14 - Graphing Quadratic Functions - Max & Min Values - Part 1 - By Math and Science
Transcript
| 00:00 | Hello . Welcome back . I'm Jason with math and | |
| 00:02 | science dot com . The title of this lesson is | |
| 00:04 | we're going to find the maximum and the minimum values | |
| 00:07 | of a parabola . This is part one . We | |
| 00:09 | also have a part to in the skill . So | |
| 00:12 | I'm very excited about teaching this lesson because probably the | |
| 00:14 | most , the most interesting thing that I like doing | |
| 00:18 | when I teach these classes and the fun thing for | |
| 00:20 | me is explaining to you where things come from . | |
| 00:23 | So there's gonna be a simple equation in here that | |
| 00:25 | I'm going to present , that will be in every | |
| 00:27 | algebra textbook that you have to learn and it has | |
| 00:29 | to do with figuring out where the minimum value of | |
| 00:31 | the parabola is or if it goes upside down where | |
| 00:34 | the maximum value of that parabola is , it's usually | |
| 00:37 | just given to you and you memorize it and you | |
| 00:39 | use it but you usually don't know where it comes | |
| 00:42 | from . So what we're gonna do is introduce the | |
| 00:44 | concept of maximum minimum value of a parabola , which | |
| 00:46 | is not hard to understand . We're gonna draw a | |
| 00:48 | couple of pictures . I'm gonna show you how to | |
| 00:50 | calculate the maximum and the minimum value of a problem | |
| 00:53 | , whether it's in vertex form or if it's in | |
| 00:55 | standard form and then if you stick with me to | |
| 00:57 | the end of the lesson , I'm going to actually | |
| 00:59 | show you where the equation comes from , which is | |
| 01:01 | more important than almost anything else because these equations , | |
| 01:04 | they don't just fall out of thin air . They're | |
| 01:06 | very easy to understand if we just take the second | |
| 01:08 | , just a few seconds to go through it . | |
| 01:11 | So let's first talk and go down memory lane uh | |
| 01:14 | , when it comes to what a maximum and a | |
| 01:17 | minimum value is in a parabola . So first let | |
| 01:20 | me give you a parabola in vertex form because the | |
| 01:24 | goal of this lesson is to really talk about the | |
| 01:26 | max and men when we have either form vertex form | |
| 01:29 | or standard form of a parabola . And by the | |
| 01:32 | way , if you don't know what vertex form of | |
| 01:33 | a parabola is , if you don't know what standard | |
| 01:35 | form is , then you need to go back to | |
| 01:36 | my previous lessons because we've covered that extensively . All | |
| 01:40 | right . So we have a vertex form of a | |
| 01:42 | problem . And the reason we like vertex form so | |
| 01:44 | much is because it makes graphing and understanding what what | |
| 01:49 | the problem looks like . Very , very simple . | |
| 01:51 | So , here's an example , concrete example why -1 | |
| 01:55 | is equal to x minus two quantity squared . So | |
| 01:58 | this is a parabola . We know that it opens | |
| 02:01 | up . The reason it opens up is because the | |
| 02:03 | coefficient in front of the X term here , in | |
| 02:05 | front of the princess is just a positive number , | |
| 02:07 | it's a positive one . So we know it opens | |
| 02:09 | up and we know it's the regular shape of an | |
| 02:11 | X . Squared parabola . Now we know this thing | |
| 02:13 | is not centered at the origin because of the two | |
| 02:15 | . And the one here we know it's shifted so | |
| 02:18 | we can read the vertex directly off of this graph | |
| 02:21 | . That's why it's called the vertex form . And | |
| 02:23 | so the vertex , we've talked about all this stuff | |
| 02:26 | before , the vertex is two units shifted over an | |
| 02:29 | X . And one unit shifted up . And why | |
| 02:32 | ? So the vertex of this problem is it two | |
| 02:34 | comma one . Also as I said , because there's | |
| 02:36 | a one here , we know this problem opens up | |
| 02:40 | . All right , we know that it opens up | |
| 02:43 | . So now what I want to do is take | |
| 02:44 | a break from that just a second and explain what | |
| 02:46 | I mean . When I talk about the maximum value | |
| 02:48 | or the minimum value of a problem , you can | |
| 02:51 | think of a parabola , like a roller coaster , | |
| 02:53 | you know , it goes to the top and then | |
| 02:55 | it comes down . The maximum value is just what's | |
| 02:57 | at the very top there . If it think of | |
| 02:59 | a roller coaster coming down like this , the minimum | |
| 03:02 | value is the value at the very bottom of the | |
| 03:04 | trough if it opens in that direction . So let's | |
| 03:07 | just draw a couple of images to make sure that | |
| 03:10 | we're all on the same page so that when I | |
| 03:12 | tell you maximum maximum value , minimum value , you | |
| 03:15 | know exactly what I'm talking about . So for this | |
| 03:18 | problem right here I'm not gonna graph it , but | |
| 03:20 | I know the vertex is two units to the right | |
| 03:23 | and one unit up . So the vertex is here | |
| 03:25 | and I know that this one opens up so this | |
