You will learn how to calculate the zeros of a quadratic function. But what are zeros in a quadratic function? You know?
Firstly, do you know what a quadratic function is? If you don't already know, I'll start the class by giving you an indication of the previous quadratic function class, right? So when you say it like this, the statement says it like this: Determine the zeros of the quadratic function, the zeros of the quadratic function are the roots.
How about the roots? I find the zeros of the quadratic function when I take my function, which is given by fx and I equal to zero. So these are the values of x that I will find by equating my function to 0.
What is my function in this exercise? my function is minus x squared plus 3x plus 28 so I'm going to take this function which is going to be minus x squared + 3x + 28 and I'm going to equate it to zero, ok, function is equal to zero, function is equal to zero, so I'm going to find out here which values of x satisfy this equation, I will now find the roots. You heard what I said equation, that's right, here when I equate my function to zero I have here what a quadratic equation and a quadratic equation have several resolution methods so you choose the one that is most appropriate, which one do you like best?
I know that most students like to solve a quadratic equation by doing what, applying Bhaskara's solving formula, so here we will remove what the coefficients are, who is the coefficient A in this function of mine? It's minus one, right, because here, when there's no one, it's a coefficient, since it has a negative sign, it ends up being minus one. The coefficient B is the one on the side of X, which is three.
I almost already say the answer, I ask, I can say the answer, and the coefficient c is 28 so coefficient C is 28. Now we are going to do the delta, the calculation of Delta B squared will be 3 squared minus 4 times the value of A, which is - 1 times the value of C, which is 28, okay, I'm going to have Delta = 9 here because 3 squared is 9, 4 x1= 4, 4 x 28, 28 x 2 That's 56 times two again are 112 so guys I do minus with minus here which will be more soon I will find a Delta = 121 which even though I know when I have a Delta greater than zero I know that the two results for x which are the two roots are two different values square of Delta the delta is 121 the square root of 121 will be 11, so more or less 11 / twice the A, which is negative two because it is twice one minus one ok here let me do a little tidying up to stay on the line right x = now what and I will find here is X1 and X2 are already here X1 will be minus three plus 11 / minus 2 and X 2 which is the second root will be - 3 - 11 / 2 and then I already managed to find it, right people, here minus three plus 11 are what are 8 positive / - 2, 8 by two are four plus and minus -4 is in the other root I will find it then - 3 - 11 They are minus 14 Why should I three I owe 11 / minus two here, right, I had forgotten about it - 2 - with minus will be more and 14 by two is seven, so notice that I managed to find what are the two roots of our equation and finding what are our roots of equation I find the zeros of my function, ok because it means that I equated my function to zero, I will find what the values of x are when I equate this function to zero. But is that all Gis does?
When the exercise talks about determining the zero of the function, that's all we do, and now mark a utility for you to use in the next class. When we find the zero, we find the zeros of the function, which would be these two values, here, these two values will show Guys, where is the parabola, because when I graph a 2nd degree function, a quadratic function, this graph has a drawing of a parabola and these two values here will show where this parabola cuts the X axis, so symbolizing here for you When you already have this class, you already know the scheme, more or less the graph, more or less like this, I know that it will cut at negative four, so suppose it is here and at the seventh, suppose it is here, right, this on the axis X so these two points show where the parabola cuts the X axis. And then I know that my parabola has more or less this design and this behavior, okay?
So it does more or less this and goes through these two points but stay calm, there will be the next class in which we will explain how we are going to assemble the graph, okay? But I like to highlight these two points that we find here which are the zeros of the function, what are they for when I put together a graph, did you mark them well then? It is used to cut the parabola to cut the X axis, it is the point of intersection with the the next classes on quadratic function.
Goodbye. . .
