 Hi, I'm Zor. Welcome to Inezor Education. This is yet another lecture which is dedicated to quadratic functions. In this particular case, it's a kind of practical exercise. I have a set of six different quadratic functions which I would like to basically analyze. Analysis of a function is very easily accomplished by basically building its graph. So I would probably concentrate mostly on how to build the graph of these quadratic functions. And in particular, where is the vertex of the parabola? We know it's parabola, right? So where is the vertex of the parabola? What's the direction of its two branches, two horns, as I called them once, upwards or downwards, and what's the steepness of these particular branches? This is basically how you investigate the behavior of the function. Now, all of these functions are represented in their general format. Now, general format is px squared plus qx plus r, where p, q and r are real numbers coefficients. Now, I cannot really just looking at this particular form of the function tell how the graph actually looks like. I can if the function is represented in this format. Now, this is a full square format of the function. Now, how does this parabola look? But that's very easy. You start from the regular parabola, which is something like this, y is equal to x squared. You shift it to the left by b. If b is positive, it's to the left. If b is negative, it would be to the right by the absolute value of b. But in any case, you shift it to the left by b, so it will be something like this. If this point is minus b. Then you multiply this function by a. And that means if a is positive, the branches remain pointing upwards. If a is negative, it will be changed to pointing downwards. If a is greater than 1, the steepness would be greater. Let's say something like this. If a is less than 1 in absolute value, the steepness would be less, something like this. And then finally, you shift it up by c. That would be your new vertex. So I know how to build the graph of this function, whatever coefficients a, b and c are. So my most important problem is to convert the function from this representation to this representation. Last lecture, which was about this generic format of the function, I actually converted this into this in general variables a, b and c, and p, q and r. Now, I actually wrote a couple of formulas here, which basically relate this p, q and r to the new coefficients a, b and c. From this, we can always derive a, b and c. Obviously, p is not equal to 0. Now, and the formulas are, I will just write it down, a is equal to p, b is equal to q divided by 2p, and c is equal to r minus q square divided by 4p. Now, these are formulas, and I can actually use these formulas to convert this into this in any case which I have. Now, before doing this, I would like to say that I don't want to do it, and I don't really want you, I don't want to encourage you to do it this way. Why? Because you have to remember these formulas. I myself don't remember. That's why I put it here as a kind of reference point. And I don't think it's the right approach. I mean, we are here not to achieve certain transformation, but to get smart about how to do this. So these logical and analytical skills would help us to do something else. The development of these analytical skills is the most important part, which we would like actually to accomplish here. So that's why I don't want to use these formulas. And I will do it differently. So to demonstrate how you can do it without knowing these formulas. However, however, I will do the first one just for fun. I will do using these formulas, and we will see whether it will give me the same result if I will do it more analytically, more logically, if you wish. So in the first case, I have the function y is equal to x squared plus x plus 1. Now, what does it mean? It means p is equal to 1, right? q is equal to 1, and r is equal to 1. Now, using these formulas, I will have a is equal to 1, b is equal to q divided by 2p, 1 divided by 2, so it's 1 half. And c would be equal to r, which is 1, minus 1 fourth, which is 3 quarters. So I can say that in this particular case, I can rewrite this function as y is equal to 1 times x plus b, which is 1 half, square plus c plus 3 quarters. Okay, now, let's check it out. I don't really need this anymore. I don't need this formula because I don't want to use it anymore. I will do something else. And let's check this out. Well, x plus 1 half square, so it's x square, 2 times x times 1 half. 2 times 1 half is 1 times x, so it's x, plus 1 half square is 1 quarter, and plus 3 quarters, and this is 1. So it's exactly the same with this. So this is actually a correct representation of our generic function in the generic format, and this is the correct representation in the full square format. Now, I know how to build the graph of this function. This is the regular parabola shifted to the left by 1 half and up by 3 quarters, so it would be something like this. This is minus 1 half, and this is 3 quarters. This would be the vertex of my parabola, and it will have the same direction and steepness of the branches as the regular y is equal to x square parabola. So that's my analysis, but I was using these formulas which I don't want to remember and I don't want actually you to remember. So how can I come up with the same result like this one using some more human-like, natural, analytical abilities which everybody called full test? Well, here is the point. What's most important right now is to find this particular coefficient, y, because this we can always adjust whatever will be left. Now, let's just think about it. Now, x square is the result of obviously squaring of the x, and then we have a double product. So if you have