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The axis of symmetry for the graph of the function f(x) = 3x^2 + bx + 4 is x = 3/2 . What is the value of b?

User KevenDenen
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Answer:

b = -9

Explanation:

Axis of symmetry of a parabola = -b / 2a.

Where y = ax² + bx + c.

Given f(x) = 3x^2 + bx + 4,

a = 3, and c = 4.

If the axis of symmetry is 3/2, than b can be found by substitution and algebra.

[ x = -b / 2a ] → [ 3 / 2 = -b / 2a ] →

[ 3 / 2 = -b / 2 (3) ] → [ 3 / 2 = -b / 6 ] →

[ 3 / 2 × 6 = -b / 6 × 6 ] → [ 18 / 2 = -b ] →

[ 9 = -b ] → [ -b = 9 ] → [ (-)(-b) = (-)(9) ] →

b = -9

User Carloluis
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