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### Course: Get ready for Algebra 2 > Unit 3

Lesson 10: Graphing quadratics in standard form# Finding the vertex of a parabola in standard form

Sal rewrites the equation y=-5x^2-20x+15 in vertex form (by completing the square) in order to identify the vertex of the corresponding parabola. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

- why is it that to find a vertex you must do -b/2a? is there a separate video on it?(48 votes)
- A parabola is defined as

𝑦 = 𝑎𝑥² + 𝑏𝑥 + 𝑐 for 𝑎 ≠ 0

By factoring out 𝑎 and completing the square, we get

𝑦 = 𝑎(𝑥² + (𝑏 ∕ 𝑎)𝑥) + 𝑐 =

= 𝑎(𝑥 + 𝑏 ∕ (2𝑎))² + 𝑐 − 𝑏² ∕ (4𝑎)

With ℎ = −𝑏 ∕ (2𝑎) and 𝑘 = 𝑐 − 𝑏² ∕ (4𝑎) we get

𝑦 = 𝑎(𝑥 − ℎ)² + 𝑘

(𝑥 − ℎ)² ≥ 0 for all 𝑥

So the parabola will have a vertex when (𝑥 − ℎ)² = 0 ⇔ 𝑥 = ℎ ⇒ 𝑦 = 𝑘

𝑎 > 0 ⇒ (ℎ, 𝑘) is the minimum point.

𝑎 < 0 ⇒ (ℎ, 𝑘) is the maximum point.(46 votes)

- Is there a video about vertex form?(12 votes)
- Not specifically, from the looks of things. When Sal gets into talking about graphing quadratic equations he talks about how to calculate the vertex. On the other hand, there are several exercises in the practice section about vertex form, so the hints there give a good sense of how to proceed.(16 votes)

- In which video do they teach about formula -b/2a(14 votes)
- This is not a derivation or proof of -b/2a, but he shows another way to get the vertex: https://www.khanacademy.org/math/algebra/quadratics/features-of-quadratic-functions/v/quadratic-functions-2

There are some answers to the derivation of -b/2a here:

https://math.stackexchange.com/q/709/592818(6 votes)

- Why does x+4 have to = 0?(7 votes)
- Because then you will have a y coordinate for a given x. When x-4 = 0 (i.e. when x =4) you are left with just y=21 in the equation: because

4-4=0

0^2=0

-3(0)=0

This leaves the equation looking like y=0+21

Then you know that when x=4 that y=21. Then you have solved for x and y.

If you want to think about it a different way you could use y=f(x). Then f(4)=21. Some people might find the f(x) way easier to understand.(15 votes)

- @0:49he mentions it's mentioned in multiple videos but I made a search and watched a few videos trying to find it in vain. Would someone kindly reply with URLs or page titles of the videos he referred to?(4 votes)
- I don't know there those videos are, but I think is quite easy to realize it. See: the x coordinate of the vertex is the average point between the two roots of the quadratic (the two points where the graph of the parabola intersects with the horizonal axis. So if we know that the formula for one of those roots, as per the quadratic formula, is -b+square root of -bsquared-4ac, all divided by 2a, and the formula for the other root i the same but with a minus sign on the numerator instead (this is, -b - squareroot of bsquared -4ac, all divided by 2a). Then the average between those two roots is obtained by adding those two formulas and dividing by two.

So: x coordinate of the vertex as an average point between the two roots->

((-b+sqrt(bsquared-4ac))/2a + (-b-sqrt(bsquared-4ac))/2a) / 2 =

(-b/2a + (sqrt(bsquared-4ac))/2a -b/2a -(sqrt(bsquared-4ac))/2a) / 2=

(2*-b/2a) /2 =

-b/2a

See how sqrt(bsquared-4ac)/2a and -sqrt(bsquared-4ac)/21 get cancelled out of the expression.

So the key seems to be

1) the x coordinate of the vertex is equal to the average point between the two roots of the parabola -the points where the parabola intersects the horizontal axis

2) the roots of the parabola can be found via the quadratic formula.

So applying the arithmetic average formula (a+b)/2 where a is -b+sqrt(bsquared-4ac)/2a and b is -b-sqrt(bsquared-4ac)/a gives -b/2a as solution for x coordinate of vertex. This formula also works if the parabola has only one root. In that case, the vertex would lie on the horizontal axis.-(10 votes)

- Can someone explain why the x-vertex formula is -b/2a?(4 votes)
- We know we can find the x-intercepts of the parabola by using the quadratic formula.

1st x-intercept: x = [-b + sqrt(b^2-4ac)]/(2a)

2nd x-intercept: x = [-b - sqrt(b^2-4ac)]/(2a)

The x-value of the vertex is located midway between these 2 points. If you average the two x-intercepts, you get their midpoint. This means you add the 2 points then divide by 2.

ADD 1st: [-b + sqrt(b^2-4ac)]/(2a) + [-b - sqrt(b^2-4ac)]/(2a)

-- Notice, we already have a common denominator, so we add the numerators.

-- Also notice, the square roots add to 0 because one is positive and the other negative.

-- This leaves: (-b-b)/(2a) = (-2b)/(2a) = -b/a

DIVIDE BY 2: -b/a divided by 2 = -b/a * 1/2 = -b/(2a)

Hope this helps.(9 votes)

- At3:38how does Sal get x=4? Wouldn't the expression -3(x-4)^2 have to equal - 21 for the whole equation to equal zero?(5 votes)
- You want that term to be equal to zero and to do that x has to equal 4 because (4-4)^2 is equal to zero.(4 votes)

- Why is x vertex equal to -b/2a ?(3 votes)
- This is not a derivation or proof of " -b/2a", but he shows another way to get the vertex: https://www.khanacademy.org/math/algebra/quadratics/features-of-quadratic-functions/v/quadratic-functions-2

There are some answers to the derivation of -b/2a here:

https://math.stackexchange.com/q/709/592818(6 votes)

- This video is not about the equation y=-3x^2+24x-27

It is about completing the square to solve 4x^2+40x-300=0

Can anyone help me?

