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### Course: Algebra 1 > Unit 14

Lesson 11: Features & forms of quadratic functions- Forms & features of quadratic functions
- Worked examples: Forms & features of quadratic functions
- Features of quadratic functions: strategy
- Vertex & axis of symmetry of a parabola
- Finding features of quadratic functions
- Features of quadratic functions
- Graph parabolas in all forms
- Interpret quadratic models: Factored form
- Interpret quadratic models: Vertex form
- Interpret quadratic models
- Graphing quadratics review
- Creativity break: How does creativity play a role in your everyday life?

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# Vertex & axis of symmetry of a parabola

Sal rewrites a quadratic equation in vertex form and shows how it reveals the vertex of the corresponding parabola. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

- He doesn't explain the formula very well here, I guess like he said, it's gained from completing the square. But here is a better and easier way for you to understand how to find the symmetry of parabola: think of a line, you have 2 coordinate points on each end, since you want to find the symmetry of parabola, you have to find the midpoint of that line.... how would you find it? Well one way would be to find the average value of the X or that is: simply add the two values where the symmetry is (in parabola case, the x axis), so how do you find the average? (X1+X2)/2. Doesn't matter if the values are negative either, you 're simply taking 2 distances from origin which are also coordinates of that line and dividing it by 2, effectively finding the midpoint coordinate. In any case, I'd like to see Sal derive this formula to make things a bit clearer in the future.(33 votes)
- Let 𝑥₁, 𝑥₂ be distinct points on the 𝑥-axis, equidistant to the vertex of the parabola 𝑎𝑥² + 𝑏𝑥 + 𝑐.

Because a parabola is symmetrical about the vertex, we then have

𝑎𝑥₁² + 𝑏𝑥₁ + 𝑐 = 𝑎𝑥₂² + 𝑏𝑥₂ + 𝑐

⇒ 𝑎𝑥₁² − 𝑎𝑥₂² + 𝑏𝑥₁ − 𝑏𝑥₂ + 𝑐 − 𝑐 = 0

⇒ 𝑎(𝑥₁² − 𝑥₂²) + 𝑏(𝑥₁ − 𝑥₂) = 0

⇒ 𝑎(𝑥₁ + 𝑥₂)(𝑥₁ − 𝑥₂) + 𝑏(𝑥₁ − 𝑥₂) = 0

⇒ (𝑥₁ − 𝑥₂)(𝑎(𝑥₁ + 𝑥₂) + 𝑏) = 0.

𝑥₁ ≠ 𝑥₂ ⇒ 𝑎(𝑥₁ + 𝑥₂) + 𝑏 = 0

⇒ 𝑎(𝑥₁ + 𝑥₂) = −𝑏

⇒ 𝑥₁ + 𝑥₂ = −𝑏∕𝑎

⇒ 𝑥₂ = −𝑥₁ − 𝑏∕𝑎.

Since 𝑥₁, 𝑥₂ are distinct points on the 𝑥-axis and equidistant to the vertex, their midpoint must lie on the same vertical line as the vertex.

In other words,

(𝑥₁ + 𝑥₂)∕2

= (𝑥₁ − 𝑥₁ − 𝑏∕𝑎)∕2

= −𝑏∕(2𝑎)

must be the 𝑥-coordinate of the vertex.(1 vote)

- I get that the parabola will open upward when a>0 and downward when a<0, but what if a=0?(14 votes)
- Good question.

It is not a parabola anymore if a = 0. It is just a horizonal line because when you multiply the x by zero it is eliminated and your just have y=?(23 votes)

- I got lost at1:54because idk where the -4 comes from. then we all of a sudden get to y= -2((x-2)^2 +8) like how did we get there?(11 votes)
- x was divisible by -2 as well as the rest of the numbers on that side of the equal sign. So, the -4 is the result of dividing the 8 on the right end by -2 . then he added another 4 to get a perfect square, but to avoid adding 4 to y as well he had to subtract 4 without making them cancel each-other out then he has a perfect square: X^2 -4X +4= [x-2]^2 the entire equation from here being y= -2([x-2]^2 -8) because the two subtractions of 4 you now have combine to make - 8 at the end now if you multiply the -8 by -2 you can get it out of the parenthesis for the equation: y = -2[x-2]^2 + 16 you never should have ended up with a +8 in this equation . (part of your question) comes from. then we all of a sudden get to y= -2((x-2)^2 +8) like how did we get there?(7 votes)

- I don't get why the formula Sal talks about at6:40(
`-b/2a`

) works.

