A7, an important[br]thing to notice here is that you 0:01:56.974,0:02:01.748 don't need every single power of[br]X all the way up to 7. The 0:02:01.748,0:02:05.840 important thing is that the[br]highest power of X is 7, and 0:02:05.840,0:02:07.545 that's why it's a polynomial 0:02:07.545,0:02:13.394 degree 7. And the final example[br]F of X equals. 0:02:13.960,0:02:19.344 Four X squared minus two[br]X minus 4. 0:02:19.870,0:02:23.445 So polynomial of degree 2[br]because the highest power of X 0:02:23.445,0:02:26.862 is a 2. Less often called a 0:02:26.862,0:02:30.553 quadratic. Now it's important[br]when we're thinking about 0:02:30.553,0:02:34.612 polynomials that we only have[br]positive powers of X, and we 0:02:34.612,0:02:38.302 don't have any other kind of[br]functions. For example, square 0:02:38.302,0:02:40.147 roots or division by X. 0:02:40.740,0:02:42.636 So for example, if we had. 0:02:43.230,0:02:46.550 F of X equals 0:02:46.550,0:02:52.586 4X cubed. Plus the square[br]root of X minus one. 0:02:53.120,0:02:58.411 This is not a polynomial because[br]we have the square root of X 0:02:58.411,0:03:04.109 here and we need all the powers[br]of X to be positive integers to 0:03:04.109,0:03:07.365 have function to be a[br]polynomial. Second example. 0:03:07.390,0:03:14.482 If we had F of X equals 5X to[br]the Power 4 - 2 X squared plus 3 0:03:14.482,0:03:19.604 divided by X. Once again this is[br]not a polynomial and this is 0:03:19.604,0:03:24.726 because, as I said before, we[br]need all the powers of X should 0:03:24.726,0:03:30.242 be positive integers and so this[br]3 divided by X does not fit in 0:03:30.242,0:03:33.000 with that. So this is not a 0:03:33.000,0:03:38.264 polynomial. OK, we've already[br]met some basic polynomials, and 0:03:38.264,0:03:44.237 you'll recognize these. For[br]example, F of X equals 2 is 0:03:44.237,0:03:50.210 as a constant function, and this[br]is actually a type of 0:03:50.210,0:03:57.269 polynomial, and likewise F of X[br]equals 2X plus one, which is a 0:03:57.269,0:04:03.242 linear function is also a type[br]of polynomial, and we could 0:04:03.242,0:04:06.000 sketch those. So the F of X. 0:04:06.640,0:04:12.820 And X and we know that F of X[br]equals 2 is a horizontal line 0:04:12.820,0:04:19.000 which passes through two on the[br]F of X axis and F of X equals 0:04:19.000,0:04:20.236 2X plus one. 0:04:21.290,0:04:26.250 Is straight line with gradient[br]two which passes through one. 0:04:28.400,0:04:34.418 On the F of X axis, stuff of X[br]equals 2 and this is F of X. 0:04:35.240,0:04:38.315 Equals 2X plus 0:04:38.315,0:04:43.238 one. Important things[br]remember is that all 0:04:43.238,0:04:46.256 constant functions are[br]horizontal straight lines 0:04:46.256,0:04:50.280 and all linear functions[br]are straight lines which 0:04:50.280,0:04:51.789 are not horizontal. 0:04:52.880,0:04:56.624 So let's have a look now at some 0:04:56.624,0:05:02.697 quadratic functions. If we had F[br]of X equals X squared, we know 0:05:02.697,0:05:06.767 this is a quadratic function[br]because it's a polynomial of 0:05:06.767,0:05:11.651 degree two and we can sketch.[br]This is a very familiar curve. 0:05:11.651,0:05:14.907 If you have F of X on the 0:05:14.907,0:05:19.774 vertical axis. An X on[br]our horizontal axis. We 0:05:19.774,0:05:25.296 know that the graph F of[br]X equals X squared looks 0:05:25.296,0:05:26.802 something like this. 0:05:29.090,0:05:32.822 This is F of X equals 0:05:32.822,0:05:38.480 X squared. So what happens as we[br]vary the coefficient of X 0:05:38.480,0:05:42.528 squared? That's the number that[br]multiplies by X squared, so we 0:05:42.528,0:05:47.312 had, for example, F of X equals[br]2 X squared. How would that 0:05:47.312,0:05:51.728 affect our graph? What actually[br]happens is you can see that all 0:05:51.728,0:05:56.880 of our X squared values are the[br]F of X values have now been 0:05:56.880,0:06:00.928 multiplied by two and hence[br]stretched in the F of X 0:06:00.928,0:06:02.032 direction by two. 0:06:02.640,0:06:04.439 So you can see we actually get. 0:06:07.270,0:06:09.120 Occurs that looks something like 0:06:09.120,0:06:16.140 this. And this is F[br]of X equals 2 X squared. 0:06:17.340,0:06:22.428 And likewise, we could do that[br]for any coefficients. So if we 0:06:22.428,0:06:25.820 had F of X equals 5 X squared. 0:06:26.450,0:06:29.783 It would be stretched five[br]times from the value it was 0:06:29.783,0:06:33.116 when it was X squared, and it[br]would look something like 0:06:33.116,0:06:33.419 this. 0:06:36.410,0:06:42.900 So this would be F of[br]X equals 5 X squared. 