SOLUTION: Find b such that f(x)=-2x^2+bx-30 has a maximum value of 2 and the vertex is located in the second quadrant.

Question 159196This question is from textbook College Algebra
: Find b such that f(x)=-2x^2+bx-30 has a maximum value of 2 and the vertex is located in the second quadrant. This question is from textbook College Algebra

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Find b such that f(x)=-2x^2+bx-30 has a maximum value of 2 and the vertex is located in the second quadrant.
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You have a quadratic equation with a = -2, b = b, and c = -30
Maximum value for f(x) occurs when x = -b/2a = -b/(-4) = b/4
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Then f(b/4) = -2(b/4)^2 + b(b/4) -30 = 2
-b^2/8 + b^2/4 - 30 = 2
b^2 - 2b^2 + 240 = -16
b^2 = 256
b = +16 or b = -16
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If b = 16 the vertex is at x= b/4 = 16 and x = 64 which is in the 1st quadrant
If b = -16 the vertex is at x=b/4 = -16 and x = -64 which is in the 2nd Quad.
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Answer b = -16
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f(x)=-2x^2-16x-30

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Cheers,
Stan H.
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