To determine the value of [tex]\( b \)[/tex] in the quadratic function [tex]\( f(x) = 3x^2 + bx + 4 \)[/tex], we need to use the fact that the axis of symmetry of a parabola described by the equation [tex]\( ax^2 + bx + c \)[/tex] is given by the formula:
[tex]\[ x = -\frac{b}{2a} \][/tex]
In our given function [tex]\( f(x) = 3x^2 + bx + 4 \)[/tex], the coefficient [tex]\( a \)[/tex] is 3. According to the problem, the axis of symmetry is [tex]\( x = \frac{3}{2} \)[/tex]. Let's set up the equation using the axis of symmetry formula:
[tex]\[ \frac{3}{2} = -\frac{b}{2 \cdot 3} \][/tex]
Simplify the denominator on the right hand side:
[tex]\[ \frac{3}{2} = -\frac{b}{6} \][/tex]
To solve for [tex]\( b \)[/tex], we multiply both sides of the equation by 6:
[tex]\[ 6 \cdot \frac{3}{2} = -b \][/tex]
[tex]\[ 3 \cdot 3 = -b \][/tex]
[tex]\[ 9 = -b \][/tex]
So, multiplying both sides by -1 gives:
[tex]\[ b = -9 \][/tex]
Hence, the value of [tex]\( b \)[/tex] is [tex]\( -9 \)[/tex].
The correct answer is:
[tex]\[ \boxed{-9} \][/tex]
