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Math Help - determing quadratic rules

  1. #1
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    determing quadratic rules


    Between e, and f how do I know whether to use y= a(x-h)^2+k form, or ax^2 +bx +c form? And how do I work it out?


    And how do I work out this one step by step?
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  2. #2
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    Quote Originally Posted by delicate_tears View Post

    Between e, and f how do I know whether to use y= a(x-h)^2+k form, or ax^2 +bx +c form? And how do I work it out?


    And how do I work out this one step by step?
    to #3:

    All points on the x-axis have the y-coordinate zero. Thus you know 2 points of the parabola:

    P(-1, 4), Q(6, 0) . Plug in these coordinates into the given equation. You'll get a system of linear equations. Solve for a and b:

    \left|\begin{array}{l}4 = a - b\\0 = 36a+6b\end{array}\right. ...... I've got (a, b) = \left(\frac47, -\frac{24}{7}\right)
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  3. #3
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    Quote Originally Posted by delicate_tears View Post

    Between e, and f how do I know whether to use y= a(x-h)^2+k form, or ax^2 +bx +c form? And how do I work it out?

    ...
    If you know the coordinates of the vertex the equation y= a(x-h)^2+k would be the best.

    In both cases you know the coordinates of the vertex and the coordinates of one additional point. I'm going to do f so you can try to do e:

    You know: V(2, 2) and P(0, 6). Plug in these coordinates:

    6 = a (0 - 2)^2 + 2~\implies~ 6 = 4a+2~\implies~a = 1 . Thus your equation becomes:

    y = (x-2)^2+2
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  4. #4
    rbell35
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    Question Quadratic Rules - h & k values

    Does this mean that h & k will always be the x & y co-ordinates of the turning point, if the quadratic is written in the form y = a(x-h) squared + k??? Please clarify..Thanks, Rob
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