| 03:28 | parable opens up so I'm just going to sketch it | |
| 03:30 | . This is not an exact graph at all , | |
| 03:32 | but you can see that it opens up , so | |
| 03:34 | why am I doing this ? Because every Parabola opens | |
| 03:37 | up like this , there's always a minimum value . | |
| 03:41 | So this is called a minimum value . So in | |
| 03:47 | oftentimes in a lot of problems , the question will | |
| 03:50 | say tell me the minimum value of the Parabolas . | |
| 03:52 | So in order for it to have a minimum value | |
| 03:55 | , it has to open up like that in order | |
| 03:57 | for there to be a bottom to it , right | |
| 03:59 | ? There is no maximum value of this parabola because | |
| 04:02 | the probably goes up forever and ever it goes to | |
| 04:04 | infinity . So there is no maximum , the maximum | |
| 04:06 | is just infinity , but there is a minimum value | |
| 04:09 | and it's right here , we're gonna talk more about | |
| 04:11 | about that in just a second . Now let me | |
| 04:13 | draw the alternative . This guy opened up so there | |
| 04:15 | is a minimum value here . But let me just | |
| 04:17 | kind of sketch another possibility . So this will be | |
| 04:20 | X and Y . As always . And let's just | |
| 04:24 | draw a problem over here , here's a vertex and | |
| 04:27 | this this problem might go upside down like this . | |
| 04:30 | So for problems that opened down like this , they | |
| 04:33 | have what we call a maximum value , yeah , | |
| 04:38 | value . And that maximum value is right at the | |
| 04:41 | top . So you see parable isn't open like this | |
| 04:44 | , they don't have any minimum value because the thing | |
| 04:47 | goes on down to negative infinity . They only have | |
| 04:49 | a maximum value for Parabolas that open like this , | |
| 04:52 | they don't have any maximum value , they only have | |
| 04:54 | a minimum value . So the very first step is | |
| 04:57 | when a question says , tell me the maximum maximum | |
| 05:00 | or minimum value , you have to figure out if | |
| 05:02 | the Parabola opens up or if it opens down because | |
| 05:05 | that's going to tell you if you have a maximum | |
| 05:07 | or if you have a minimum you can't have a | |
| 05:09 | maximum and a minimum at the same time in any | |
| 05:12 | given problem because it's only going to open one direction | |
| 05:16 | . All right . So when you have the equation | |
| 05:19 | in vertex form like this , it's very easy to | |
| 05:21 | figure out what the maximum and minimum value of the | |
| 05:23 | problem is because it's all given to you by the | |
| 05:26 | vertex because in this case the vertex of this problem | |
| 05:29 | is right here , which was at uh two comma | |
| 05:33 | one , right , two comma one . And if | |
| 05:36 | I asked you give me the maximum or the minimum | |
| 05:37 | value of this guy , you just look on the | |
| 05:39 | graph and you would say , well the minimum values | |
| 05:41 | right here . But when I ask you what the | |
| 05:43 | minimum value is , What I'm asking you is what | |
| 05:46 | is the lowest value and why that this thing really | |
| 05:48 | has . Now you can look at the vertex and | |
| 05:51 | you can see that this point is to comma one | |
| 05:53 | . Right ? And so the lowest value and why | |
| 05:57 | that the thing gets to is just one because this | |
| 05:59 | is one right here . Right ? The X value | |
| 06:01 | doesn't matter . The X value is where the problem | |
| 06:04 | is this direction . When I'm asking you for the | |
| 06:05 | minimum value or the maximum value , I'm asking you | |
| 06:08 | what value of why is or how big or small | |
| 06:12 | can y B for this problem ? In this case | |
| 06:14 | the minimum value is one . So you can read | |
| 06:16 | it directly out of the vertex but I want to | |
| 06:18 | show you something because we're gonna have to do this | |
| 06:19 | kind of thing when we solve our problems . All | |
| 06:23 | right . If you want to know the minimum value | |
| 06:26 | of this , I know it's given to you here | |
| 06:27 | . But you you know where the vertex is ? | |
| 06:30 | The X component of the vertex . The X component | |
| 06:32 | of the vertex is at two . X . Is | |
| 06:34 | equal to two . So I can take the value | |
| 06:36 | of two and stick it in here . Why minus | |
| 06:38 | one is two ? Because that's the X value where | |
| 06:42 | the vertex is minus two quantity squared . And I | |
| 06:46 | can solve for the value of why ? So why | |
| 06:48 | minus one is equal to this is zero square , | |
| 06:50 | so zero . So why is equal to one ? | |
| 06:52 | So you see all I've done here is kind of | |
| 06:54 | silly really . I've said hey the vertex is at | |
| 06:56 | two comma one . So if I take and I | |
| 06:58 | know that the vertex is at the bottom . Right | |
| 07:01 | ? So if I take the X . Value of | |
| 07:03 | the vertex and stick it in here , then I'm | |
| 07:05 | gonna get the y value of the vertex which is | |
| 07:07 | one which is exactly what the vertex is . So | |
| 07:10 | you can read it off . So you see I've | |