0:06:43.610,0:06:47.482 So this is all for positive[br]coefficients of X squared. But 0:06:47.482,0:06:50.298 what would happen if the[br]coefficients were negative? 0:06:50.800,0:06:53.130 Well, let's have a look. 0:06:53.160,0:06:57.840 It starts off with F of X[br]equals minus X squared. Now, 0:06:57.840,0:07:02.520 compared with F of X equals X[br]squared would taken all of 0:07:02.520,0:07:06.030 our positive values and we've[br]multiplied them all by 0:07:06.030,0:07:09.150 negative one. So this[br]actually results in a 0:07:09.150,0:07:13.050 reflection in the X axis[br]because every single value is 0:07:13.050,0:07:17.730 now, which was positive has[br]become negative. So F of F of 0:07:17.730,0:07:21.630 X equals minus X squared will[br]look something like this. 0:07:24.210,0:07:30.330 So this is F of X[br]equals minus X squared. 0:07:31.120,0:07:36.184 Likewise, F of X equals minus[br]two X squared will be a 0:07:36.184,0:07:42.092 reflection in the X axis of F[br]of X equals 2 X squared, and 0:07:42.092,0:07:45.468 so we get something which[br]looks like this. 0:07:48.810,0:07:55.674 F of X equals 2 X squared 3 -[br]2 X squared and Leslie are going 0:07:55.674,0:08:02.109 to look at F of X equals minus[br]five X squared and you may well 0:08:02.109,0:08:08.115 have guessed by now. That's a[br]reflection of F of X equals 5 X 0:08:08.115,0:08:13.263 squared in the X axis, which[br]gives us a graph which would 0:08:13.263,0:08:14.550 look something like. 0:08:15.480,0:08:16.180 This. 0:08:17.510,0:08:23.738 We F of X equals[br]minus five X squared. 0:08:24.550,0:08:29.503 And in fact, it's true to say[br]that for any polynomial, if you 0:08:29.503,0:08:33.694 multiply the function by minus[br]one, you will always get the 0:08:33.694,0:08:35.599 reflection in the X axis. 0:08:36.240,0:08:41.557 I will have a look at some other[br]quadratics and see what happens 0:08:41.557,0:08:46.874 if we vary the coefficients of X[br]as opposed to the coefficient of 0:08:46.874,0:08:51.782 X squared. And to do this I'm[br]going to construct the table. 0:08:52.610,0:08:54.346 So we'll have RX value at the 0:08:54.346,0:08:59.670 top. And the three functions I'm[br]going to look at our X squared 0:08:59.670,0:09:03.174 plus X. X squared 0:09:03.174,0:09:09.110 plus 4X. And X[br]squared plus 6X. 0:09:10.040,0:09:17.400 And for my X file use, I'm going[br]to choose values of minus 5 - 4 0:09:17.400,0:09:24.760 - 3 - 2 - 1 zero. I'm[br]going to go all the way up to 0:09:24.760,0:09:26.600 a value of two. 0:09:27.480,0:09:31.176 Now for each of these functions,[br]I'm not necessarily going to 0:09:31.176,0:09:35.544 workout the F of X value for[br]every single value of X. I'm 0:09:35.544,0:09:39.240 just going to focus around the[br]turning point of the quadratic, 0:09:39.240,0:09:40.920 which is where the quadratic 0:09:40.920,0:09:46.289 dips. So just draw some lines in[br]on my table. 0:09:46.290,0:09:52.926 On some horizontal[br]lines as well. 0:09:55.350,0:09:59.172 And now we can fill this 0:09:59.172,0:10:05.902 in. So for RF of X equals[br]X squared plus X, I'm just going 0:10:05.902,0:10:13.110 to focus on minus 2 + 2. So when[br]I put X equals minus two in here 0:10:13.110,0:10:19.046 we get minus 2 squared, which is[br]4 takeaway two which gives Me 2. 0:10:19.540,0:10:24.430 When we put X is minus one in we[br]get minus one squared, which is 0:10:24.430,0:10:27.120 one. Take away one which gives 0:10:27.120,0:10:34.715 us 0. Report this X is zero in[br]here we just got 0 + 0 which is 0:10:34.715,0:10:41.008 0. I put X is one in here.[br]We've got one plus one which 0:10:41.008,0:10:42.226 gives Me 2. 0:10:42.890,0:10:47.024 And finally, when I pay taxes 2[br]in here, the two squared, which 0:10:47.024,0:10:52.572 is 4. +2, which gives me 6 and[br]in fact just for symmetry we 0:10:52.572,0:10:57.265 could put X is minus three in[br]here as well, which gives us 0:10:57.265,0:10:58.709 minus 3 squared 9. 0:10:59.350,0:11:02.990 Plus minus three. Sorry, which[br]gives us 6. 0:11:03.630,0:11:09.840 OK, for my next function,[br]X squared plus 4X. 0:11:10.590,0:11:15.582 I'm going to look at values from[br]minus four up to 0. 0:11:16.160,0:11:23.167 So if I put minus four in[br]here, we get 16 takeaway 16 0:11:23.167,0:11:25.323 which gives me 0. 0:11:26.060,0:11:27.680 Put minus three in here. 0:11:28.430,0:11:35.040 We get 9 takeaway 12 which[br]gives us minus three. 0:11:35.690,0:11:42.307 For put minus two in here, we[br]get four takeaway 8 which is 0:11:42.307,0:11:49.386 minus 4. If I put minus one[br]in here, we get one takeaway 4 0:11:49.386,0:11:56.031 which is minus three, and if I[br]put zero in here 0 + 0 just 0:11:56.031,0:11:57.360 gives me 0. 