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

About James Carlton

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15 thoughts on “Completing the square for vertex form | Quadratic equations | Algebra I | Khan Academy

  1. Was anyone else's mind blown when he just added 16 and then subtracted 16, just so he can get a perfect square? This is sooo useful when you have an equation that does not have a perfect square.

  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

  3. 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.

  4. 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.

  5. 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

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