anonymous
  • anonymous
determine the polynomial function whose graph passes through the given points (-1,-1) (0,0) (1,1) (2,4)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
@amistre64
amistre64
  • amistre64
we have 4 reference points, what do you consider we can do with them?
amistre64
  • amistre64
4 equations, with 4 unknowns ... can we do matrix stuff yet?

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anonymous
  • anonymous
ax^2 + bx +c ?
amistre64
  • amistre64
thats only 3 unknowns :) lets go another degree
anonymous
  • anonymous
oooohhhh ax^3 +bx^2+cx
amistre64
  • amistre64
ax^3 + bx^2 + cx + d = y
amistre64
  • amistre64
but yeah, since 0,0, then d=0 by default
anonymous
  • anonymous
so then |dw:1444009179944:dw|
anonymous
  • anonymous
is this right then
amistre64
  • 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)}}
amistre64
  • 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}}
anonymous
  • anonymous
so my matrix table is right then
amistre64
  • amistre64
yes
amistre64
  • 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
anonymous
  • anonymous
but i have 4 column
amistre64
  • amistre64
|dw:1444009601928:dw|
amistre64
  • amistre64
your last column is a y column .. you had no d variable column
amistre64
  • amistre64
|dw:1444009738187:dw|
amistre64
  • amistre64
|dw:1444009782699:dw|
amistre64
  • amistre64
do you follow that?
anonymous
  • anonymous
nah really
amistre64
  • 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
amistre64
  • amistre64
IF we can already determine a variable, then it is reduced to 3 unknown ... 3 columns and a 4th for y
amistre64
  • 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
amistre64
  • amistre64
do you see that d=0 already? if not we can simply work the matrix with these columns as is
anonymous
  • anonymous
yes
anonymous
  • anonymous
alright thanks ill go ahead and work on it now
amistre64
  • amistre64
ok
anonymous
  • anonymous
so now how do i graph it

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