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Determining the second order polynomial from the intersection points

  1. Oct 13, 2012 #1
    1. The problem statement, all variables and given/known data
    Let's say that we have a second order polynomial function, and we know all of the points where it intersects with the x and y axis. Ex: (-2; 0), (0; 2), (1; 0)

    How does on determine the ax^2+bx+c polynomial form based on that?

    2. Relevant equations
    -


    3. The attempt at a solution

    Tried searching for it on Google without any luck.
     
  2. jcsd
  3. Oct 13, 2012 #2
    What is a second order polynomial? What are both coordinates of the intersections?
     
  4. Oct 13, 2012 #3

    HallsofIvy

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    I assume that you are saying that this is a general second order polynomial so you have answered sbj-2812's first question. You should also know that the point (a, b) lies on the graph of y= f(x) if and only if b= f(a). If (-2, 0), (0, 2), and (1, 0) are on the graph of [itex]y= ax^2+ bx+ c[/itex] then we must have [itex]0= a(-2)^2+ b(-2)+ c[/itex] or 4a- 2b+ c= 0, [itex]2= a(0)^2+ b(0)+ c[/itex] or [itex]c= 2[/itex], and [itex]0= a(1)^2+ b(1)+ c[/itex] or [itex]a+ b+ c= 0[/itex]
    Solve the equations a+ b+ c= 0, c= 0, and 4a- 2b+ c= 0 for a, b, and c.

     
  5. Oct 13, 2012 #4

    Ray Vickson

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    Do you know the relation between factors of a polynomial and the roots of the polynomial? If you do not, see http://www.sosmath.com/algebra/factor/fac02/fac02.html . Using the relationship makes your problem very easy. That is material well worth knowing.

    RGV
     
  6. Oct 13, 2012 #5

    SammyS

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    A second order (second degree) polynomial having the form ax2+bx+c and having two real root may be written as a(x-D)(x-F).

    From this it should be easy to solve your problem.
     
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