0:11:58.030,0:12:04.180 And Lastly, I'm going to look at[br]F of X equals X squared plus 6X. 0:12:04.710,0:12:08.182 And I'm going to take this from[br]minus five all the way up to 0:12:08.182,0:12:13.590 minus one. So we put minus five[br]in first of all minus 5 squared 0:12:13.590,0:12:17.200 is 25. Take away 30 gives us 0:12:17.200,0:12:25.064 minus 5. Minus four in[br]here 16 takeaway 24 which 0:12:25.064,0:12:28.256 gives us minus 8. 0:12:29.150,0:12:35.940 For X is minus three in. Here we[br]get 9 takeaway 18, which is 0:12:35.940,0:12:38.500 minus 9. For X is minus two in. 0:12:39.000,0:12:45.230 We get minus two square root of[br]4 takeaway 12, which is minus 8. 0:12:45.780,0:12:52.530 And we put minus one in. We get[br]one takeaway 6 which is minus 5. 0:12:53.050,0:12:56.710 And you can actually see the[br]symmetry in each row here, to 0:12:56.710,0:12:59.760 show that we've actually focused[br]around that turning point we 0:12:59.760,0:13:02.810 talked about. So let's draw the[br]graph of these functions. 0:13:03.400,0:13:07.180 So in a vertical scale will 0:13:07.180,0:13:13.335 need. F of X and that's going to[br]take us from all those values 0:13:13.335,0:13:15.885 minus nine up 2 - 6. 0:13:16.430,0:13:20.084 So. If we do minus 0:13:20.084,0:13:26.187 9. Minus 8 minus Seven[br]6 - 5. 0:13:26.840,0:13:30.578 For most 3 - 2 - 0:13:30.578,0:13:37.721 1 zero. 12345 and we[br]can just about squeeze 6 in 0:13:37.721,0:13:39.488 at the top. 0:13:40.430,0:13:43.240 On a longer horizontal axis. 0:13:44.480,0:13:46.928 We've gone from minus five to 0:13:46.928,0:13:50.550 two. So we just use one 0:13:50.550,0:13:57.641 2. Minus 1 -[br]2 - 3 - 0:13:57.641,0:14:01.150 4. And minus 5. 0:14:01.660,0:14:06.462 So first of all, let's look at[br]our function F of X equals X 0:14:06.462,0:14:09.549 squared plus X. We've got minus[br]three and six. 0:14:10.250,0:14:11.459 Which is appear. 0:14:11.960,0:14:15.698 We've got minus two and two. 0:14:16.550,0:14:20.030 Which is here. 0:14:20.260,0:14:22.748 Minus one and 0. 0:14:23.680,0:14:25.880 We've got zero and zero. 0:14:26.800,0:14:28.099 One and two. 0:14:28.660,0:14:32.239 And two 16. 0:14:33.650,0:14:36.750 And you can see clearly[br]that this is a parabola 0:14:36.750,0:14:39.850 which, as we expected, we[br]can draw a smooth curve. 0:14:42.260,0:14:43.640 Through those points. 0:14:45.770,0:14:49.698 So this is F of X equals X 0:14:49.698,0:14:52.170 squared. Plus X. 0:14:53.070,0:14:58.110 Next, we're going to look at the[br]function F of X equals X squared 0:14:58.110,0:15:02.100 plus 4X. And the points we had[br]with minus four and zero. 0:15:02.800,0:15:06.128 Minus three and minus 0:15:06.128,0:15:12.510 three. Minus two[br]and minus 4. 0:15:14.130,0:15:18.010 Minus one and minus three. 0:15:19.570,0:15:23.730 And zero and zero which is[br]already drawn. And once 0:15:23.730,0:15:28.722 again you can see this is a[br]smooth curve is a parabola 0:15:28.722,0:15:29.970 as we expected. 0:15:34.200,0:15:42.100 And this is F of[br]X equals X squared plus 0:15:42.100,0:15:48.336 4X. And the final function we're[br]going to look at F of X equals X 0:15:48.336,0:15:52.940 squared plus 6X. And so we've[br]got minus 5 - 5. 0:15:53.800,0:15:57.776 So over here minus 4 - 8. 0:15:58.770,0:16:00.110 Which is down here. 0:16:00.780,0:16:02.670 Minus three and minus 9. 0:16:03.480,0:16:06.980 Minus 2 - 8. 0:16:07.530,0:16:14.225 And minus one and minus five and[br]once again you can see smooth 0:16:14.225,0:16:16.800 curve in the parabola shape. 0:16:21.010,0:16:28.050 And this is F of X[br]equals X squared plus 6X. 0:16:28.860,0:16:34.320 So we can see that as the[br]coefficient of X increases from 0:16:34.320,0:16:40.235 here to one to four to six, the[br]curve the parabola is actually 0:16:40.235,0:16:42.965 moving down and to the left. 0:16:44.170,0:16:48.548 But what would happen if the[br]coefficients of X was negative? 0:16:48.548,0:16:54.518 Well, let's have a look and will[br]do this in the same way as we've 0:16:54.518,0:16:59.692 just on the previous examples.