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## Quadratic vertex form

Current time:0:00Total duration:3:16

# Graphing quadratics: vertex form

CCSS Math: HSF.IF.C.7, HSF.IF.C.7a

## Video transcript

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.