16 - Domain and Range of a Quadratic Function - Part 1 (Graphing Quadratics) - By Math and Science
Transcript
| 00:00 | Hello . Welcome back up Jason with math and science | |
| 00:02 | dot com . And today the lesson that we're going | |
| 00:04 | to conquer is finding the domain and range of these | |
| 00:08 | quadratic parabolas basically . And so for every one of | |
| 00:11 | these problems we're gonna actually do four things . We're | |
| 00:14 | gonna find the vertex of the parabola . We've done | |
| 00:16 | that many times . We're gonna find the domain of | |
| 00:18 | the problem . We haven't done that really much at | |
| 00:20 | all . We'll find the domain , we'll find the | |
| 00:22 | range of the parable of the function . We haven't | |
| 00:24 | really done that much at all and we'll find the | |
| 00:26 | zeros of the problems . So we've actually found the | |
| 00:29 | zeros several times when we were graphing them . We're | |
| 00:32 | not really gonna graph these problems . We're just gonna | |
| 00:34 | find those four pieces of information specifically focusing on domain | |
| 00:38 | and range . So the only thing I need you | |
| 00:39 | to know going into this lesson is that the domain | |
| 00:42 | of any function . It's just the allowable values of | |
| 00:45 | X . That are allowed to go into the function | |
| 00:48 | basically . And the range are the are the values | |
| 00:52 | that come out of the function . So you can | |
| 00:54 | think of the domain as being the input values that | |
| 00:56 | are allowed to go into the function . And the | |
| 00:59 | range are the set of values that come out of | |
| 01:02 | that function after you've plugged all the domain into it | |
| 01:05 | . So domain is the X values that the function | |
| 01:07 | is valid for . Range are the y values that | |
| 01:11 | come out of the function . And so we're going | |
| 01:13 | to use that information as we go along here . | |
| 01:15 | So we're gonna find the vertex , the domain , | |
| 01:16 | the range and the zeros of every one of these | |
| 01:18 | functions . So we're putting a lot of different skills | |
| 01:21 | together essentially . So what if I give you f | |
| 01:23 | of X is equal to X squared minus four , | |
| 01:27 | X minus three . We're gonna find the vertex first | |
| 01:32 | . That's the very first thing we're gonna do and | |
| 01:34 | we've done this many times before . So in the | |
| 01:37 | part A we're gonna find the vertex . So because | |
| 01:42 | this is not in vertex form we have to of | |
| 01:43 | course use our formula that we have derived and and | |
| 01:47 | have used in the last lesson and that is the | |
| 01:48 | X value of the vertex negative B over two . | |
| 01:51 | A . In this case A B and C are | |
| 01:54 | given here . So it's negative , but B is | |
| 01:57 | negative four . So it goes in the top and | |
| 01:59 | then two times A . But is positive , one | |
| 02:02 | goes right here . So what you have is the | |
| 02:04 | negative times the positive gives you a positive four . | |
| 02:07 | And on the bottom two times one gives you two | |
| 02:09 | and you all know that four divided by two is | |
| 02:11 | two . So the X value of the vertex is | |
| 02:13 | given to be it too . Now how do you | |
| 02:16 | find the corresponding why value of the vertex ? Because | |
| 02:19 | vertex is a point has X comma Y Well if | |
| 02:22 | we know the X coordinate of the vertex , we | |
| 02:23 | just stick it into the function and we figure out | |
| 02:26 | that f of evaluated at x is equal to two | |
| 02:30 | is sticking into the function two squared -4 times X | |
| 02:35 | . Which is now 2 -3 . And so what | |
| 02:37 | we'll have here two times two is four , this | |
| 02:39 | is gonna be minus eight , this is gonna be | |
| 02:41 | minus three . And we'll just subtract four minus eight | |
| 02:43 | is negative four . Then we have minus uh website | |
| 02:47 | got ahead of myself minus the three . And so | |
| 02:49 | we're gonna have negative seven . So the Y value | |
| 02:52 | here is going to be a negative seven . So | |
| 02:54 | we have the X . Value in the Y . | |
| 02:56 | Value . And so what we're trying to drive that | |
| 02:58 | is the vertex . Right ? So the vertex okay | |
| 03:02 | , is drum roll please ? Two comma negative seven | |
| 03:07 | . So that would be kind of the first step | |