[br]And we look at a table, and this 0:16:59.692,0:17:04.468 time we'll look at how negative[br]values of our coefficients of X 0:17:04.468,0:17:09.642 affect the graph. So we look at[br]X squared minus XX squared minus 0:17:09.642,0:17:15.690 4X. And the graph of F of X[br]equals X squared minus 6X. 0:17:18.030,0:17:19.998 Now as before, will. 0:17:20.590,0:17:25.495 Put a number of values in for X,[br]but we won't use. All of them, 0:17:25.495,0:17:30.073 will only use the ones which are[br]near to the turning point of the 0:17:30.073,0:17:33.997 quadratic, but the values will[br]put in the range will be from 0:17:33.997,0:17:36.286 minus two all the way up to 0:17:36.286,0:17:40.528 five. Four and[br]five. 0:17:44.910,0:17:48.676 I just put in the lines for 0:17:48.676,0:17:55.378 our table. OK, so for the first[br]one, F of X equals X squared 0:17:55.378,0:18:01.734 minus X. We look at what happens[br]when we substitute in X is minus 0:18:01.734,0:18:07.386 2. So minus 2 squared is 4[br]takeaway, minus two is the same 0:18:07.386,0:18:12.906 as a plus two, which gives us[br]six when X is minus 1 - 1 0:18:12.906,0:18:17.690 squared, is one takeaway minus[br]one is the same as a plus one 0:18:17.690,0:18:19.530 which just gives us 2. 0:18:20.300,0:18:23.690 Zero is just zero takeaway 0:18:23.690,0:18:29.961 00. When we put one in, we've[br]got 1. Take away, one which is 0:18:29.961,0:18:36.010 0. And two gives us 2 squared[br]four takeaway two which gives us 0:18:36.010,0:18:38.625 2. And again, for symmetry will 0:18:38.625,0:18:44.639 do access 3. So 3 squared is 9[br]takeaway, three is 6. 0:18:45.710,0:18:51.222 So our second function F of X[br]equals X squared minus 4X. We're 0:18:51.222,0:18:55.462 going to look at going from zero[br]up to five. 0:18:55.980,0:19:01.230 And we put zero and we just get[br]zero takeaway 0 which is 0. 0:19:01.850,0:19:03.410 When we put X is one. 0:19:04.220,0:19:07.755 We get one takeaway 4 which is 0:19:07.755,0:19:14.935 minus 3. When we put X[br]is 2, two squared is 4 takeaway 0:19:14.935,0:19:17.785 eight, which gives us minus 4. 0:19:18.760,0:19:23.912 When X is 3, that gives us 3[br]squared, which is 9 takeaway 12 0:19:23.912,0:19:27.960 which gives us minus three and[br]you can see the symmetry 0:19:27.960,0:19:32.744 starting to form here. Now a[br]Nexus 4 gives us 4 squared 16 0:19:32.744,0:19:36.424 takeaway 16 which as we expect[br]it is a 0. 0:19:37.140,0:19:41.928 And for our final function, FX[br]equals X squared minus six X, 0:19:41.928,0:19:47.913 we're going to look at coming[br]from X is one all the way up to 0:19:47.913,0:19:53.100 X equals 5, X equals 1, gives[br]US1 takeaway six, which gives us 0:19:53.100,0:19:59.190 minus 5. X equals 2 gives us[br]2 squared. Is 4 takeaway 12 0:19:59.190,0:20:01.190 which gives us minus 8. 0:20:02.320,0:20:07.585 I mean for taxes, three in three[br]squared is 9 takeaway 18, which 0:20:07.585,0:20:09.205 gives us minus 9. 0:20:09.900,0:20:13.716 X is 4 + 4 squared. 0:20:14.360,0:20:21.100 Take away 24. Which 16[br]takeaway 24 which is minus 0:20:21.100,0:20:27.083 8. I'm waiting for X is 5 in[br]we get 25 takeaway 30, which 0:20:27.083,0:20:31.494 is minus five and once again[br]symmetry as expected. So as 0:20:31.494,0:20:35.504 with our previous examples[br]we want to draw this graphs 0:20:35.504,0:20:40.316 of these functions so we can[br]see what's going on as our 0:20:40.316,0:20:44.326 value of the coefficients of[br]X or Y is changing. 0:20:46.260,0:20:52.014 So if we got F of X on our[br]vertical axis, this time we're 0:20:52.014,0:20:56.535 going from Arlo's values, minus[br]nine. Our highest value is 6. 0:20:57.070,0:21:00.902 So we're going from minus 9 - 8 0:21:00.902,0:21:08.060 - 7. 6 - 5[br]- 4 - 3 - 2 0:21:08.060,0:21:09.920 - 1 zero. 0:21:10.630,0:21:18.460 12345 I will just squeeze[br]in sex and are horizontal 0:21:18.460,0:21:21.592 axis. We've got one. 0:21:23.180,0:21:29.838 2. 3. Four[br]and five letter VRX axis, and we 0:21:29.838,0:21:36.078 go breakdown some minus 2 - 1 -[br]2. So first of all, let's look 0:21:36.078,0:21:41.902 at the graph F of X equals X[br]squared minus X. So minus two 0:21:41.902,0:21:47.310 and six was our first point, so[br]minus two and six, which is 0:21:47.310,0:21:51.760 here. And we've got minus one[br]and two which is here. 0:21:52.940,0:21:54.840 We've got the origin 00. 0:21:55.630,0:21:57.290 And we've got 10. 0:21:58.150,0:22:00.790 22 0:22:01.440,0:22:07.322 I'm 36 And[br]as we expected, you can see we 0:22:07.322,0:22:10.611 can join the points of here.[br]Let's make smooth curve which 0:22:10.611,0:22:11.807 will be a parabola. 0:22:16.740,0:22:22.658 So this is F of X equals[br]X squared minus X. 0:22:23.520,0:22:29.019 Our second function was F of X[br]equals X squared minus 4X. So 0:22:29.019,0:22:31.557 first point was 00, which we've 0:22:31.557,0:22:34.480 got. One and negative 3. 0:22:35.430,0:22:38.948 Say it. We've got[br]two and minus 4. 0:22:40.030,0:22:42.874 Just hang up three and negative 0:22:42.874,0:22:45.138 3. Which is here. 0:22:46.150,0:22:51.330 And four and zero, which is here[br]and straight away. We can see a 0:22:51.330,0:22:53.180 smooth curve which is a 0:22:53.180,0:22:56.430 parabola. Coming through[br]those points. 