12 - Writing Quadratic Functions in Vertex Form - Part 1 (Graphing Parabolas) - By Math and Science
Transcript
00:01 | Hello . Welcome back to algebra . I'm Jason with | |
00:02 | math science dot com . And today we're going to | |
00:05 | tackle the topic of writing these quadratic functions in vertex | |
00:09 | form . And this is part one . We have | |
00:10 | a couple of lessons here to get this topic of | |
00:12 | cross . So if you've been with me and these | |
00:14 | lessons up until this point we have talked about parameters | |
00:17 | for quite a long time . And most recently we've | |
00:20 | talked about parabolas in this vertex form , which is | |
00:22 | just a different way of writing uh these parabola equations | |
00:26 | down these quadratic equations that make it very easy to | |
00:29 | graph it because the vertex is basically right there in | |
00:32 | front of you in the equation . And then we've | |
00:34 | shown in the last few lessons how to plot the | |
00:36 | vertex and how to plot the rest of the problem | |
00:39 | . So now what we wanna do is tie everything | |
00:41 | together . All right , we want to talk about | |
00:43 | uh the idea that the quadratic equations can be written | |
00:46 | really in two ways . And specifically I want to | |
00:49 | be able to give you any quadratic equation that I | |
00:51 | might write down and I want you to be able | |
00:53 | to then convert it into vertex form because right now | |
00:56 | I've been telling you , hey , here is quadratic | |
00:58 | equations and regular form . Here is quadratic equations in | |
01:01 | vertex form and we've been kind of dealing with them | |
01:03 | separately . Now I'm going to actually teach you how | |
01:05 | to take any quadratic that we have and how to | |
01:08 | convert it into right in terms of vertex form so | |
01:11 | that you can then plotted or graphic or sketch it | |
01:13 | very easily . So we have in general two ways | |
01:16 | to write these quadratic equations . The first one is | |
01:19 | the kind of the first one we learn in algebra | |
01:20 | . I'm gonna call it . Most books , will | |
01:22 | call it the general general role for . I'm gonna | |
01:25 | spell general general form also , this is called in | |
01:33 | some books Standard Form and this is the form of | |
01:39 | a parable that you're basically taught when you very first | |
01:42 | very first time that you're introduced to the concept of | |
01:44 | what a parable is . So the general form or | |
01:47 | the standard form , she'll look really familiar to F | |
01:49 | of X . So some function right , is gonna | |
01:51 | look like this , A X squared plus Bx plus | |
01:56 | C . Now A and B and C are just | |
01:58 | numbers , but specifically A cannot be equal to zero | |
02:02 | because if A is actually equal to zero , then | |
02:05 | it kills this first term completely . And then it's | |
02:07 | not a quadratic equation anymore . Because if A zero | |
02:10 | then all you have is a line Mx plus B | |
02:12 | or B , X plus C . This is just | |
02:13 | a line , right ? But when you add this | |
02:15 | term to it , it changes it into that very | |
02:17 | famous familiar parabolas shape . Now , you know that | |
02:21 | A and B and C can basically be anything that | |
02:23 | could be negative numbers , that can be positive numbers | |
02:26 | . They can be fractions , they can be decimals | |
02:28 | and the different coefficients A B and C when you | |
02:31 | dial them in are going to basically produce a parabola | |
02:33 | which can live anywhere in the xy plane . Might | |
02:36 | be up here , might be over here , might | |
02:38 | be upside down , might be right side up but | |
02:40 | it's very hard to look at these coefficients on paper | |
02:43 | and really be able to look at it and say | |
02:45 | well yeah that's over here because it's just all wrapped | |
02:47 | up in an expanded form basically which is more typically | |
02:52 | called general former standard form . But there's another way | |
02:55 | to write these uh parabolas so I can take any | |
03:00 | parabola in this form and by the way this is | |
03:02 | the general form . So A X . Squared plus | |
03:04 | bx plus C . But of course you know be | |
03:05 | could be zero . So a very a perfectly valid | |
03:09 | . Um Let me just go off off the rails | |
03:12 | here and give you a couple of examples . Okay | |
03:13 | , A parable . It can be in standard form | |
03:15 | , it probably could be a . Would be maybe | |
03:17 | two X squared plus three eggs plus . See that's | |
03:21 | valid . That's a parabola . You can also have | |
03:23 | a parabola which would be maybe you know four X | |
03:26 | . Squared and be can be zero . So there's | |
03:28 | no exit all plus two . This is a parabola | |
03:30 | . So it doesn't have to have all three terms | |
03:32 | . It just has to have the X squared term | |
03:34 | . In fact equally good parabola as you all know | |
03:37 | will be just X squared . So in the case | |
03:40 | of X squared , this the term is zero . | |
03:42 | And also this term is zero . So really A | |
03:44 | , B and C . Or any number you want | |
03:46 | A is the only thing that cannot be zero . | |
03:48 | The other terms can all three of these are parabolas | |
03:50 | in different parts of the xy plane . But by | |
03:53 | looking at these equations , you cannot really tell where | |
03:56 | the parabola is unless you're just a human computer . | |
03:58 | Right ? So we have another way of writing these | |
04:01 | things which we have already learned and it's called vertex | |
04:05 | form , vertex form of a parabola . Another term | |
04:10 | that you might see in books or teachers might tell | |
04:12 | you it's also called the completed square form . Now | |
04:21 | you might say , oh my gosh completing square , | |
04:23 | what does that mean ? What we've already learned in | |
04:25 | previous lessons how to complete the square . It's a | |
04:28 | topic that we've covered in the past , I've covered | |