What's the technical proof.(5 votes)- Look by using the completing the square method we essentially turn the equation into vertex form which I suppose you know. Then the result seems as follows: A (x+b)^2+C. Here you know how you derived up to this. Then see the part ( x + B ). Here, you know that B = b/2a. And Sal told that to obtain the vertex form the Part A( x + B )^2 should be equal to zero in both the cases. And for that (x+(b/2a)) should be equal to zero. And now we can derive that as follows:

x + (b/2a) = 0

=> x = -b/2a.

And here your formula is whose deriving seems pretty daunting but is based on just simple logical reasoning.

Hope that helps you ...

Well it took me more than half a hour to derive this formula and I had the same doubt too.(8 votes)

- At about2:25, Sal adds and subtracts four to the equation. Wouldn't he have to add 8 because the 4 is in the parenthesis being multiplied by 2?(2 votes)
- You're right, maybe you did it the way I did. However, he did the problem differently, since he factored the 8 out, it changed how he had to solve it. He worked it out as 4 was added instead of -8, he ignored the -2. He multiplied the -2 at the end. I prefer to multiply the number on the outside of the parentheses before the end so I don't forget, but either way is fine.

Example of how I did it:

I factor out the -2 like he did because it's not good to have a negative in front of the x^2 term.

-2(x^2+4 ) +8

What is half of 4? 2.

2^2=4

I don't factor the 8, I leave it alone and deal with it later

I then solve

-2(x^2+4+4)+8

-2(x+2)^2

-8 was added to the equation (because -2*4=-8, we can't ignore the -2 in front of the parentheses) so to ensure the equation stays the same, I have to add 8 to the equation (-8+-8=0, so the equation does not change).

Add 8+8=16

-2(x^2+4+4)-8+8+8

8+8=16

-2(x+2)^2 16(9 votes)

- Kind of a general question, what are quadratic formulas and how are they used? I know they are a complex reverse form of factoring, however that doesn't correspond with it's name 'quadratics'. Wouldn't quadratics have to do something dealing with '4' as suggested by it's name?

From roughly1:40to2:50, Sal 'completes the square' using a quadratic formula. The process seems to make sense once I understood the formula used for it , which was simply required to find the vertex of the parabola however why is it called 'completing a square'? Sal is completing a parabolic curve so is 'completing the square just unfamiliar terminology?(4 votes)- Quadratic equations are any equation that can be written in the form: Ax^2 + Bx + C = 0, where A is not = 0. As to why they are called quadratics, see this link: http://mathforum.org/library/drmath/view/52572.html

Quadratic equations can be solved using 4 different methods. You are mixing up the name of the methods with the type of equations. The 4 methods for solving are:

1) Factoring. This does not work for all quadratics because not all are factorable. But, it is usually easier than other methods when it works.

2) Square root method. This also only works for some quadratics. Specifically, it can be used for any quadratics in the form of "x^2 = a number" or "(ax+b)^2 = a number"

3) Completing the square. This works for all quadratics. This method is called completing the square because you create a perfect square trinomial, which factors into a binomial-squared.

4) Quadratic formula. This also works for all quadratics.

Note: Completing the square and the quadratic formula are**not**the same method.

For more info on these methods, see the lessons at this link. These are sections for each method: https://www.khanacademy.org/math/algebra/quadratics#quadratics-square-root(0 votes)

- how would u get 16 as the maximum point by using the simple and fast formulae?(4 votes)
- I asked myself the same question, as Sal seems to have more than a little contempt for the "mindless application" method ;p 'Sokay! I do understand where he's coming from, and am glad he takes the time to go through the math-intuition/derivation form he does in these videos ;)

I took the quick-'n'-dirty method result (`x=2`

), plugged it back into the original formula (`y = -2x^2 + 8x + 8`

), and solved for y. Result was (surprise!)`y = 16`

. Since I haven't seen any further responses to your question since you had asked lo, these many ages ago, I will simply assume that that's the way it's done and leave it at that :)

Anyone else know any better? 'Cuz believe me, I'm all ears! \o/(3 votes)

- I don't understand why he suddenly said x-2=0(1 vote)
- Well, let's look at4:22.

y = -2(x-2)² + 16

So, how does this tell us where the vertex is? Well, the vertex will be at the point where y is a minimum or maximum. Let's look at the equation in parts:

The last term is 16. It doesn't matter what x is, this term isn't going to change, so it tells us nothing about where the minimum or maximum is, so we can ignore it. That is, if we find the minimum for:

y = -2(x-2)²

it will occur at the same value of x as the minimum of our original equation.

