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Urgent help needed for quadratic graphs

  1. Jun 2, 2009 #1
    can someone pls tell me how you can get the value of 'a' or the gradient in a quadratic equation by analysing the quadratic graph


    as a reminder the equation is y=ax[tex]^{}2[/tex]+ bx + c
     
    Last edited: Jun 2, 2009
  2. jcsd
  3. Jun 2, 2009 #2

    danago

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    Are you able to find three points on the curve by looking at the graph?
     
  4. Jun 2, 2009 #3
    i'm not sure what you mean by the 3 points on the graph. can you please elaborate on that.
     
  5. Jun 2, 2009 #4

    danago

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    Well each position along the curve has an x and y coordinate -- When i refer to a point, im just talking about an ordered pair (x,y) that lies somewhere on the curve.

    If you know three such points, you could plug them each into the equation for the graph and solve for a,b,c simultaneously.
     
  6. Jun 2, 2009 #5
    i see what you mean. well yes, i know the points tht lie on the graph. but how do you plug these coordinates into the equation to find the value of 'a'
     
  7. Jun 2, 2009 #6

    danago

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    Well lets say you have three points, (p,q), (r,s) and (t,u). What these are saying is that when the x coordinates are p,q or r, the y coordinates are q,s and u respectively, so the must fit the following three equations:

    q = ap^2 + bp + c
    s = ar^2 + br + c
    u = at^2 + bt + c

    Which is essentially just a linear system of three equations in three variables (a,b and c).
     
  8. Jun 2, 2009 #7
    If you choose one of the points at x=0, the problem is slightly easier.

    The following equations result, for the points (0,Y_0), (X_1,Y_1) and (X_2, Y_2)

    [tex] c=Y_0 [/tex]

    [tex] a {X_1}^{\;2}+b X_1=Y_1-Y_0 [/tex]

    [tex] a X_2^{\;2}+b X_2=Y_2-Y_0 [/tex]

    You now have two linear equations with two unknowns.

    The problem is quick to solve either way, but often the X=0 point is well-known. So, you might as well take advantage of it when this is true.
     
    Last edited: Jun 2, 2009
  9. Jun 2, 2009 #8
    i sooort of see what you mean, but could you please give me an example with this method to find the value of a 'a'/gradient.
     
  10. Jun 2, 2009 #9
    (0,0) (1,1) (2,4)

    c=0
    a+b=1
    4a+2b=4 (or 2a+b=2 is better)

    subtract third equation from second.

    (2a+b=2)
    -(a+b=1)
    ----------------
    a=1

    Use second equation

    a+b=1

    b=0
     
  11. Jun 2, 2009 #10

    danago

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    Consider a quadratic graph that passes through the points (-1,3), (0,1) and (1,3)

    The general form of thw quadratic is y = ax^2+bx+c, so we can form the three equations:

    3 = a(-1)^2 + (-1)b + c = a-b+c
    1 = a(0)^2 + (0)b + c = c
    3 = a(1)^2 + (1)b + c = a+b+c

    Clearly, we already have c=1.

    Add the first and last equations together and we get:
    6 = 2a + 2c = 2a + 2 (since c=1)

    Hence a=2.

    You can then plug in a=2 and c=1 into the first equation and solve for b to find that b=0, hence the equation for the graph is y = 2x^2 + 1


    EDIT: aha elect_eng beat me to it :P ah well i guess two examples is better than 1 :smile:



    Also, is there any reason you are referring to 'a' as the gradient of the graph? For a linear equation, the coefficient of x is the gradient because the slope does not change along the whole line, however for a quadratic, the slope does change, so it not always equal to 'a'.
     
  12. Jun 2, 2009 #11
    Ha! The OP can easily see that I'm much more lazy than you. I took the easier case. :smile:
     
  13. Jun 2, 2009 #12
    i think i understand it now.

    thanks a lot for the help danago and elect_eng!!!!!!!!!!!! :smile:
     
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