| 07:12 | done a little bit of work but I don't really | |
| 07:14 | need to because when you are in vertex form you | |
| 07:17 | already know the maximum or the minimum value . You | |
| 07:20 | just read it right off the vertex and you do | |
| 07:22 | a little sketch tell yourself if it's a maximum or | |
| 07:25 | minimum value . All right . I had to go | |
| 07:28 | over that stuff because it's important to contrast it with | |
| 07:33 | what I'm going to talk about next . And that | |
| 07:35 | is what if the parable is not given to you | |
| 07:38 | in vertex form . And what if I give you | |
| 07:41 | a parabola in general form or standard form ? And | |
| 07:44 | I ask you what is the maximum or the minimum | |
| 07:46 | value of this problem ? Now , of course you | |
| 07:48 | can take that parable and you can convert it to | |
| 07:50 | vertex form . We actually did that in the last | |
| 07:52 | section . Lots and lots of times . So you | |
| 07:54 | could do that . That's a lot of work . | |
| 07:56 | It turns out there's a pretty easy way to figure | |
| 07:59 | out the maximum or the minimum value of a parabola | |
| 08:01 | , even if it's not in this form and that's | |
| 08:04 | what we're gonna focus on . But I wanted to | |
| 08:05 | go through this first . So you understand what maximum | |
| 08:08 | and minimum is and how you can read it directly | |
| 08:10 | off of the vertex form . But what if I | |
| 08:13 | give uh give you a parabola ? What if the | |
| 08:19 | parabola is given in this form ? Uh Let's see | |
| 08:27 | here , Y equals a x squared plus bx plus | |
| 08:33 | seat . In other words this is a standard form | |
| 08:35 | , right ? So if it's given to you in | |
| 08:37 | standard form , you can tell if it opens up | |
| 08:40 | or down , you can look at the at the | |
| 08:42 | coefficient if it's positive and the one in front of | |
| 08:45 | X squared , if it's positive , it will open | |
| 08:47 | up and if it's negative it'll open down . But | |
| 08:50 | this doesn't really directly tell you what by just looking | |
| 08:53 | at it what the vertex is or if it hasn't | |
| 08:56 | what maximum or minimum value there is . So a | |
| 08:59 | question typically will say here is a parabola like this | |
| 09:03 | in this form , tell me the maximum value of | |
| 09:05 | it . Now you could convert that to vertex form | |
| 09:08 | of course and then you can read it off but | |
| 09:10 | that's a lot of work . So what we're gonna | |
| 09:11 | do is I'm going to present to you a rule | |
| 09:14 | or an equation to figure it out and then we're | |
| 09:17 | gonna solve a quick problem to show you how it | |
| 09:18 | works . And then at the end of it I'm | |
| 09:20 | gonna actually derive and show you where the equation comes | |
| 09:23 | from , which is not hard to understand . So | |
| 09:24 | stick with me to the end and we'll get there | |
| 09:27 | . So if I give you an equation in this | |
| 09:30 | form like this , then there's a couple of choices | |
| 09:34 | because it depends on if it opens up or opens | |
| 09:36 | down , right ? So what we say is if | |
| 09:39 | A is less than zero , then we say F | |
| 09:43 | of X , right ? Which I did . I | |
| 09:46 | put why here ? Right ? But I could say | |
| 09:48 | I could call it F . Of X because everything | |
| 09:50 | is a function , right ? So if this value | |
| 09:53 | of A is less than zero , then this function | |
| 09:55 | has a max value . And why does it have | |
| 10:01 | a max value will do a little sketch right there | |
| 10:04 | . Because if the thing if the thing has a | |
| 10:05 | negative value in front , we know it opens down | |
| 10:08 | so it has to have a max value here at | |
| 10:11 | the top . There's no minimum value in that case | |
| 10:14 | there's only a maximum value . Now . The other | |
| 10:17 | case you might guess is if a is bigger than | |
| 10:20 | zero a positive number , then ffx has a minimum | |
| 10:27 | value , right ? Why does it have a minimum | |
| 10:30 | value ? Because if this number here is bigger than | |
| 10:33 | zero , like if it's three or four or five | |
| 10:35 | halves or one half anything bigger than zero , then | |
| 10:38 | we know the problem opens up . So then the | |
| 10:40 | problem , it looks like this , and that means | |
| 10:43 | that always has a minimum value here , a minimum | |
| 10:46 | value . All right . So , we know that | |
| 10:50 | if I give it to you in standard form , | |
| 10:51 | I can definitely figure out very easily if it's a | |
| 10:54 | maximum or minimum value that we're dealing with . Now | |
| 10:57 | the next part of this little theorem is going to | |
| 10:58 | show you how to calculate that so I can find | |
| 11:02 | the vertex the vertex is located and this is something | |
| 11:11 | we haven't learned before is located at an X . | |
| 11:14 | Value of negative B over two . A . This | |
| 11:20 | should sort of look familiar because B over two A | |
| 11:24 | . It's something that we run into before and we | |
| 11:26 | were doing completing the square . It's something we ran | |
| 11:28 | into before with the quadratic formula . You know , | |