04:30 | it extensively . So if you have no idea what | |
04:32 | completing the square is and you probably should go back | |
04:35 | to my previous lessons because I've covered it extensively . | |
04:37 | But this topic , this type of parable , it's | |
04:40 | called vertex form in almost every book . But some | |
04:43 | books call it the completed square form . The reason | |
04:45 | will become obvious in a minute , but it's because | |
04:47 | we have to use completing the square in order to | |
04:50 | write the vertex form . So that's why it's called | |
04:52 | that . Um And just to refresh your memory , | |
04:56 | the vertex form of a parabola looks something like this | |
04:59 | F . Of X is equal to some constant A | |
05:03 | X minus H quantity squared plus K . Some books | |
05:08 | will write it like this right ? I actually or | |
05:11 | you might instead of F of X , you might | |
05:12 | see it as Y . Right ? So you might | |
05:15 | see it like I can leave F of X so | |
05:16 | that's fine . But I could also write it as | |
05:17 | Y . Okay . But I actually prefer to do | |
05:20 | it like some other books that I use that right | |
05:23 | like this , I just moved the K to the | |
05:24 | other side . So what you have is why minus | |
05:26 | K . Is a X minus H quantity squared . | |
05:31 | So basically these are the same equations . This one | |
05:33 | just has the shift in the Y direction , written | |
05:36 | on the right hand side of the equal sign . | |
05:37 | And this one has the shift in the Y direction | |
05:40 | here . The advantage to writing it like this is | |
05:43 | because the shift in Y is written right next to | |
05:45 | the Y variable and the shift in X has written | |
05:47 | next to the X variable . So it's very easy | |
05:51 | to recognize that this guy is basically easy to graph | |
05:55 | , this is easy to graph and we've done it | |
06:00 | many times before . We had the entire lessons on | |
06:02 | this . The reason it's easy is because the vertex | |
06:06 | is written right into the equation is at the coordinate | |
06:09 | h comma K . In other words , this tells | |
06:12 | you that from the origin this parable is shifted eight | |
06:15 | units to the right and it shifted up K unit | |
06:18 | . So this would be the point H comma K | |
06:20 | . So the vertex is written right there . So | |
06:22 | basically when you have a problem in vertex form , | |
06:25 | you can write you can put a dot on the | |
06:27 | X Y plane showing its vertex which is its maximum | |
06:30 | or minimum value just right away . You don't have | |
06:32 | to calculate anything , you just read it . So | |
06:35 | then the question is , if we're given an equation | |
06:38 | of a parabola or quadratic any quadratic we want , | |
06:42 | how do I take it from this form and somehow | |
06:45 | come up with this form ? Because so far I've | |
06:47 | given you some problems in the past with parable is | |
06:50 | written like this , we've done things . And then | |
06:52 | I've also given you lots of problems recently where the | |
06:55 | problem looks like this and we've of course graph them | |
06:57 | . But how can I give you this guy ? | |
06:59 | And then you calculate this ? That is the topic | |
07:01 | of this lesson . So what we're gonna be doing | |
07:03 | is I'm gonna give you a quadratic like this , | |
07:05 | we're gonna do some mathematical manipulation and we're gonna get | |
07:08 | this and here's a big hint we're gonna end up | |
07:10 | having to complete the square , which is a skill | |
07:13 | that we've learned in the past . So let's go | |
07:14 | over here and kind of inch are way forward into | |
07:17 | that . Let's take a specific example . What if | |
07:20 | I have uh the parabola or the quadratic F . | |
07:24 | Of X is equal to X squared plus two X | |
07:28 | uh minus three . What can I ascertain from this | |
07:32 | ? Uh as far as what the graph looks like | |
07:35 | without doing anything else ? Well , pretty much the | |
07:37 | only thing that I can tell is number one that | |
07:39 | it is a quadratic because I have a square term | |
07:41 | . But also I can look at the coefficient in | |
07:43 | front of the X squared . And I see that | |
07:45 | that is just the number one . So the first | |
07:48 | thing you can tell is that it's a positive value | |
07:50 | . So this parabola , even though you don't know | |
07:52 | exactly where it is , you know that it opens | |
07:54 | up because positive values in front of the X , | |
07:57 | we're always opened up . So you can tell from | |
08:00 | this right away that it opens up instead of opening | |
08:04 | upside down , right ? You can also tell that | |
08:08 | since it's just a one X square , that it's | |
08:10 | gonna have that standard parabolas shape . The basic graph | |
08:13 | is what I've been trying to tell you if it | |
08:14 | were two X squared or three X squared for the | |
08:16 | first time , you know that it would be much | |
08:18 | , much more steep and closed up because that's what | |
08:20 | happens when you have larger and larger coefficients in front | |
08:24 | . But honestly , other than this opening up , | |
08:26 | you really can't tell anything at all about this problem | |
08:28 | . I don't have any idea if it's way down | |
08:30 | in the bottom of the plane or if it's up | |
08:32 | here or up here . I don't have any idea | |
08:34 | where the crossing points are . I can't tell anything | |
08:36 | from really looking at this . So it's hard to | |
08:38 | graph to graph . So ultimately , I want to | |
08:46 | get this thing into the form of this . I | |
08:50 | want to get into form that looks like this . | |
08:57 | Why minus K A x minus H quantity squared and | |
09:03 | some books , again , I tell you , has | |
09:05 | the K value over there . But this is how | |
09:06 | I generally right , this this guy down now this | |
09:09 | why is this thing called completed square form ? Or | |
