anonymous
  • anonymous
Find a cubic function with the given zeros. √6 , -√6, -3
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
@amistre64
amistre64
  • amistre64
what do we know about zeros of a function, what do they do to the function?
anonymous
  • anonymous
they mark the intercepts?

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amistre64
  • amistre64
yeah, they do mark the x intercepts. but something more useful to is is that the make the function, equal out to zero.
amistre64
  • amistre64
now, another thing we should know, is that anything times 0, is equal to 0 ... right?
anonymous
  • anonymous
yep
amistre64
  • amistre64
so, would you agree that if x=a, then x-a = 0 ?
anonymous
  • anonymous
yes
amistre64
  • amistre64
given any number of roots, we can then create a function: say the roots are a,b,c let f(x) = (x-a)(x-b)(x-c) will f(a)=0? f(b)=0? and f(c)=0?
anonymous
  • anonymous
okay, so we have to find which one ?
amistre64
  • amistre64
we simply need to expand (multiply thru, distribute) this into the cubic format.
amistre64
  • amistre64
let f(x) = (x-a)(x-b)(x-c) for the stated zeros; a,b, and c how would we expand this into a cubic format?
anonymous
  • anonymous
multiply it out
amistre64
  • amistre64
can you show me how it looks when we multiply it out?
anonymous
  • anonymous
http://www.wolframalpha.com/share/clip?f=d41d8cd98f00b204e9800998ecf8427elr1dfihajj is that right?
amistre64
  • amistre64
that link is not working for my browser :/ (x-c)(xx-bx-ax+ba) xxx-bxx-axx+bax -cxx+bcx+cax-abc x^3 -(a+b+c)xx^2 +(ab+ac+bc)x -abc
amistre64
  • amistre64
let a=sqrt(6), b=-sqrt(6), and c=-3
anonymous
  • anonymous
ok
anonymous
  • anonymous
btw the link says x^3 + 3x^2 -6x -18
amistre64
  • amistre64
a+b+c = -3, so --3 = 3 is fine ab+ac+bc = -6, thats good and -(--6.3) is -18
anonymous
  • anonymous
so now we just combine them to get the function
amistre64
  • amistre64
that IS the function ..
amistre64
  • amistre64
f(x) = x^3 +3x^2-6x-18
amistre64
  • amistre64
any scaled form of this also works ... so this is not a unique solution
anonymous
  • anonymous
i see, thanks so much!
amistre64
  • amistre64
goodluck

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