| 03:09 | of graphing . This thing is I would want to | |
| 03:10 | figure out where the vertex was . I put it | |
| 03:12 | on the xy plane and then I'd start looking at | |
| 03:14 | the remaining parts of the problem . Now part B | |
| 03:18 | . We said we're going to find the domain and | |
| 03:19 | the range . So I told you many times before | |
| 03:23 | and also at the beginning of this lesson that the | |
| 03:25 | domain of a function is just the allowable values of | |
| 03:29 | X that can go into the function because remember a | |
| 03:32 | function is like a black box , you don't know | |
| 03:34 | what's inside of it . But it's a calculating machine | |
| 03:36 | inside right ? It takes inputs into one side . | |
| 03:40 | It calculates on those inputs and it sticks values out | |
| 03:44 | . So the input values come in one side . | |
| 03:46 | The output values come out . So we say the | |
| 03:48 | X values go in and the f of X or | |
| 03:51 | the y values are the ones that come out the | |
| 03:52 | other side . So the input values of extra , | |
| 03:55 | the allowable values in the domain of the function . | |
| 03:58 | So for every Parabola you're going to basically have the | |
| 04:01 | same answer because I don't really know how this thing | |
| 04:03 | looks yet but I know that this parabola is either | |
| 04:06 | going to be a smiley face parabola or it's going | |
| 04:10 | to be a frowny face parabola and in this case | |
| 04:12 | I can look and see this one actually has a | |
| 04:14 | positive one here . So it opens up so this | |
| 04:17 | Parabola opens up . If I had a negative sign | |
| 04:22 | up here then it would open down and it would | |
| 04:23 | be a frowny face problem . What I call a | |
| 04:25 | frowny face problem . Alright , of course I haven't | |
| 04:28 | drawn it proper . The vertex is at two comma | |
| 04:30 | negative seven , so two comma negative seven means the | |
| 04:32 | vertex would barely be down here somewhere . That's not | |
| 04:35 | the point of what I'm saying . I'm just trying | |
| 04:36 | to get you to realize that parable is either open | |
| 04:38 | up or open down . Those are the only valid | |
| 04:41 | . Um Those are the only ways it can go | |
| 04:44 | . So you have to ask yourself for the domain | |
| 04:45 | , which is what we're trying to figure out here | |
| 04:48 | . What are the allowable values of X . That | |
| 04:51 | I can stick into this function when you think about | |
| 04:53 | it ? I can put 0123456789 10 . All the | |
| 04:57 | way to positive infinity . On my my output is | |
| 04:59 | going to get larger and larger . I can go | |
| 05:01 | here negative all the way to negative infinity . And | |
| 05:04 | my output still gonna get larger and larger and larger | |
| 05:06 | . So really I can put any value I want | |
| 05:09 | for X . Into this uh polynomial into this quadratic | |
| 05:13 | , I can put any value of X in . | |
| 05:14 | There's no restrictions on the value of X . So | |
| 05:17 | when you have a function , which a lot of | |
| 05:19 | functions behave this way , where you have no restrictions | |
| 05:22 | on the the domain of the function , you just | |
| 05:26 | say that the domain is all real numbers . Right | |
| 05:32 | ? So we're not we're not including imaginary numbers here | |
| 05:35 | because this is a graph of real numbers on the | |
| 05:36 | X axis . But every real number on this axis | |
| 05:39 | can go into this function without any problems at all | |
| 05:43 | . Alright , now I'm gonna circle back at the | |
| 05:44 | end of this problem and tell you kind of remind | |
| 05:47 | you because we talked about and we talked about rational | |
| 05:49 | functions when you might have problems with the domain of | |
| 05:52 | the function is not all real numbers , but basically | |
| 05:54 | for every Parabola , you're gonna see it's always going | |
| 05:56 | to be a domain of all real numbers . So | |
| 05:58 | for all of these problems , you're gonna say all | |
| 06:00 | real numbers for every one of these problems . So | |
| 06:02 | the next part part C is asking what is the | |
| 06:04 | range of values with a range of values ? If | |
| 06:08 | you look at this Parabola here , in fact , | |
| 06:10 | let's go over here to the range of values . | |
| 06:12 | Let me just say we're going to find the range | |
