## satellite73 3 years ago @amistre64 please refresh my feeble memory snap way to find polynomial through (-2,4),(0,-6),(4,70)

1. amistre64

hmm, my method is to construct it such that each x value causes the unknown constants to zero out

2. satellite73

i thought i had this, but i resorted to a two by two system then i tried writing \[4-5(x+2)+4x(x+2)\] which worked but it was agony finding those constants i though you had a snap way of finding them

3. amistre64

if you can do a matrix augment, thats pretty snappy

4. satellite73

yeah but i wanted the snappy method you used

5. Hero

....also interested in "snappy" method :D

6. amistre64

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)

7. joemath314159

Expanding this should work as well

8. amistre64

if there was another method that was snappier; you might have to refresh me memory how it looked to you

9. amistre64

joes looks kinda legendre to me

10. Hero

@joemath314159, your method looks like some numerical methods stuff

11. satellite73

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

12. joemath314159

It is legendre, learned it in Linear Algebra

13. amistre64

its these lighntning fast clickers of mine, just makes it appear rico y suave lol

14. satellite73

(0,-2) (1,1) (2,6) (3,19)

15. Hero

@satellite73, are you taking a course?

16. satellite73

yes, at the school of amistre

17. Hero

lol

18. satellite73

\[a+bx+cx(x-1)+cx(x-1)(x-2)+dx(x-1)(x-2)\] actually i think this is slightly different then legrange, but maybe it is identical

19. amistre64

(0,-2) (1,1) (2,6) (3,19) y = -2 + bx +cx(x-1) + dx(x-1)(x-2) 1 = -2 + b; b= 3 y = -2 + 3x +cx(x-1) + dx(x-1)(x-2) 6 = -2 +6 + 2c ; c=1 y = -2 + 3x +x(x-1) + dx(x-1)(x-2) 19 = -2 + 9 + 6 + d(2); d=3 y = -2 + 3x +x(x-1) + 3x(x-1)(x-2) if i mathed it right

20. amistre64

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

21. satellite73

yeah legendre looks like what @joemath314159 wrote newtons i am not sure about

22. satellite73

now i can sleep better. thanks!

23. amistre64