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 of a parabola by competing the square, so whenever we're completing the square, what we have to do is first get our coefficient on our xÂ² to be 1.
So in this particular problem I have to factor out a 1/2 .At the same time I want to somehow isolate my constant term from the xs, so we can either subtract it over the other side or just bring it outside the parenthesis. For this one I tend to just bring it to the outside, you can bring it to the other side as well it doesn't really matter.
So what we end up with is we factor out a 1/2. Where people tend to go right is when we factor out a fraction trying to figure out what this coefficient is going to be. And the easiest way is to just check, whatever goes here times 1/2 has to equal 3. What times 1 half is equal to 3? That's 6, you can always check your work. For some reason people always wonder if 3/2 or something like that, but if you have 3/2, you multiply it out, you wouldn't end up with the same thing. And then our +4 gets moved outside of our parenthesis.
We now have to complete the square, so we can have a different color and we end up with 1/2 and we need to go figure out what goes here it's always half of our middle term, so 6 divided by 2 is 3, so we end up with +3 here and that squared ends up going inside of our parenthesis.
So we've added 9 inside of our parenthesis, but before we subtract 9 to keep everything balanced, we have to remember that we had this 1/2 on the outside. So when we added 9 inside of our parenthesis, but we really just added 9/2 as a whole because of this 1 half, so we added 9/2 which means we have to subtract 9/2 to keep it balanced.
If you brought the 4 over to the other side, you would add 4/2 to this side to keep everything balanced, if you added to one side, you have to add it to the other because I'm adding to one side I want to subtract to keep it balanced. So then all we have here is 4 minus 9/2, 4 minus 4 and 1/2, so this just ends up with a minus one half and our f(x) is equal to.
So don't let that fraction out in front scare you, the main thing you want to be careful of is when you factored out, you get the right coefficient on your x term, other than that just completing the square as you know it making sure to distribute your coefficient through once you've completed the square.
Unit
Equations of Lines, Parabolas and Circles