Question 423981: Consider the quadratic function; f(x)=r^2 + (x+s)(x-s) here r,s are arbitary constants, find the min possible value for f and also the value of x for which the min is achieved.
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Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Consider the quadratic function; f(x)=r^2 + (x+s)(x-s) here r,s are arbitary constants, find the min possible value for f and also the value of x for which the min is achieved.
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f(x)=r^2 + (x+s)(x-s)
f(x) = x^2-s^2+r^2
a = 1 ; b = 0 ; c = (r^2-s^2)
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The minimum value occurs when x = -b/(2a) = 0/(2) = 0
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f(0) = r^2-s^2
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Min: (0, r^2-s^2)
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Cheers,
Stan H.
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