Question

domain:-1 0 1 2 3 4
range: -2 0 0 1 2 3 what is the rule?

Answer 1

what does the domain differ from the range?

Answer 2

domain is the x and range is the y'

Answer 3

a rather lengthy way is to simply setup a 5th degree poly that hits all those points

Answer 4

and i think i mixed up some x and y parts in that so its prolly not the correct stuff; but the general idea is sound :)

Answer 5

-1 0 1 2 3 4

(x+1)
(x-0)
(x-1)
(x-2)
(x-3)
(x-4)


+a(x-0)(x-1)(x-2)(x-3)(x-4)
+b(x+1)(x-1)(x-2)(x-3)(x-4)
+c(x+1)(x-0)(x-2)(x-3)(x-4)
+d(x+1)(x-0)(x-1)(x-3)(x-4)
+e(x+1)(x-0)(x-1)(x-2)(x-4)
+f(x+1)(x-0)(x-1)(x-2)(x-3)

that might prove more exacting

Answer 6

when x=-1, y=-2; everything zeros out but the a term, giving us
\[-2=a(-1-0)(-1-1)(-1-2)(-1-3)(-1-4)\]
\[\frac{-2}{(-1)(-2)(-3)(-4)(-5)}=a\]
\[\frac{1}{60}=a\]

Answer 7

I believe newton had a similar method for hitting all the points; just as lengthy tho

setup so that each term zeros out to build up the coeffs as you go

y = a + b(x+1) + c(x+1)(x-0) + d(x+1)(x-0)(x-1)
+e(x+1)(x-0)(x-1)(x-2) + f(x+1)(x-0)(x-1)(x-2)(x-3)

now, when x=-1, y=-2 such that

-2 = a

y = -2


when x=0, y=0

0 = -2 + b(1) ; b=2


y = -2 + 2x+2 + c(x+1)(x-0) + d(x+1)(x-0)(x-1)
+e(x+1)(x-0)(x-1)(x-2) + f(x+1)(x-0)(x-1)(x-2)(x-3)


y = 2x + c(x+1)(x-0) + d(x+1)(x-0)(x-1)
+e(x+1)(x-0)(x-1)(x-2) + f(x+1)(x-0)(x-1)(x-2)(x-3)


etc ...

Answer 8

(1,0)

0 = 2 + 2c ; c = -1

y = -x^2 + x + d(x+1)(x-0)(x-1)
+e(x+1)(x-0)(x-1)(x-2) + f(x+1)(x-0)(x-1)(x-2)(x-3)


(2,1)

1 = -4 + 2 + 6d

3 = 6d ; d=2


y = 2x^3 - x^2 + 3x + e(x+1)(x-0)(x-1)(x-2)
+ f (x+1)(x-0)(x-1)(x-2)(x-3)

etc

Answer 9

err, d=1/2 :)