x plus b square, this is x square plus 2bx plus b square, right? Everybody knows this formula. And if you know it, you can always multiply it yourself. So how can I find b such that 2bx would be equal to whatever my element with x to the first degree is? Well, obviously, if this coefficient is 1 in this particular case, then this b should be half of it because we double it, right? So if I have this, it means b must be equal to 1 half. So after doubling, I will have exactly x. So this is the first and most important part. So I know that it should be x plus 1 half squared. Now, let's just think about it. x plus 1 half squared will give me x squared. It will give me 2 times x times 1 half, which is x. And it will always give me something else, which is 1 quarter. This is a free member, 1 quarter, but I need 1. So I have to compensate. I have to add another 3 quarters to get to the 1. So this is the logic. And I will repeat this logic another five times so you will kind of familiarize yourself better. Now, the only thing is a small nuance, which does not really exist in this case. The coefficient in this case is 1 with x squared, right? Now, what if it's not 1? Well, obviously, we have to really factor it out because, again, that is something which is very easily to accomplish. And then the rest will have a coefficient of 1 with x squared. And I will use this approach. So let me demonstrate it for the second example. Now, I'm not using the formula. I'm just using the second example straight. And I'll just demonstrate the logic. So you have y is equal to x squared plus x plus 1. So the first step in converging this into full square mode is to factor out the coefficient at x squared. So it's equal to 2 times x squared plus 1 half x plus 1 half, right? So now I have x squared here. Now, what I'm thinking is I have to find out such a b that x plus b squared would have the first numbers like this. Now, what is it? Obviously, b should be equal to 1 quarter because then 2xb would be equal to 2x and 1 quarter. So it would be 2 and 1 quarter. So it's 1 half. That's exactly what I mean. So next is, I should say this is supposed to be equal to x plus 1 quarter square. Now, that would satisfy x squared and it would satisfy 1 half x. Now, 1 quarter square would be 1 sixteenths, but I need 1 half. So I have to add whatever I need to add to 1 sixteenths to get 1 half. 1 half is 8 sixteenths. I have 1 sixteenths. So I have to add 7 sixteenths, right? Basically, that's almost it. The only thing which remains actually to do is multiply separately two by these two things using the distributive law. So it's 2 times x plus 1 quarter squared plus 2 times 7 sixteenth is 7 eighths. So this is my answer. This is my full square for this particular thing. Just in case, let me check if I'm right. Okay, it was 2. 2 times x squared plus 2 times x times 1 quarter. So it's 1 half x plus 1 quarter squared which is 1 sixteenths plus 7 eighths. Okay, multiply. 2 times x squared, it's 2x squared plus 2 times 1 half x is x. 2 times 1 sixteenths is 1 eighths plus 7 eighths. 1 eighths and 7 eighths is 1, so everything is fine. So this is just checking, but most important is that I have transformed or converted my original general representation of quadratic function into full square. And now I can say very definitively how this particular graph looks like. So you have to shift my x square, y is equal to x square to minus 1 fourth, so it's this way. So then the parabola should be multiplied by 2 which means it should be 2 times steeper. So it's something like this. And then shift it upwards by 7 eighths. So this would be the new vertex. And that's how it goes. So this is my graph. This is how parabola goes. It's twice as steep as the regular y is equal to x square. And its vertex has the x coordinate minus 1 fourth and y coordinate 7 eighths. That's the end of this particular analysis. So full square is the purpose of all these transformations and I was trying to do it using any kind of a formula but just some kind of logical transformations which led me to this particular case. Now next one, 3x square plus 2x plus 5. Again, the first step, factor out the coefficient at x square. So it's 3 times x square plus 2 third x plus 5 third second. How can I get this? Well, it's x plus plus how much to get 2 third x? Well, this is supposed to be double x this b, whatever the b is supposed to be and obviously b is supposed to be 1 third, right? Because if I will do this, it will be x square plus 2 times x times 1 third which is 2 third x. Now I have 1 ninths square. I need 5 third. 1 ninths and this is 15 ninths. So I need 14 ninths to add. That's it. 1 ninths plus 14 ninths would be 15 ninths factor 3 5 third. Yes, seems to be okay. Now, more I can rewrite it as 3 times x plus 1 third square plus 3 times 14 ninths is 14 third. So this is my full square and now I can say that this is the regular parabola y is equal to x square. Then it should be shifted to the left by 1 third made 3 times steeper and then shifted up by 14 third. Just in case, let me check that I'm right. So it's 3 times x square plus 2 times x times 1 third so it's 2 third x plus 1 ninths plus 14 third. So it's 3x square plus 3 times 2 third is 2x 3 times 1 ninths is 1 third plus 14 third. So it's 15 third which is 5. Yeah, I'm right. So checking is fine. All right, so that's it for this particular example. So let me just continue doing exactly the same thing after a few more we will all be very comfortable with this particular transformation. y is equal to minus 4x square plus x minus 1. Okay, equals minus 4 outside factor out. What's left? x square. This is minus 4. This is plus. So it would be minus 1 fourth x and plus 1 fourth. Right? Minus 4 times minus 1 fourth would be 1. Minus 4 times plus 1 fourth would be minus 1 equals minus 4. Now, what is it? It's x plus b square, right? Now, x square. Now, we will have 2xb and we will have to have minus 1 fourth. So what is b? Well, b should be equal to minus 1 eighths. Right? 