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Math Help - gradient, minimum point

  1. #1
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    gradient, minimum point

    The curve C has the equation
     y = x^2 + ax + b
    where a and b are constants.
    Given that the minimum point of C has coordinates (−2, 5), find the values of a and b.
    I don't get how to do this question, cause I have missed classes when we were doing this and even after looking at examples in my book still can't understand
    can anyone explain ?

    thanks
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Tweety View Post
    I don't get how to do this question, cause I have missed classes when we were doing this and even after looking at examples in my book still can't understand
    can anyone explain ?

    thanks
    The minimum occurs when:

    \frac{d}{dx}\left[x^2+ax+b\right]=2x+a=0

    that is when x=-a/2. At x=-a/2: y=a^2/4 - a^2/2 +b=-a^2/4+b.

    But you are told that the point (-a/2, -a^2/4+b) is (-2,5) , so solve for a abd b.

    CB
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  3. #3
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    Without using calculus, completing the square:

    x^2+ ax+ b= x^2+ ax+ \frac{a^2}{4}- \frac{a^2}{4}+ b[/tex]= (x- a/2)^2+ b- \frac{a^2}{4}[/tex]
    Since a square is always positive, that has its minimum value, b- \frac{a^2}{4} when x= a/2.

    Your are told that x= a/2= -2 and y= b- \frac{a^2}{4}= 5, the same equations Captain Black got.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by HallsofIvy View Post
    Without using calculus, completing the square:

    x^2+ ax+ b= x^2+ ax+ \frac{a^2}{4}- \frac{a^2}{4}+ b[/tex]= (x- a/2)^2+ b- \frac{a^2}{4}[/tex]
    Since a square is always positive, that has its minimum value, b- \frac{a^2}{4} when x= a/2.

    Your are told that x= a/2= -2 and y= b- \frac{a^2}{4}= 5, the same equations Captain Black got.
    The question is in the calculus sub-forum.

    CB
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