| 11:31 | those kinds of the quadratic formula is over to a | |
| 11:34 | negative B plus or minus B squared minus four A | |
| 11:37 | . C over two A . So we've run into | |
| 11:39 | like things with to A . On the bottom , | |
| 11:40 | but this is kind of uh different from all of | |
| 11:44 | that . This is an equation that tells you if | |
| 11:47 | I have this guy in standard form , I cannot | |
| 11:49 | read the vertex directly off of it . But I | |
| 11:52 | will prove it to you in a minute that the | |
| 11:53 | X value of that vertex is this quantity negative B | |
| 11:57 | . Over two A . So if this is 12 | |
| 12:00 | and three then it would I would just stick the | |
| 12:02 | value and for be the value in for A . | |
| 12:04 | I calculate it , that would be the X . | |
| 12:06 | Value of the vertex . Now before like the vertex | |
| 12:09 | here is that to common one . If I were | |
| 12:11 | to calculate the X value the vertex by this equation | |
| 12:15 | , I would be getting the first number here . | |
| 12:16 | This is the X value of the vertex . And | |
| 12:20 | if you remember when you have the X value of | |
| 12:22 | the vertex , like we knew here once we know | |
| 12:24 | the X value of the vertex . If you stick | |
| 12:27 | it into your equation for X . Of course then | |
| 12:29 | you're gonna find the corresponding why value of the vertex | |
| 12:31 | which is exactly what we did here . So once | |
| 12:34 | you know the X . Value of the vertex then | |
| 12:37 | the Y . Value of the vertex or the uh | |
| 12:40 | yeah the Y value the vertex is you can get | |
| 12:43 | it from here . You take this value of X | |
| 12:46 | . And you stick it into the function negative B | |
| 12:48 | . Over two . A . This stuff here is | |
| 12:51 | exactly what we did here . If you know the | |
| 12:53 | X value of the vertex , stick it into the | |
| 12:55 | function and you get the Y . Value of the | |
| 12:57 | vertex . If you know the X value of the | |
| 12:59 | vertex , stick it into the function , that's what | |
| 13:01 | this means . I just stick that value of X | |
| 13:03 | in you know and that's why I say F . | |
| 13:06 | Of negative B . Over two . A . What | |
| 13:08 | this means is that calculate the X . Value let's | |
| 13:10 | say for this I got the value of two , | |
| 13:12 | that's the X . Value the vertex . Then I | |
| 13:14 | just stick into the function a value of two . | |
| 13:17 | And I calculate the corresponding why value and you'll have | |
| 13:20 | a vertex in X comma y once you have X | |
| 13:23 | come alive for the vertex , you know what the | |
| 13:24 | minimum value is . Because here is the vertex here | |
| 13:27 | the minimum value is one for this one . And | |
| 13:30 | so it will basically be self evident once you have | |
| 13:33 | the vertex . So this allows you to calculate the | |
| 13:35 | vertex of a parabola . Even if you can't read | |
| 13:38 | it so easily from the parabola . When it's in | |
| 13:41 | standard form , when it's in vertex form you just | |
| 13:43 | read it directly off . You don't have to do | |
| 13:44 | any calculations . But here you have to do one | |
| 13:47 | little calculation . You got to find the X value | |
| 13:48 | . Then you stick it in to the function and | |
| 13:50 | you get the y . Value back . All right | |
| 13:53 | . And then from this you find the maximum or | |
| 13:56 | minimum value . The minimum or maximum value is gonna | |
| 13:58 | be uh the Y value . So this is gonna | |
| 14:01 | be the max or minimum value of parabola because the | |
| 14:12 | y value is always the maximum or minimum value . | |
| 14:15 | So this whole thing kind of is its own theorem | |
| 14:19 | . And usually you will see this in books . | |
| 14:21 | And uh literally , they'll just tell you if you | |
| 14:24 | have a parabola , calculate the stuff and you'll get | |
| 14:27 | maximum or minimum value of it . Great . And | |
| 14:29 | you just do it . So what we're gonna do | |
| 14:30 | now is solve a quick little problem is very simple | |
| 14:33 | . And then at the end of it , I'm | |
| 14:34 | gonna show you where this comes from . And then | |
| 14:36 | you'll know not only how to use it , you'll | |
| 14:37 | know why why we can do this . So you | |
| 14:40 | won't think it's black magic or something because I don't | |
| 14:44 | want you to think that . All right . So | |
| 14:46 | let's take a look at a real equation . Let's | |
| 14:48 | say I have a function F of X . It's | |
| 14:51 | equal to two X squared plus eight X . Now | |
| 14:55 | , in this particular case it's a X squared plus | |
| 14:59 | B X plus C . But there is no C | |
| 15:02 | C A zero , basically . So , what we | |
| 15:04 | really have in this situation for this equation is we | |
| 15:08 | know that A . Is too and we know that | |
| 15:11 | B is eight and we know that C . Is | |
| 15:13 | zero because it's not there . Right . And so | |
| 15:16 | I go back to my theory and I say , | |
| 15:18 | and I'm asking you the question , what is the | |
| 15:20 | minimum value of that function or the maximum value of | |
