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    The least value of the function
    x^2 + px + q is 3, and this occurs when x = -2.
    Find the values of p and q.

    I first substituted -2 into the equation replacing x,
    and I got
    -2p + q = -1

    But I'm not sure what to do next?
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    (Original post by Sakura-Sama)
    The least value of the function
    x^2 + px + q is 3, and this occurs when x = -2.
    Find the values of p and q.

    I first substituted -2 into the equation replacing x,
    and I got
    -2p + q = -1

    But I'm not sure what to do next?
    If you say (x+\frac{p}{2})^2-\frac{p^2}{4}+q then substituting x=-2 will make the square term 0 (it has to in order for the function to take min value), so we need p such that (-2+\frac{p}{2})^2=0

    Get it?
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    You know the turning point of the function because you know the minimum value, 3, which occurs whey x = -2. Therefore the turning point is (-2,3).

    You can then use this if you complete the square of the function and sub in the values.
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    (Original post by Sakura-Sama)
    The least value of the function
    x^2 + px + q is 3, and this occurs when x = -2.
    Find the values of p and q.

    I first substituted -2 into the equation replacing x,
    and I got
    -2p + q = -1

    But I'm not sure what to do next?
    Two ways to do this:

    1) You need another equation as you have two unknowns.
    You know the minimum point and the coordinates of where this happens.

    Can you use differentiation and take it from there?

    2) (My preferred method), as the other posters above this post have stated, you can use complete the square.

    Re-write the equation  x^{2} + bx + c (note a=1) in form:

     f(x) = \big( x + (b/2) \big)^{2} + d,

    (d is found after you have completed the square) and using your knowledge of graph transformations, you have two equations to use to find p and q!

    -----------------------------------------------------------------------------

    Can you take it from here? :-)
 
 
 
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