0:22:57.970,0:23:00.580 And we can label up this is F of 0:23:00.580,0:23:06.790 X. Equals X[br]squared minus 4X. 0:23:08.340,0:23:13.870 Lost function to look at is the[br]function F of X equals X squared 0:23:13.870,0:23:21.625 minus 6X. So our first point[br]was one and minus five, which is 0:23:21.625,0:23:24.010 here. We are two and negative 8. 0:23:25.090,0:23:28.190 Yep. Three and minus 9. 0:23:28.890,0:23:31.920 4 - 8. 0:23:32.480,0:23:38.187 And five and minus five, which[br]is hit, and once again we can 0:23:38.187,0:23:40.821 just draw a smooth curve through 0:23:40.821,0:23:48.660 these points. To give us the[br]parabola we wanted, this is F 0:23:48.660,0:23:52.500 of X equals X squared minus 0:23:52.500,0:23:57.960 6X. So what's happening as the[br]coefficients of X is getting 0:23:57.960,0:24:02.772 bigger in absolute terms. So for[br]instance, we can from minus one 0:24:02.772,0:24:07.985 to minus four to minus six, and[br]we can see straight away that 0:24:07.985,0:24:12.797 the actual graph this parabola[br]is moving down and to the right 0:24:12.797,0:24:16.807 as the coefficients of X gets[br]bigger in absolute terms. 0:24:18.220,0:24:20.938 And that's where the[br]coefficients of X is negative. 0:24:21.710,0:24:25.970 OK, so we know what happens when[br]we varied coefficients of X 0:24:25.970,0:24:30.230 squared and we know what happens[br]when we vary the coefficients of 0:24:30.230,0:24:34.490 X and that's both of them. For[br]quadratics, what happens when we 0:24:34.490,0:24:38.750 vary the constants at the end of[br]a quadratic well? Likewise table 0:24:38.750,0:24:42.655 of values is a good way to see[br]what's going on. 0:24:42.700,0:24:48.776 So this time I'm going to use X[br]again. I'm going to go from 0:24:48.776,0:24:56.154 minus two all the way up to +2,[br]so minus 2 - 1 zero one and two, 0:24:56.154,0:25:01.362 and for my functions I'm going[br]to use X squared plus X. 0:25:01.940,0:25:05.284 X squared plus X 0:25:05.284,0:25:11.930 plus one. X[br]squared plus X 0:25:11.930,0:25:18.398 +5. And X squared[br]plus X minus 4. 0:25:19.190,0:25:22.542 Now this table is 0:25:22.542,0:25:28.115 particularly. Easy for us to[br]workout compared with the other 0:25:28.115,0:25:33.510 ones because we have a slight[br]advantage in that we have a Head 0:25:33.510,0:25:38.490 Start because we already know[br]the values for X squared plus X 0:25:38.490,0:25:43.470 that go here and we can just[br]take them directly from Aurora, 0:25:43.470,0:25:48.035 the table and you'll remember[br]that they actually gave us the 0:25:48.035,0:25:50.110 values 200, two and six. 0:25:50.730,0:25:55.168 And in fact, we can use this[br]line in our table to help us 0:25:55.168,0:25:59.289 with all of the other lines. The[br]second line here, which is X 0:25:59.289,0:26:00.874 squared plus X plus one. 0:26:01.500,0:26:05.100 His only one bigger than every[br]value in the table before. So 0:26:05.100,0:26:06.600 all we need to do. 0:26:07.330,0:26:14.285 Is just add 1 answer every value[br]from the line before, so this 0:26:14.285,0:26:18.030 will be a 311, three and Seven. 0:26:18.970,0:26:24.010 Likewise, for this line, when[br]we've got X squared plus X +5. 0:26:24.570,0:26:28.962 All this line is just five[br]bigger than our first line, so 0:26:28.962,0:26:30.426 we can just 7. 0:26:30.940,0:26:36.340 557[br]and 0:26:36.340,0:26:42.680 11. And likewise, our[br]last line is minus four in the 0:26:42.680,0:26:46.880 end with exactly the same thing[br]before hand, so it's just take 0:26:46.880,0:26:51.080 away for from every value from[br]our first line, which gives us 0:26:51.080,0:26:58.274 minus 2. Minus 4 -[br]4 - 2 and 0:26:58.274,0:27:03.054 2. So let's put this information[br]on a graph now so we can 0:27:03.054,0:27:06.378 actually see what's happening as[br]we vary the constants at the end 0:27:06.378,0:27:12.135 of this quadratic. So you put F[br]of X and are vertical scale so 0:27:12.135,0:27:17.760 that we need to go from minus[br]four all over to 11 so it starts 0:27:17.760,0:27:19.635 at minus 4 - 3. 0:27:20.260,0:27:22.280 Minus 2 - 1. 0:27:22.870,0:27:29.932 0123456789, ten and[br]we can just 0:27:29.932,0:27:33.463 about squeeze already 0:27:33.463,0:27:39.396 left it. I no[br]longer horizontal axis. We're 0:27:39.396,0:27:46.548 going from minus 2 + 2[br]- 1 - 2 plus one 0:27:46.548,0:27:52.468 and +2. So first graph is the[br]graph of the function F of X 0:27:52.468,0:27:56.142 equals X squared plus X, so it's[br]minus two and two. 0:27:57.230,0:28:02.450 Which is a point here, minus[br]one and 0. 