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!
To unlock all 5,300 videos, start your free trial.
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!
Graphing a quadratic equation is easy when we have our equation in vertex form. The problem is that we don't always get our equation in such a form, we often get it just in standard form which we don't know how to interpret.
So what we're going to have to do is turn this equation into vertex form and we do that by completing the square. The process is exactly the same, remember when we did it for solving an equation except instead of dealing with a zero over on the side we're just dealing with a y. The y can stay there and everything else behaves just the same.
First step was to get our xs by themselves, so we subtract this 6 over giving us y minus 6 is equal to xÂ² minus 4x. We then want to figure out what we need to add to this right side in order to make it a perfect square, x plus or minus something and what ends up going here is this middle term divided by 2. so we have -4 divided by 2, this just turns into -2. When we square this out, we would end up with a +4 over here, so what we've really done by rewriting this as a square, we've added 4 to the right side to keep everything balanced we have to add 4 to the other side as well.
So simplifying what I have now, I have y, -6 plus 4 is -2 is equal to x minus 2 quantity squared. Solving this into vertex form, add the 2 over, y is equal to x minus 2 quantity squared plus 2, a lot of 2s in this problem.
So what we know is by completing the square, how to find the vertex. The plus 2 on the outside makes it go up 2, the minus 2 on the inside makes it go over to the right 2 and we don't have any stretches or anything like that. So what we've done is complete the square to find the vertex of this graph.
I do to want to note that there is another way to complete the square for this particular problem. And what we do in this case instead of bringing this 6 around to the other side, we just isolate it by itself on the same side, so what we have is xÂ² minus 4x and then just draw my parenthesis and throw the +6 all the way on the outside.
We still want to complete the square inside the parenthesis, so we need to add 4 here. Instead of adding 4 to the other side, I just want to keep everything balanced by subtracting 4 from the same side. We add 4, we subtract 4 everything cancels and works out to be the same, then by factoring we end up with x minus 2Â², +6 plus 4 is +2 we end up with the same answer.
It's a little excessive to do it both ways because you always are going to end up with the same way, but just showing two different ways of completing the square to find the vertex of a parabola.
Unit
Quadratic Equations and Inequalities