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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.

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• 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.
• 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?
• 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=?
• I got lost at because 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?
• 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?
• I don't get why the formula Sal talks about at (`-b/2a`) works.
What's the technical proof.
• 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.
• At about , 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?
• 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).
-2(x^2+4+4)-8+8+8
8+8=16
-2(x+2)^2 16
• 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 roughly to , 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?
• 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.

• how would u get 16 as the maximum point by using the simple and fast formulae?
• 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/
• I don't understand why he suddenly said x-2=0
(1 vote)
• Well, let's look at .
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).