0:28:03.030,0:28:07.706 So first graph is the graph of[br]the function F of X equals X 0:28:07.706,0:28:10.712 squared plus X, so it's minus[br]two and two. 0:28:11.800,0:28:15.712 Which is a .8 - 1 0:28:15.712,0:28:23.069 and 0. Because the origin[br]next 00 at the .1 two 0:28:23.069,0:28:26.124 and we got the .26. 0:28:27.620,0:28:30.797 For our next graph, actually[br]sorry, now we should. 0:28:31.430,0:28:34.027 Draw a smooth curve for this one 0:28:34.027,0:28:39.278 first. And Labor Lux F of X[br]equals X squared plus X. 0:28:40.340,0:28:45.530 For an X graph which is F of X[br]equals X squared plus X plus 0:28:45.530,0:28:49.336 one, we need to put the[br]following points minus two and 0:28:49.336,0:28:54.530 three. So minus one and one[br]you can see straight away 0:28:54.530,0:28:58.886 that all of these points[br]are just want above what we 0:28:58.886,0:29:00.866 had before zero and one. 0:29:01.970,0:29:03.869 One and three. 0:29:04.910,0:29:07.808 Two and Seven. 0:29:07.980,0:29:11.630 So you can imagine our next[br]parabola that we draw 0:29:11.630,0:29:12.725 through these points. 0:29:15.260,0:29:20.096 Is exactly the same but one[br]above previous, that's F of X 0:29:20.096,0:29:24.025 equals. X squared plus X plus 0:29:24.025,0:29:30.152 one. So what about F of X[br]equals X squared plus X +5? 0:29:30.790,0:29:33.886 Well, let's have a look minus[br]two and Seven. 0:29:35.480,0:29:37.316 Gives us a point up here. 0:29:38.280,0:29:40.120 Minus one and five. 0:29:41.260,0:29:43.990 Zero and five. 0:29:44.730,0:29:47.649 One and Seven. 0:29:48.460,0:29:51.728 And two and 11. 0:29:53.530,0:29:57.157 So we draw a smooth curve[br]through these points. 0:30:02.060,0:30:09.032 See, that gives us a parabola[br]and this is F of X equals X 0:30:09.032,0:30:11.024 squared plus X +5. 0:30:12.320,0:30:17.832 So finally we look at the[br]function F of X equals X squared 0:30:17.832,0:30:22.920 plus X minus four. So we've got[br]minus two and minus 2. 0:30:23.450,0:30:26.938 Minus 1 - 4. 0:30:27.960,0:30:29.598 0 - 4. 0:30:30.240,0:30:33.928 One 1 - 2. 0:30:33.930,0:30:35.010 Two and two. 0:30:36.510,0:30:40.074 I once again we can[br]draw a smooth curve. 0:30:41.300,0:30:48.463 Through these points and this is[br]F of X equals X squared plus 0:30:48.463,0:30:50.116 X minus 4. 0:30:50.900,0:30:54.200 So we can see that what's[br]actually happening here is from 0:30:54.200,0:30:58.400 our original graph of F of X[br]equals X squared plus X, and if 0:30:58.400,0:31:02.300 you like, you could have put a[br]plus zero there. And when we 0:31:02.300,0:31:06.200 added one, it's moved one up[br]when we added five, it's moved 5 0:31:06.200,0:31:11.546 or. And when we took away four,[br]it moved 4 down. So it's quite 0:31:11.546,0:31:15.530 clear to see the effect that the[br]constant has on our parabola. 0:31:16.880,0:31:23.198 When looking at the graph of a[br]function, a turning point is the 0:31:23.198,0:31:28.058 point on the curve where the[br]gradient changes from negative 0:31:28.058,0:31:30.974 to positive or from positive to 0:31:30.974,0:31:34.916 negative. And we're thinking[br]about polynomials. Are 0:31:34.916,0:31:40.317 polynomial of degree an has at[br]most an minus one turning 0:31:40.317,0:31:44.736 points. So for example, a[br]quadratic of degree 2. 0:31:45.470,0:31:47.220 Can only have one turning 0:31:47.220,0:31:51.510 points. And if we draw just a[br]sketch of a quadratic, you can 0:31:51.510,0:31:54.360 see this point here would be[br]at one turning point. 0:31:55.570,0:32:00.058 We think about cubics, obviously[br]cubic as a polynomial of degree 0:32:00.058,0:32:05.770 3, so that can have at most two[br]turning points, which is why the 0:32:05.770,0:32:09.442 general shape of a cubic looks[br]something like this. 0:32:10.700,0:32:14.150 We see the two turning points[br]are here and here. 0:32:14.750,0:32:20.075 But as I said, it can have at[br]most 2. There's no reason it has 0:32:20.075,0:32:25.400 to have two, and a good example[br]is this of this is if you look 0:32:25.400,0:32:27.885 at F of X equals X cubed. 0:32:28.470,0:32:31.460 And in fact, that would look if[br]I do a sketch over here. 0:32:32.130,0:32:38.440 F of X&X that would come[br]from the bottom left. 0:32:39.870,0:32:42.198 Through zero and come up here 0:32:42.198,0:32:47.980 like this. That's F of X equals[br]X cubed, and so we can see that 0:32:47.980,0:32:52.140 the function F of X equals X[br]cubed does not have a turning 0:32:52.140,0:32:56.846 point. Another example, if[br]we were looking at a quartic 0:32:56.846,0:33:01.086 curve I a polynomial of[br]degree four, we know can 0:33:01.086,0:33:05.750 have up to at most three[br]turning points, which is why 0:33:05.750,0:33:09.566 the general shape of a[br]quartic tends to be 0:33:09.566,0:33:10.838 something like this. 0:33:12.910,0:33:18.240 With the three, turning points[br]are here, here and here. 0:33:19.010,0:33:25.346 Now let's suppose I had the[br]function F of X equals. 