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    Say you have the graph of a quadratic equation, and you have all the x/y values but not the equation of the line, is there a way you go about finding the equation without trial & error?
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    well if any of your co-ordinates have x=0 or y=0 it will help in terms of working out y-intercept etc... apart from that i actually have no idea
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    you have to...quadra....BOOOOOOOOBS......
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    Well where it intercepts the x axis would be the solutions, so then you could write it as a multiple of that factorised form, and then another point would enable you to deduce the exact equation.
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    All the x/y values?
    If you have 3 different sets, you can form simultaneous equations and solve them for a, b and c(for a quadratic being ax^2+bx+c).
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    In other news:

    Put 3 points through y = ax^2 + bx + c, getting you 3 equations in terms of a, b and c. use your favourite method to solve them, and write your quadratic.
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    can you upload a pic of the graph??

    it helps if everyone can see it
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    (Original post by Slumpy)
    All the x/y values?
    If you have 3 different sets, you can form simultaneous equations and solve them for a, b and c(for a quadratic being ax^2+bx+c).
    Tried that, just took me round in a circle.

    http://tutor2u.net/economics/revisio...p_image001.gif

    The graph looks a bit like that (the TR line, disregard everything else), but the problem is it doesnt cut the x-axis. I was thinking maybe using the vertex and trying to complete the square?
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    (Original post by Zweihander)
    Tried that, just took me round in a circle.

    http://tutor2u.net/economics/revisio...p_image001.gif

    The graph looks a bit like that (the TR line, disregard everything else), but the problem is it doesnt cut the x-axis
    That looks like it cuts the x-axis at x=0, and also like it's not a quadratic, are you certain it is?
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    (Original post by Slumpy)
    That looks like it cuts the x-axis at x=0, and also like it's not a quadratic, are you certain it is?
    yeah, but x=0 doesnt count because you cant use it to determine the equation, and yes it's a quadratic in the form y = -ax^2 + bx + c, where y = revenue, and x = quantity sold
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    (Original post by Zweihander)
    yeah, but x=0 doesnt count because you cant use it to determine the equation, and yes it's a quadratic in the form y = -ax^2 + bx + c, where y = revenue, and x = quantity sold
    You certainly can.
    If x=0 is a root of a quadratic, you can factorise the quadratic to x(x-a)
    And, it looks like a circle arc, but I'll take your word for it.
    All you need now is any other point (x,y) to put into the equation
    y=x(x-a).
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    (Original post by Slumpy)
    You certainly can.
    If x=0 is a root of a quadratic, you can factorise the quadratic to x(x-a)
    And, it looks like a circle arc, but I'll take your word for it.
    All you need now is any other point (x,y) to put into the equation
    y=x(x-a).
    Wait, maybe I'm missing something but surely

     y = x(x-a) = x^2 - ax, how does that relate to this graph? and the picture I posted isnt the actual graph, it's just a graph that looks like the graph in my textbook, which is definitely a quadratic equation.
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    (Original post by Zweihander)
    Wait, maybe I'm missing something but surely

     y = x(x-a) = x^2 - ax, how does that relate to this equation? and the picture I posted isnt the graph, it's just a graph that looks like the graph in my textbook, which is definitely a quadratic equation.
    You said you have all values of x and y.
    Take a point (x,y), put the values of x and y in  y =  x^2 - ax, get  a =  \frac{x^2 - y}{x}, and you've found a.
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    (Original post by Slumpy)
    You said you have all values of x and y.
    Take a point (x,y), put the values of x and y in  y =  x^2 - ax, get  a =  \frac{x^2 - y}{x}, and you've found a.
    what I dont get is how you've gone from

     y = -ax^2 + bx + c, to

     y = x(x-a)
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    (Original post by Zweihander)
    what I dont get is how you've gone from

     y = -ax^2 + bx + c, to

     y = x(x-a)
    In the graph you've shown, x=0 is a point where y is 0. If this isn't the case on your actual graph, disregard it.
    Do you know every quadratic can be expressed as (x-a)(x-b) where a and b are the x values for the x-axis intercepts?
    Using that knowledge, and the knowledge that (0,0) is an intercept, we can set b=0, and get the quadratic=x(x-a)

    Edit-the two aren't the same a, the a in the first is the coefficient of x^2, and the second it's one of the roots of the quadratic. In general they won't be the same.
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    well if one of the x=0 and y=0 at the same point... y=x(x-a) proves that either x=0 or x-a=0 which would make y=0... from there its easy
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    (Original post by Zweihander)
    what I dont get is how you've gone from

     y = -ax^2 + bx + c, to

     y = x(x-a)
    he's just saying that since this quadratic has 2 roots and one of them is 0, then you can factorise it as y = (x - 0)(x - k) ie y = x(x-k)
    now find a point x,y, sub it in so that you get a value for k, and you've found your quadratic
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    (Original post by Slumpy)
    In the graph you've shown, x=0 is a point where y is 0. If this isn't the case on your actual graph, disregard it.
    Do you know every quadratic can be expressed as (x-a)(x-b) where a and b are the x values for the x-axis intercepts?
    Using that knowledge, and the knowledge that (0,0) is an intercept, we can set b=0, and get the quadratic=x(x-a)

    Edit-the two aren't the same a, the a in the first is the coefficient of x^2, and the second it's one of the roots of the quadratic. In general they won't be the same.
    yeah, I get it now, but like the graph in the picture there's only one x-axis intercept - (0,0). In this particular case there's a maximum quantity of goods sold (i.e. a maximum value for x) so the graph cuts off at that point. Am I right in guessing that unless the graph has 2 roots this is impossible?
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    (Original post by Zweihander)
    yeah, I get it now, but like the graph in the picture there's only one x-axis intercept - (0,0). In this particular case there's a maximum quantity of goods sold (i.e. a maximum value for x) so the graph cuts off at that point. Am I right in guessing that unless the graph has 2 roots this is impossible?
    The graph does have two roots, you just aren't shown where the other is.
    And as stated above, if you're given the values of x and y at any other point you can solve to find the equation.
 
 
 
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