| 06:15 | the range of values . All right . What you | |
| 06:17 | need to do is think about the vertex this Parabola | |
| 06:19 | opens up in the vertex is at two comma negative | |
| 06:22 | seven . So , let me go over here and | |
| 06:23 | just sketch it . I don't care about making a | |
| 06:25 | perfect graph , but it's at 212 -1234567 . Let | |
| 06:30 | me go ahead and go down like this . Right | |
| 06:32 | ? So two comma negative seven . There's the vertex | |
| 06:35 | and I already know that this thing opens up . | |
| 06:37 | So I know that this Parabola does something like this | |
| 06:41 | . So you have to ask yourself what is the | |
| 06:42 | range of values mean ? It means what are this | |
| 06:45 | , what are the set of numbers that come out | |
| 06:48 | of the function ? Remember the functions of box input | |
| 06:50 | values of X . Output values are F of X | |
| 06:53 | . Or why ? The range is What are the | |
| 06:55 | values of why that come out ? But you can | |
| 06:57 | see that the value uh the minimum of this problem | |
| 07:01 | is down here at negative seven . So the very | |
| 07:04 | lowest value possible that can come out of this function | |
| 07:07 | is just negative seven because this is the value of | |
| 07:09 | negative seven down here in the Y direction . So | |
| 07:11 | domain is concerned with X . Range is concerned with | |
| 07:14 | why But of course the output of this function could | |
| 07:17 | be negative six . Negative five . Negative four could | |
| 07:20 | be zero . And if you continue to graph on | |
| 07:22 | , the output could be 12345 on and on and | |
| 07:25 | on to positive infinity . So for the range of | |
| 07:28 | values , the very very lowest minimum value is a | |
| 07:32 | negative seven . But anything bigger than that is totally | |
| 07:35 | fine to graph as an output of this function . | |
| 07:38 | So what we say is the range Is why greater | |
| 07:42 | than or equal to -7 . Anything bigger than -7 | |
| 07:47 | or equal to -7 is a valid output of this | |
| 07:50 | function . So the domain is associated with all the | |
| 07:55 | inputs I can put in for the X values along | |
| 07:57 | X . Any real number I can think of , | |
| 07:59 | I can plug into this function , but the output | |
| 08:01 | is not all real numbers . The output has a | |
| 08:03 | minimum here and everything larger than that , but you'll | |
| 08:06 | never get a negative eight or negative 10 or negative | |
| 08:09 | 12 or negative 14 out of this function . So | |
| 08:12 | it's not part of the range . So it's really | |
| 08:14 | helpful when you're asked to figure out the domain and | |
| 08:16 | range of the function to do a quick little sketch | |
| 08:18 | . You don't have to make it perfect . You | |
| 08:20 | don't have to find the intercepts but just do a | |
| 08:21 | little sketch so that you in your brain can see | |
| 08:24 | what the values of the output might be . All | |
| 08:28 | right . So that was part A , B and | |
| 08:29 | C . The last part of this problem is we're | |
| 08:31 | gonna figure out what the zeros are . Zeros are | |
| 08:36 | . The zeros of the function is exactly the same | |
| 08:39 | thing as saying what the roots of the function are | |
| 08:41 | , which are exactly the same thing as what we | |
| 08:43 | say . The x intercepts of the function are the | |
| 08:45 | different ways of saying the same thing . The zeros | |
| 08:47 | of the function are the values of the function where | |
| 08:50 | the Y is equal to zero . And that means | |
| 08:53 | that it's the crossing points . Wherever this function crosses | |
| 08:56 | , whether it's here and here or wherever it is | |
| 08:58 | , our values of X where the y output of | |
| 09:01 | the function is equal to zero . That's why it's | |
| 09:04 | called a zero of the function . When you say | |
| 09:06 | find the zeros , you're telling me all right , | |
| 09:09 | set the output to zero and figure out where that | |
| 09:11 | happens in the X direction . So to figure it | |
| 09:14 | out , what you have to do is go back | |
| 09:15 | to your equation and you have to set the output | |
| 09:18 | of the function equal to zero . And so the | |
| 09:21 | function that we were given in the problem statement was | |
| 09:23 | x squared minus four , X minus three . That's | |