0:33:26.070,0:33:32.310 X minus a multiplied by[br]X minus B. 0:33:33.270,0:33:37.014 Now to find the roots of this[br]function, I want to know the 0:33:37.014,0:33:39.606 value of X when F of X equals 0. 0:33:40.270,0:33:46.752 So when F of X equals 0, we[br]have 0 equals X minus a 0:33:46.752,0:33:49.067 multiplied by X minus B. 0:33:49.790,0:33:55.460 So either X minus[br]a equals 0. 0:33:55.700,0:34:03.056 Or X minus B equals 0,[br]so the roots must be X 0:34:03.056,0:34:06.780 equals A. Or X equals 0:34:06.780,0:34:13.688 B. Now we can use the[br]converse of this and say that if 0:34:13.688,0:34:19.764 we know the roots are A&B, then[br]the function must be F of X 0:34:19.764,0:34:25.840 equals X minus a times by X[br]minus B or a multiple of that, 0:34:25.840,0:34:31.482 and that multiple could be a[br]constant. Or it could be in fact 0:34:31.482,0:34:36.690 any polynomial we choose. So for[br]example, if we knew that the 0:34:36.690,0:34:38.860 roots were three and negative. 0:34:38.890,0:34:46.208 2. We would say that F[br]of X would be X minus 3 0:34:46.208,0:34:50.924 multiplied by X +2 or[br]multiple so that multiple 0:34:50.924,0:34:57.212 could be three could be 5, or[br]it could be any polynomial. 0:34:58.600,0:35:05.620 OK, another example. If my[br]roots were one 2, three 0:35:05.620,0:35:13.252 and four. Then my function would[br]have to be F of X equals X 0:35:13.252,0:35:15.607 minus One X minus 2. 0:35:16.300,0:35:18.499 X minus three. 0:35:19.090,0:35:24.640 At times by X minus four, or it[br]could be a multiple of that, and 0:35:24.640,0:35:28.340 as I keep saying that multiple[br]could be any polynomial. 0:35:28.960,0:35:36.201 Right so Lastly, I'd like to[br]think about the function F of X. 0:35:36.240,0:35:39.695 Equals X minus two all 0:35:39.695,0:35:44.990 squared. Now if we try to find[br]the roots of this function I 0:35:44.990,0:35:49.680 when F of X equals 0, look what[br]happens. We had zero equals and 0:35:49.680,0:35:54.370 I'll just rewrite this side as X[br]minus two times by X minus 2. 0:35:55.040,0:35:59.549 Which means either X minus two[br]must be 0. 0:36:00.190,0:36:08.146 Or X minus 2 equals 0[br]the same thing. So X equals 0:36:08.146,0:36:14.461 2. Or X equals 2. So[br]there are two solutions and two 0:36:14.461,0:36:20.915 routes at the value X equals 2[br]as what we call a repeated root. 0:36:21.740,0:36:27.968 OK, another example. Let's[br]suppose we have F of 0:36:27.968,0:36:31.428 X equals X minus 2 0:36:31.428,0:36:37.018 cubed. Multiply by X +4 to[br]the power 4. 0:36:37.930,0:36:42.670 Now, as before, we want to find[br]out what the value of X has to 0:36:42.670,0:36:45.514 be to make F of X equal to 0. 0:36:46.170,0:36:51.495 And we can see that if we put X[br]equals 2 in here, we will 0:36:51.495,0:36:55.755 actually get zero in this[br]bracket. So X equals 2 is one 0:36:55.755,0:36:59.305 route and actually there are[br]three of those because it's 0:36:59.305,0:37:03.565 cubed. So this has a repeated[br]route, three of them, so three 0:37:03.565,0:37:06.760 repeated roots. So three roots[br]for X equals 2. 0:37:07.400,0:37:12.020 But also we can see that if X is[br]minus four, then this bracket 0:37:12.020,0:37:16.970 will be equal to 0, so X equals[br]minus four is a second route and 0:37:16.970,0:37:20.930 there are four of them there. So[br]there are four repeated roots 0:37:20.930,0:37:24.828 there. Now what we say is that. 0:37:25.340,0:37:26.618 If a route. 0:37:27.140,0:37:31.440 Has an odd number of repeated[br]roots. For instance, this 0:37:31.440,0:37:35.740 one's got 3 routes, then it[br]has an odd multiplicity. 0:37:36.810,0:37:41.144 If a root, for instance X[br]equals minus four, has an 0:37:41.144,0:37:44.690 even number of roots, then it[br]hasn't even multiplicity. 0:37:45.840,0:37:49.520 Why are we interested in[br]multiplicity at all? Well, for 0:37:49.520,0:37:53.400 this reason. If the[br]multiplicities odd, then that 0:37:53.400,0:37:57.910 means the graph actually crosses[br]the X axis at the roots. 0:37:58.450,0:38:02.949 If the multiplicity is even,[br]then it means that the graph 0:38:02.949,0:38:07.857 just touches the X axis and this[br]is very useful tool when 0:38:07.857,0:38:11.400 sketching functions. For[br]example, if we had. 0:38:12.100,0:38:15.108 F of X equals. 