Easy to Hard Find the Vertex of a Quadratic

in this video I want to work through a progression of finding the vertex of a quadratic I want to start with some basic problems that are in vertex form and then slowly work our way up to completing the square to rewrite the equation from a standard form to vertex form so we can identify the vertex but it's very important to understand if we want everything to be in vertex form and if vertex form is how we find the vertex then what is vertex form so vertex form of a quadratic is a times x minus h quantity squared plus K and just remember what we're trying to find is basically either the max or the Min of a quadratic okay so if we have like a nice little paraba here's an upside down one you can see it has a max point right here that is what we call the vertex and that vertex is going to be in the form of H comma K now we're not going to be working with graphs but what I want to do is help you identify how you can find that point when you're given an equation so let's go ahead and work on an example here and what if I have a function here f of x and this one's going to be a x squared minus 2. right so what is the vertex in this case so just remember the vertex is going to be your H represents your y or your x squared I'm sorry and K represents the uh the y coordinate right so these are going to be your transformations of your graph right because typically if we look at a quadratic graph you're going to have your vertex is going to be at 0 0 right and it's going to open up okay so in this case you can see my vertex is at 0 comma zero or the origin so what's happening with H and K is that's going to be shifting My Graph left to right or up or down so it's important to recognize which transformation is which well when you have a transformation inside of your parentheses or inside of your function that's going to be a horizontal shift and when you have this transformation of this adding K outside of the function that is going to be your vertical shift so you can see in this case that this is a vertical shift down two units or also that my K is equal to a negative two so the vertex in this example is just going to be a zero because I didn't do anything horizontally comma a negative 2. now what about if we do something on the inside so for instance what if I had an f of x is equal to a x minus two quantity squared right so obviously you know that it's not the same transformation right the negative 2 shifted the graph down to units that's what was my K now a lot of students will say oh H is negative 2 right so therefore this vertex is negative two comma zero don't do that okay so what's really really important about this is is to understand we have to write it in this format so what I can do is I can rewrite this using parentheses I can rewrite this as x minus a negative 2. or x minus yeah negative 2 because what's x minus negative 2 No No's I'm sorry about that why am I doing that that's for the plus ones why am I doing it like that what you want to do is rewrite it like this x minus a 2 right but notice when I put these in parentheses notice how my 2 is actually positive right it's x minus 2. we're looking for our x minus H we're looking for what H is right so you can see this is x minus 2. so H is actually equal to a 2. so my vertex in this case is going to be a 2 and that's a really bad 2. comma zero why is it zero in this case because again I'm not adding or subtracting anything outside of the function so again just as a quick review let's just kind of do one now where we're going to have both of them in the same set so in this case I could have a X plus one quantity squared and then we could do a plus three okay so this is what I was trying to kind of explain to you in that last one whereas you can now rewrite this I got if I got to write this as x minus I'm not going to use a positive um one in this case because that would be that would give me the same result but in this case what I need to do is rewrite this as an x minus a negative 1 quantity squared and then plus 3. all right and that was my point that I was trying to make or I mistakenly made in the previous example because minus the negative is the same thing as adding right so now hopefully you can see here that this vertex is actually going to be a negative one and then we can just take K exactly as it is as a 3. okay so what that's all that's doing is that's just shifting the graph one unit to the right and then three units up so that is going to be your new vertex now these are all fine and dandy when we have something in vertex form and that's why we like vertex form because once you get a little bit of practice with this you can just instantly look at an equation and you can be able to identify what the vertex is but sometimes we're going to get equations that are not that obvious so for instance like what if I had an equation let's just do a y equals and in this case I have an x squared plus 4X right so now it's not so obvious what's your a and your K are or your H and your K so what we need to do now is we need to complete the square and the process of completing the square is basically taking a quite is basically taking a quadratic equation in standard form ax squared plus BX plus C and rewriting it into vertex form which is y is equal to a times x minus h quantity squared plus K okay so there's a couple other variables we could throw in there but we'll just keep this as the very basic but that is our goal we're trying to go from this to this so how do we do that well the way that we do that is we're going to create a binomial squared because what I want you to recognize is x minus H is a binomial x minus H squared is a binomial squared right so to create a binomial squared we have to create a perfect to our trinomial how do you create a perfect square trinomial you find the value C that makes a perfect square trinomial how do you find that value C well all you simply do is you're going to take to do this you're going to take C is equal to a b divided by 2. quantity squared so let's go and see this in practice here what I'm going to do is I'm going to take my middle term which is my B divided by 2 and square it so 4 divided by 2 is going to be a 2. 2 2 squared is equal to a 4. all right so now what I'm going to do is I'm going to add that value to this quadratic so I'll say Y is equal to an x squared plus a 4X plus 4. all right now with the main thing though you have to do though isn't in addition to like adding four you just can't randomly add 4 to an equation without subtracting the 4 right because we've got to keep these equations balanced either add it to both sides or add and subtract it on the same side now what's important about this though is I now just created a perfect square trinomial right so now I have y is equal to an X let's see this can be factored down into an X plus two quantity squared minus four and guess what now I can quickly identify my vertex my vertex in this case is going to be a negative 2 negative 4. see all that practice we did up there see how that made everything go by so much quicker right now sometimes we get some quadratics that are actually pretty cool because sometimes we have a quadratic that's actually already in that form like in this case I have an x squared minus an 8X Plus 16. right so sometimes students will start completing the square but you also want you to think about this this is already a perfect square trinomial right because if you take your B right negative eight divided by two that's just four four squared is 16. so what I can do is I can automatically already Factor this down so this one would be a I can Factor this down into an x minus 4 quantity squared and guess what that's now in standard form the vertex in this case is 4 comma zero kind of one of those that we you know already kind of looked at but sometimes guys it does take a little bit of work so in this example um was that we had a Y is equal to a x squared minus a2x plus 1. all right so in this case um no sorry 2x plus 2. that's a perfect square trinomial I don't want to do another one of those all right so this is um this is not a perfect for our channel so in this case we got to go back through the process again so what was that process you're going to take your middle term divided by 2 and square it negative 2 divided by 2 is negative 1. negative 1 squared is going to be a positive one now you're going to take that term and you're going to add it to your first two terms so remember up here I just kind of like added it to the equation right well now we have an extra number here so we got to be kind of like careful how we're doing this so I'm going to use um some different colors so we can kind of represent this so Y is equal to an x squared minus a2x now we're going to do is I'm going to say all right let's do plus one minus one okay so we're going to add one and subtract one like we did before now I'm going to group though these two and then don't forget we have a 2 over here right there's still that 2 there so that 2 needs to make sure you get transformed make sure you add and subtract and then we're going to group these um these first two that you the the first three terms and why do we want to group those because that's going to perform our perfect square trinomial so Y is equal to a x minus a 1 quantity squared negative 1 plus 2 is going to be a positive one now we have a Vertex guys of one comma one and that is the progression of finding the vertex of a quadratic I hope this video is helpful for you if it was I look forward to seeing you in the next video