09:13 | sometimes it's called the perfect square form . It's because | |
09:16 | this guy right here differs from this quite , quite | |
09:19 | obviously because of this term is a perfect square , | |
09:28 | it's a perfect square . In other words , I'm | |
09:30 | able to to manipulate this equation into a form where | |
09:33 | I have a binomial squared like that , right ? | |
09:36 | You can always do that with every one of these | |
09:39 | quadratic that can give you any quadratic and I'm gonna | |
09:41 | show you how to manipulate it so that on one | |
09:43 | side of the equal sign , I'm going to have | |
09:44 | a perfect square like that . When I say perfect | |
09:46 | square . What I mean is In terms of binomial | |
09:49 | , it's just a binomial squared , but the analog | |
09:51 | back to numbers would be a perfect square , would | |
09:53 | be like three square or a perfect square be like | |
09:56 | four square . Well this can be written in such | |
09:58 | a way where I can have this perfect binomial square | |
10:00 | . In other words , I don't have a bunch | |
10:01 | of terms running around , expand it out . I | |
10:04 | can compact if I it into something that's basically just | |
10:07 | nice and square and that's why it's called uh sometimes | |
10:11 | you see this uh called vertex form obviously , but | |
10:13 | you might see it's written as perfect square form or | |
10:16 | completed square form . So now we're ready to actually | |
10:20 | jump into it . Let me go . That was | |
10:22 | kind of uh an introduction more than anything . We're | |
10:25 | gonna work with the same exact equation . But I | |
10:26 | want a fresh board and I want to teach you | |
10:29 | how to go from point A to point B . | |
10:31 | So if I give you the function F . Of | |
10:33 | X X squared plus two , X minus three . | |
10:37 | And I want to write it in that other form | |
10:40 | . The first thing you want to do , if | |
10:41 | you see it written in function notation F . Of | |
10:43 | X is I want you to replace F . Of | |
10:46 | X with Y . Because it just ends up becoming | |
10:48 | a lot easier to have wise and excess . You | |
10:51 | could carry the ffx down . It would be fine | |
10:53 | , but then it would just look a little cumbersome | |
10:54 | after a while . All right . So just first | |
10:57 | thing you do replace it with , why ? That's | |
10:58 | all I've done , The next thing I wanna do | |
11:01 | is I want to pretend the why isn't here ? | |
11:03 | I want to complete the square over here . Remember | |
11:05 | we did that a long time ago . And remember | |
11:07 | the very first step of completing the square was take | |
11:10 | the number the constant and move it to the other | |
11:12 | side of the equal side . Right ? So here | |
11:14 | I have a minus three , so I'm gonna add | |
11:15 | three to both sides . So I'll have y plus | |
11:17 | three on the left , an x squared plus two | |
11:20 | X on the right . So all I've done is | |
11:23 | add three to this side and add three to this | |
11:25 | side . Right ? So what I'm trying to do | |
11:27 | is complete the square over here . And if you | |
11:28 | remember the steps , first thing you have to do | |
11:30 | is move the constant term over to the other side | |
11:33 | . Then what you have to do is check . | |
11:35 | It's very important for the next step to check , | |
11:37 | make sure the coefficient is a one in front of | |
11:41 | the x square term . The coefficient must b one | |
11:48 | in this case . This coefficient already is one . | |
11:50 | So we don't have anything else to do in the | |
11:52 | next problem . I'm gonna show you what happens when | |
11:54 | we don't have that coefficient there . But if you | |
11:56 | remember back to completing the square , we've handled that | |
11:59 | many times before , so for now it's already done | |
12:02 | . So the next step we have to do , | |
12:03 | remember if you're just completing the square here is we | |
12:06 | have to add one half of this term , The | |
12:09 | coefficient in front of the X . We have to | |
12:10 | add one half of this square , it and we're | |
12:12 | gonna add it to this side and then we have | |
12:14 | to add it to this side . Why do we | |
12:15 | add it to both sides ? Because this is an | |
12:17 | equation , so we can add anything we want to | |
12:20 | both sides of the equation that keeps it balanced . | |
12:23 | So we have to do whatever we're gonna do to | |
12:24 | both sides . So let's just work on the right | |
12:26 | hand side uh for the time being let's go and | |
12:30 | use let's continue with the purple . So we're gonna | |
12:33 | have X . Squared plus two X . But what | |
12:36 | I'm going to add now is another term and that | |
12:39 | term is going to be written like this , you | |
12:41 | open it up and you take this coefficient your divided | |
12:43 | by two and you square this is just a number | |
12:46 | , I know that you know that this is just | |
12:47 | one squared . So it means you're just adding one | |
12:50 | . But I don't want to just put a one | |
12:52 | here because if I just put a one there then | |
12:53 | you don't really know how I got the one . | |
12:55 | So I'm writing it down exactly as I would on | |
12:57 | my paper , I'm saying take this coefficient divided by | |
13:00 | two and square it . And if I'm going to | |
13:02 | add it to this side then of course I have | |
13:04 | to add it to the other side . So it's | |
13:05 | going to be y plus three and you have to | |
13:08 | add exactly the same thing . Take that coefficient divided | |
13:12 | by two quantity squared . Yeah . So now that | |
13:15 | we're in a position where we can see that we're | |
13:17 | just adding some number two both sides . What's gonna | |
13:20 | end up happening is adding this very special number on | |
13:23 | the right hand side . Makes this thing a perfect | |
13:25 | square . Try no meal that I can factor very | |
13:28 | nicely . So on the left I have y plus | |
13:30 | three plus , this is one square which means one | |
