In this class you will learn how to construct the graph of a quadratic function, so take a look here, I will tell you a script on how you will make this graph, so the function I brought is fx equal to x squared minus 2X minus 3 and I will give you some guidance For you, it is important that you have seen classes previous to this one which are base content for this class here so you will find the links to these classes in the description, agreed? So, first of all, for me to build the graph of the 2nd degree function or quadratic function, I know that this drawing of the graph will be a parabola, right, well, looking at the function, I can already identify one thing, I can already identify if my parabola will be happy or it will be sad, so you don't know that I'm talking about a happy or sad parable, that's why I talked about the importance of going back to previous classes, okay? So we are going to look here at the coefficient A, the coefficient A of my function is 1 it is a positive value when A is the positive value my parabola is happy it has the concavity facing upwards, so here today I identified a point So I already know how my parable will behave.
The other thing I can look at here is this coefficient C, which is the independent term. What this coefficients C shows me it shows where the parabola cuts the Y axis so here I'm going to mark it, it cuts Y just to summarize so it will pass at minus three, I was going to mark it here, it will pass at minus three ok now There's a point where she 's going to pass and I already more or less know that she's going to do this drawing here just because she's happy, okay ? Now I need to find out more, I already know that it cuts the Y axis in the independent term so I can find out where it will cut the X, the X axis, I just need to find what the zeros of the function are.
So how do I find the zero of the function, you will take your function and you will equate it to zero. So look, the function is x squared minus 2X minus 3, no, I'm going to take this function and equate it to zero and once again I say, I've also explained in previous classes the zeros of the quadratic function, if you have any questions, come back to class, ok, good here so I can solve it by doing it for Bhaskara, right, the resolving formula for báskara or on the product you choose which is better, I'm going to do it for Bhaskara here because I'm going to need some moment of the Delta of the discriminant. So let's start with delta = B squared, let's do it a little straight, right?
You're used to Bhaskara, right? Negative 2 squared will be what will be four then minus four times the coefficient A which is 1 times the coefficient C which is minus three, well here I will find that Delta will be equal to four leave the four here and I will multiply four times 1 = 4, 4 times 3= 12 so it will be 12 - with - you have to do the sign rule that it will be + so I will find that my Delta will be 16 which I know even when the delta gives a greater value that zero I know that this parabola when drawing the graph it will cut the X axis at two distinct points because when Delta greater than zero X1 is different from X2, right the roots of the equation. Now continuing, so it's going to be my X2 which are the roots of the equation and who will be the X1?
So I'm going to add two first, I'm going to use the plus sign here in the middle, okay, so it's going to be two plus four. What is two plus four is six then divide by 2, 6 divided by two is three and who will be X2 now I will do one minus now 2 - 4 will be - 2 / 2 will be minus one because I already I'm doing it a little straight here guys because like I said, when you get to this class on the graph of the quadratic function, you've already gone through all the others, so if you don't remember, you'll go back to understand everything, okay then? I identified that these two values here which are x 1 and x 2 which are the zeros of the function are where my parabola will cut the X axis because when I equate the function to zero I find out where it cuts the X axis so let's mark it now It's going to cut the X axis here at least one and here at three, at these two points, so I've already identified it.
I can trace. No, there's one little thing missing, what's missing? Every time I draw the graph, you know, the parabola, I have to identify if its extreme point is at the bottom or if it is at the top.
As I already know her behavior, she will be happy. I know that the extreme point is at the bottom. So I have a minimum point and how do I find what this minimum value is?
I need to find what the coordinates of the vertex are and to find the coordinates of the vertex I will find who is the x of the vertex which is minus B so it is the opposite of B divided by 2 a, eoy of the vertex will be who Y of the vertex because when I'm finding the coordinates of the vertex I need to find who is this ordered pair that makes this curve, right that gives us this extreme, so it will be minus Delta, I'm going to delete my happy thing here now - Delta / 4 A, so these are the two formulas for me to find the coordinates of the vertex so starting x from the vertex will be x equal to minus B. Where are you guys? Bb is not negative two, I do the opposite of two of negative two, which is positive two divided by two times A, who is A?
