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satellite73
Group Title
@amistre64 please refresh my feeble memory
snap way to find polynomial through (2,4),(0,6),(4,70)
 one year ago
 one year ago
satellite73 Group Title
@amistre64 please refresh my feeble memory snap way to find polynomial through (2,4),(0,6),(4,70)
 one year ago
 one year ago

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amistre64 Group TitleBest ResponseYou've already chosen the best response.3
hmm, my method is to construct it such that each x value causes the unknown constants to zero out
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
i thought i had this, but i resorted to a two by two system then i tried writing \[45(x+2)+4x(x+2)\] which worked but it was agony finding those constants i though you had a snap way of finding them
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.3
if you can do a matrix augment, thats pretty snappy
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
yeah but i wanted the snappy method you used
 one year ago

Hero Group TitleBest ResponseYou've already chosen the best response.0
....also interested in "snappy" method :D
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.3
i believe the method you posted is what i used at first (2,4),(0,6),(4,70) y = a + bx + cx(x+2) ; (0,6) 6 = a +0+0 y = 6 + bx + cx(x+2) ; (2,4) 10 = 2b + 0 ; b=5 y = 6 5x + cx(x+2) ; (4,70) 70+6+20 = c4(4+2) 96/24 = c = 4 y = 6 5x + 4x(x+2)
 one year ago

joemath314159 Group TitleBest ResponseYou've already chosen the best response.1
Expanding this should work as well
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.3
if there was another method that was snappier; you might have to refresh me memory how it looked to you
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.3
joes looks kinda legendre to me
 one year ago

Hero Group TitleBest ResponseYou've already chosen the best response.0
@joemath314159, your method looks like some numerical methods stuff
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
no that was what i was looking for i guess that is what i did, somehow it looked easier when you did it. i guess the grass is always greener hello @Hero hello @joemath314159
 one year ago

joemath314159 Group TitleBest ResponseYou've already chosen the best response.1
It is legendre, learned it in Linear Algebra
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.3
its these lighntning fast clickers of mine, just makes it appear rico y suave lol
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
(0,2) (1,1) (2,6) (3,19)
 one year ago

Hero Group TitleBest ResponseYou've already chosen the best response.0
@satellite73, are you taking a course?
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
yes, at the school of amistre
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
\[a+bx+cx(x1)+cx(x1)(x2)+dx(x1)(x2)\] actually i think this is slightly different then legrange, but maybe it is identical
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.3
(0,2) (1,1) (2,6) (3,19) y = 2 + bx +cx(x1) + dx(x1)(x2) 1 = 2 + b; b= 3 y = 2 + 3x +cx(x1) + dx(x1)(x2) 6 = 2 +6 + 2c ; c=1 y = 2 + 3x +x(x1) + dx(x1)(x2) 19 = 2 + 9 + 6 + d(2); d=3 y = 2 + 3x +x(x1) + 3x(x1)(x2) if i mathed it right
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.3
legendre i believe zeros out each constant term and omits one zero so that just a single constant is exposed at any time. Newtons method is similar to mine in that new information can be added as needed without have to reconstruct the whole equation
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
yeah legendre looks like what @joemath314159 wrote newtons i am not sure about
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
now i can sleep better. thanks!
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.3
http://nptel.iitm.ac.in/courses/Webcoursecontents/IITKANPUR/mathematics2/node109.html
 one year ago
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