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Family of quadratic functions?

  1. Sep 26, 2006 #1
    This question has been killing me for days, :
    Give a family of quadratic functions of the form y=ax^2+bx+c, that passes through the following points:
    (1,1) and (2,0)
    I see how we can find the family, but how do we find the specific functions that pass through both those points? I tried making two separate equations (1 for each point, like(1,1) I would sub in for x and y) but that didn't give me anything useful. Even if I use elimination for the 2 equations, I don't know where it will get me....what should i do?
  2. jcsd
  3. Sep 26, 2006 #2
    Of course it does. You get the two equations

    [tex]1 = a + b + c[/tex]


    [tex]0 = 4a + 2b + c,[/tex]

    which can be solved to get [itex]b = -1 - 3a[/itex] and [itex]c = 2(a+1)[/itex] with no conditions on [itex]a[/itex]. So what's your family of functions?
  4. Sep 26, 2006 #3
    Really? Well first of all I put down the wrong 2 points (aorry, my fault) they should be (1,) and (-1,-2). I got 0=a+b+c and -2=a-b+c. I solved to get b=-1 and i think c=-1-a. I don't know how to continue!
  5. Sep 26, 2006 #4
    woops, points= (1,0) and (-1,-2)
  6. Sep 26, 2006 #5
    does anyone know?
  7. Sep 26, 2006 #6
    well, if b=-1 and c=-1-a in ax^2 + bx + c then what do the polynomials that go through those points look like?
  8. Sep 26, 2006 #7
    i don't know, now i'm lost:uhh: do you mean like a parabola? a linear line with a slope? how do i define the family? why can't i just sub in the known values of b and c:
  9. Sep 26, 2006 #8
    you can! that is exactly the "family" it's looking for: all the quadratics with the form [itex]a^2 + x - 1 - a[/itex] for nonzero constants [itex]a[/itex] (you actually made an error earlier. You should have found [itex]b=1[/itex], not [itex]b=-1[/itex]). :tongue2:
  10. Sep 28, 2006 #9
    should it be [tex]a^2 + x - 1 - a[/tex]? or [tex]ax^2 + x - 1 - a[/tex]?
  11. Sep 28, 2006 #10
    [itex]ax^2 + x - 1 -a.[/itex] I'm not very good at typing!
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