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. Let's first look at a. Now we want to consider 2 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. Now if a is negative, it is instead going to open downward.
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, that tells us that we are undergoing a vertical stretch. Now if instead the absolute value of a is less than 1, we are undergoing a vertical compression. So that's a. Let's go ahead and move on to h. So our value for h, recognize that this is x minus h, so we need to be careful with our signs there. Our h 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 minus 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.
Now k is going to tell me my vertical shift. Here, I have (x−1)2−4. This minus 4 tells me my k value. So this tells me that I'm going to shift vertically by negative 4 units. 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. 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 (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 upward. Now when my parabola opens up, that tells me that I am definitely dealing with a minimum here.
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 is equal to h, so just that point in our vertex. So here, my axis of symmetry is simply going to be x equals 1. 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) is equal to 0. We know that we need to be using the square root property. So let's go ahead and solve this using the square root property. (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 minus 1 is equal to plus or minus the square root of 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 is equal to 3 and -1. So here I have my 2 x intercepts, 3 and -1.
Moving on to step 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. Now -1 squared is 1 minus 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. First, starting with my vertex (1, -4), let's go ahead and plot that. 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, 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. 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.