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Question 1111593: The minimum value of f(x) = x^2 + bx +4 is equal to the maximum value of g(x) = b + 2x -x^2. Find the value of b. Is It possible for the vertices of the two parabolas to coincide?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! The minimum value of f(x) = x^2 + bx +4 is equal to the maximum value of g(x) = b + 2x -x^2. Find the value of b. Is It possible for the vertices of the two parabolas to coincide?
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In each case the vertex occurs when x = -b/(2a)
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Examine f(x) = x^2 + bx + 4
Minimum value and Vertex occur at x = -b/(2*1) = -(1/2)b
Minimum value is f(-b/2) = (b^2/4)-b^2/2+4 = (-b^2/4)+4
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Examine g(x) = -x^2 + 2x + b
Maximum value and Vertex occur at x = -2/(2*-1) = 1
Maximum value is g(1) = -1+2+b = 1+b
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Since min of f(x) = max of g(x), (-b^2/4)+4 = 1 + b
-b^2/4 - b + 3 = 0
b = [1 +- sqrt(1-4(-1/4)(3)]/(2(-1/4)) = [1+-2]/(-1/2) = -6 or 1
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Can they coincide ?::
For f(x), vertex is at (3,-5) or (-1/2,15/4)
For g(x), vertex is at (1,-5) or (1,2)
Ans: No
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Cheers,
Stan H.
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