| 09:26 | f . Of X is equal to this . But | |
| 09:27 | now we're going to set the whole thing is equal | |
| 09:28 | to zero . That's gonna tell us the crossing points | |
| 09:31 | on the X axis because that's where y is equal | |
| 09:33 | to zero . How do you solve it ? We | |
| 09:35 | can use the quadratic formula . But the first thing | |
| 09:37 | you do is you try to factor it because it's | |
| 09:39 | easier if it works X and X one times three | |
| 09:43 | gives you three and then you try to make it | |
| 09:45 | work and you think you can but you really can't | |
| 09:47 | get it to factor properly because if I choose positive | |
| 09:50 | and positive , I'll never get a negative here . | |
| 09:53 | If I choose negative and negative , I'll also never | |
| 09:57 | get a negative here . So I have to have | |
| 09:59 | different signs . So if I do positive negative or | |
| 10:02 | negative positive , when I do it all , you | |
| 10:04 | know , let's just choose this one here . If | |
| 10:05 | I do positive negative it will be this will be | |
| 10:07 | X . And this will be negative three X . | |
| 10:09 | I can add those together and it's not gonna be | |
| 10:11 | negative for if I flip the signs around , I'm | |
| 10:13 | gonna have the same problem . If I put a | |
| 10:15 | negative here in a positive here , I have a | |
| 10:16 | negative X . Positive three X . I add those | |
| 10:19 | together . I'm not going to get a negative four | |
| 10:21 | . So even though it kind of looks like it's | |
| 10:22 | factory ble you really can't factor this thing so you | |
| 10:25 | can't factor Yes , but fortunately , you know how | |
| 10:30 | to solve things . Even when you can't factor them | |
| 10:32 | , you can complete the square if you want . | |
| 10:34 | But it's just always easier to use the quadratic formula | |
| 10:36 | negative B plus or minus B squared minus four A | |
| 10:41 | . C . Uh There's a radical right here and | |
| 10:44 | then you're driving the whole thing divided by two times | |
| 10:46 | a right two times A . So we just have | |
| 10:50 | to plug the values in negative B . But B | |
| 10:52 | itself is negative . So you say negative negative for | |
| 10:56 | plus or minus B is again negative four . So | |
| 10:58 | I'm squaring that negative four squared minus four times A | |
| 11:03 | . Which is one time C . Which is negative | |
| 11:05 | three . And there's a radical around all of this | |
| 11:08 | stuff and then two times A . But A was | |
| 11:11 | one because A was one here . So I have | |
| 11:14 | to do that calculation and it's not fun but it's | |
| 11:17 | not that bad either negative times negative is positive for | |
| 11:20 | they have plus or minus negative . Four times negative | |
| 11:23 | four is 16 . Then I have negative times positive | |
| 11:26 | . That's going to give you 12 . I have | |
| 11:29 | a radical around all this on the bottom . It's | |
| 11:31 | just too , so then I have four plus or | |
| 11:34 | minus square root . Uh 16 . Just double check | |
| 11:37 | my math here . Uh 16 plus 12 is 28 | |
| 11:41 | . And then on the bottom I'm gonna have a | |
| 11:43 | two now at this point you need to do a | |
| 11:47 | radical of 28 here . So I can go to | |
| 11:49 | do 28 I can say seven times four is 28 | |
| 11:52 | 2 times two is four . So I can pull | |
| 11:54 | this out of the radical . So what I'm going | |
| 11:56 | to have is four plus or minus for this guy | |
| 11:59 | . It's going to be a two coming out in | |
| 12:01 | the square root of seven . And we've covered all | |
| 12:03 | of this stuff when we did radicals in detail . | |
| 12:05 | And you have a two on the bottom four plus | |
| 12:08 | or minus two . Route 7/2 . Now . In | |
| 12:10 | order to uh continue I say I have a four | |
| 12:14 | or two in the two you can cancel things how | |
| 12:15 | you want but I think it's cleaner to split this | |
| 12:18 | thing up . So I'm gonna write it as four | |
| 12:20 | over to the plus or minus comes along for the | |
| 12:22 | ride . Two square root of seven over two is | |
| 12:26 | how I'm gonna split it up . Why does this | |
| 12:28 | work ? Because if I go backwards I have a | |
| 12:29 | common denominator here . Four plus or minus this in | |
| 12:33 | the numerator . So I'm just taking this thing and | |
| 12:35 | breaking it apart . Why am I doing like this | |