OA is not 1, twice 1= 2 so the x of the vertex will be formed by 1, now I need to find out who the Y of the vertex is to assemble the ordered pair, let's go, the y of the vertex will be equal to, here we needed to write x from the vertex, minus Delta that's why I said I was going to do it for Bhaskara guys because I already had the value of Delta here, the delta is here so it's 16. So it's going to be minus 16 / 4 times A, four times A , 4 x 1 right so it gives 4 - 16 divided by four we will then find one - 4. What do these two values that I just found here mean these two values here they will form an ordered pair and this ordered pair formed here it's going to give that extreme, where that curve is, okay, so let's connect it here on our graph, 1 of X, 1 of X is here and - 4 of Y, 1 of Seeing here, this here will be the minimum point, the lowest possible value that our parabola will go through and then it goes up again.
Just remembering if you wanted to find the value of this x of the vertex thinking about symmetry because when I draw a parabola it has a Symmetry and when I have the two values from zero of the function from - 1 to 3 I have how many spaces one two three four take half, half of four is 2, 1 2, 2 spaces would be 1, that's not what I do, this average is right in the middle, so I already know that X is 1 so I can find who this Y is. vertex here you know I could have done this by taking this x that I found here doing half the spacing and putting it in its place here, then if I put it here in its place I would find the - 4 ok, that's another way for you to do it but I wanted to show for you, using the formulas for x at the vertex and y at the vertex, well, what do I do now ? to draw us, it passes through the zero of the function, cuts the Y axis and reaches its extreme point, right, here I have a minimum point and now what does it do now it will go all the way again in a symmetrical way but not You're going to laugh at me, if my parable isn't symmetrical, I'm doing it here live for you, how to make the parable.
So see that here I have the graph of the quadratic function that I brought you Which is where the design of our parabola shows if it is happy or if it is sad, right, if it is positive, our parabola is happy, as happened here, the coefficient C, which is where our parabola intersects the Y axis and then the coefficient B, people he will determine for me if the parabola will intersect the Y axis on the decreasing increasing branch or at the vertex, in this case as it is negative two then he will tell me that it is the negative value so he says that my parabola will intersect the Y axis the decreasing branch, it's going down until it intersects the Y axis, so mark it there, it's very important, okay guys, in the other case we'll see what will happen here with the B coefficient, let's go to the next one. Now I want to see who is already an expert and brought a second function for us to build the graph of the function. So now I have minus 2X squared plus x plus 3, how do I start doing it, there's that tip I gave you, coefficient A, look at it, coefficient A is now minus two, it's negative and every time A is less than zero which happens our parabola will be sad as it has the concavity facing downwards the coefficient C it determines for me where the parabola will be right a Y occurs I already know that it will cut here at the cloverleaf right here it will cut my Y axis and b here guys, so mark it well.
You can see that b is positive 1. So this means that as B is a number greater than zero it will indicate to us that the parabola will intersect the Y axis on the growing branch, so mark it right there and now that I start doing it I already know more or less that it is a parabola will do this right and so what I have to think about now I have to think about the roots of the function in the zeros of the function and to determine the zero of the function which are those values where the parabola intersects the X axis I will take the function and I'm going to equate the function to zero so it will be minus 2 x + x + 3 I'm going to equate it to zero and now we're going to calculate the Delta which will be Delta equal to b squared we know that B is 1, so 1 squared it's 1 - 4 x o A x C, why did I put parentheses, I didn't even need it, right people, but leave the parentheses there, now what's going to happen? Delta equals 1, I do the multiplication first 4x 2= 8, 8 x 3= 24 - with less it gives more so I arrive at Delta being worth 25 OK now let's find the roots x equals minus b so it's the opposite of B plus or minus the square root of Delta which will be root of 25 which is five divided by two times A, well now it is two times negative two and it will be = -4 let's find the two values now x 1 x 2, so X1 How will it happen, it will be minus 1 plus, I'm going to do it in my head, right?
Better - 1 + 5, - 1 + 5 = how many people? minus 1 plus 5 = 4 = 4, 4 / - 4 which gives minus 1, and here the other one - 1 - 5 gives minus six I will leave it marked - 6 / - 4 - with minus I know it will give more and it gives I simplify here by two, right, and then it will be equal to 3 - , so I can identify that now my parabola intersects the X axis at those two points, which would be minus 1 and 3/2, 3/2, whatever Is it 3/2 people? three just divide three by two 3/2 is the same thing as 1.
