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Quadratic Equations

  1. Mar 5, 2008 #1
    [SOLVED] Quadratic Equations

    How would I find what 'a' and 'q' is in the equation of y=a(x)^2+q when I'm given 2 plots?

    Example:Find 'a' and 'q' so that a parabola y=ax^2+q passes through each pair.
    a)(-3,11)and(4,18)

    if you could give me step by step explanation that would be much appreciated.
     
  2. jcsd
  3. Mar 5, 2008 #2
    What does it mean for the parabola y = ax2 + q to pass through a point (m, n)? It means that: n = am2 + q, because we substitute m and n for x and y. Do that for the two points you are given (-3, 11) and (4, 18). What equations do you get?
     
  4. Mar 5, 2008 #3
    well if i did substitute then i would get 11=a(-3)^2+q
    11=9a+q
    18=16a+q

    hmmm now i have a feeling i need to cancel out variable by elimination right?

    well yeah... i end up with 2 varibales 'a' and 'q'
     
    Last edited: Mar 5, 2008
  5. Mar 5, 2008 #4
    Yes, that is exactly what you would do. You have two equations, and two unknowns, which is exactly what you need.

    This is an interesting thing to notice. This quadratic equation is in a special form y = ax2 + c. However, if it was in the general form y = ax2 + bx + c, you will find that, after substituting points for x and y, you will have equations in three unknowns, namely a, b, and c! That means you will need three total equations, and three points. In general, three points completely determine a parabola.

    You can extend this further. Since a line is always y = ax + b, in two unknowns a and b, you only need two points to define a line. Since a cubic is y = ax3 + bx2 + cx + d is in 4 unknowns (a, b, c, d) you will need four equations, or four points to define a cubic. And so on.

    In your question, the quadratic was missing a bx term (or you can think of it as you already knowing that b = 0), so you only have two unknowns (a and c), and you only needed two points.
     
  6. Mar 5, 2008 #5
    oook i got it now thnx alot
     
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