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Family of quadratic functions

  1. Feb 17, 2007 #1
    A family of quadratic functions passes through the points (3,0). Find the family of quadratic functions

    err i have no idea hwo to do it except substituting those values in ... 0=9a+3b+c

    what does it meant the family of quadratic functions?
  2. jcsd
  3. Feb 18, 2007 #2
    And then find a in terms of b and c
    b in terms of a and c
    c in terms of a and b
    then put them back into the original quadratic.
  4. Feb 18, 2007 #3


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    From two equations, in general, you at most can eliminate only one of the unknowns. So from the original y=ax^2 +bx +c, you can remove only one of a or b or c.
    Write, a in terms of b and c (or, b in terms of a and c; or, c in terms of a and b) and put in the original.... that is your family of equations for any choice of the existing two parameters.
  5. Feb 18, 2007 #4

    and put in the original what
  6. Feb 18, 2007 #5
    so c =-12x
    is that right?
  7. Feb 18, 2007 #6


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    No, c = -(9a + 3b).

    So the family is,

    [tex]y(x) = a x^2 + b x -(9a + 3b)[/tex]

    The one given condition only lets you eliminate one unknown parameter. So you end up with a quadratic function that still has two free parameters, that's why it's referred to as a "family", there's lot of 'em. Get it?
    Last edited: Feb 18, 2007
  8. Feb 18, 2007 #7


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    The point is that any "quadratic function" can be written in the form
    y= f(x)= ax2+ bx+ c. You want to write a formula that describes the "family" (i.e. set) of all those that pass through (3,0)- that is, all those for which y= 0 when x= 3. Putting y= 0 and x= 3 into that original formula,
    0= 9a+ 3b+ c so c= -(9a+3b). The answer to the question is that the family of all quadratic functions that pass through (3, 0) are those of the form f(x)= ax2+ bx- (9a+ 3b).

    (That's one way to write the answer: we could also, of course, have solved 9a+ 3b+ c= 0 for a, in terms of b and c, or for b, in terms of a and c, and replaced that parameter instead of c.)
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