Graphing quadratics: vertex form

We're asked to graph the equation y is equal to negative 2 times x minus 2 squared plus 5. So let me get by scratch pad out so we could think about this. So y is equal to negative 2 times x minus 2 squared plus 5. So one thing, when you see a quadratic or a parabola graph expressed in this way, the thing that might jump out at you is that this term right over here is always going to be positive because it's some quantity squared. Or I should say, it's always going to be non-negative. It could be equal to 0. So it's always going to be some quantity squared. And then we're multiplying it by a negative. So this whole quantity right over here is going to be non positive. It's always going to be less than or equal to 0. So this thing is always less than or equal to 0, the maximum value that y will take on is when this thing actually does equal 0. So the maximum value for y is at 5. The maximum value for y is 5. And when does that happen? Well, y hits 5 when this whole thing is 0. And when does this thing equal 0? Well, this whole thing equals 0 when x minus 2 is equal to 0. And x minus 2 is equal to 0 when x is equal to 2. So the point 2 comma 5 is the maximum point for this parabola. And it is actually going to be the vertex. So if we were to graph this, so the point 2 comma 5. So that's my y-axis. This is my x-axis. So this is 1, 2, 1, 2, 3, 4, 5. So this right here is the point 2 comma 5. This is a maximum point, it's a maximum point for this parabola. And now I want to find two more points so that I can really determine the parabola. Three points completely determine a parabola. So that's 1, the vertex, that's interesting. Now, what I'd like to do is just get two points that are equidistant from the vertex. And the easiest way to do that is to maybe figure out what happens when x is equal to 1 and when x is equal to 3. So I could make a table here actually, let me do that. So I care about x being equal to 1, 2, and 3, and what the corresponding y is. We already know that when x is equal to 2, y is equal to 5. 2 comma 5 is our vertex. When x is equal to 1, 1 minus 2 is negative 1, squared is just 1. So this thing is going to be negative 2 plus 5, so it's going to be 3. And when x is equal to 3, this is 3 minus 2, which is 1 squared is 1 times negative 2 is negative 2 plus 5 is 3 as well. So we have three points. We have the point 1 comma 3, the point 2 comma 5, and the point 3 comma 3 for this parabola. So let me go back to the exercise and actually put those three points in. And so we have the point 1 comma 3, we have the point 2 comma 5, and we have the point 3 comma 3. And we have now fully determined our parabola.