The next detail is the exponent. Any number squared is a positive number (OK, there are exceptions, but I like to keep it real!) So (x-2)² is a positive number (or zero), no matter what x is. So if we multiply it by a negative, it must give us a negative number.

So, -2(x-2)² is negative, or zero. The bigger we make x, the more negative this term gets. The more negative we make x, the more negative this term gets. The term vanishes off the bottom of the graph paper to the left and to the right. So we know we must be looking for a maximum. We know -2(x-2)² is never positive, so it must be when it is least negative (or zero). But when is this?**Here is where it happens**

Well, the -2 doesn't change - it just makes the graph steeper. We know the maximum must occur when (x-2)² is smallest. So, x-2 must be neither very positive nor very negative. Well, when x = 2, (x-2) = 0, which is the smallest we can possibly get without going negative, so (x-2)² = 0² = 0. The maximum occurs when x = 2. This is where he got the mysterious x-2 = 0.

Let's do some sanity checking. What if x = 3? Then (x-2) = 1, 1² = 1, -2*1 = -2.

What if x = 1? Then (x-2) = -1, (x-2)² = 1, and we are back where we were, but on the other side.

So how do we get the y? Just plug the x back in and we see the vertex is at (2, 16).(10 votes)

- Also, I need some advice on where I can go to get a deeper understanding of parabola's. I know how to do the equations the long way in the video provided, but I don't know WHY I just see a pattern and know how to get the right answer.(4 votes)
- Mathematical intuition is not enjoyed by all - and you may have it in this area and maybe even more. Other than playing with equations and their graphs, which you can do here: https://www.desmos.com/calculator, the next step is to begin investigating other conic sections, of which the parabola is one. Khan has some: https://www.khanacademy.org/math/algebra2/conics_precalc

Keep Studying!(1 vote)

- At3:36, Sal said (x-2)^2 is "always positive" but I think it could be zero too. So shouldn't it be always non-negative?(3 votes)
- Yes, it should be "always non-negative".(5 votes)