| 12:37 | ? Because two divided by two is 14 divided by | |
| 12:40 | two is two and then these two is just canceled | |
| 12:43 | . So what I'm going to have is the to | |
| 12:46 | from here plus or minus . This is just a | |
| 12:48 | one square root of +72 plus or minus square to | |
| 12:51 | seven . These are the zeros . So the answers | |
| 12:55 | to the answer to the problem is that the crossing | |
| 12:58 | points of this function , the points where this function | |
| 13:00 | would cross on the X axis there are X is | |
| 13:04 | equal to two plus the square root of seven . | |
| 13:06 | That's one value and two minus the square root of | |
| 13:08 | seven . That's another value . So really all we're | |
| 13:10 | doing in this lesson is putting a lot of things | |
| 13:12 | together . But I'm doing a little bit more work | |
| 13:14 | on explaining domain and range . We already know how | |
| 13:16 | to find the vertex . We've already found the zeros | |
| 13:19 | of these functions when we were graphing them . I'm | |
| 13:20 | just doing it again to give you some more practice | |
| 13:22 | and we're calculating the domain and range . One final | |
| 13:26 | thing I want to talk about as I said for | |
| 13:28 | every one of these parabolas , the domain is gonna | |
| 13:30 | be all real numbers . So for every one of | |
| 13:31 | these problems you're gonna be able to um put down | |
| 13:35 | that the domain is all real numbers . That only | |
| 13:38 | works because this parable is very well defined , right | |
| 13:42 | ? I can put in any value negative or positive | |
| 13:44 | . It's gonna go in and calculate a value . | |
| 13:46 | No problem . So then you might ask yourself and | |
| 13:48 | we have covered it a little bit in the past | |
| 13:50 | . What kind of function would ever give you a | |
| 13:52 | problem with the domain ? Like when would you ever | |
| 13:54 | have a function where the domain was not all real | |
| 13:57 | numbers ? Let me give you just one quick example | |
| 13:59 | . Uh I don't want to get into you know | |
| 14:02 | uh ton of detailed math because that's not the point | |
| 14:07 | here . But if I graft and we talked about | |
| 14:10 | this when we did rational functions . If I had | |
| 14:12 | a graph that didn't look like this but look like | |
| 14:15 | um where I had an Asem tope like something like | |
| 14:18 | this over here . Like let's say this is that | |
| 14:21 | negative to write . And then the function looks something | |
| 14:26 | like this . Um Yeah the function looks something like | |
| 14:31 | this and then down like this , something like this | |
| 14:38 | , we did this kind of thing . We did | |
| 14:39 | rational functions , right ? You see if you have | |
| 14:42 | a graph that goes to either positive or negative infinity | |
| 14:44 | like this , then the domain of the function is | |
| 14:46 | not all real numbers because the function it's perfectly fine | |
| 14:50 | to calculate values here closer and closer and closer to | |
| 14:52 | negative two . But as they get closer to negative | |
| 14:55 | , I'm sorry not negative to this would be like | |
| 14:57 | a positive view on this side as they get closer | |
| 14:59 | and closer to positive to the value would get really | |
| 15:01 | really , really big negative . And then as soon | |
| 15:03 | as I crossed to the other side , it would | |
| 15:04 | get really , really big uh in the positive infinity | |
| 15:08 | way . So basically , if you ever have a | |
| 15:09 | function that when you graph it goes to infinity , | |
| 15:12 | then the spot where it goes to infinity cannot be | |
| 15:16 | in the in the domain of the function . It's | |
| 15:18 | an invalid thing . I mean if I put the | |
| 15:20 | value of positive to in this function , the answer | |
| 15:22 | is going to be basically infinity either positive or negative | |
| 15:25 | infinity , depending on which way I'm approaching the thing | |
| 15:28 | . Right ? So the bottom line is Parabolas have | |
| 15:30 | a domain of all real numbers because there's no infinities | |
| 15:33 | . I mean there is an infinity , it goes | |
| 15:34 | up forever and ever , but that's not a that's | |
| 15:37 | not a , it doesn't go to infinity at a | |
| 15:38 | specific value . It's just that as I go off | |
| 15:41 | to infinity and X then the thing goes off to | |