| 15:22 | that function ? Well , what you do is you | |
| 15:25 | say , let me find the vertex So to find | |
| 15:27 | the vertex it's going to be negative B over two | |
| 15:30 | way I just calculated . So for the vertex I | |
| 15:32 | go down here to find the vertex I say this | |
| 15:36 | , I say the X value of the vertex is | |
| 15:38 | negative B over two . A . So b is | |
| 15:42 | eight and A . Is too . So it's two | |
| 15:45 | times two to a right . And so I get | |
| 15:48 | don't forget that negative sign 8/4 which is negative to | |
| 15:51 | this is the X value . So the X value | |
| 15:53 | , the vertex is negative two . So I label | |
| 15:55 | vertex X . Value of the vertex is negative two | |
| 15:59 | . All right . And just like we did before | |
| 16:02 | when we had it over here , once we have | |
| 16:04 | the X value of the vertex , we can just | |
| 16:06 | plug it into the function for X and calculate the | |
| 16:09 | corresponding why value of the vertex . And by the | |
| 16:12 | way that's exactly what this is saying . Take the | |
| 16:14 | value you get , stick it into the function to | |
| 16:16 | get the Y . Value of the vertex . So | |
| 16:19 | let me switch colors and do that right now . | |
| 16:21 | And so in order to find the Y value , | |
| 16:23 | it's going to be taking that function and evaluating it | |
| 16:27 | at negative two . Why negative two ? Because that's | |
| 16:29 | the X . Value of the vertex . So I | |
| 16:31 | go back to my function and see what it is | |
| 16:33 | . It's two X squared . But X is now | |
| 16:35 | negative too . Don't forget to square it plus eight | |
| 16:39 | times X . Eight of course uh evaluated at negative | |
| 16:42 | two . And so what I'm gonna get here is | |
| 16:45 | um Two times negative , two times negative two . | |
| 16:49 | When you square it is going to be positive for | |
| 16:51 | . And then here you have a negative 16 . | |
| 16:54 | So I'm gonna erase this and make it a negative | |
| 16:56 | 16 . And so what you have here is 8 | |
| 16:59 | -16 . And so what you have here is -8 | |
| 17:02 | . And so what you have here is I wanted | |
| 17:05 | to run it over there . I'll put it right | |
| 17:06 | here . Why is equal to negative eight ? So | |
| 17:09 | , what you figured out from this is we've calculated | |
| 17:11 | the vertex the vertex is located at an X . | |
| 17:16 | Value of negative two and a Y value of negative | |
| 17:19 | eight . Negative two comma negative eight . All right | |
| 17:22 | . So if the question said , what is the | |
| 17:24 | vertex ? I would just circle . This is my | |
| 17:25 | answer . But the question says usually for these problems | |
| 17:28 | is what is the maximum or the minimum value of | |
| 17:30 | this function ? Now , just knowing the vertex doesn't | |
| 17:33 | tell you that . Because remember the vertex is going | |
| 17:35 | to be either at the top or the bottom of | |
| 17:37 | the problem of the vertex is here or here . | |
| 17:40 | But in order to figure out if it's a maximum | |
| 17:42 | or minimum value , you have to know does the | |
| 17:44 | problem open up or down ? So the Parabola has | |
| 17:47 | a coefficient of x squared of positive to that means | |
| 17:51 | that this Parabola is gonna open up , you know | |
| 17:54 | the vertex is at negative two right here and then | |
| 17:58 | you have to go negative 12345678 So it's way down | |
| 18:02 | here and you know that it opens up . So | |
| 18:05 | basically you don't have to do all the sketching , | |
| 18:07 | but I'm showing you the vertex is down here . | |
| 18:09 | It opens up so this is a minimum value . | |
| 18:12 | So this uh Parabola has a minimum value , not | |
| 18:16 | a maximum value value of negative eight . It's always | |
| 18:20 | the y value . You're looking for a minimum value | |
| 18:22 | and why ? So this is what you would circle | |
| 18:24 | . The minimum value is negative eight . Or you | |
| 18:27 | could say the minimum value is y equals -8 . | |
| 18:30 | You can also circle the vertex if it's asked or | |
| 18:32 | required as well . So now you can see the | |
| 18:35 | value of the vertex form of the equation is very | |
| 18:38 | very even more powerful because now you can see that | |
| 18:42 | you can pull the vertex out and from that you | |
| 18:44 | can figure out if it's a combination with if it | |
| 18:47 | opens up or down . You know if it's a | |
| 18:49 | maximum or minimum value when you just read it directly | |
| 18:51 | off . If the equation is given to you in | |
| 18:54 | this form with any coefficients at all , then I | |
| 18:57 | know if it opens up or down just by looking | |
| 18:59 | at it by knowing the value of A . And | |
| 19:01 | I can calculate the vertex with these equations here , | |
| 19:04 | basically all you have to remember is this one because | |
| 19:06 | once you have the X . Value , you just | |
| 19:08 | stick it into the function to calculate the y value | |
| 19:11 | , why it works right , you stick it in | |
| 19:13 | there and you get the corresponding x comma y value | |
| 19:15 | of the vertex right now what I want to do | |
