# Completing the square for vertex form | Quadratic equations | Algebra I | Khan Academy

Use completing the square to

write the quadratic equation y is equal to negative 3x squared,

plus 24x, minus 27 in vertex form, and then

identify the vertex. So we’ll see what vertex form

is, but we essentially complete the square, and we

generate the function, or we algebraically manipulate it so

it’s in the form y is equal to A times x minus B

squared, plus C. We want to get the equation

into this form right here. This is vertex form

right there. And once you have it in vertex

form, you’ll see that you can identify the x value of the

vertex as what value will make this expression equal to 0. So in this case it would be B. And the y value of the vertex,

if this is equal to 0, then the y value is just

going to be C. And we’re going to see that. We’re going to understand why

that is the vertex, why this vertex form is useful. So let’s try to manipulate

this equation to get it into that form. So if we just rewrite it, the

first thing that immediately jumps out at me, at least, is

that all of these numbers are divisible by negative 3. And I just always find it

easier to manipulate an equation if I have a 1

coefficient out in front of the x squared. So let’s just factor

out a negative 3 right from the get-go. So we can rewrite this as y is

equal to negative 3 times x squared, minus 8x– 24 divided

by negative 3 is negative 8– plus 9. Negative 27 divided by negative

3 is positive 9. Let me actually write the

positive 9 out here. You’re going to see in a second

why I’m doing that. Now, we want to be able to

express part of this expression as a perfect

square. That’s what vertex

form does for us. We want to be able to express

part of this expression as a perfect square. Now how can we do that? Well, we have an x

squared minus 8x. So if we had a positive 16

here– because, well, just think about it this way, if we

had negative 8, you divide it by 2, you get negative 4. You square that, it’s

positive 16. So if you had a positive

16 here, this would be a perfect square. This would be x minus

4 squared. But you can’t just willy-nilly

add a 16 there, you would either have to add a similar

amount to the other side, and you would have to scale it by

the negative 3 and all of that, or, you can just subtract

a 16 right here. I haven’t changed

the expression. I’m adding a 16, subtracting

a 16. I’ve added a 0. I haven’t it changed it. But what it allows me to do is

express this part of the equation as a perfect square. That right there is

x minus 4 squared. And if you’re confused, how

did I know it was 16? Just think, I took negative

8, I divided by 2, I got negative 4. And I squared negative 4. This is negative 4 squared

right there. And then I have to subtract that

same amount so I don’t change the equation. So that part is x

minus 4 squared. And then we still have this

negative 3 hanging out there. And then we have negative 16

plus 9, which is negative 7. So we’re almost there. We have y equal to negative 3

times this whole thing, not quite there. To get it there, we just

multiply negative 3. We distribute the negative 3 on

to both of these terms. So we get y is equal to negative

3 times x, minus 4 squared. And negative 3 times negative

7 is positive 21. So we have it in our vertex

form, we’re done with that. And if you want to think about

what the vertex is, I told you how to do it. You say, well, what’s the

x value that makes this equal to 0? Well, in order for this term to

be 0, x minus 4 has to be equal to 0. x minus 4 has to be equal to 0,

or add 4 to both sides. x has to be equal to 4. And if x is equal to 4, this is

0, this whole thing becomes 0, then y is equal to 21. So the vertex of this parabola–

I’ll just do a quick graph right here– the

vertex of this parabola occurs at the point 4, 21. So I’ll draw it like this. Occurs at the point. If this is the point 4, if this

right here is the– so this is the y-axis, that’s

the x-axis– so this is the point 4, 21. Now, that’s either going to be

the minimum or the maximum point in our parabola, and to

think about whether it’s the minimum or maximum point, think

about what happens. Let’s explore this equation

a little bit. This thing, this x minus 4

squared is always greater than or equal to 0. Right? At worst it could be 0, but

you’re taking a square, so it’s going to be a non-negative

number. But when you take a non-negative

number, and then you multiply it by negative 3,

that guarantees that this whole thing is going to be

less than or equal to 0. So the best, the highest, value

that this function can attain, is when this

expression right here is equal to 0. And this expression is equal

to 0 when x is equal to 4 and y is 21. So this is the highest value

that the function can attain. It can only go down

from there. Because if you shift the x

around 4, then this expression right here will become, well,

it’ll become non-zero. When you square it, it’ll

become positive. When you multiply it

by negative 3, it’ll become negative. So you’re going to take a

negative number plus 21, it’ll be less than 21, so your

parabola is going to look like this. Your parabola is going

to look like that. And that’s why vertex

form is useful. You break it up into the part

of the equation that changes in value, and say,

well, what’s its maximum value attained? That’s the vertex. That happens when

x is equal to 4. And you know its y value. And because you have a negative

coefficient out here that’s a negative 3, you know

that it’s going to be a downward opening graph. If that was a positive 3, then

this thing would be, at minimum, 0 and it would be

an upward opening graph.

why did you divide 8 by 2??

I would use h and k instead of b and c because in completing the square ax^2+bx+c the b and c don’t equal b and c in vertex form

does anyone else find his voice sounding like nails on a chalkboard

Thanks so much! Now I finally understand

It would be nice to see an example where the "a" value isn't factorable from the constant term, as this is more commonly the case students see, where you have to partially factor "a" from the variable terms and leave the constant out altogether until the end.

okay he lost me when he started talking about max and min. SMH

Useful but man you speak very fast…

kahn Academy be out here saving lives

My teacher keeps giving us problems where C isn’t divisible by the A coefficient. You wouldn’t be able to use this exact method in that case.

I get this and wrote notes and everything but my equation is 5x^2+10x+6 and I got stuck. Can someone explain to me please

Thank you… You are a life saverrrr

I’m failing math omg