13:33 | and then I have X squared plus two X Plus | |
13:37 | . again this comes out to be one on the | |
13:40 | left hand side , I have Y plus four and | |
13:43 | on the right hand side I'm going to try to | |
13:44 | factor this . If I'm unable to factor it , | |
13:47 | I've already messed up because completing the square makes it | |
13:50 | so that I can always factor the thing , that's | |
13:52 | the whole point of it . So if I cannot | |
13:54 | factor this , I've already messed up . So I'm | |
13:56 | going to open up to parents season and put an | |
13:57 | X right here in an X right here and I | |
13:59 | have a one . The only way that can work | |
14:01 | is if I have one times one and trying to | |
14:03 | make a positive too . So the only way it | |
14:05 | works is with positive sides , check yourself X times | |
14:08 | X is X squared inside terms gives you X . | |
14:11 | Outside terms give you X . Those add to give | |
14:13 | you two X . And then these multiply to give | |
14:15 | you the one . Now you see what's happened . | |
14:17 | I was able to factor it , but I have | |
14:18 | carbon copy exact duplicates of the terms here . So | |
14:22 | I have Y plus four is X plus one quantity | |
14:25 | squared because they're just multiplied together . I should always | |
14:29 | be able when I factor the right hand side after | |
14:31 | I complete the square . Remember we did this a | |
14:33 | lot when we completed the square in those lessons , | |
14:36 | when you factor it , it should always be duplicate | |
14:39 | terms because you should always be able to write it | |
14:41 | as a square . That's the whole point . You'll | |
14:44 | always be able to do it that way and we | |
14:45 | prove that mathematically back then . But if you notice | |
14:49 | , look what we ended up with were trying to | |
14:51 | get into a form that looks like this , we | |
14:55 | want y minus some number is some constant , A | |
14:59 | X minus some number squared . And we got y | |
15:03 | minus some number equals a constant , which is one | |
15:05 | X plus or minus some number squared . So of | |
15:09 | course we don't have minus signs here . But that | |
15:11 | just means that this parable is shifted in a different | |
15:14 | , you know , opposite lee than than the normal | |
15:16 | uh guy . And so you can kind of see | |
15:18 | that whenever you try to plot this thing . So | |
15:20 | this is the answer . This is what we're trying | |
15:22 | to get to , we're trying to get to vertex | |
15:24 | form , you can read the vertex directly off the | |
15:27 | X shift and the Y shift and the vertex in | |
15:30 | this case Is going to be what shifted from the | |
15:33 | origin . If it was X -1 , it would | |
15:35 | be shifted to the right , but it's X plus | |
15:38 | one . So that means it shifted to the left | |
15:39 | one unit . And if it was why minus the | |
15:42 | number it will be shifted up for . But it's | |
15:44 | why plus this . So it shifted down for which | |
15:46 | means the vertex is at negative one comma negative four | |
15:50 | . Also , we've already done in the past lessons | |
15:52 | when we graph these guys , we look at the | |
15:54 | coefficient in front again , it's positive . So this | |
15:57 | means it opens up and because it's just a positive | |
16:01 | one , we know that it opens up with the | |
16:03 | same shape as y is equal to X squared because | |
16:05 | A is one . If it were three as a | |
16:08 | coefficient out there , it would be much more closed | |
16:10 | off . If it were negative out in front here | |
16:12 | , like negative two or something , it would be | |
16:14 | upside down opening frowny face and if it were negative | |
16:17 | two would be closed in a little bit more because | |
16:20 | of what we've talked about when we graph these things | |
16:22 | . So now we're in a position to attempt to | |
16:25 | sketch this thing . So we're gonna go over here | |
16:28 | and we're gonna try to sketch it . So the | |
16:30 | first thing you want to do when you sketch these | |
16:32 | things obviously is you want to put the vertex down | |
16:34 | . Negative one comma negative four . So here's negative | |
16:36 | one comma one negative two negative three negative four . | |
16:39 | Of course , this is X . And this is | |
16:41 | why so negative one comma negative four means the problem | |
16:44 | is the vertex of it , is there , you | |
16:46 | know , it opens up so this is the bottom | |
16:49 | most point of the problem . So I could just | |
16:51 | sketch it here and say you're done . But really | |
16:53 | when you sketch parable is you want to figure out | |
16:55 | where the crossing points are so that you can plot | |
16:57 | them , You know , effectively there . So let | |
17:00 | me go ahead and try to say to this is | |
17:03 | three , this is four 1 , 2 , 3 | |
17:05 | , 4 , put a couple tick marks all over | |
17:07 | the place , you can kind of see it . | |
17:09 | So when we draft these things in the past , | |
17:11 | what we always needed to do is um find these | |
17:15 | intercepts . So here is the equation . We want | |
17:18 | to find the intercept . We want to figure out | |
17:19 | where are these points that are crossing points ? They're | |
17:22 | probably somewhere around here . But I want to figure | |
17:25 | that out . So I have to set Y equal | |
17:27 | to zero to do that . We've done that skill | |
17:28 | before . So just use this equation and set Y | |
17:31 | equal to zero , Y plus four X plus one | |
17:35 | quantity squared I have to solve for X . So | |
17:38 | what do I have ? I have four equals I | |
17:40 | can expand this out by foil . And this one | |
17:42 | is so easy , it's just gonna be X squared | |
17:45 | plus two times X times one . So two X | |
17:48 | plus one squared like this , whoops forgot plus down | |
17:52 | here . So it's plus one . And then I | |
17:55 | need to subtract the four . So it's going to | |
17:57 | be zero equals X squared plus two X . And | |