5 so here is 1 half here will be 1. 5 this point is 3/2 so the parabola goes through here, you can get an idea then it will do this here and now what I do now I need to find where the maximum point is because in this case the extreme will be up here I need to identify my maximum point. So you can even increase it here and it will be a little bit upwards, put your Y axis here.
Come on, do you remember how I find the maximum point there, what is the coordinate of the vertex, the coordinates of the vertex that will be x of the vertex equal to minus B / 2A, what will it look like ? two times the negative two of - 4 so I got that it will be a quarter because a quarter minus with minus gives more, 1/4 that if you divide 1 by 4 to 0. 25 now the y of the vertex who will be oy will be minus Delta divided by 4 A, the delta is here so it will be minus 25 divided by 4 A, 4 times the A are 4 x - 2 which will be - 8 - with - people, less with less gives more so it will be 25/8 .
Wow 25/8 how do I think how much is 25/8? you must divide 25 by 8 and by dividing 25 by 8 you will find 3. 125, okay?
And here, dividing 1 by four you find 0. 25 so I need to unite these two here because they will form the ordered pair that will give me the coordinate of the vertex. So let's go, I'm going to take 0.
25 of the a little above 3. Remembering that we always do it by approximation here, it's good and now what I do now is just draw the parabola I have so that it is sad, right for concavity facing downwards. So let's draw it here and it won't make you laugh, just draw it crookedly, it also comes from below, cut the X axis intersects the Y axis, goes to its maximum point and goes down again, until it looks pretty, right guys, so here I have the graph of the quadratic function, right?
Mark right here, now I'm going to summarize for you that when you start graphing the quadratic function, you'll already know that you're doing it right or won't you go there then? Here I brought three situations, the first, second and third, here in the first situation we are going to assume that the delta is greater than zero, what does it mean for the delta to be greater than zero? when you are there finding the zero of the quadratic function then you find a Delta greater than zero that is a positive Delta you will find an X1 different from an So now we have two other situations.
So, taking as a basis that the delta is greater than zero, here I will represent a being greater than zero and here a being less than zero, what did I say, even when A is greater than 0, the coefficient A of the function is greater than zero You have a parabola that is happy, so this will happen here, when your Delta is greater than zero and A is greater than zero, your parabola will intersect the X axis at two different points, which would be X1 and X2, the two values two zeros of the function and remember as it is greater than zero it is happy, even making a happy face so you don't forget. Now if I'm still working in the same situation where the delta is greater than zero then it intersects the intercept at the two different points, but it will be a sad concavity facing downwards and here the concavity facing upwards, look at its little face here. Now if I am working with Delta equal to zero, when you are there solving the zeros of the quadratic function, your Delta was equal to zero, which means that the two zeros of the function are equal values, the same value, so your parabola will intersect the axis X in just a single point.
And then you take into consideration again if your a is greater than zero the parabola is happy if your a is less than zero your parabola will be sad but then the difference is that it will not pass below the axis here it will hit a single point here and it goes back up again, that's it when A is greater than zero if A is less than zero it is sad so it does this here, it hits a single point and goes down again ok . If the drawings are more closed here, more open, they don't take into consideration people, what matters is the summary that I want to give you so you can analyze whether the graph you are making is correct or not or also other activities that can be done analyze the quadratic function and in the latter case I now have a Delta less than zero when you are there solving the zero of the quadratic function and you find in the Delta less than zero you do not have the roots In other words, the parabola does not intersect the X axis but then you don't even need to do it, right Gis , it won't intersect the greater than zero she is happy so she does this here she flies like this she doesn't intersect the X axis but she intersects the Y axis and if her a is less than zero what will happen if her is less than zero she will keep the concavity facing downwards and does not intersect the Alright guys, I hope you understood today's class, look, you also liked it. If you liked it, leave a like for Gis and subscribe to the channel and we will have many other new classes for you and also activate the bell so you can receive the notifications for all classes and then you will rock your activities.
And I'll see you in the next class! Goodbye. .
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