| 15:43 | infinity and why there's no discontinuities like exist in the | |
| 15:47 | graph here . So if you have a graph of | |
| 15:49 | function and you have a discontinuity like that where the | |
| 15:51 | thing just runs off to infinity , that part of | |
| 15:54 | the function is not in the domain , it doesn't | |
| 15:55 | exist because there is no number here . It's just | |
| 15:58 | a infinity concept . And we did talk about that | |
| 16:01 | when we talk about rational functions . All right , | |
| 16:04 | So for Parabolas , they're always well behaved and and | |
| 16:08 | by the way , I should say one more thing | |
| 16:09 | , we've talked about equations of lines , the lines | |
| 16:12 | are well behaved to most lines , I should say | |
| 16:14 | , lines that go like this are well behaved , | |
| 16:17 | they're going to have any value of X . I | |
| 16:19 | want to put into this line is completely fine . | |
| 16:21 | I'm just going to get a new value of the | |
| 16:23 | output , right ? If I were to have a | |
| 16:25 | a line that goes straight up and down , it's | |
| 16:27 | a different story . But for oblique lines like this | |
| 16:29 | , the domain of the line is all real numbers | |
| 16:32 | as well . Lots and lots and lots of real | |
| 16:34 | functions in real life have a domain of all real | |
| 16:37 | numbers . It's just if you have a function that | |
| 16:38 | goes off to infinity at a sharp disk continuity like | |
| 16:41 | that , that's when you're gonna run into problems with | |
| 16:43 | the domain . All right . One more problem . | |
| 16:49 | Uh And it's gonna look like this . What if | |
| 16:51 | I have F of X . The function to find | |
| 16:53 | to be eight minus two X minus X squared . | |
| 16:57 | And we're gonna find those four things . We're gonna | |
| 16:59 | find the vertex , we're gonna find the domain , | |
| 17:01 | we're gonna find the range , we're gonna find the | |
| 17:03 | zeros of this function . Uh And so we're just | |
| 17:05 | gonna crank through it . First thing is you always | |
| 17:07 | want to rearrange , you never want to have it | |
| 17:09 | written like this . You always want to have it | |
| 17:11 | with your biggest power first this term . So it | |
| 17:14 | has to be negative X . Squared . Then you | |
| 17:16 | have this term which is -2 X . Then you | |
| 17:20 | have eight . Always rearrange it into something that is | |
| 17:22 | familiar to you . All right . How do we | |
| 17:26 | find the vertex when it's in standard form like this | |
| 17:30 | , the vertex X . Value is just negative . | |
| 17:34 | B . Over two . A . Always B . | |
| 17:37 | Is itself negative ? So you have negative to their | |
| 17:39 | two times A . But don't forget A . Is | |
| 17:42 | negative as well . So you have negative one . | |
| 17:43 | So , I really want you to write all the | |
| 17:45 | negatives everywhere because if you forget to write them down | |
| 17:47 | , you're gonna make a sign error . So the | |
| 17:49 | way I'm gonna handle this is this negative will stay | |
| 17:52 | out front . But when I divide all of this | |
| 17:54 | away , what's going to happen is negative divided by | |
| 17:56 | negative is positive . Two divided by two is one | |
| 17:59 | . So it's really negative one . That's what's left | |
| 18:01 | over . And you can do the math however you | |
| 18:03 | like differently than me . But the answer that you | |
| 18:05 | get as negative one because these negatives go away . | |
| 18:07 | Two divided by two is one . This negative is | |
| 18:10 | what is left over . So that's the X . | |
| 18:13 | Value of the vertex . Mhm . All right now | |
| 18:17 | the next thing we need to do once we have | |
| 18:19 | the X . Value of the vertex is to figure | |
| 18:21 | out the corresponding why value . And we do that | |
| 18:24 | by sticking this into the function . So the x | |
| 18:27 | value we're putting in is negative one in the function | |
| 18:30 | that already has a negative sign here . So we | |
| 18:32 | put a negative one in for the X . Value | |
| 18:34 | that we then have to square it . I have | |
| 18:36 | a minus sign from here . And I put in | |
| 18:39 | for this X value again , negative one . And | |
| 18:42 | then I have a plus eight . Now the negative | |
| 18:45 | sign stays here . But this square negative one squared | |