18:00 | then one minus four is negative three . So let | |
18:03 | me just double check myself . I have X squared | |
18:05 | plus two X minus three . Right ? So now | |
18:07 | I need to try to factor this and I'm gonna | |
18:11 | factor this guy with an X . Here in an | |
18:12 | X . Here and I have , the only thing | |
18:15 | I can do to make three is one times three | |
18:16 | is three and I have a negative . So it | |
18:18 | needs to be opposite signs and a plus sign here | |
18:20 | . The only way it's gonna work is a plus | |
18:21 | here in the mine is here double check yourself X | |
18:24 | times X is X squared . This gives you negative | |
18:27 | X . This gives you positive three X . I | |
18:29 | add those together . I get the two X . | |
18:31 | And then this gives me negative three . So then | |
18:33 | I have these two guys here . If I set | |
18:35 | this equal to zero , I'll find that X . | |
18:37 | Is equal to one . Move the one over . | |
18:39 | If I set this equal to zero , X . | |
18:41 | Is equal to negative three . So what I've done | |
18:43 | is I've taken the equation of the problem , I | |
18:45 | found I forced Y to be equal to zero and | |
18:48 | I figured out the X values that correspond to that | |
18:51 | and it's at one and three , so uh or | |
18:55 | one and negative three . So here's the first one | |
18:56 | at one and the negative 123 is the other one | |
18:59 | right here . And you got to remember this is | |
19:00 | a free hand parabola . So my tick marks aren't | |
19:02 | exact , but you can see basically what happens is | |
19:05 | this is the problem . Well , that's a very | |
19:07 | bad problem , I'm sorry about that , but you | |
19:09 | get the idea goes down like this , something like | |
19:14 | this , and of course it's a little too crunch | |
19:16 | at the bottom , whatever , but that's the general | |
19:17 | idea . These intercepts should always be equal distant on | |
19:21 | both sides of the vertex . You can see this | |
19:23 | is negative one . So we have to tick marks | |
19:25 | for that one and two tick marks over for that | |
19:27 | one . And the axis of symmetry . If you | |
19:29 | ever asked that The axis of symmetry goes right down | |
19:32 | through this guy , the axis of symmetry has got | |
19:34 | to be access equal to negative one because that's the | |
19:37 | vertical line negative one , that slices the whole thing | |
19:39 | in half . So this is the axis of cemetery | |
19:42 | . I'm not gonna write the cemetery park down . | |
19:44 | So every one of these problems is basically going to | |
19:47 | proceed the same way you're going to be . Given | |
19:49 | some kind of quadratic equation , you're gonna replace this | |
19:53 | with . Why ? Then you need to complete the | |
19:55 | square on the right hand side . So you're gonna | |
19:56 | take the constant term , move it over . You're | |
19:59 | going to check that the coefficient in front of X | |
20:00 | square is one . If it's not , we'll fix | |
20:03 | it , I'll show you how later . But for | |
20:04 | now you proceed , you take the coefficient of the | |
20:08 | X term divided by two and square it . You | |
20:10 | add it to both sides . Then you simplify everything | |
20:13 | . Factor . You should always get a perfect square | |
20:15 | on the right . And then basically you're done , | |
20:17 | this is the vertex form , it should be y | |
20:19 | plus or minus a number X plus or minus a | |
20:22 | number quantity squared . Then after that you just graph | |
20:25 | it and we already practiced graphing these things many many | |
20:28 | times before . Now in the next problem , the | |
20:32 | process is the same . But completing the square part | |
20:34 | becomes a little bit um uh trickier because there's an | |
20:40 | extra step with the coefficient , I'm gonna have to | |
20:42 | show you . But let's go and actually start this | |
20:45 | problem . Let's start this problem over here so I | |
20:48 | can make sure you have enough room . So the | |
20:49 | problem that we're gonna solve is the following ffx is | |
20:53 | equal to two X squared minus four X plus one | |
20:58 | . Now , if I asked you , what does | |
21:00 | this problem look like ? The only thing you can | |
21:02 | tell me is that it opens up because it's got | |
21:04 | a positive number in front of the X squared because | |
21:06 | it's a positive to , you know , that it's | |
21:08 | closed in on itself a little more than the regular | |
21:10 | X squared parabola . And that's really all , you | |
21:13 | know , you don't have any idea where the thing | |
21:14 | is located or anything . So what we want to | |
21:17 | do is a take this guy and transform it into | |
21:20 | a vertex form . So the thing we have to | |
21:22 | do is the very first thing . Take the F | |
21:24 | of X and replace it with . Why ? So | |
21:26 | two X squared minus four X plus one . Next | |
21:31 | step , you take the constant term on the right | |
21:33 | and you move it over by subtraction or addition or | |
21:35 | whatever you have to do . So in this case | |
21:36 | I have to subtract two , X squared minus four | |
21:39 | X . So that's exactly what we did in the | |
21:42 | previous problem . We just moved the constant term over | |
21:44 | now . In the last problem remember I told you | |
21:47 | but after you do that you have to check that | |
21:50 | the coefficient there is a one in this case it | |
21:52 | was a one . So we were we could proceed | |
21:54 | . But in this case the coefficient in front of | |
21:57 | the x squared term is actually a two . So | |
21:59 | you cannot do that . You can't proceed without doing | |
22:02 | the following steps . So you have y minus one | |
22:04 | . If the coefficient in front is not a one | |
22:06 | , then all you have to do is factor to | |
22:08 | make it a one . So you have x squared | |
22:09 | minus two eggs . So I've just factored out of | |
22:12 | two there and notice that if I cover everything up | |
