Hey everyone. Let's put everything that we just learned together in this example. So we want to graph the given quadratic function and then identify all of the possible information about our parabola. And the function I have here is f(x)=-2x2-4x+6. But seeing that this function is in standard form, I want to go ahead and convert it to vertex form by completing the square so that it's much easier for us to graph. So let's go ahead and do that here.
The first thing I want to do is identify a, b, and c before I get started with my steps. So a here is -2, b is -4, and c is 6. Starting with step 1, we want to factor a out of just the first two terms. We just identified a as -2. So, if I pull -2 out of my function out of those first two terms, that leaves me with x2 and then a x. Remember, only your first two terms and do not forget to tack that constant on the end. Step 1 is done here.
Now I can move on to step 2, which is to both add and subtract b2×a squared inside my parentheses. So, let's compute b2×a first. b is -4. 2×a, which is -2. This is -4 over -4, which is really just 1. Now, if we square 1, that just leaves us with 1. So we're going to add and subtract 1 here. -2x2+2x, and then +1-1 inside those parentheses, with my constant on the end. That's the first part of step 2 done. We can move on to the second part of step 2, which is going to be to move my subtraction×a outside of my parentheses. So, here, this -1 and then times this -2, I want to move it to the outside here. So this becomes -1×-2, and that cancels it inside of my parentheses. Let's go ahead and rewrite this to get it a little more compact. So this -2 and then x2+2x+1. I moved my subtraction outside, so this just becomes +6, and then -1×-2 is going to give me positive 2. So looking at all of this, I have finished step 2. I can move on to step 3, which is to factor 2x+b2×a squared. We already calculated b2×a, and we know that it's just 1. So this just becomes -2. And then inside my parentheses, x+1. That's what it factors down to. And then combining this 6 and 2 to simplify is going to give me a +8 on the end. And looking at this, I'm done. I'm in vertex form, and this is the function that I am now going to graph.
So I can take my function here, and step 3 is done. I can move on to step 4 and go ahead and graph it right down here. So copying my function I have f(x)=-2×(x+1)2+8. And we are ready to go. Let's get graphing.
Starting with step 1, we want to identify our vertex here, which is just h,k. Now, looking at this, since I have x+1, I know that this plus is really minus a negative because, remember, we have x-h. So this is x--1. So our value for h is -1 and then k is just positive 8. So that's my vertex, -1,8.
Is this a minimum or a maximum point? Looking at my function here, I have this negative on the front, which tells me my parabola is opening downwards, which means my vertex is at the top. It's at a maximum. Then identifying our axis of symmetry here, x is equal to h, we just identified h as -1. So this is simply the line x=-1. Now moving on to steps 3 and 4, we know we're going to do some calculations, which again I'm going to do right down here. Let's go ahead and start with step 3 and solve f(x)=0.
Setting up our equation down here, my function is -2×(x+1)2+8, and of course, equal to 0. So, let's go ahead and solve this. I'm going to move my 8 over to the other side. Remember, we're going to be using the square root property with our vertex form here. Subtracting 8 from both sides, I have -2×(x+1)2=-8. Then dividing both sides by -2 to isolate that squared expression, I have (x+1)2=-8÷-2, which is positive 4. Now I can go ahead and apply the square root property by squaring both sides, leaving me with x+1=±4, which we know is 2. Now from here, I can go ahead and isolate x by moving my 1 over to the other side by subtracting, canceling out, and leaving me with x=-1±2. Now here is, of course, where we want to split it into our 2 possible answers. So this is really -1+2 and -1-2. Now -1+2 is going to give me a positive 1, and the -1-2 is going to give me a -3. So these are my 2 x intercepts here, 1 and -3, which I can go ahead and fill in on my table up here, -1 and -3.
Moving on to our y intercept, we can go ahead and compute f(0) by plugging 0 into our function. So this is -2×(0+1)2+8. Now simplifying this, I have -2×12+8. One squared is just 1, so this is -2×1+8. Negative 2 times 1 is, of course, just