posted 14 years ago
Given two points on a graph (you know, each with a xcoordinate and a ycoordinate) which method could I use to determine the slope of a line between those two points?
Which method could I use to determine the distance between those two points?
Given three integer coefficients, A, B, C, which method could I use to solve a quadratic function of the form: Ax^2 + Bx + C = 0 ?
Given four integer coefficients, A, B, C, D which method could I use to find the derivative of a cube polynomial function: Ax^3 + Bx^2 + Cx + D?
Which method could I use to determine the distance between those two points?
Given three integer coefficients, A, B, C, which method could I use to solve a quadratic function of the form: Ax^2 + Bx + C = 0 ?
Given four integer coefficients, A, B, C, D which method could I use to find the derivative of a cube polynomial function: Ax^3 + Bx^2 + Cx + D?
posted 14 years ago
If point one has coordinates x1,y1 and point two has coords x2,y2
then the slope is
(y2  y1) / (x2  x1)
Pythagoras's Theorem:
distance = sqrt[ (y2y1)^2 + (x2  x1)^2 ]
Oh dear how embarrassing that I had to look that one up
One root is x = [ B + sqrt( B^2  4*A*C) ] / (2*A)
The other is x = [ B  sqrt( B^2  4*A*C) ] / (2*A)
Yikes... er ... .... hey what's that thing behind you !? .... *RUNS* (hee hee hee)
[ March 20, 2004: Message edited by: Carlos Failde ]
Given two points on a graph (you know, each with a xcoordinate and a ycoordinate) which method could I use to determine the slope of a line between those two points?
If point one has coordinates x1,y1 and point two has coords x2,y2
then the slope is
(y2  y1) / (x2  x1)
Which method could I use to determine the distance between those two points?
Pythagoras's Theorem:
distance = sqrt[ (y2y1)^2 + (x2  x1)^2 ]
Given three integer coefficients, A, B, C, which method could I use to solve a quadratic function of the form: Ax^2 + Bx + C = 0 ?
Oh dear how embarrassing that I had to look that one up
One root is x = [ B + sqrt( B^2  4*A*C) ] / (2*A)
The other is x = [ B  sqrt( B^2  4*A*C) ] / (2*A)
Given four integer coefficients, A, B, C, D which method could I use to find the derivative of a cube polynomial function: Ax^3 + Bx^2 + Cx + D?
Yikes... er ... .... hey what's that thing behind you !? .... *RUNS* (hee hee hee)
[ March 20, 2004: Message edited by: Carlos Failde ]
Will Carpenter
Greenhorn
Posts: 26
posted 14 years ago
There's a proof in number theory that there's no formula for the zeroes of an order5 polynomial (ie, p(x) = ax^5 + bx^4 + cx^3 + dx^2 + ex + f), or any polynomial of greater order.
OK, when I say "formula", I mean a function f(a, b, c, d, e, f) which yields the zeroes to p(x) above.
IIRC there are formulae for the zeroes of polynomials of orders 3 and 4, but they're sufficiently complex that in almost all circumstances an approximation is better.
Tim
OK, when I say "formula", I mean a function f(a, b, c, d, e, f) which yields the zeroes to p(x) above.
IIRC there are formulae for the zeroes of polynomials of orders 3 and 4, but they're sufficiently complex that in almost all circumstances an approximation is better.
Tim
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