| 18:48 | is a positive one . So you get a negative | |
| 18:49 | one for that first part this becomes a positive too | |
| 18:53 | . And then you have eight . So you have | |
| 18:55 | negative one plus two is just going to give you | |
| 18:57 | one plus eight . And so what you get is | |
| 18:59 | nine . The corresponding why value of the vertex is | |
| 19:02 | nine . So what does that tell you the vertex | |
| 19:07 | ? Is that an x . value of -1 and | |
| 19:09 | a Y . Value of nine . That's the vertex | |
| 19:13 | that's part a negative one comma nine . All right | |
| 19:16 | . So the second part is what is the domain | |
| 19:18 | of this guy ? So be the domain the domain | |
| 19:23 | . Um What I like to do whenever I find | |
| 19:27 | the domain and range is I just do a quick | |
| 19:29 | little sketch . Not a crazy amazing sketch . Just | |
| 19:31 | a quick little sketch because I know where the vertex | |
| 19:34 | is . It's a negative one , comma 123456789 way | |
| 19:42 | up here . -1 09 . And this Parabola has | |
| 19:48 | a negative sign for X squared . So it opens | |
| 19:50 | down and you can write that down if you want | |
| 19:52 | . You can say it opens down . But what | |
| 19:55 | that basically means is that the Parabola starts here and | |
| 19:58 | then goes down and crosses the X axis in two | |
| 20:00 | locations somewhere . All right . Why did I draw | |
| 20:03 | that ? Well , because number one , it just | |
| 20:05 | helps you visualize domain and range . But I already | |
| 20:07 | told you in just a minute ago that for the | |
| 20:09 | domain of this function , I can put any value | |
| 20:11 | I want for X into the function . The function | |
| 20:13 | is well behaved whether or not to put zero in | |
| 20:16 | or negative five , positive 10 or positive 10 million | |
| 20:19 | or negative 32,000 or whatever . I'm still gonna get | |
| 20:22 | a number back . Right . It's just gonna be | |
| 20:24 | a bigger and bigger numbers that go this way and | |
| 20:26 | a bigger and bigger number in the negative sense as | |
| 20:27 | I go this way . So the domain is as | |
| 20:33 | it always is for Parabolas . All real numbers . | |
| 20:38 | Yes . All real numbers . Okay . And then | |
| 20:43 | the next thing I need to do is figure out | |
| 20:45 | what the range is . So the range you can | |
| 20:51 | kind of see that this has a maximum here . | |
| 20:53 | The vertex , is that negative one comma nine . | |
| 20:55 | That's exactly what we said . The vertex was . | |
| 20:57 | That's the maximum value . So the maximum value in | |
| 21:00 | terms of why is nine every other output of this | |
| 21:04 | function in the Y direction is going to be smaller | |
| 21:06 | than nine . So what you say is you say | |
| 21:08 | that why is less than or equal to nine ? | |
| 21:11 | Why equal to Well because the function goes all the | |
| 21:14 | way up to nine including nine . And the outputs | |
| 21:16 | of this function are always going to be smaller than | |
| 21:18 | nine . So we say that why is less than | |
| 21:20 | or equal to nine ? So we found the vertex | |
| 21:23 | of the function , the domain of the function . | |
| 21:24 | The range of the function . And then the final | |
| 21:26 | thing we're gonna do is try to find the zeros | |
| 21:28 | of this function of negative x squared minus two X | |
| 21:30 | plus eight . So we're gonna go up to this | |
| 21:32 | last board and do that . Four D . Zeros | |
| 21:39 | is going to be negative X squared minus two X | |
| 21:43 | Plus eight . We're gonna set it equal to zero | |
| 21:45 | . That's what the Zeros are . When I set | |
| 21:47 | it equal to zero I'm finding these crossing points because | |
| 21:50 | that's when Y is equal to zero on this function | |
| 21:53 | . And so you can try to factor it as | |
| 21:55 | it sits . There's nothing wrong with that . But | |
| 21:57 | this is what I like to do when I see | |
| 21:59 | a negative sign in the front or especially just a | |
| 22:03 | negative one like this . What I'm gonna do is | |
| 22:04 | the following Equals zero . What I'm gonna do is | |
| 22:08 | I'm going to multiply this equation on both sides by | |
| 22:12 | negative one on the left . And when I do | |
| 22:14 | it to the left I got to multiply by negative | |
| 22:16 | one on the right . You see I can do | |