The transcript is going but it is different words!(3 votes) - How can we find the domain and range after compeleting the square form?(0 votes)
- The Domain of a function is the group of all the x values allowed when calculating the expression.

In this exercise all x values can be used so the domain is the group of all the Real numbers.

Examples to functions that would limit the domain would contain operations like:

Division - Because division by 0 is not allowed

Square root - Because Square root of a negative number is not a real number

As you can see there are no such operations in this exercise.

The Range of a function is the group of all the y values that result from calculating the function for all the x values allowed (the Domain).

As Sal explains in the last part of the video when you bring the parabola to its vertex form it is easier to see the Range.

The free coefficient, i.e., the C in the video, is either the minimum or the maximum point of the Range.

The sign of the leading coefficient, i.e., the A in the video, determines whether it is the minimum or the maximum.

If A>0 the parabola open upwards (we call it smiling :-) and all other values of y will be greater than C, i.e., C is minimum and the Range is y>=C

If A<0 the parabola open downwards (we call it weeping :-) and all other values of y will be smaller than C, i.e., C is maximum and the Range is y<=C

In this exercise A is (-3) and it is negative, so 21 is the maximum and the Range is y<=21

Hope it helps :-)(15 votes)

## Video transcript

I have an equation right here. It's a second degree equation. It's a quadratic. And I know its graph is
going to be a parabola. Just as a review, that means it
looks something like this or it looks something like that. Because the coefficient on the
x squared term here is positive, I know it's going to be an
upward opening parabola. And I am curious about the
vertex of this parabola. And if I have an upward
opening parabola, the vertex is going to
be the minimum point. If I had a downward
opening parabola, then the vertex would
be the maximum point. So I'm really trying
to find the x value. I don't know actually where
this does intersect the x-axis or if it does it all. But I want to find
the x value where this function takes
on a minimum value. Now, there's many
ways to find a vertex. Probably the easiest,
there's a formula for it. And we talk about where that
comes from in multiple videos, where the vertex of a
parabola or the x-coordinate of the vertex of the parabola. So the x-coordinate
of the vertex is just equal to
negative b over 2a. And the negative b, you're just
talking about the coefficient, or b is the coefficient
on the first degree term, is on the coefficient
on the x term. And a is the coefficient
on the x squared term. So this is going to be
equal to b is negative 20. So it's negative
20 over 2 times 5. Well, this is going to
be equal to positive 20 over 10, which is equal to 2. And so to find the y
value of the vertex, we just substitute
back into the equation. The y value is going
to be 5 times 2 squared minus 20 times 2 plus 15,
which is equal to let's see. This is 5 times 4, which is 20,
minus 40, which is negative 20, plus 15 is negative 5. So just like that, we're able
to figure out the coordinate. This coordinate right over here
is the point 2, negative 5. Now it's not so
satisfying just to plug and chug a formula like this. And we'll see where
this comes from when you look at the
quadratic formula. This is the first term. It's the x value that's
halfway in between the roots. So that's one way
to think about it. But another way to do
it, and this probably will be of more lasting
help for you in your life, because you might
forget this formula. It's really just try to
re-manipulate this equation so you can spot
its minimum point. And we're going to do that
by completing the square. So let me rewrite that. And what I'll do is out
of these first two terms, I'll factor out a 5, because I
want to complete a square here and I'm going to leave
this 15 out to the right, because I'm going to have
to manipulate that as well. So it is 5 times x
squared minus 4x. And then I have
this 15 out here. And I want to write this
as a perfect square. And we just have
to remind ourselves that if I have x plus
a squared, that's going to be x squared
plus 2ax plus a squared. So if I want to turn something
that looks like this, 2ax, into a perfect
square, I just have to take half of this coefficient
and square it and add it right over here in order
to make it look like that. So I'm going to do
that right over here. So if I take half of negative
4, that's negative 2. If I square it, that is
going to be positive 4. I have to be very careful here. I can't just willy nilly
add a positive 4 here. I have equality here. If they were equal
before adding the 4, then they're not going to
be equal after adding the 4. So I have to do proper
accounting here. I either have to add 4 to both
sides or I should be careful. I have to add the same
amount to both sides or subtract the
same amount again. Now, the reason why I
was careful there is I didn't just add 4 to the right
hand side of the equation. Remember, the 4 is
getting multiplied by 5. I have added 20 to the right
hand side of the equation. So if I want to make
this balance out, if I want the equality
to still be true, I either have to
now add 20 to y or I have to subtract 20 from
the right hand side. So I'll do that. I'll subtract 20 from
the right hand side. So I added 5 times 4. If you were to distribute
this, you'll see that. I could have literally, up
here, said hey, I'm adding 20 and I'm subtracting 20. This is the exact same
thing that I did over here. If you distribute the 5, it
becomes 5x squared minus 20x plus 20 plus 15 minus 20. Exactly what's up here. The whole point of
this is that now I can write this in
an interesting way. I could write this as y is equal
to 5 times x minus 2 squared, and then 15 minus 20 is minus 5. So the whole point of this is
now to be able to inspect this. When does this equation
hit a minimum value? Well, we know that this
term right over here is always going to
be non-negative. Or we could say
it's always going to be greater than
or equal to 0. This whole thing is going
to hit a minimum value when this term is equal
to 0 or when x equals 2. When x equals 2, we're going
to hit a minimum value. And when x equals
2, what happens? Well, this whole term is 0
and y is equal to negative 5. The vertex is 2, negative 5.