Another simple quadratic question!!!

[Panphobia]

Are you in grade 10? If so you won't be using it soon. Also if you haven't been taught calculus or haven't taken physics at least I do not know how it makes sense to you really, do you know what taking a derivative is?

[Nathan4102]

Yeah, greade 10, and crap. I dont know what a derivative means, and what shouldnt make sense to me?

[Panphobia]

I don't know how -b/2a makes sense to you if you don't know what a derivative is

[Nathan4102]

I suppose "underatand" was the wrong word to use. I see the math behind it, but I dont actually understand why or how this works.

[Panphobia]

Basically a derivative is the Rate of change at any point in a graph, so rate of change is the same thing as slope on a line. Since we know at a maximum or a minimum the slope of that point is 0 you set the derivative to 0 and solve for x

[Nathan4102]

So like a tangent line drawn on a point of a parabola? I think I understand the concept, one thing though. Where does 2AX + B come from in "d/dx (Ax^2 + Bx + C) = 2AX + B"?

[Panphobia]

You haven't learned functions yet but basically if y = ax^2+bx+c. And f(x) =y then to get the derivative using first principles which will work always is f'(x) = lim h->0 (f(x+h)-f(x))/h that is basically function composition and you have to arrange such that it won't be divided by h because h is approaching 0. So the rule with polynomials is called the power rule and this is it if you have a function x^n the derivative function is nx^(n-1) so Ax^2 + Bx^1 + Cx^0 would be 2Ax + (1)Bx + (0)Cx^-1 and anything multiplied by 0 is 0 so you leave it out...and there you go

[Nathan4102]

I got some of that, didn't get some of that... Neat stuff though, grade 12 calculus sounds pretty interesting. Thanks for the explanations, I think I should stick to perfecting grade 10 quadratics before taking on calculus

[Raknarg]

You'll actually kindof touch on it in Grade 11.

Lets say you wanted to find out what the slope was around the point. The simplest way is to take two points around the point you're looking at, and find the slope between them.

Ex. y = x^2, and you wanted to know the slope at 2.

Therefore: m (or slope) = (y2 - y1) / (x2 - x1)

We can replace these values:

m = (f (x + 1) - f (x)) / (x2 - x1)

Where f (x) is the formula for the graph (same thing as saying y). Now lets replace those f's:

m = ((2 + 1)^2 - (2)^2) / ((2 + 1) - 2)
m = ((3)^2 - (2)^2) / (3 - 2)
m = (9 - 4) / 1
m = 5

This is inaccurate though. This isnt the slope of the point, this is the slope of the point x and one above. Lets get lower.

m = ((2 + 0.1)^2 - (2)^2) / ((2 + 0.1) - 2)
m = ((2.1)^2 - (2)^2) / (2.1 - 2)
m = (4.41 - 4) / 0.1
m = 0.41 / 0.1
m = 4.1

This is better, but not good enough. What we're heading for is the same point though, so whjy don't we try that?

m = ((2)^2 - (2)^2) / ((2) - 2)

You can see right now that we're going to get an error. The slope is obviously not 0 / 0. The trick around that is to do this:

m = [f (x + h) - f (x)] / (x + h - x)

h is the number we want to add to x. Our goal is zero, but if we put in zero we'll run into a problem. Instead, lets try to get rid of h and see what happens:

m = [(2 + h)^2 - (2)^2] / (2 + h - 2)
m = [(2 + h)^2 - 4] / h
m = [(4 + 4h + h^2 - 4] / h
m = [4h + h^2] / h

Now lets remove an h from each polynomial:

m = [4 + h] / 1

But wait... h is zero! That leaves us with:

m = 4

This is what we're looking for. The slope is 4. Now lets extrapolate this to the entire funtion of x^2:

m = [(x + h)^2 - x^2] / (x + h - x)
m = [x^2 + 2xh + h^2 - x^2] / h
m = [2xh + h^2] / h
m = [2x + h] / 1

Now we alway set h to 0, so it leaves us with this:

m = 2x

The slope is 2x at any point. Lets try this with 2x^3 + 3x^2 + x + 5

m = [2(x + h)^3 + 3(x + h)^2 + (x + h) + 5 - (2x^3 + 3x^2 + x + 5)] / (x + h - x)

now it looks ugly but bear with me...

m = [2(x^3 + 3hx^2 + 3xh^2 + h^3) + 3(x^2 + 2xh + h^2) + (x + h) + 5 - (2x^3 + 3x^2 + x + 5)] / (x + h - x)
m = [2x^3 + 6hx^2 + 6xh^2 +2h^3 + 3x^2 + 6xh + 3h^2 + x + h + 5 - 2x^3 - 3x^2 - x - 5] / h

Now we remove anything we can right now

m = [6hx^2 + 6xh^2 +2h^3 + 6xh + 3h^2 + h] / h

new remove an h

m = [6x^2 + 6xh +2h^2 + 6x + 3h + 1]

Set all h's to 0

m = 6x^2 + 6x(0) +2(0)^2 + 6x + 3(0) + 1
m = 6x^2 + 6x + 1

Look how nicely that cleaned up! But this is gross, and we want a simpler way right? Lets look at the two formulas:

y = 2x^3 + 3x^2 + x + 5
m = 6x^2 + 6x + 1

There is a pattern here: In each case, we put the exponent down by one, and multiplied the coefficient by the previous exponent. Notice how the 5 is now gone? We can rewrite that as 5x^0. 5 * 0 is 0, so we chuck it out. The x turned into 1 because 1 * 1 = 1.
Now we can do cool stuff with this.

Lets take a random equation like y = 4x^4 - x^3 + 7x

y = 4x^4 - x^3 + 7x
m = 16x^3 - 3x^2 + 7

But anyways, this is basically caluclus. There's rules and stuff for different situations, but this is basicallt the most important one. In fact all the rules you need to know can be proved logically by this basic formula.

[Nathan4102]

Calculus doesn't look that bad... More algebra, new equations, more variables... Obviously, I assume it gets a fair bit harder, but this is a neat intro. Ill take time later to try to understand all this, thanks!

[Raknarg]

Its actually basically stuff you know writ in a different way and not really, calculus is pretty easy if you understand math in general. well it has been for me anyways...

[Panphobia]

Yea and you will not have to use the (f(x+h)-f(x))/h for all of grade 12 calculus, for most of it you will get to use rules such as the power rule which I touched on, the chain rule, the quotient rule, and the product rule. This will make getting derivatives so easy it will take literally seconds.

[Raknarg]

I remember I got so annoyed with trying to remember the procedure for finding the vertex form equation of a parabola that i just memorized the formula

[Nathan4102]

Sweet! I dont understand why they teach us all this stuff then, like if theres a faster, more efficient way, and ill never use this after I learn the better way, why teach me it???

[Panphobia]

This is because you always have to learn and understand the long and tedious way to do things before you can go on to the quick and elegant way. This is how it is these days, but like I said all your math classes, except for data management have been setting you up for calculus. In my opinion calculus is too easy in grade 12, they took out too much of actual calculus, like what you learn in grade 12 is around 30-40% of what calculus is. You are not supposed to learn integration, you are not supposed to learn any limit rules, oh yea and with limits you basically do 3-4 days and you get a test on them and you're done. It really amazes me how crappy it is. If you have a love for math I suggest doing the Calculus AP exam.