 After many years of teaching algebra, I've learned many things, right? One of those is that students don't like to complete the square. We learn about quadratic equations. We see that it's a very important tool, but still students are very apprehensive to complete the square, even though it's a process we can follow every, every time. Now, in one example, like if we're trying to solve a quadratic equation, we could factor it. Sometimes that works, sometimes it doesn't. But we could also always complete the square, but instead of complete the square, we could use the quadratic formula, right? And many students, if they're not gonna factor, really like the quadratic formula because you don't have to go through all the mechanics to complete the square, you just have to plug them into the formula and go from there. But now we have this new problem, right? Finding the vertex of a quadratic function. I keep on telling you that if your function's in the standard AX squared plus BX plus C form, you're gonna have to complete the square to find the vertex. But we then might ask, well, could we do the same thing that we did to find the quadratic formula? Can we vertex of the general quadratic function AX squared plus BX plus C? Now, if you ask that, I'm like, well, that's actually, if we did that, what if we complete the square in this situation? Doing so, we'd factor out the A from the Xs right here. So we factor it out and we would get A times X squared plus B over AX, leave a space plus C. And because again, factoring A away from AX squared is easy, just gives you an AX squared, at least you just did X squared, excuse me. Factoring away A from BA. Well, in that situation, just factoring A away from B just leaves a B over A. I don't know if that simplifies or not, but you get that fraction. And so then looking at this number right here, this B over A, let me do a different color for that. You get this B over A. We need to take half of that, B over A, half of that, that's going to give you just B over 2A. We're gonna square that and that gives us B squared over 4A squared, so far so similar to what we did with the quadratic formula, B squared over 4A squared. But then we have to subtract this, so cancel out the addition we had, in which case you're gonna get B squared over 4A squared times by A. Again, all of this looks very, very familiar to how we were doing this with the quadratic formula. So next, F of X would equal A times the middle will now factor since it's a perfect square trinomial. You get A, or you'll get A times X plus B over 2A, quantity squared, because I just grabbed, let's grab this number right here. And then we have, again, we have this plus C minus B squared over 4A. Kind of like with the quadratic formula, we're gonna have to find a common denominator times top and bottom by 4A, 4A. We then end up with F of X equals A times X plus B over 2A, quantity squared. And then we have this negative B squared minus 4AC over 4A. And so this right here is now that the square has been completed. And so this right here is a form of the quadratic, the A times X minus H squared plus K. And so there's some things we can see here. So first of all, the coefficient, the leading coefficient at the very start is the same as the transformation, the vertical stretch you do. And so if we have to determine like, is this vertex a maximum or a minimum of the graph, we actually can see that from the very beginning of the problem, right? It's the exact same A we had at the very beginning. It's just the leading coefficient. This right here is the observation I really want. And this is the main takeaway from this calculation here. Note that the H coordinate of the vertex the vertex is gonna be a negative B over 2A. Notice here that the mismatch of signs, we have a plus versus a minus. So in order to compensate, we would have to do something like the following. We really should think of this as X minus a negative B over 2A. And therefore we get this great formula right here, H equals negative B over 2A. This will find the X coordinate of the vertex. Now, if you want to, we could also get a formula for K, which K is gonna look like negative the discriminant over 4A. You can use that if you want to. That's not what I necessarily tell anyone to memorize because they're sort of like a simpler approach here. If I know what H is, then K is just F of H. I could just plug H into my quadratic function and compute that. And that would give me the K value. This one's a lot simpler here because after all, when you think of the quadratic formula, X equals negative B plus or minus the square root of B squared minus 4AC all over 2A. Some things to notice here is the discriminant is the thing inside of the square root here, but the vertex formula is bum, bum, bum, bum, bum, bum. It's right here, negative B over 2A. So it's actually hidden inside the quadratic formula to begin with. And so this is usually pretty easy to memorize because again, it's already in the quadratic formula. And then we can use that to find the vertex of many quadratic functions. And so I have these summarized here on the screen. H equals negative B over 2A. If you want to, you can do negative discriminant over 4A or more likely just F of H gives us the K value. So let's locate the vertex of this quadratic function without actually completing the square. So using the formula right here, H equals negative B over 2A. Since the function's already in the standard form, we get negative six over two times negative three. Simplifying, we get negative six over negative six and we get H equals one. That gives you the H value. Then to find K, we're just gonna compute F of one. We evaluate the function at one. That'll give you the corresponding H coordinate. That's negative three times one squared plus six times one plus one. That'll be negative three plus six plus one. Six take away three is a three plus one is a four. And so we see that the vertex, the vertex here, which is H comma K will be one comma four. And we were to find that we could, we could find that point without having to complete the square. That's really attractive in terms of finding the vertex here. People, many students really much like this. And so we can find the vertex very easily. We can also find the vertex form here because the coefficient in front is just going to be the same coefficient. So this is gonna look like negative three times X minus one squared plus four. So I can switch to the vertex form if I have the vertex and I have the leading coefficient. We'll always have the leading coefficient. We can find the vertex like this. So we can actually get the vertex form without completing the square just like we can solve quadratic equations without completing the square by.