| 19:18 | is focus on where does this come from , because | |
| 19:19 | I just threw it at you and I said hey | |
| 19:22 | that is a equation , you know enjoy it , | |
| 19:25 | but I want you to know where things come from | |
| 19:27 | that . It's very easy to understand and actually build | |
| 19:29 | your algebra skills as well because you have every bit | |
| 19:33 | of knowledge to understand that . So if I'm given | |
| 19:38 | a parabola in standard form , I'm always going to | |
| 19:40 | start with this A X squared plus bx plus C | |
| 19:46 | . Now I have to keep it general like this | |
| 19:48 | because I have to keep it , I have to | |
| 19:50 | do the proof for for any value of any parabola | |
| 19:53 | . So I I don't want to pick certain values | |
| 19:55 | . I'm gonna leave it as a B and C | |
| 19:57 | because it could be any problem anywhere in the plane | |
| 20:00 | . Right ? So what we're gonna do is essentially | |
| 20:02 | convert this thing to vertex form as we have done | |
| 20:05 | in previous lessons , but we're gonna do it with | |
| 20:08 | these variables . So we have to be careful when | |
| 20:09 | , when there's numbers , it's easy to convert to | |
| 20:11 | vertex form when there's variables that are A B and | |
| 20:14 | C . I should say placeholders for the constants . | |
| 20:17 | Then you have to be a little bit more careful | |
| 20:18 | . But it's the same process we followed in the | |
| 20:20 | past . The first thing you do is you take | |
| 20:22 | the constant , we've done this before you move it | |
| 20:24 | to the other side . So you say why minus | |
| 20:26 | C . Is a X squared plus bx so far | |
| 20:30 | . So good . That's the first thing we did | |
| 20:32 | and we convert to vertex form . The next thing | |
| 20:34 | we did is we take a look over at the | |
| 20:35 | coefficient in front of X squared and make sure it's | |
| 20:38 | a one . It's not a one , it's a | |
| 20:40 | . So I don't know what A . Is , | |
| 20:42 | it's possible that A is one , of course , | |
| 20:44 | but I have to keep it general , I don't | |
| 20:45 | know that A . Is one . So I'm going | |
| 20:47 | to pretend that it's not . So in order to | |
| 20:49 | get rid of that , what I'm gonna do is | |
| 20:52 | factor it out , which is what I have to | |
| 20:53 | do , so it's gonna be X squared plus . | |
| 20:57 | Now , here's the tricky part . How do you | |
| 20:59 | factor at an A . From this term ? And | |
| 21:01 | a lot of students look at that and say I | |
| 21:03 | can't do that because here's A and here's a B | |
| 21:05 | . How do I factor out ? Well , when | |
| 21:07 | you start doing things , uh you kind of quote | |
| 21:10 | unquote , grow up a little bit right , and | |
| 21:11 | you start doing more proof you have to use and | |
| 21:14 | become very comfortable with fractions . If you're it is | |
| 21:17 | true that it basically looks like it's impossible to factor | |
| 21:20 | out an A . From that . So you don't | |
| 21:21 | even think you can do it . However , look | |
| 21:22 | at how I can write this . I can write | |
| 21:24 | this as follows . I can write this as this | |
| 21:27 | term as be over A times X . I'm gonna | |
| 21:31 | close the parentheses , make sure you understand what happened | |
| 21:33 | here . It doesn't look like you can factor it | |
| 21:35 | out . However , if I multiply back in , | |
| 21:37 | that's going to be a X . Squared . And | |
| 21:39 | if I multiply this in it'll be A times B | |
| 21:42 | over A . But the A's would then cancel and | |
| 21:44 | it would just leave B . X . Which is | |
| 21:46 | my term right here . So the truth is that | |
| 21:50 | you can't factor out numbers unless you kind of see | |
| 21:52 | the number there or a factor of it . But | |
| 21:54 | the truth is I can factor anything I want because | |
| 21:57 | I can change the second term into a fraction . | |
| 22:00 | So that when I back multiply it back in , | |
| 22:02 | the cancellation , gives me what I started with . | |
| 22:05 | That's all that matters when it comes to factoring , | |
| 22:08 | right ? So um when I multiply this times this | |
| 22:12 | , I get A X . Squared . When I | |
| 22:13 | multiply this times this , I get A times B | |
| 22:16 | over A the A's cancel . Giving me this . | |
| 22:19 | This little trick of leaving it as a fraction here | |
| 22:21 | is not really a trick . It's just making the | |
| 22:23 | backwards motion of the multiplication work is something that you | |
| 22:26 | will see uh as you get into more advanced math | |
| 22:29 | . Okay . But it's not so hard to understand | |
| 22:31 | , it's just like a times B overhead and they | |
| 22:33 | see the cancellation . Okay , So now we have | |
| 22:36 | a coefficient in front of this of one , that's | |
| 22:38 | what we always wanted . Now we have to complete | |
| 22:40 | the square over here . Now it's ugly , right | |
| 22:42 | ? Because the coefficient , remember to complete the square | |
| 22:45 | , you look at the coefficient that's in front of | |
| 22:46 | X , whatever the coefficient in front of exes to | |
