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[Kirbypowered]

Just take N and substitute all Ts with T(t)s, then substitute T(t).

Example: N = (T+1)², T = t²+t, then N(T(t)) = (T(t)+1)² = (t²+t+1)².

Hrm. I'm fairly certain that that's what I did. Mayhaps it's an issue in my calculations.

N = 25T²-134T+72
T = 8t+1.6
I got:
N(T(t)) = 1600t²-432t+350.4
I've gone through it a couple times, so if the issue's in my calculations than I'm just doing something wrong. Bleh.

Actual calculations:
25(8t+1.6)²-134(8t-1.6)+72
25(8t+1.6)(8t+1.6)-1072+214.4+72
25(64t²+25.6t+2.56)-1072t+286.4
1600t²+640t+64-1072t+286.4
1600t²-432t+350.4

[MagmaMcFry]

25(8t+1.6)²-134(8t-1.6)+72

You have a sign change there.

[Kirbypowered]

GOSH DANGIT I'M A BUTT.

Thanks for point that out.

[Neonivek]

Now I would go to the internet for this, but unfortunately I only know it by description.

You know how (X+Y)^3 = whatever

And there is a pyramid that quickly explains how you come up with these numbers... What is that? and how do you do it?

Personally I like to write it out, but that is too slow for testing.

Afterall... (X+Y)^3 = (X+Y)(X+Y)(X+Y)... No matter how high the exponent is I can eventually compute it.

[Virex]

You're thinking of Pascal's triangle I think. Don't know how you would go about computing polynominals with that.

[Neonivek]

You're thinking of Pascal's triangle I think. Don't know how you would go about computing polynominals with that.

Yeah Wikipedia is often not very good for "elementary uses" of certain mathematical concepts.

I like to remember that basically every combination gets used.

2 can be obtained by... 2 or 1+1 or 2 (since there are two variables)...

So 3 can be obtained by... 3, 2 + 1, 1 + 2, 3...

So 4 can be obtained by 4, 1 + 3, 2 + 2, 3 + 1, 4

5 can be obtained by 5, 1 + 4, 2 + 3, 3 + 2, 4 + 1, 5

Is this correct?

[MagmaMcFry]

Oh, you want the generalized binomial formula: (a+b)ⁿ = sum(k=0..n, a^n-kb^kB(n,k)), where B(n,k) is the binomial coefficient of n over k (the number of ways to pick k elements from n), and all binomial coefficients happen to appear in Pascal's triangle.

For example: (a+b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴, and the fourth row of Pascal's triangle is (1, 4, 6, 4, 1).

[Lightningfalcon]

Does anyone know any good quick tutorials or guides on how to sketch rational functions?  Not just x/x^2+1, I get that, but how to do things like (x^2-1x+2)^7(x-4)^9/x^11(4x-1)^4   ?  Might not of been paying full attention in my Pre-Calculus class due to being tired and not having my glasses. 

[MagmaMcFry]

Does anyone know any good quick tutorials or guides on how to sketch rational functions?  Not just x/x^2+1, I get that, but how to do things like (x^2-1x+2)^7(x-4)^9/x^11(4x-1)^4   ?  Might not of been paying full attention in my Pre-Calculus class due to being tired and not having my glasses.

Input some values into the function, get some results, connect the dots? All you have to worry about is near the poles of the function, and the poles are easy to figure out, because every pole is a root of the denominator (although not vice versa). For the poles, you have to figure out the left and right limit at those poles, and draw the corresponding vertical asymptotes in the correct direction.

Example: (x³-1)/(x³+x²-x-1) = (x-1)(x²+x+1)/(x+1)²(x-1), so you have a pole at x=-1 of order 2 and a reducible singularity (no pole) at x=1. Since the order of the pole is even, both limits go in the same direction, and you can always figure out the direction of the right asymptote from the sign of lim(x->-1, f(x)*(x+1)²), which is incidentally 1.

So (x³-1)/(x³+x²-x-1) has no zeros (the pole and zero at x=1 cancel each other out) and one pole of order 2 at x=-1 with positive divergence from both sides.

EDIT: The horizontal asymptote is 0 if the degree of the denominator is greater than the degree of the numerator, undefined if the degree of the denominator is smaller, and equal to a/b when the degrees are both equal to n, where a and b are the coefficients of x^n for the numerator and denominator respectively.

[ZetaX]

EDIT: The horizontal asymptote is 0 if the degree of the denominator is greater than the degree of the numerator, undefined if the degree of the denominator is smaller, and equal to a/b when the degrees are both equal to n, where a and b are the coefficients of x^n for the numerator and denominator respectively.

There is also the concept of a nonhorizontal asymptote, e.g. that of y=x+1/x (given by y=x). You get those if the difference of degrees is <=1. More generally, this corresponds to (a x^m + lower degrees)/(b x^n + lower degrees) to behave like a/b x^(m-n) [+lower terms] for |x| large.
This actually is just a more conventional way of saying that there is a pole of order m-n at infinity.

[Descan]

Can someone tell me how

Code: [Select]

2(x+8)(x-2)
__________

4(x-2)(x+2) turns into

Code: [Select]

(x+8)
_____
2(x+2) ?


Tried to recreate it as it appears in the book, those are fractions to be clear.

[Helgoland]

(x-2)/(x-2)=1 for x unequal 2, or am I missing someting?

[Descan]

Well the question series is designed to find out which answers for x end up with the fraction becoming undefined, so... I have no idea.

I just want to know how 2 and x-2 cancel out, and how dividing 4 by 2 means that the (x-2) cancels out. I suspect it's related to dividing both the denominator and numerator by 2, but I don't know how (x-2)/2 = 0.

[Helgoland]

(x-2)/(x-2)=1, and 2/4=1/2. Two seperate steps

[Descan]

... What?