Hey, everyone. By now, we've graphed some basic functions and we've even transformed those graphs by maybe shifting them some number of h units to the right or k units up or maybe even stretching it by a factor of a. And the vertex form of a quadratic function takes all of those possible transformations that we could do to the square function f(x)=x2 and puts them all together in one equation that is then going to tell us exactly how to graph our quadratic function. So I'm going to walk you through step by step how to use this vertex form in order to easily graph our quadratic function using a bunch of stuff we already know. Seriously, I'm not going to teach you anything new here. I'm just going to take a bunch of stuff we already know and bring it together. So let's go ahead and get started here and take a look at our vertex form. So here we have f(x)=a⋅(x-h)2+k and we have our a, h, and k that all represent transformations that are happening to our x2 function. So let's first look at a. Now we want to consider two things when looking at a, both the sign and the value of it. So first, considering the sign, if a is positive, that tells us that our parabola is going to open upward. So like our square function, it's just going to open straight up. Now if it has a negative, if my a is negative, it is instead going to open down. So I'm going to have a parabola that is opening downward like this. Now the other thing we want to consider when looking at a is the actual value of it. So we're only considering the absolute value because we already took care of the sign. So if the absolute value of a is greater than 1 so some number like 5 or 20 or anything greater than 1, that tells us that we are undergoing a vertical stretch. So if I take my x2 function and I stretch it upward it's going to look something like this, much skinnier because I have stretched it. Now if instead the absolute value of a is less than 1, so something like 1/2 or 2/3 or any number less than 1, I am instead undergoing a vertical compression. So instead of stretching my x2 function, I've instead compressed it and now it will look a little bit wider here. So that's a. Let's go ahead and move on to h. So our value for h, recognize that this is x-h, so we need to be careful with our signs there. X-h. Our h is going to tell us what we are horizontally shifting by. So h, horizontal, it's a good way to remember it, is going to tell us our horizontal shift by some number h units. So if I have a quadratic function in vertex form, if I have my value h here, this is x-1, so one is my value for h. That tells me that I'm just going to shift 1 unit over and now my parabola is here. So it's shifted by some number h to the right. Now k is going to tell me my vertical shift. So instead of my horizontal, now it's dealing with a vertical shift by some number k units. Now here, I have x-1 2-4. So this minus 4 tells me my k value. So this tells me that I'm going to shift vertically by negative 4 units, paying attention to that sign there. So I'm just taking my parabola and I'm shifting it vertically by some number of units here by negative 4. So we've looked at all of these different transformations. Let's go ahead and go through, step by step, how to graph a function in vertex form. So the first thing that we want to consider is our actual vertex. It's in vertex form. Let's go ahead and find our vertex. So our vertex is actually always simply going to be the ordered pair h, k. So here, we've already identified what our h and our k are, so we know that our vertex is simply going to be 1, -4. Now the other thing we want to consider here is whether our vertex is going to be a minimum or a maximum point, which we can do by looking at our value for a. Here, I don't actually have a number for a, which tells me that I'm dealing with an invisible positive one here because there's nothing else going on. So I know that a is positive, which tells me that my parabola is going to open up. Now when my parabola opens up, that tells me that I am definitely dealing with a minimum here. So I know that that vertex will be a minimum. Now the next thing we want to consider for step 2 is our axis of symmetry. Now our axis of symmetry is actually always going to be the line x=h, so just that point in our vertex. So here, my axis of symmetry is simply going to be x=1 and I'm done. I can move on to step 3. Now step 3 is to find the x-intercepts. Now this is where we're going to have to do a little bit more math, but don't worry, it's using something that we already know, how to solve a quadratic equation. So here we're going to want to solve the equation f(x)=0. So if I take my function here and I set it equal to 0 and I think about how to solve that, you might recall that we have a bunch of different methods. So if I consider everything that's happening here and I'm looking for a clue of how I should actually solve this, I can see that this has the form x+somenumbers2=aconstant, or rather I can make it in that form. So I know that I need to be using the square root property. So let's go ahead and solve this using the square root property. So I have x-1 2-4. I'm going to go ahead and move that 4 over to the other side, canceling it out, leaving me with x-1 2=4. Now using the square root property, I'm going to go ahead and square root both sides, leaving me with x-1=±4, which is just 2. So from here, I can go ahead and move my one over by simply adding 1 to both sides, leaving me with just x=±2+1. So now I can just go ahead and split it into my 2 possible answers, positive 2 + 1 and negative 2 + 1. So going ahead and solving those, 2+1 I know is 3 and negative 2+1 is simply -1. So here I have my 2 x-intercepts, 3 and -1, and I've completed step number 3. My x-intercepts are 3 and -1. Moving on to step number 4, which is to find our y-intercepts. Now we're going to have to do a little more math here, but all we're doing is computing f(0), which means I'm just plugging in 0 for x.\ So down here, I have f(0)=0-1 2-4. So I've really just taken my vertex form and plugged in a 0 for x. So let's go ahead and calculate that. Now 0 - 1 is simply -1, so this is just -1 2-4. Now -1 2 is 1 - 4. This will leave me with -3. So this is my y-intercept, -3. Okay. So I found a bunch of points here. Now my last step to actually graph this is going to be to plot all of these points and then connect them with a smooth curve. We already know what the shape of our quadratic function should be, a parabola, so we know what it should look like. Let's go ahead and plot all these points. So first starting with my vertex 1, -4, let's go ahead and plot that. So 1, -4, right here, and then my axis of symmetry, which we're going to represent with a dotted line, is simply the line x=1 that goes all the way through my vertex there. And then my x-intercepts, which are 3 and -1, plotting those on my x-axis, 3 and -1, and then finally my y-intercept at -3. Okay,\ now connecting this with a smooth curve, I know it needs to be in the shape of a parabola and here is my parabola. I want to put arrows on the end to represent that it doesn't just stop there, it keeps going. And looking at this graph so we're done, we've completely graphed it. I know that I have my x2 function here. You'll notice that this is just the same thing but shifted one over and down 4 as was shown in our vertex form. So even though we can always just sketch it from the beginning we want to make sure and calculate these extra points just so that we can draw it accurately. So that's all you need to know in order to graph a function from vertex form, let's get some practice.