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## anonymous one year ago determine the polynomial function whose graph passes through the given points (-1,-1) (0,0) (1,1) (2,4)

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1. anonymous

@amistre64

2. amistre64

we have 4 reference points, what do you consider we can do with them?

3. amistre64

4 equations, with 4 unknowns ... can we do matrix stuff yet?

4. anonymous

ax^2 + bx +c ?

5. amistre64

thats only 3 unknowns :) lets go another degree

6. anonymous

oooohhhh ax^3 +bx^2+cx

7. amistre64

ax^3 + bx^2 + cx + d = y

8. amistre64

but yeah, since 0,0, then d=0 by default

9. anonymous

so then |dw:1444009179944:dw|

10. anonymous

is this right then

11. amistre64

the 0 0 0 0 is not required, but yeah, its fine. then row reduce it to echelon form rref{{(-1)^3,(-1)^2,(-1)^1,(-1)},{(1)^3,(1)^2,(1)^1,(1)},{(2)^3,(2)^2,(2)^1,(4)}}

12. amistre64

for checking .. if the link posts properly http://www.wolframalpha.com/input/?i=rref {{%28-1%29^3%2C%28-1%29^2%2C%28-1%29^1%2C%28-1%29}%2C{%281%29^3%2C%281%29^2%2C%281%29^1%2C%281%29}%2C{%282%29^3%2C%282%29^2%2C%282%29^1%2C%284%29}}

13. anonymous

so my matrix table is right then

14. amistre64

yes

15. amistre64

only thing that may be off is that you dont have your column for d, but d=0 due to the point 0,0

16. anonymous

but i have 4 column

17. amistre64

|dw:1444009601928:dw|

18. amistre64

your last column is a y column .. you had no d variable column

19. amistre64

|dw:1444009738187:dw|

20. amistre64

|dw:1444009782699:dw|

21. amistre64

do you follow that?

22. anonymous

nah really

23. amistre64

each column is an unknown, and the last is the y ... we have 4 unknown so we need 4 columns, and a 5th one for y

24. amistre64

IF we can already determine a variable, then it is reduced to 3 unknown ... 3 columns and a 4th for y

25. amistre64

ax^3 + bx^2 + cx + d = y -1a + 1b + -1c + 1d = -1 0a + 0b + 0c + 1d = 0 1a + 1b + 1c + 1d = 1 8a + 4b + 2c + 1d = -1

26. amistre64

do you see that d=0 already? if not we can simply work the matrix with these columns as is

27. anonymous

yes

28. anonymous

alright thanks ill go ahead and work on it now

29. amistre64

ok

30. anonymous

so now how do i graph it

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