 Hi, I'm Zor. Welcome to Inezor Education. We continue solving different problems which are somehow not exactly typical problems which usually are solved in school. The typical problems are those which basically are verifying how well you know the theory. These problems which I am putting in the course called Mass Plus and problems, these problems are not really typical in this sense. They require you some kind of ingenuity, creativity, thinking basically to come up with a solution which is not part of whatever you were presented in a theoretical part of the course. Now, this particular course, Mass Plus and Problems, is presented on Unizor.com and I do suggest you to watch all these lectures from that website. First of all, there are prerequisites on the website. There is a Mass 14th course which basically gives you all the theory pertaining to regular high school course of mathematics. Also, there is a course called Physics for Teens and Relativity for Teens. I mean, if you are interested. So, there are some other courses on the same website. Now, what's important about any course presented on Unizor.com is, well, there is basically some kind of a table of contents. It's hierarchically organized. Lectures are ordered properly in logical sequence. So, whatever I'm presenting in one lecture will be used in subsequent. Also, every lecture has detailed notes which basically are like textbook, if you wish. There are exams which you can take as many times as you want. Now, the site is totally free. There are no advertisements, so nothing really distract you from the process of studying. Okay. Now, back to the problem which I wanted to present today. It's related to a certain property of certain functions. So, that's why I categorize it as algebra and the number of the program is 0-1. So, this is algebra 0-1. It's a one problem. Relatively simple one, but it really gives you some flavor that there are problems which are not really properly covered in the regular course. And they do require you some kind of thinking about. Okay. So, what's the problem? Now, there is a function x square minus a times x plus a. So, a represents any real number. Now, what does it mean that there is a function which depends on some kind of a which can be any real number? Well, this equation actually defines a set of functions. For a is equal to 0, it's one function. It's y is equal to x square. For a is equal to 1,000, let's say it's another function. It's y equals x square minus 1,000 x plus 1,000. So, this is a family of functions, as we are saying. Okay. Now, what's important is that if you will graph a few of these functions, representatives of this family, let's say we start with a is equal to 0, what do we have? Well, we have a parabola. So, let's do a parabola. Okay. Now, let's have a is equal to 2. You have y is equal to x square minus 2x plus 2. Which is the same as x minus 1 square plus 1, right? This is x square minus 2x plus 1 and another one gives me 2. Now, what is this? This is the same kind of parabola, but shifted to the right by 1 and up by 1. So, it will be somewhere here. So, they're crossing in point 1, 1. Let's take another parabola. Let's say a is equal to 3. So, we have y is equal to x square minus 3x plus 3. Which is what? x minus 1.5 square. This is what? x minus 3x plus 225 plus 0.75, something like this. If I'm not mistaken. Which means 1 and a half shifted to the right and 0.75 somewhere here. So, it's parabola something like here. What's interesting, you see, this is a point which is kind of common to these three parabolas. And now the problem. First problem, actually, there are two problems. The first problem is find the point on the plane where all these parabolas are going through. Regardless of the value of A. Like in this case, we have A is equal to 0, 2 and 3 and they look like on my very imperfect picture. They look like they're crossing at point 1, 1. Now, you have to prove it, basically. You can formulate it this way. Is there a point on the plane which is common for all these parabolas? And if it is, what is it? So, that's basically the point. Well, how can we find this particular point? Well, very easy. Let's just take a couple of parabolas for a couple of different values of A and see if there are any points which they cross. And then we will check it for some others in general way. So, in this particular case, let's just concentrate on these two. So, these two seem to be crossing at point 1, 1. So, let's just check what if x is equal to 1 and check the value of the function y for any A. So, y in this case is equal to 1 square minus A times 1 plus A, right? Well, as we see A and A is and y is equal to 1. So, what we actually have proven that the point x is equal to 1 and y is equal to 1, which is this point, is common for any parabola from this family. No matter what A is, if x is equal to 1, y will be equal to 1, which means all these parabolas, all these functions, graphs of these functions, they are going through the point 1, 1. Okay, now this is kind of maybe not exactly direct way of proving it. We can prove it many different ways. For example, instead of two concrete values and then checking if this works for anything else, let's just take two completely different values. Let's say y is equal to x square minus A1, x plus A2, and we are looking for, sorry, A1, and we are looking for another function from the same family. We are looking for such an x, which produces the same result for y in both cases. Well, if x is the same and y is the same, let's subtract them. What do we have? 0 equals 0 minus A1 minus A2, x plus A1 minus A2, from which follows 0 goes out, 0 goes out. We can cancel A1 and we will have x is equal to 1. It's a solution to this equation for x. So as we see, we can put x is equal to 1, we can find x is equal to 1, even from some kind of a general consideration, not exactly two concrete values of A. So it all goes to the same way. So all