| 22:49 | complete the square . We've done it many times , | |
| 22:51 | you divided by two , then you square it and | |
| 22:54 | then you add it to both sides . But the | |
| 22:56 | coefficient that's in front is a fraction , so it | |
| 22:58 | makes it tricky . So we're gonna have to be | |
| 23:00 | careful , but that's okay , that's what we're gonna | |
| 23:02 | do here . So we're gonna say why minus C | |
| 23:04 | . I'm gonna leave myself some space , I'm gonna | |
| 23:06 | have A and I'm gonna have X squared plus B | |
| 23:10 | over a times X . And then I'm gonna add | |
| 23:13 | a term to both sides , Right ? What am | |
| 23:16 | I gonna add ? I have to take the coefficient | |
| 23:18 | of X . Which is going to be I'm gonna | |
| 23:19 | write it like this , I'm gonna write it as | |
| 23:21 | be over a right ? But then I have to | |
| 23:24 | divide that coefficient by two , and then I have | |
| 23:28 | to wrap parentheses around it , square it . And | |
| 23:31 | that whole thing is living inside of the princes . | |
| 23:34 | This isn't exactly the same thing as we've done every | |
| 23:37 | time we've completed the square . If the coefficient in | |
| 23:40 | front of X were six , I would say six | |
| 23:42 | divided by two Would be three square . It it | |
| 23:45 | would be nine , I wouldn't add nine to both | |
| 23:47 | sides . But uh in this case the coefficient is | |
| 23:50 | wrapped up in these be in a variable placeholders . | |
| 23:54 | So I have to leave at exact , I have | |
| 23:56 | to say , well this thing's gonna be divided by | |
| 23:57 | two and then one change I'm gonna make is when | |
| 23:59 | you divide by two , this two on the bottom | |
| 24:02 | can be written as to over one . Now , | |
| 24:05 | what I'm gonna do is you got to be a | |
| 24:07 | little careful when you add it to both sides . | |
| 24:09 | Remember we did this in the past lessons on this | |
| 24:12 | , I can just add it to this side . | |
| 24:14 | However , I can't just let me let me just | |
| 24:17 | start by saying that this whole term is multiplied by | |
| 24:19 | A . So when I added it to the side | |
| 24:22 | , I really added a times that so to to | |
| 24:25 | keep it balanced , I actually have to add a | |
| 24:27 | larger term over here . It's gonna be a times | |
| 24:31 | b over a over to over one quantity squared . | |
| 24:37 | Close the bracket , I add that to the left | |
| 24:38 | hand side . Why did I multiply by A . | |
| 24:40 | Because when I added it to the right , when | |
| 24:43 | you multiply a in here , you're really adding eight | |
| 24:45 | times this , that's the value you're adding , you're | |
| 24:47 | not just adding one of them , you're adding eight | |
| 24:49 | times it to the right hand side . So to | |
| 24:51 | balance it , I have to add eight times on | |
| 24:53 | the left . Now the truth is I'm not going | |
| 24:55 | to continue simplifying this on the left because it doesn't | |
| 24:58 | matter and you'll see why it doesn't matter in a | |
| 25:00 | second . So I'm gonna say this is why minus | |
| 25:02 | blah because I'm not gonna go through simplifying all that | |
| 25:05 | stuff . You'll see why in a second . Over | |
| 25:08 | here I can try to factor , remember completing the | |
| 25:12 | square the goal of it is to factor . So | |
| 25:14 | it's gonna be X . Times X . To give | |
| 25:16 | me the X . Squared . And then over here | |
| 25:19 | um I think probably an easier way to do it | |
| 25:21 | . Let me I should have probably done one more | |
| 25:23 | step . This this thing right here reduces to the | |
| 25:26 | following , it reduces to when you take the 2/1 | |
| 25:31 | and flip it upside down . And multiply it becomes | |
| 25:34 | B . Over two . A . That's where to | |
| 25:36 | be over to A . Uh comes from . So | |
| 25:39 | be over a divided by two . You can think | |
| 25:41 | of it as flipping over , multiplying it . So | |
| 25:43 | it's B . Over two A . That's what this | |
| 25:44 | is squared . So that's what I really added to | |
| 25:47 | both sides . And so that's what this term really | |
| 25:50 | is . And so when I come in here to | |
| 25:51 | do the factoring , see it squared . So I'm | |
| 25:53 | gonna call this B . Over two A . And | |
| 25:58 | I think I'm going to you and I'll leave it | |
| 25:59 | like that be over two eggs . And then of | |
| 26:02 | course there's a plus here . So I'm gonna factor | |
| 26:04 | it like this now . I know it looks confusing | |
| 26:06 | but just go backwards with me , X times X | |
| 26:08 | . Is X squared . The inside term is B | |
| 26:12 | over two A . Times X . And then this | |
| 26:14 | is B over two A . Times X . So | |
| 26:15 | you can think of it as one half B . | |
| 26:17 | A . One half B . A . You add | |
| 26:19 | them together , you just get to be over A | |
| 26:21 | . X . Which is what this is multiply the | |
| 26:24 | last terms together . You get the term that squared | |
| 26:26 | . So this is the factored form of it . | |
| 26:28 | And so what I get is why minus blah , | |
| 26:30 | I'm not going to mess with that anymore . Doesn't | |