## Video transcript

We need to find the vertex
and the axis of symmetry of this graph. The whole point of doing this
problem is so that you understand what the vertex
and axis of symmetry is. And just as a bit of a
refresher, if a parabola looks like this, the vertex is the
lowest point here, so this minimum point here, for an
upward opening parabola. If the parabola opens downward
like this, the vertex is the topmost point right like that. It's the maximum point. And the axis of symmetry is
the line that you could reflect the parabola around,
and it's symmetric. So that's the axis
of symmetry. That is a reflection of the
left-hand side along that axis of symmetry. Same thing if it's a
downward-opening parabola. And the general way of telling
the difference between an upward-opening and a
downward-opening parabola is that this will have a positive
coefficient on the x squared term, and this will have
a negative coefficient. And we'll see that in a little
bit more detail. So let's just work on this. Now, in order to figure out the
vertex, there's a quick and dirty formula, but I'm not
going to do the formula here because the formula really tells
you nothing about how you got it. But I'll show you how to apply
the formula at the end of this video, if you see this on a math
test and just want to do it really quickly. But we're going to do it the
slow, intuitive way first. So let's think about how we can
find either the maximum or the minimum point of
this parabola. So the best way I can think
of doing it is to complete the square. And it might seem like a very
foreign concept right now, but let's just do it one
step at a time. So I can rewrite this as y is
equal to-- well, I can factor out a negative 2. It's equal to negative 2 times
x squared minus 4x minus 4. And I'm going to put the
minus 4 out here. And this is where I'm going
to complete the square. Now, what I want to do is
express the stuff in the parentheses as a sum of a
perfect square and then some number over here. And I have x squared minus 4x. If I wanted this to be a perfect
square, it would be a perfect square if I had a
positive 4 over here. If I had a positive 4 over
there, then this would be a perfect square. It would be x minus 2 squared. And I got the 4, because I said,
well, I want whatever half of this number is,
so half of negative 4 is negative 2. Let me square it. That'll give me a positive
4 right there. But I can't just add
a 4 willy-nilly to one side of an equation. I either have to add it to the
other side or I would have to then just subtract it. So here I haven't changed
equation. I added 4 and then
I subtracted 4. I just added zero to this little
expression here, so it didn't change it. But what it does allow me to do
is express this part right here as a perfect square. x
squared minus 4x plus 4 is x minus 2 squared. It is x minus 2 squared. And then you have this
negative 2 out front multiplying everything, and
then you have a negative 4 minus negative 4, minus
8, just like that. So you have y is equal to
negative 2 times this entire thing, and now we can multiply
out the negative 2 again. So we can distribute it. Y is equal to negative 2 times
x minus 2 squared. And then negative 2 times
negative 8 is plus 16. Now, all I did is algebraically arrange this equation. But what this allows us to do
is think about what the maximum or minimum point
of this equation is. So let's just explore
this a little bit. This quantity right here, x
minus 2 squared, if you're squaring anything, this
is always going to be a positive quantity. That right there is
always positive. But it's being multiplied
by a negative number. So if you look at the larger
context, if you look at the always positive multiplied by
the negative 2, that's going to be always negative. And the more positive that this
number becomes when you multiply it by a negative, the
more negative this entire expression becomes. So if you think about it,
this is going to be a downward-opening parabola. We have a negative coefficient
out here. And the maximum point on this
downward-opening parabola is when this expression right here
is as small as possible. If this gets any larger, it's
just multiplied by a negative number, and then you subtract
it from 16. So if this expression right here
is 0, then we have our maximum y value, which is 16. So how do we get x is
equal to 0 here? Well, the way to get x minus
2 equal to 0-- so let's just do it. x minus 2 is equal to 0,
so that happens when x is equal to 2. So when x is equal to 2,
this expression is 0. 0 times a negative number,
it's all 0, and then y is equal to 16. This is our vertex, this
is our maximum point. We just reasoned through it,
just looking at the algebra, that the highest value this
can take on is 16. As x moves away from 2 in
the positive or negative direction, this quantity right
here, it might be negative or positive, but when you square
it, it's going to be positive. And when you multiply it by
negative 2, it's going to become negative and it's going
to subtract from 16. So our vertex right here
is x is equal to 2. Actually, let's say each
of these units are 2. So this is 2, 4, 6,
8, 10, 12, 14, 16. So my vertex is here. That is the absolute maximum
point for this parabola. And its axis of symmetry is
going to be along the line x is equal to 2, along the
vertical line x is equal to 2. That is going to be its
axis of symmetry. And now if we're just curious
for a couple of other points, just because we want to plot
this thing, we could say, well, what happens when
x is equal to 0? That's an easy one. When x is equal to 0,
y is equal to 8. So when x is equal to 0, we have
1, 2, 3, 4-- oh, well, these are 2. 2, 4, 6, 8. It's right there. This is an axis of symmetry. So when x is equal to 3, y is
also going to be equal to 8. So this parabola is a really
steep and narrow one that looks something like this, where
this right here is the maximum point. Now I told you this is the slow
and intuitive way to do the problem. If you wanted a quick and dirty
way to figure out a vertex, there is a formula
that you can derive it actually, doing this exact same
process we just did, but the formula for the vertex, or
the x-value of the vertex, or the axis of symmetry, is x is
equal to negative b over 2a. So if we just apply this-- but,
you know, this is just kind of mindless application
of a formula. I wanted to show you the
intuition why this formula even exists. But if you just mindlessly
apply this, you'll get-- what's b here? So x is equal to negative--
b here is 8. 8 over 2 times a. a right here is a negative 2. 2 times negative 2. So what is that going
to be equal to? It is negative 8 over negative
4, which is equal to 2, which is the exact same thing we
got by reasoning it out. And when x is equal to
2, y is equal to 16. Same exact result there. That's the point 2 comma 16.