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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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• @ through Sal says "when x is equal to three it's also going to be equal to 8" ... he meant "when x is equal to 4" because we are calling each line in the graph 2, not one... I know this isn't a question, but it confused me since -2(3)^2+8(3)≠0... always check your (and the teacher's) work! • this may be a stupid question.... I understand that the vertex is (2,16) , but what exactly is the equation of the axis of symmetry? If my teacher hypothetically asks me "What is the equation for the axis of symmetry of this parabola?" , how do I find that equation? • I get that the parabola will open upward when a>0 and downward when a<0, but what if a=0? • 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. • 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?
• how would you pit x^2+56y-28.28x=0 into general form? • 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.
• 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.
Note: Completing the square and the quadratic formula are not the same method.

(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). 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 :)