22:14 | , the coefficient in front of this one is a | |
22:16 | one , that's what you're really trying to do now | |
22:18 | . Yes , we have all this other stuff , | |
22:20 | we pulled it out , but if you remember the | |
22:22 | completing the square process , what we did to get | |
22:24 | rid of it as we just factor it out and | |
22:25 | we focus on what's on the inside . So the | |
22:28 | coefficient of what's inside here is now one . Okay | |
22:32 | , so let's go on and add one half of | |
22:36 | this term of this coefficient square . So we're gonna | |
22:38 | have why minus one ? I'm gonna give myself a | |
22:41 | lot of space , two X squared minus two X | |
22:47 | . And now I have to add a term . | |
22:49 | What do I add ? I'm going to add to | |
22:51 | it , 1/2 of the coefficient of the x . | |
22:54 | term . So the way you write it is -2 | |
22:57 | . That's the coefficient there , divide by two , | |
22:59 | that's one half of it . And I'm gonna square | |
23:02 | and don't forget that , this whole thing , I | |
23:04 | think I'm adding it inside of these purple parentheses . | |
23:07 | So this is what's inside of here , and I'm | |
23:09 | kind of completing the square of what's on the inside | |
23:11 | . So I do one half of this squared , | |
23:13 | and that's all living inside of here . So the | |
23:16 | biggest mistake students are going to make is that they're | |
23:18 | going to uh add this term to the right hand | |
23:21 | side . So let me go ahead and do it | |
23:23 | exactly as you probably would think to do it negative | |
23:25 | to over two quantity square . So you would think | |
23:28 | that you're just gonna like uh let's do it like | |
23:30 | this , I'm going to add this guy like this | |
23:32 | and you would be done . But the problem is | |
23:35 | the following . Remember when you're adding things to both | |
23:37 | sides of an equation , you have to add the | |
23:39 | same value to both sides . So it looks like | |
23:42 | I'm adding this over here and I'm adding this over | |
23:44 | here and you're good . But actually think about it | |
23:46 | when I evaluate this is gonna be negative one , | |
23:49 | negative 2/2 is negative one squared . So I'm gonna | |
23:51 | get a positive one . So I'm adding one over | |
23:53 | here . So you would think that you just add | |
23:55 | one over here also , which is kind of what | |
23:57 | I have . But the problem is I'm not really | |
23:59 | adding one over here , this whole thing is inside | |
24:02 | and the two is multiplying . So really I'm adding | |
24:05 | to , but I've just added it on the inside | |
24:08 | so it looks like I'm just adding this term . | |
24:10 | But really I'm adding this term times too , so | |
24:12 | to make the equation balanced to add it to this | |
24:15 | side , I have to also multi , have to | |
24:18 | multiply the thing by two and this is the term | |
24:21 | that I'm really adding to both sides . So it | |
24:23 | looks like I'm adding something unbalanced cause it looks like | |
24:26 | I'm adding two times something and it's not there . | |
24:28 | But really it is because if I were to blow | |
24:30 | this thing out of multiplying by two anyway , so | |
24:33 | I'm adding the same thing to both sides . Basically | |
24:35 | I'm adding to to both sides two times one , | |
24:37 | two times one and that's what I'm adding to both | |
24:40 | sides and you can see a little bit more clearly | |
24:43 | in the next lesson um what I have here , | |
24:49 | so I may have misspoken minute ago , I'm not | |
24:51 | sure I'm adding two to the left and I'm adding | |
24:53 | two to the right because two times one is two | |
24:55 | and two times one is two . I I may | |
24:57 | have misspoke there , I apologize if I did . | |
24:59 | So what you're going to have here is why minus | |
25:01 | one and then on the inside you're gonna have what | |
25:04 | you're going to have one . I'm sorry negative one | |
25:06 | square that gives you one And then one times 2 | |
25:09 | is two . So I'm adding just a two right | |
25:11 | there then I'm gonna have to X squared minus two | |
25:15 | X . Plus this is negative one squared , that's | |
25:17 | just adding one . But again two times one is | |
25:20 | I'm really adding to I'm really adding to so the | |
25:22 | equation is completely balanced . So then I have negative | |
25:27 | one plus two is one right and on the right | |
25:30 | hand side to have it too . And now I'm | |
25:32 | an attempt to factor the right hand side . If | |
25:34 | I'm unable to factor it , I've made a mistake | |
25:36 | . So have action X . The only way I | |
25:38 | can get a one is one and a one . | |
25:40 | And the only way I can make this negative is | |
25:41 | to have both of them negative double check yourself X | |
25:44 | times X is X squared . This gives me negative | |
25:47 | X . This gives me negative X . This gives | |
25:49 | me positive one . The negative X plus the negative | |
25:51 | X gives me negative two . X . And of | |
25:54 | course that's what I expect . Just like in the | |
25:56 | previous problem when I factored it , I told you | |
25:58 | when you factor this thing , you'll always get carbon | |
26:00 | copies and be able to write it as a square | |
26:03 | and the exact same thing happened . So what I'm | |
26:05 | able to do is that I have Y plus one | |
26:08 | equals two x minus one . Now it's squared because | |
26:12 | they're multiplied together and look at what I have . | |
26:14 | This is exactly the form that I wanted in . | |
26:17 | This is why plus or minus a number , the | |
26:20 | constant A . Out in front X plus or minus | |
26:22 | some number . It matches exactly the form of the | |
26:24 | equation that I'm trying to get . That's why sometimes | |
26:27 | it's called when I opened the lesson up , I | |
26:30 | said this is called the vertex form of a parabola | |
26:33 | but it's also called the completed square form because you | |
26:36 | have to complete the square to get . So now | |