2 times x times minus 1 eighths would be minus 1 fourth. Fine. Now, the free member would be 1 sixth fourth and we need 1 fourth. So 1 fourth is 16 sixth fourths. So I need another 15 sixth fourths to get to my right. So x square minus 1 quarter x plus 1 sixth fourth plus another 15 sixth fourths. So it's 16 sixth fourths which is 1 fourth. Seems to be okay. Now, what is this particular graph? All right? First, it's shifted by, well, minus 1 eighths to the left which is actually plus 1 eighths, right? So this is 1 eighths. That's my vertex. Now, then, by the way, it probably would be easier if I will open these parentheses. So it will be minus 4x minus 1 eighth square plus actual minus. Minus 4 and 15 sixth fourths is 15 sixteenths minus 15 sixteenths. That's the right. Well, just in case, let me check. It's minus 4x square, okay? Then 2 times x times minus 1 eighth is minus 1 quarter x times minus 4 is x and this is 1 sixth fourth times minus 4 is 4 sixth fourths which is minus 1 sixteenths and 15 sixteenths. So it's minus 1. Yeah, seems to be fine. Now, this graph is, again, as I was saying, it's shifted by 1 eighths to the right in this case because this is negative. Then the parabola should be directed with its branches with its horns downwards because this is a minus sign and it should be 4 times steeper than the regular parabola. So it's something like this. Very, very steep goes down. And then it should be minus 15 sixteenths so it should be brought down. So this is the parabola which I'm looking for. So the horns are looking down because this coefficient is negative and this is the original coefficient. And it's very steeply goes down because the coefficient is 4 here. And then 1 eighths and minus 15 sixteenths is basically where the vertex is located. Alright? Okay, so that's as much as we need for this particular case. Y equals minus 3x squared plus 6x plus 4. Okay, same thing. The first step is factor out minus 3. Now we have x squared minus 2x minus 4 third. Minus 3x squared plus 6x and plus 4 third. Okay, seems to be fine. How to convert this into full squared? Minus 3. So it's x. What should we need to get minus 2? Obviously minus 1 squared. Right? It will be x squared minus 2x plus 1. Now, we have plus 1 here but we do need minus 4 third which means we have to subtract 7 third. Right? So it should be 1 which is 3 thirds minus 7 thirds would be minus 4 thirds. That's what it is. Okay. Now, just to make it a little easier it would be this plus 7. Let me check. Minus 3x squared. This is minus 2x so it's plus 6x. Now this is plus 1 times minus 3 is minus 3 plus 7 is 4. Correct. Now, how to draw this particular graph? Well, again, obviously it's 1 to the right then direct the parabola downwards 3 times steeper and then raise it up by 7 so it would be something like this. This is 7. This is 1 and parabola is downwards 3 times steeper than original y is equal to x squared. Okay? And that's the last one. y is equal to minus 2x squared minus 2x minus 3. Okay. I'm sure you are very proficient by now with the technique. Well, factor out minus 2 we have x squared plus x plus 3 second, right? Equals minus 2 x plus x squared plus x would be x plus 1 half squared, right? It will give me x squared plus 2 times x times 1 half which is x and it will give me 1 fourth. Now, this is 1 fourth but I need 3 seconds. So I have to add this is 6 fourth. I have 1 fourth so I have to add 5 fourths. x squared plus 2x and 1 1 half so it's x plus 1 quarter and 5 quarter it's 6 quarters which is 3 seconds. That's fine. Or minus 2x plus 1 half squared minus 2 times 5 quarters would be minus 5 seconds which means my parabola would be with a vertex minus 1 half and minus 5 seconds. So it would be something like this minus 1 half and minus 5 seconds. That's the vertex. The horns will be downwards because this is the minus sign and the steepness would be twice as big as the steepness of the regular parabola. Well, that's the end of it. Basically, my purpose was to basically encourage you not to remember the formula which you will probably forget anyway but to try to logically approach what we need is a full square. How can I accomplish this transformation to a full square? Well, first of all, this definitely prevents me from doing it in a simple way. So let's just factor it out. That's the first step. Now I have the function which has 1 at the x which means we should convert it into x plus something squared and all we need to do is just to find this something that's efficient with x to the first degree. In this case, for instance, it's 1 half. And then I see what my free member is supposed to be, what I have here, so how much I have to add or subtract to get to this. So this is the logical kind of a set of conclusions which you can make, set of steps which you can make to get the result without knowing any formula. And that's personal. I prefer this way because it will develop your ability to find the way in a new situation which you are not familiar with. Nobody gives you the formula to solve some real-life problems. You have to basically come up with a way to solve it just on your own, without the help of anybody. So this is just a simple example how you can use your own logic to accomplish this without knowing any additional formulas or anything like that. That's it for now. I do suggest you to do some exercises yourself. You can do exactly the same exercises which I did. The notes for this particular lecture are on unison.com. And well, good luck. Thank you.