| 26:32 | matter . On the right hand side is going to | |
| 26:33 | be X plus B over two a quantity squared now | |
| 26:40 | , why do I care about that ? Because the | |
| 26:43 | vertex form always looks like y minus blah is something | |
| 26:48 | X minus something squared . That's exactly what I've done | |
| 26:52 | . I've done the completing the square operation to take | |
| 26:54 | this quadratic and beat it into shape . So it's | |
| 26:58 | why mine is blah something and then X . Of | |
| 27:01 | course I have a plus sign instead of a minus | |
| 27:03 | a number . The number just happens to be B | |
| 27:05 | over two . A . So the X component of | |
| 27:07 | this vertex can be written from this equation . I | |
| 27:12 | don't care about the Y value , I only care | |
| 27:14 | about finding the equation for that X value . The | |
| 27:16 | vertex has an X value . If this were in | |
| 27:19 | minus sine x minus this then it would be shifted | |
| 27:22 | to the right by that value . But it's A | |
| 27:24 | . Plus , so it's shifted to the left , | |
| 27:26 | so it's really an X . Value of negative B | |
| 27:28 | over two . A . Let me say that one | |
| 27:30 | more time . This is telling you the shift in | |
| 27:33 | the parabola in terms of X , how far over | |
| 27:36 | it's shifted . If it were a minus sign it | |
| 27:38 | would be shifted to the right and the vertex would | |
| 27:40 | have a positive sign there , but there's a plus | |
| 27:42 | sign . So it shifted to the left . That's | |
| 27:44 | why the negative where the negative comes from . So | |
| 27:46 | the vertex has an X . Value of negative B | |
| 27:48 | over two . A . Which is exactly what we | |
| 27:51 | said in the theorem . So the theorem is just | |
| 27:54 | usually thrown at you and it said what if you | |
| 27:56 | have a standard form a X squared dx and see | |
| 28:01 | , you can figure out if it goes up and | |
| 28:02 | down . The vertex has an X . Value of | |
| 28:04 | this . Once you calculate that , you stick it | |
| 28:07 | back into the function to find the Y value of | |
| 28:09 | the vertex here I'm saying where does that come from | |
| 28:13 | ? Start from the standard form of a parabola and | |
| 28:16 | complete the square . Which is what we've done when | |
| 28:18 | we've changed the vertex form , move the C . | |
| 28:20 | Value over . Then I have to factor out that | |
| 28:22 | A . But then you have to know how to | |
| 28:24 | handle that has to be be over A . So | |
| 28:25 | the multiplication works then since you have a coefficient of | |
| 28:29 | one there you complete the square . It's this plus | |
| 28:32 | this one half of this coefficient squared . But when | |
| 28:36 | you take one half of this coefficient or divided by | |
| 28:38 | two and square it it becomes when you flip over | |
| 28:40 | and multiply B over two A squared . And then | |
| 28:43 | you try to factor the right hand side . The | |
| 28:45 | is still here . This plus this and this plus | |
| 28:48 | this exactly comes back and gives you that factored form | |
| 28:51 | , which then can be written as this plus this | |
| 28:53 | , B over two A squared . And then the | |
| 28:56 | vertex is read directly from it when you have X | |
| 28:58 | plus anything , the vertex has shifted left and that's | |
| 29:01 | what it is , that's where it comes , that's | |
| 29:03 | where it comes from . So in this lesson we've | |
| 29:05 | introduced the concept of the maximum and minimum value of | |
| 29:07 | a parabola and then we've also calculated the X value | |
| 29:11 | of the vertex of a parabola . And we've done | |
| 29:13 | a problem where practically speaking , when we are given | |
| 29:17 | one of these parabolas , we can calculate the X | |
| 29:20 | value of the parabola . Once we know that x | |
| 29:22 | value , we just stick it back into the function | |
| 29:24 | to find the corresponding why value and that why value | |
| 29:28 | is either going to be a maximum or minimum value | |
| 29:31 | . And this problem , it was a minimum value | |
| 29:33 | because the thing opened upwards . But in another problem | |
| 29:37 | , if you had a problem where it was opening | |
| 29:39 | down , then that vertex uh y value would be | |
| 29:42 | the maximum value . So make sure you understand where | |
| 29:46 | I'm coming from and that you can solve these problems | |
| 29:47 | and understand mostly the concepts of what's happening . Follow | |
| 29:50 | me on to the next lesson , we're going to | |
| 29:51 | do a few problems to get your skills going and | |
| 29:54 | it's not gonna be that hard . We don't ever | |
| 29:56 | have to do this completing the square business ever again | |
| 29:59 | . I just did that to show you where this | |
| 30:01 | equation comes from , the future . Problems are just | |
| 30:03 | going to calculate the X vertex , calculate the minimum | |
| 30:06 | maximum value . And then you're basically done with what | |
| 30:09 | the problem is asking for . So let's move on | |
| 30:11 | to that lesson and learn those skills right now . |
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