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Math Help - Maximum Value???

  1. #1
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    Maximum Value???

    Find the value of a such that h(x) = ax^2 + 2ax + 4 will have a maximum value of 7?

    How do I go about doing this? I've been stuck for almost an hour with this problem.
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  2. #2
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    Quote Originally Posted by strigy View Post
    Find the value of a such that h(x) = ax^2 + 2ax + 4 will have a maximum value of 7?

    How do I go about doing this? I've been stuck for almost an hour with this problem.
    The standard way of determining a maximum value for a quadratic is to complete the square: ax^2+ 2ax+ 4= a(x^2+ 2x)+ 4 = a(x^2+ 2x+ 1- 1)+ 4= a(x^2+ 2x+ 1)- a+ 4 = a(x^2+ 2x+1)- a+4= a(x+1)^2- a+4. Since a square if never negative, as long as a itself is negative, that is -a+ 4 minus something. That is, its maximum value occurs at x= -1 and is equal to -a+ 4.
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    Quote Originally Posted by strigy View Post
    Find the value of a such that h(x) = ax^2 + 2ax + 4 will have a maximum value of 7?

    How do I go about doing this? I've been stuck for almost an hour with this problem.

    For a\neq 0\,,\,\,ax^2+2ax+4 is a parabola. A parabola has a maximum value if and only if it is a concave-down parabola, this is: if and only if a<0 (and this already is a limitation to a).
    In this case, the maximal point is at the parabola's vertex, which in this case is given by \left(-2\,,-\frac{4a^2-16a}{4a}\right) , with the y-entry being the maximal value. Well, now solve your problem.

    Tonio
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