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. Getting started 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+2x. Now remember only your first two terms and don't forget to tack that constant on the end. Step 1 is done here.

Now I can go ahead and move on to step 2, which is to both add and subtract b over 2a squared inside of my parenthesis. So let's go ahead and compute b over 2a first. So b is -4, 2 times a, which is -2. This is -4 over -4, which is really just 1. Now if we take 1 and we square it, that just leaves us with 1. So we're going to add and subtract 1 here. Now -2x2+2x, and then a +1-1 inside of those parentheses, and then with my constant on the end. So that's the first part of step 2 is done.

We can go ahead and move on to the second part of step 2 which is going to be to move my subtraction times a outside of my parentheses. So here, this negative one and then times this negative 2, I want to move it to the outside here. So this becomes negative one times negative 2, and that cancels it inside of my parentheses. So 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 negative one times negative 2 is going to give me positive 2. So looking at all of this, I have finished step 2.

I can go ahead and move on to step 3, which is to factor 2x+b2a squared. So b over 2a, we already calculated, and we know that it's just 1. So this just becomes -2(x+1)2. 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 am done. I am 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 minus h. So this is x minus negative one. 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? Well, 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 is equal to -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) is equal to 0. So 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 come down here and solve this.

So I am going to go ahead and move my 8 over to the other side. Remember, we're going to be using the square root property with our vertex form here. So 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=4. Now I can go ahead and apply the square root property by square rooting 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 one 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, negative one and negative 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(1)2+8. One squared is just 1, so this is -2 times one plus 8. -2 times one is, of course, just -2 plus 8. And then finally, our last step, -2 +8 gives me positive 6. So 6 is going to be our y intercept here, filling that in on my table.

Now we are done. We can go ahead and plot and connect everything with a smooth curve. So coming right back up to my graph here, let's start by graphing symmetry right through that vertex. Then our x intercepts at 1 and at -3. So positive 1, negative 3, and then my y intercept at 6. So right up here. Okay. Now I can go ahead and connect all of this with a smooth curve. We know that this is a parabola facing downwards because we identified our maximum point. So here is my parabola and now I can go ahead and look at all of this remaining information and look at my graph to fill that in.

Of course, our domain still the same, still negative infinity to infinity, nothing changed there. And then our range since we're dealing with a maximum point we know goes from negative infinity up into that y max, which in this case is 8, including that 8 in a square bracket, of course. Then finally, we have increasing and decreasing. Increasing is for what x values is that going up, which from negative infinity until I reach that vertex point at x equals -1, it is increasing. So from negative infinity to -1, it is increasing. And then on the other side from -1 to positive infinity, it is falling back down, so it is decreasing. So decreasing from -1 to positive infinity, those increasing and decreasing intervals always separated by that vertex. So that's all for this one. I'll see you in the next one.