| 22:17 | that . I can multiply the left and right by | |
| 22:18 | negative one . Why am I doing that ? Because | |
| 22:21 | when I multiply by this I'm going to just get | |
| 22:22 | positive X squared . I multiply this , I'm gonna | |
| 22:25 | get positive two X . And multiply this . I | |
| 22:27 | get negative eight on the right hand side , it's | |
| 22:30 | just zero . So you see what I've done is | |
| 22:32 | I now no longer have a negative sign out in | |
| 22:34 | front . Because I find when I try to factor | |
| 22:36 | things , which is what I'm gonna try to do | |
| 22:38 | if I have a negative sign in the front , | |
| 22:40 | it just makes it where my signs get a little | |
| 22:42 | bit hard to figure out . So when I see | |
| 22:44 | a negative sign I just multiply through by a negative | |
| 22:46 | one to get rid of it . And now I | |
| 22:48 | can try to factor this . Mhm . And it's | |
| 22:52 | gonna be easier for me to think about . You | |
| 22:54 | guys may not have that problem . You can certainly | |
| 22:56 | factor this expression , There's nothing wrong with it , | |
| 22:58 | I just prefer to do it this way . So | |
| 22:59 | now I know I have X times X . And | |
| 23:02 | then for eight you can do uh there's one times | |
| 23:05 | eight of course . But then you have two times | |
| 23:07 | for you're trying to make a positive two on the | |
| 23:09 | inside . So the only way to really make that | |
| 23:11 | happen is a positive and a negative right this right | |
| 23:14 | there and to check yourself with it , this is | |
| 23:16 | going to give you negative two X . And this | |
| 23:18 | is going to give you four X . When you | |
| 23:20 | add these together you get the positive two X . | |
| 23:22 | Of course . The first terms give you this in | |
| 23:24 | the last terms to give you this . Yeah , | |
| 23:27 | so then you have X minus two is equal to | |
| 23:31 | zero X . Being when you move it over equal | |
| 23:33 | to two and then X plus four is equal to | |
| 23:36 | zero and X of course can be equal to whoops | |
| 23:39 | negative four Like this . So there's two values just | |
| 23:43 | like there was two values before there's two crossing points | |
| 23:46 | at two and also at negative four At two . | |
| 23:50 | And also at -4 . Now this was just a | |
| 23:52 | sketch . I just sketched this thing . Freehand . | |
| 23:54 | Okay ? But you can see that one of them | |
| 23:55 | is positive and one of them is negative . And | |
| 23:58 | if I had done a proper graph with everything spaced | |
| 24:00 | out exactly right and got the shape right , it | |
| 24:02 | would it would look a little closer . Actually not | |
| 24:04 | too bad . I mean that's that's two units . | |
| 24:06 | That's two units , that's two years . So that's | |
| 24:07 | about -4 . That's about four units as well . | |
| 24:11 | Uh And so these are the values of the zeros | |
| 24:14 | two and x is equal to negative four . So | |
| 24:17 | these are the zeros . All right , so we're | |
| 24:20 | not doing anything particularly difficult . We're just wrapping the | |
| 24:23 | concept up of of these quadratic functions . A lot | |
| 24:25 | of times , students will understand how to find the | |
| 24:27 | vertex of a parabola because , you know , we | |
| 24:30 | learned that and then when they there has to find | |
| 24:31 | the range of the problem , they have no idea | |
| 24:33 | what to do . So I'm just showing you that | |
| 24:35 | by doing a quick sketch and looking at if the | |
| 24:37 | thing opens up or opens down . Uh this problem | |
| 24:40 | , it opened down in the previous problem . It | |
| 24:43 | opened up . Then you immediately know what the domain | |
| 24:47 | and range is going to do . Once you know | |
| 24:48 | , if the thing opens up or down and you | |
| 24:49 | know what the vertex is , You know what the | |
| 24:52 | domain and range are . So we're just getting practice | |
| 24:53 | with that . We have one more lesson with finding | |
| 24:57 | this domain and range of problems . And then we're | |
| 24:59 | gonna wrap up this topic . So make sure you | |
| 25:01 | understand all of these problems and then follow me on | |
| 25:04 | to the next lesson where we will continue the topic | |
| 25:06 | of finding the domain and range of quadratic functions . |
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