Hey, everyone. We just learned that whenever we have a quadratic function in vertex form, we can quickly find our vertex and our x and y intercepts in order to easily graph it. But what if I'm given a quadratic function in standard form instead? How am I supposed to find my vertex if I don't know what h and k are? Well, the good news is that we can actually just take our standard form function and put it right back in vertex form by looking to a familiar friend: completing the square. So here, I'm going to show you how you can take your standard form function, rewrite it, and then add some number and subtract that same number in order to get back to vertex form by using something that you already know how to do: completing the square. So let's go ahead and get started here.
Now, something important to consider is that we're now working with a function instead of an equation, so our steps are going to change slightly and you're going to need to pay attention and be careful with some of the changes that are happening. So let's go ahead and look at our example here. We have \( f(x) = x^2 + 6x + 7 \), so we want to take this standard form and put it into vertex form. Let's go ahead and identify a, b, and c here before we get started. So here I just have \( x^2 \). I just have an invisible one in front of that \( x \) so \( a = 1 \), \( b = 6 \), and \( c = \text{this positive } 7 \). Let's go ahead and get started with step 1, which is going to be to factor out \( a \) of my first two terms. So since here \( a \) is just 1, really if I factor out 1, I'm just left with \( x^2 + 6x \) because that \( 6x \) isn't going to change if factoring out a 1. So this is going to leave me with the form \( a(x^2 + \frac{b}{a}x + c) \) because I've simply factored out an \( a \) there. So I've completed step number 1.
Let's move on to step 2, which is going to be to add and subtract \(( \frac{b}{2a})^2 \) inside of our parentheses. So this \( \frac{b}{2a} \) is going to show back up. Let's go ahead and calculate that here. So \( \frac{b}{2a} \), I know that \( b \) here is just 6, \( 2 \times a \) is just 1, and this is equal to \( \frac{6}{2} \) which is just 3. Now that's my \( \frac{b}{2a} \). I want to make sure and square it here before I add it. That's going to give me 9. So, here, I want to add 9 and subtract 9 inside of my parentheses. Now I can do that because I'm really just adding nothing if I add 9 and then take it right back away. So I've completed that first part of step 2. Let's look at the second part. Now for the second part of step 2, I want to move my subtraction times \( a \) to the outside of my parentheses. So I want to take this negative 9, and I want to move it. But since it's inside a parenthesis being multiplied by a factor of \( a \) I need to make sure I account for that. Now here \( a \) is just 1 so all I'm doing is taking minus 9 times 1 to the outside in order to cancel it on the inside there. Now that one isn't going to do anything this is really just minus 9 but remember that if your \( a \) isn't 1 it's going to get a little bit more complicated. So we've completed step 2.
Let's go ahead and rewrite this to make it a little bit cleaner. So I still just have my 1 on the outside, but that's not doing anything. So I just have \( x^2 + 6x + 9 \). I moved that negative 9 away. So this is \( +7-9 \) on the outside. Okay. So moving on to step 3. Now I want to factor this to \( (x + \text{this } \frac{b}{2a})^2 \) and then simplify it. So this \( \frac{b}{2a} \) we've seen it before and we know that it's just this 3 right here so this simply factors to \( (x+3)^2 \) because it is a perfect square trinomial as we know happens with completing the square. So we want to take this \( +7 -9 \) into account and I add my negative 2 to the outside of those parentheses leaving me with \( (x + 3)^2 - 2 \) and I've completed step 3.
Now, this form of an equation or form of a function might look familiar to you because it is now in vertex form and we can simply graph from vertex form. Now, I'm going to leave this for practice that we'll do in just a little bit, but let's take one more look at our function here. \( (x + 3)^2 - 2 \). Now, I can easily extract \( h \) and \( k \) and graph it from there. That's all you need to know to take standard form back to vertex form. And now you can graph it just like you know how. Let's go ahead and get some practice.