What do you mean by "compute"? Do you want to graph it? Factor it? Calculate it's function given a set of points that lie on it? If you're looking to compute the function given three points that fall on the parabola, then I have just the code for you. If you're given three points, (x1, y1), (x2, y2) and (x3, y3), then you can compute the coefficients of your quadratic equation like this: a = (y1 * (x2 - x3) + y2 * (x3 - x1) + y3 * (x1 - x2)) / (x1 * x1 * (x2 - x3) + x2 * x2 * (x3 - x1) + x3 * x3 * (x1 - x2)) b = (y1 - y2) / (x1 - x2) - a * (x1 + x2); c = y1 - (x1 * x1) * a - x1 * b; You now can calculate the y co-ordinate of any point given it's x co-ordinate by saying: y = a * x * x + b * x + c;
i think its pretty much the same thing because matrix X1 X2 IS ACTUALLY X1 X2
The equation for the slope between the points A = (x1, y1) and B = (x2, y2) = (y2 - y1)/(x2 - x1), provided x1 is different from x2. If x1 and x2 are the same then the slope is not defined.
If (x1, y1) and (x2, y2) are two points on the line, then the formula for the slope is (y2-y1)/(x2-x1) provided x2 ≠x1. If x2 = x1 then the line is vertical and the slope is not defined.
Suppose that you have simple two variable model: Y=b0+b1X1+e The least squares estimator for the slope coefficient, b1 can be obtained with b1=cov(X1,Y)/var(X1) the intercept term can be calculated from the means of X1 and Y b0=mean(Y)-b1*mean(X1) In a larger model, Y=b0+b1X1+b2X2+e the estimator for b1 can be found with b1=(cov(X1,Y)var(X2)-cov(X2,Y)cov(X1,X2))/(var(X1)var(X2)-cov(X1,X2)2) to find b2, simply swap the X1 and X2 terms in the above to get b2=(cov(X2,Y)var(X1)-cov(X1,Y)cov(X1,X2))/(var(X1)var(X2)-cov(X1,X2)2) Find the intercept with b0=mean(Y)-b1*mean(X1)-b2*mean(X2) Beyond two regressors, it just gets ugly.
the function (x),sory I can`t use the sign of the function because it is not available. the function of (x)=4x+4 is one to one function assume function(x1)= function(x2) then 4(x1)+4 =4(x2)+4 4(x1)=4(x2) (x1)=(x2) hence,the function is one to one
this is the increasing function theorem, hope it helps "If F'(x) >= 0 , and all x's are and element of [a,b], Then F is increasing on [a,b]" use Mean Value Theorem (M.V.T) Let F'(x)>=0 on some interval Let x1< x2 (points from that interval) by M.V.T there is a point C which is an element of [x1,x2] such that F(x2)-F(x1) / X2- X1 = F'(C) this implies: F(x2)-F(x1) = F'(C) X [x2-x1] F'(C)>=0 [x2-x1]>0 therefore: F(x2)>=F(x1) Therefore: F is increasing on that interval.
Need two points. m = slope. (X1, Y1) and (X2, Y2) m = Y2 - Y1/X2 - X1 ==============Or, if function is in this form...... Y =mX + b ======== Read off of function, or get function is this form.
What do you mean by "compute"? Do you want to graph it? Factor it? Calculate it's function given a set of points that lie on it? If you're looking to compute the function given three points that fall on the parabola, then I have just the code for you. If you're given three points, (x1, y1), (x2, y2) and (x3, y3), then you can compute the coefficients of your quadratic equation like this: a = (y1 * (x2 - x3) + y2 * (x3 - x1) + y3 * (x1 - x2)) / (x1 * x1 * (x2 - x3) + x2 * x2 * (x3 - x1) + x3 * x3 * (x1 - x2)) b = (y1 - y2) / (x1 - x2) - a * (x1 + x2); c = y1 - (x1 * x1) * a - x1 * b; You now can calculate the y co-ordinate of any point given it's x co-ordinate by saying: y = a * x * x + b * x + c;
If the variables are x1 & x2 the solution is : 1) x1=x1+x2; 2) x2=x1-x2; 3) x1=x1-x2; EX: x1=1 , x2=6; 1) x1= 1+6 = 7 2) x2= 7-6 =1 3 x1=7-1 =6 ============================================
y=x2+3 x1=1 x2=2 y(x1) = 1*1+3 = 4 y(x2) = 2*2+3 = 7 x2/x1 = 2, While y2/y1 = 7/4 !=2, and thus the function is nonlinear.
i think its pretty much the same thing because matrix X1 X2 IS ACTUALLY X1 X2
This deck contains. Monsters Flamvell firedog x2 flamvell magician x3 Flamvell guard x2 masked dragon x3 dread dragon x2 majestic dragon x2 worm caratos x2 worm dimikles x2 worm gulse x1 mystic tomato x3 changer synchron x2 reptilianne naga x2 drill barnicle x1 spells pot of avarice x1 pot of benevolence x2 monster reborn x1 cards of consonence x1 scapegoat x1 stardust shimmer x2 dark hole x1 heavy storm x1 burden of the mighty x1 crashbug road x2 swords of revealing light x1 spirit burner x1 Traps parallel selection x2 bottomless trap hole x2 trap hole x2 threating roar x2 call of the huanted x1 shadow spell x1 starlight road x2 urgent tuning x2 magic cylinder x1
The formula: distance=sqrt(((x1-x2)*(x1-x2))+((y1-y2)*(y1-y2))+((z1-z2)*(z1-z2))) In DarkBASIC it's: function distance3D(x1,y1,z1,x2,y2,z2) x=x1-x2 y=y1-y2 z=z1-z2 result=sqrt((x*x)+(y*y)+(z*z)) endfunction result In classic BASIC I think it's: FUNCTION distance3D(x1,y1,z1,x2,y2,z2) x=x1-x2 y=y1-y2 z=z1-z2 result=SQRT((x*x)+(y*y)+(z*z)) RETURN result END FUNCTION
okay... took me ages to figure out how to finish it but... up(X3), Right(X5), Up(X2), Left(X1), Down(X4), Right(X2), Down(X1), Left(X1), Right(X1), Down(X1), Left(X1), Down(X1), Right(X1), Down(X1), Left(X1), Down(X3), Right(X2), Down(X1), Right(X1), Down(X2), Left(X2), Up(X1), Right(X1), Down(X4), Right(X1), Up(X2), Left(X2), Down(X2), Right(X1) and finally Up(X1) :) code for level 30 is 676534
It is pretty much impossible
If this is in the context of finding a root of an equation, the answer is to make some guesses. Find value x1 and x2 such that f(x1) and f(x2) have opposite signs. Then, provided that f is a continuous function over (x1, x2), the bisection method will find its root.