these parabolas are going through the point 1, 1. x is equal to 1, y is equal to 1. So that's my first problem related to this particular thing. A little bit more, maybe an unusual result would be the result of the second problem. So the second problem for the same kind is as follows. Consider the vertices of all these parabolas. In this case it's this one, this one, this one and all other. Proof that all these vertices are actually lying on some parabola, which should probably look something like this. Now this is a very non-trivial result, which can be very simply obtained, but personally I would not expect that all the parabolas of this particular family would have the vertices forming another parabola. Well, let's try and prove it. Let's start from the general case. You have the quadratic function, any quadratic function. Where is its vertex? Well, for those who remember it from the course of mathematics, its x is equal to minus b divided by 2a. For those who don't remember, and that actually includes myself, I do know, the only formula which I know, which I remember actually about quadratic equation, is the formula for its roots. And the vertex is actually in the middle between the roots, right? So if you have some kind of parabola, these are two roots, x crossing. The vertex is also always in between, right in the middle between these two points. And they do remember the formula for roots, which is minus b plus minus square root of b square minus 4ac divided by 2a. So the middle point is you just add these two things and divide it by 2, right? If you will add these two things and divide it by 2, this thing disappears and you will have minus b divided by 2a. So that's the formula for x coordinate of the vertex. Now, what's the y coordinate? Well, we don't really need to go with the y coordinate in general case, where we can say, we can only consider the family of parabolas which we have. In this particular case, a coefficient at x is equal to 1, coefficient at b is equal to minus a, the coefficient at c is equal to a. So minus b divided by 2a is minus minus a divided by 2, which is a divided by 2. Okay, now, this is a coordinate, let's call it x0. It's a coordinate of the vertex of any of these parabolas, the x coordinate. Now, what's the y coordinate? Well, let's substitute this into, we will have a square divided by 4 minus a times this a square divided by 2 and plus a, and it's equal to, this is minus a square divided by 4 plus a. So we have x and y coordinates of the vertex of this parabola. This is x coordinate and this is y coordinate. So as we see, x and y coordinates are parametrically given by the value of a. But if you would like to know what kind of a curve it produces, this type of parametric definition of our point. x is equal to some function of a and y is equal to some function of a. We actually need to convert it into form as a function, like y is equal to some kind of function of x. Then we can judge what exactly this particular curve is. Well, how to find from parametric definition of x and y, how to find the dependency from them? Well, we'll just exclude a from these two equations and resolve a, y in terms of x. How can we do it? Well, I will find a from here as function of x and substitute it to this. So a is equal to, from this, 2x0. And therefore, y0 is equal to minus a squared by the before 2x0 squared, that's 4x0 squared, divided by 4, that's minus x0 squared. And plus a, which is to the plus 2x0. So basically the equation, sorry, that's y. Okay, so this is the equation of the dependency between y and x, where y and x are vertices of parabolas which belong to our family. And as you see, this is a quadratic dependency. This is definitely a parabola, which may look actually like this. Now, y, it looks like this. Well, first of all, since this is minus, so the horns of parabola are downwards, obviously. Now, we can find where are the x crossing, the roots of this parabola. Well, the roots of this parabola is when y is equal to 0, if y is equal to 0, we will have minus x0 squared plus 2x0 is equal to 0. So obviously the first root is x is equal to 0. Now we can take x out, and that would be another root, would be 2. So it goes through points 0 and 2. Now the vertex will be in between at a point x is equal to 1, so this will be actually the vertex. And the value of parabola at this point is exactly that common value which we had. It's 1, 1. It's x is equal to 1, y is equal to 1. If you will substitute 1 here, you will have minus 1 plus 2, which is 1. So this is a parabola which has vertex at 1, 1. Two roots at 0 and 2. Horns are down. This is a parabola on which all the vertices of these parabolas, family of parabolas, are located. And as I say, that's a little bit unusual result. I quite frankly did not expect just looking at this. You don't really realize that all the vertices of all these different parabolas belong to one particular parabola which looks like this. And that's an equation of this parabola. Well, that's the whole problem which I, couple of problems actually which I wanted to present to you. Are they unusual? Well, I think it's not really like part of the general course of algebra. In the general course, they will probably talk different kinds of equations and how to solve these equations. That's basically the standard course of mathematics which is presented at school. Now, this property of this particular kind of parabolas is, as I said, unusual and it requires certain research. So that's why I presented this problem as part of this course called Mass Plus and Problems. Okay, that's it for today. I do suggest you to read the notes for this lecture. They have a much better graph than I presented. So you go to Unisor.com, you choose the course Mass Plus and Problems. And within that course you have Algebra, Geometry, etc. You choose Algebra and Algebra 01 will be on the next menu. Okay, that's it. Thank you very much and good luck.