Graph Quadratic Inequalities - Example 2
Solve Quadratic Equations Algebraically - Example 2
Solve Quadratic Equationsby Factoring - Example 2
Solve Quadratic Equationsby Graphing - Example 2
Solving Quadratic Equationsby Completingthe Square - Example 2
Liberty University
Graph Quadratic Functions - Example 2


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Video Transcript

G of X equals X squared minus six X plus three. Here we're trying Thio graph this quadratic function. Well, let's start with a few key points that we can go ahead and find those points because otherwise we don't want to continually just put numbers in place for X. So the first thing we can go ahead and we could find is our Y intercept. We get our Y intercept from our seater, and in this case, that's going to be three. So that tells me when x zero my wives wannabe three. So that's gonna be our first one we're gonna use. The second thing is, we're gonna go ahead and Bonner, Axes of symmetry and our Vertex. So this is gonna be our axis of symmetry is when X equals Arbet term divided by negative to a RB terminus. Negative six. We have negative two times one. So that's negative. Six over negative too. So that means that our axes of symmetry is at three. And if I put three in place of X in my equation, so they'd be three squared minus six times three plus three, I would get negative six. So let's start with these two points, and then we'll go from there to discover some more points. So let's go ahead and start at zero three. That's gonna be my wife Intercept. And then we said, three negative six. And this right here is also going to be what we call my axes of symmetry. So at this point, this is going to be kind of the center of the parabola. So I go ahead and I can kind of start my parabola to this point so I can see that I'm really want to get to come on around in this direction right here. But that means that I need some larger numbers for my ex values. Well, we've got our y intercept here, and I can tell that this right here is where X equals zero. And right here is where X equals three. Well, there's a difference of three between those two spots. So let's come back to our table and let's do another point of X and let's go ahead. And we would let's create another difference of three. So three units to the right of X would be six. So let's find out what that would be that would be six squared minus six, Tom six plus three. So I'd be 36 minus 36 plus three, which would be zero plus three. So that would be Excuse me, That would be three. So that would be a perfect point to you. So that would be 3456 So right there would be my 63 And I could curve this back around to that point. So there, you see, that is how I can find my parabola using just three points.

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