Unit
Quadratic Equations and Inequalities
University of Michigan
Runs his own tutoring company
Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
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University of Michigan
Runs his own tutoring company
Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
Finding the vertex by completing the square; so what we have behind me is a parabola in standard form and I want to find the vertex of it and the only way I can do that is by completing the square.
So rules for completing the square are to isolate our xs together and to make sure that our coefficient on our x² term is 1. So what we have to do is isolate the xs from this 3 and there's two ways of doing that. We can either bring the 3 over to the other side, or just bring it outside of our parenthesis. I'm going to do the latter, I'm just going to bring it to the outside, but if you want to bring it around, you can do that as well.
The other thing we have to do is get the coefficient on our x² to be a 1, so what I have to do is then factor out a -2 from everything that has an x. Let's go ahead and do that one.
So what we end up with is f(x) stays the same and we end up with -2, x² and we're taking a -2 out of 8 so this turns to -4x and my +3 I just brought that side of the parenthesis.
So we now what to complete the square, so we want to figure out what goes in here to make this a perfect square, divide the middle term by 2, -4 divide by 2 is -2, and square that so we are adding 4 inside of our parenthesis.
Now be careful, we're adding 4 inside of the parenthesis, but what we're actually doing is that plus 4 is getting multiplied by -2, so we've actually subtracted 8 by putting that 4 in here, subtracted 8 from that part, so in order to keep it balanced, I have to add 8 outside the parenthesis. So we add 8 out here, so what we actually end up with is +11 on the outside and we have f(x) equals.
I was able to find my vertex, my vertex is going to be (2,11). The graph is going to be shifted up 11 to the right 2 and this -2 is going to tell me that my graph is going to be a little bit steeper and actually flipped upside down because of that negative.
So finding the vertex of an equation by completing the square; always make sure you have your x² coefficient to be 1 and make sure that when you add or subtract something inside the parenthesis you have to distribute whatever you factored out.