>> In some problems involving quadratic functions, the primary goal is to find the maximum value that the function that can take on or the minimum value that the function can take on. Now, we know that if the graph is kind of shaped like this, where it turns downward and it's cupped downward that has a maximum value and that maximum value is at the vertex. On the other hand, if the graph is shaped like this, then it has a minimum value and that minimum value is at the vertex. Consider this problem. And so, let's find the maximum value of f of x equals this function. Now, we know that it has a maximum because the leading coefficient here is negative and so we know the graph has this kind of shape. And so it has a maximum value and the maximum value at the vertex, so let's find that maximum value. Now, the maximum value of the function means the maximum value of f of x equals y, the maximum y, the maximum vertical distance here. And we know that the vertex is found by first finding the value of x. The x component of the vertex is minus b over 2a, which in this case is, let's see, b is 4, 2a means 2 times negative 2, gee this would end up being 4 over negative 4, the negative of that gee this would be 1. Alright, now if x is 1 the y component would be, let's see, f of 1, which is negative 2 times 1 squared plus 4 times 1 plus 3. So, that y component would be negative 2 plus 4 plus 3 and this is 5, y then is 5. So, the maximum value that the function can take on is 5 and it's found as one of the components, the y component for the vertex. Now, this kind of situation arises in word problems like this. The height in feet designated s of a ball thrown straight up is given by s height in terms of t. You see this is the function notation is equal to this expression where t is time in seconds. Now, we want to find a couple of things. First of all, we want to find the time it takes the ball to reach its maximum height. We know that maximum height corresponds with the vertex of the graph of this relationship. And so, here's how we might proceed. Now, we're talking about time and height as our two variables. Now, the time here can be found as, let's see, now understand that we're looking for the vertex and the t part of the vertex is negative b over 2a. And let's see for our equation the b part is the 64 and the a part is negative 16. So, this becomes negative 64 over negative 32 or 2. So, t, the time is 2 seconds to reach that maximum height and that's what we're looking for, the time it takes the ball to reach its maximum height. Then it says, what is the maximum height? Well, the maximum height is the height after 2 seconds. Well, we can find that easily by plugging into the original equation and what we're finding is s of 2, you see. The height associated with 2 seconds and we find that by negative 16 times 2 squared plus 64 times 2 and that turns out to be 64 and that then would be 64 feet and this is 2 seconds. I should put the units of measure here in answering these questions.