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 my scratch pad out so we can think about this so it's y is equal to negative 2 times X minus 2 squared plus 5 so one thing when you see a quadratic or a parabola expressed in this way the thing that might jump out at you is that this term right over here 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 zero so it's always going to be some quantity squared and then we're multiplying it by a negative so this whole quantity this whole quantity right over here is going to be non positive it's always going to be less than or equal to zero less than or equal to zero so if this thing is always less than or equal to zero the maximum value that this that Y will take on is when this thing actually does equal zero so the maximum value for Y is at five so Y the maximum value for Y is five and when does that happen well Y hits five when this whole thing is zero and when does this thing equal zero well this whole thing equals zero when X minus two is equal to zero and X minus 2 is equal to zero when X is equal to two so the point 2 comma five 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 one two one two three four five so this right over here is a point 2 comma five 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 determinate of the completely determined parabola so that's one the vertex that's interesting now what I like to do is just get two points that are equidistant from the vertex and the easiest way to do that is to maybe to figure out what happens when X is equal to when X is equal to 1 and when X is equal to 3 so I can make a table here actually let me do that so I care about X being equal to one two and three and what the corresponding y is we already know that when X is equal to two 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 3 points we have the point 1 comma 3 1 comma 1 comma 3 the point 2 comma 5 and the point 3 3 comma 3 for this parabola so let me go back to the exercise and actually put those put those three points in so we have the point 1 comma 3 through the point 2 comma 5 and we have the point 3 comma 3 and we have now fully determined our parabola