26:39 | what I want to do to finish the problem out | |
26:41 | , let me re copy this on the right hand | |
26:43 | side . This is actually the answer to the problem | |
26:45 | . So the answer is we checked myself , Y | |
26:47 | plus one is equal to two times x minus one | |
26:49 | quantity squared . If you're asked to take this quadratic | |
26:52 | and change it to a vertex form , you would | |
26:55 | leave it like this and of course if your teacher | |
26:58 | prefers it or your book you could also write it | |
27:00 | with the constant term over here . So I'd have | |
27:02 | a -1 over here and it would be -1 . | |
27:05 | Move to the right hand side like that . But | |
27:07 | I like to leave it in this form . So | |
27:09 | now let's go ahead since we're getting practice with all | |
27:12 | the stuff , we're gonna sketch this problem . So | |
27:14 | let me rewrite it . It's Y plus one is | |
27:18 | two X minus one quantity squared . We want to | |
27:21 | graph this guy . So the first thing you do | |
27:23 | is you say well what's the vertex , what's the | |
27:27 | vertex The X . Value is going to be shifted | |
27:30 | one unit to the right from this , so it's | |
27:32 | gonna be one unit over and it's gonna be one | |
27:35 | unit and why ? But if it was why minus | |
27:37 | it would be up and it's why plus , so | |
27:39 | it's down so it's negative light . So it's one | |
27:41 | common minus one . That's the vertex . It also | |
27:44 | opens up why ? Because it's a positive to if | |
27:49 | this were a negative to , it would open down | |
27:51 | but it's a positive to . It opens up and | |
27:53 | also again it's a little steeper , it's a little | |
27:55 | closed in because it's too as the coefficient instead of | |
27:58 | regular X squared . Parable has coefficient of one . | |
28:02 | So let's take a little sketch of this guy and | |
28:07 | see what it kind of looks like . So the | |
28:09 | vertex is at 1:00 -1 . Which is from doing | |
28:13 | the math right basically down here and then actually I | |
28:17 | want to change the color of it . Just do | |
28:19 | green . So I put the vertex down here one | |
28:22 | comma negative one . And we know it opens up | |
28:25 | so we know this problem is going to go and | |
28:27 | look something like this so we know it's going to | |
28:28 | cut through the X axis . And so in order | |
28:31 | to really graph it properly we probably should find the | |
28:33 | X intercepts . So the find the X intercepts . | |
28:37 | Um What we have to do is take this equation | |
28:40 | that we have and set y equal to zero . | |
28:43 | That's gonna find the points on the X axis when | |
28:46 | y is zero . So we're gonna take this equation | |
28:48 | , we're gonna set y equal to zero . So | |
28:50 | we're gonna set is gonna be zero plus one and | |
28:53 | then it's gonna be too X -1 quantity squared . | |
28:56 | Now we have to find the values of X where | |
28:58 | the crossing points happen . So we have 12 and | |
29:03 | then here it's x minus one . I need to | |
29:05 | expand this out , I need to blow the whole | |
29:07 | thing out again , moves the one over and then | |
29:09 | solve . Now we've already done this stuff , we | |
29:11 | know what this is equal to but let's just do | |
29:13 | it again . Uh It's going to be X squared | |
29:16 | minus X . For the interior term minus X . | |
29:19 | For the exterior term plus number one . So we're | |
29:23 | gonna have one too X squared minus two X plus | |
29:28 | one . Now we're gonna multiply the two in two | |
29:32 | X squared minus , multiply in their four X two | |
29:37 | times , one is two . And now we have | |
29:39 | to move the one over so we'll have a 02 | |
29:42 | X square uh minus four X . And then one | |
29:47 | minus two is just a one . Let me double | |
29:49 | check myself . I have two X squared minus four | |
29:51 | X plus one . And that's equal to zero . | |
29:53 | So the first thing that you're always gonna try is | |
29:56 | going to try to factor this thing , see if | |
29:58 | you can factor it and set it equal to zero | |
29:59 | . Right ? So I'm gonna have a two X | |
30:01 | . Here have a two X . And I'm gonna | |
30:03 | have an X . Here . So I multiply those | |
30:05 | two . Give me X squared . The only way | |
30:06 | I can do this is a one in a one | |
30:09 | . So then I try to fiddle with it . | |
30:10 | But you're gonna figure out that it's not gonna work | |
30:12 | because the outside terms are going to be two X | |
30:15 | . When you multiply , this is going to give | |
30:17 | you X . So I can't get four X from | |
30:20 | that . Even if I add them together . Two | |
30:22 | plus one is three . There's no way I'm gonna | |
30:24 | get a four in the middle . So this is | |
30:25 | not factory able . Now when a uh quadratic is | |
30:29 | in fact a rubble , it doesn't mean there's no | |
30:31 | solution . There's definitely a crossing point here . There's | |
30:34 | definitely a solution . But what it means is the | |
30:36 | numbers in here , the factored form of that are | |
30:39 | not whole numbers . The way that it would factor | |
30:41 | out is not gonna be a whole number , Right | |
30:43 | ? So I can't just pick whole numbers and find | |
30:46 | the crossing points , basically . So if you come | |
30:48 | up empty and you can't factor it , like in | |
30:49 | the previous problem , we were able to when we | |
30:52 | set it equal to zero and get down to this | |
30:54 | point , we could factor and solve for X . | |
30:56 | But here we are unable to do it . So | |
30:57 | you don't give up what you do is you just | |
31:00 | go and use the quadratic formula , right ? So | |
31:05 | we have to use the quadratic formula on the sky | |
31:07 | . So here we say that A . Is to | |
31:12 | B is negative for and C is one . And | |
31:17 | we have to do the quadratic formula which is negative | |
31:20 | B plus or minus B squared minus four A . | |
31:23 | C . And this is under a radical . Oops | |