0:38:16.120,0:38:19.925 X minus 3 squared multiplied 0:38:19.925,0:38:25.160 by. X plus one to[br]the power 5. 0:38:25.440,0:38:31.576 By X minus 2 cubed times by X[br]+2 to the power 4. 0:38:32.370,0:38:35.700 Now, first sight, this might[br]look very complicated, but in 0:38:35.700,0:38:39.363 fact we can identify the four[br]roots straight away. The first 0:38:39.363,0:38:43.692 One X equals 3 will make this[br]brackets equal to 0, so that's 0:38:43.692,0:38:48.021 our first roots. X equals 3 and[br]we can see straight away there 0:38:48.021,0:38:54.220 are. Two repeated roots there,[br]which means that this has an 0:38:54.220,0:38:57.330 even multiplicity. Some 0:38:57.330,0:39:02.579 even multiplicity.[br]Our second route. 0:39:03.130,0:39:07.232 He is going to be X equals minus[br]one because that will make this 0:39:07.232,0:39:09.283 bracket equal to 0. So X equals 0:39:09.283,0:39:14.248 minus one. I'm not actually five[br]of those repeated root cause. 0:39:14.248,0:39:19.876 The power of the bracket is 5,[br]which means that this is an odd 0:39:19.876,0:39:26.000 multiplicity. We look at the[br]next One X is 2 will give us a 0:39:26.000,0:39:31.720 zero in this bracket. So X[br]equals 2 is a root there and 0:39:31.720,0:39:36.560 there are three of them, which[br]means is an odd multiplicity. 0:39:37.490,0:39:44.630 And finally, our last bracket X[br]+2 here. If we put X 0:39:44.630,0:39:47.605 equals minus two in there. 0:39:48.180,0:39:52.600 That will give us zero in this[br]bracket and there are four of 0:39:52.600,0:39:55.320 those repeated roots, which[br]means it isn't even. 0:39:55.960,0:40:02.370 Multiplicity. So as I said[br]before, we can use this to help 0:40:02.370,0:40:07.050 us plot a graph or helper sketch[br]the graph. So for instance, and 0:40:07.050,0:40:11.010 even multiplicity here would[br]mean that at the roots X equals 0:40:11.010,0:40:15.330 3. This curve would just touch[br]the access at the roots, X 0:40:15.330,0:40:20.010 equals minus one. It would cross[br]the Axis and at the River X 0:40:20.010,0:40:24.330 equals 2 is an odd multiplicity,[br]so it will cross the axis. 0:40:24.880,0:40:29.131 And at the roots for X equals[br]minus two is an even more 0:40:29.131,0:40:33.055 simplicity, so it would just[br]touch the axis. So let's have a 0:40:33.055,0:40:36.652 look at a function where we can[br]actually sketch this now. 0:40:36.830,0:40:39.758 So if we look at the function F 0:40:39.758,0:40:47.600 of X. Equals X minus two[br]all squared multiplied by X plus 0:40:47.600,0:40:51.495 one. We can identify the two 0:40:51.495,0:40:57.639 routes immediately. X equals 2[br]will make this bracket zero and 0:40:57.639,0:41:01.216 we can see that this hasn't even 0:41:01.216,0:41:05.700 multiplicity. So even[br]multiplicity. 0:41:06.820,0:41:13.504 And X equals negative one would[br]make this make this bracket 0. 0:41:13.710,0:41:17.200 So this will be an 0:41:17.200,0:41:21.586 odd multiplicity. Because[br]the power of this bracket 0:41:21.586,0:41:22.804 is actually one. 0:41:24.330,0:41:28.206 Now we talked before about the[br]general shape of a cubic, which 0:41:28.206,0:41:29.821 we knew would look something 0:41:29.821,0:41:34.499 like this. So how can we combine[br]this information and this 0:41:34.499,0:41:38.109 information to help us sketch[br]the graph of this function? 0:41:38.109,0:41:40.275 Well, let's first of all draw 0:41:40.275,0:41:42.940 axes. Graph of X here. 0:41:43.480,0:41:48.466 Convert sleep X going[br]horizontally. Now we know the 0:41:48.466,0:41:55.114 two routes ones X equals 2 to[br]the X equals negative one. 0:41:55.930,0:41:59.740 Now about the multiplicity. This[br]tells us that the even 0:41:59.740,0:42:04.312 multiplicity is that X equals 2,[br]so that means it just touches 0:42:04.312,0:42:06.217 the curve X equals 2. 0:42:07.200,0:42:10.512 But it crosses the curve X[br]equals minus one. 0:42:11.280,0:42:14.928 I should add here that because[br]we've got a positive value for 0:42:14.928,0:42:18.272 our coefficients of X cubed, if[br]we multiply this out, we're 0:42:18.272,0:42:20.096 definitely going to get curve of 0:42:20.096,0:42:23.269 this shape. So we can see[br]it's going to cross through 0:42:23.269,0:42:23.711 minus one. 0:42:26.130,0:42:27.490 It's going to come down. 0:42:28.000,0:42:31.300 And it's just going to touch the[br]access at minus 2. 0:42:32.330,0:42:38.770 So this is a sketch of the[br]curve F of X equals X minus 0:42:38.770,0:42:42.450 two all squared multiplied[br]by X plus one. 0:42:43.680,0:42:46.850 We do not know precisely[br]where this point is, but 0:42:46.850,0:42:50.020 we do know that it lies[br]somewhere in this region.