31:28 | this whole kind of bought . Shut up . Sorry | |
31:30 | about that . So this whole thing is there to | |
31:32 | A So we have to substitute in for a B | |
31:36 | and C and crank through it . And I know | |
31:37 | you know how to do this . We've done it | |
31:38 | . But let's do it again together . So we | |
31:40 | have negative B . But B is itself negative . | |
31:42 | So we do negative for in parentheses plus or minus | |
31:46 | the be a squared . So it's negative four quantity | |
31:48 | squared minus four times A . Which is to times | |
31:53 | C , which is one . And that whole thing | |
31:55 | is under a radical and on top of two times | |
31:58 | a which is to All right , So then what | |
32:04 | do I have ? I have x equals I have | |
32:06 | four because this is negative times negative plus or minus | |
32:10 | on the inside . I have 16 -8 from inside | |
32:15 | here . There's a radical here and then on the | |
32:17 | bottom two times two is four . Which means that | |
32:20 | X is going to be equal to four Plus or | |
32:22 | minus the square root of eight because the 16 1988 | |
32:27 | and then I'm gonna have a four on the bottom | |
32:30 | . Now if you do a factor tree on eight | |
32:32 | , you know , we could go off to the | |
32:33 | side here . I've done it so many times . | |
32:35 | It's kind of crazy . But if you do a | |
32:37 | factor tree on eight it's two times four and two | |
32:40 | times two . So really the square root of eight | |
32:43 | is the two comes out The square root of two | |
32:46 | left over stays on the inside . So eight is | |
32:50 | two times a squared of two . So what you | |
32:51 | have is four plus or minus two times the square | |
32:55 | root of two divided by four . Now I can | |
32:57 | simplify this further by simplifying these coefficients and all that | |
33:01 | , but really I'm trying to sketch it . So | |
33:03 | what I'm gonna do is let you dump that in | |
33:05 | a calculator and say four plus two square root of | |
33:08 | two divided by four when you dump that and you're | |
33:10 | gonna get about 1.7 . This is a approximate , | |
33:14 | it's not an exact , there's decimals that go on | |
33:16 | after this thing forever . And then the other answer | |
33:19 | is going to be approximately equal two for minus two | |
33:22 | squared of 2/2 , which is 0.3 . These are | |
33:26 | both approximate answers , but they are the places where | |
33:29 | the graph has a y value of zero . So | |
33:33 | if you go back to your graph here and say | |
33:35 | this is one , this is two , this is | |
33:37 | three negative one , negative two negative three . Then | |
33:40 | the crossing points are gonna be at 0.3 Right ? | |
33:44 | Which is right here , here's here's one , here's | |
33:47 | half , so 0.3 is a little bit left of | |
33:49 | half , right , and then 1.7 here's one , | |
33:53 | here's two , so 1.7 is pretty close to two | |
33:55 | and you can see these dots are on either side | |
33:58 | of the vertex . So the problem is going to | |
34:00 | go something like this is gonna go down cross up | |
34:03 | through those points and go up . Of course it's | |
34:05 | a sketch , it's not exact , but to get | |
34:07 | the idea , it's at 0.3 and 1.7 the vertex | |
34:10 | is down here . And if you ever asked what | |
34:13 | the axis of symmetry is , that's the vertical line | |
34:15 | that goes through the whole thing and that's X equals | |
34:18 | one . Is the axis of symmetry , right ? | |
34:21 | Because it's the value here that cuts in their one | |
34:24 | common negative ones . The vertex of the axis has | |
34:26 | to be one . So that was a long lesson | |
34:29 | . But the process , it depends on so many | |
34:32 | skills that we've learned in the past . First of | |
34:35 | all , you have to understand that they're basically two | |
34:40 | ways to write parabolas are quadratic in general . The | |
34:42 | first way is that the whole thing expanded and blown | |
34:44 | up like this . Right ? The other way is | |
34:47 | a completed square form or alternately called the vertex form | |
34:50 | , which looks like this . Now this looks very | |
34:53 | different than this , but they're the same equation . | |
34:55 | If you take any of these answers that we got | |
34:57 | like this one , right ? And if you multiply | |
35:00 | out with foil multiply by two and move the one | |
35:03 | over , you're going to get what we started with | |
35:05 | , you're gonna get this . So all of these | |
35:07 | work that way . It's basically you have this form | |
35:10 | and it looks different . If you blow this out | |
35:12 | with foil , take the four and move it over | |
35:14 | , you're going to get what we started with . | |
35:16 | They're just two different ways of writing the exact same | |
35:19 | equation , right ? And so we learned that we | |
35:22 | have to complete the square in order to do that | |
35:24 | . And so we've learned the topic of completing the | |
35:26 | square in the past . Here is an application of | |
35:28 | why we care why do we care about completing the | |
35:30 | square ? That kind of rhymes ? Right ? So | |
35:33 | make sure you understand this ? I have one more | |
35:34 | lesson on this topic , It's the same sort of | |
35:37 | thing where the problem complexity to be slightly more involved | |
35:40 | . But ultimately it's all the same steps . So | |
35:42 | follow me on to the next lesson , we're going | |
35:43 | to continue getting practice with writing these quadratic in vertex | |
35:47 | form . |
Summarizer
DESCRIPTION:
Quality Math And Science Videos that feature step-by-step example problems!
OVERVIEW:
12 - Writing Quadratic Functions in Vertex Form - Part 1 (Graphing Parabolas) is a free educational video by Math and Science.
This page not only allows students and teachers view 12 - Writing Quadratic Functions in Vertex Form - Part 1 (Graphing Parabolas) videos but also find engaging Sample Questions, Apps, Pins, Worksheets, Books related to the following topics.