To find the value of \( b \) for the quadratic function \( f(x) = 3x^2 + bx + 4 \) given that the axis of symmetry is \( x = \frac{3}{2} \), we will use the formula for the axis of symmetry of a quadratic function.
The formula for the axis of symmetry for a quadratic function \( f(x) = ax^2 + bx + c \) is given by:
[tex]\[
x = -\frac{b}{2a}
\][/tex]
Here, the function is \( f(x) = 3x^2 + bx + 4 \). Therefore, \( a = 3 \) and \( b \) is the coefficient we need to find. We know the axis of symmetry is \( x = \frac{3}{2} \).
Substitute \( x \) and \( a \) into the axis of symmetry formula:
[tex]\[
\frac{3}{2} = -\frac{b}{2 \cdot 3}
\][/tex]
Simplify the right side:
[tex]\[
\frac{3}{2} = -\frac{b}{6}
\][/tex]
To isolate \( b \), we will multiply both sides of the equation by 6:
[tex]\[
6 \cdot \frac{3}{2} = -b
\][/tex]
Simplify the left side:
[tex]\[
9 = -b
\][/tex]
Therefore, we multiply both sides by -1 to solve for \( b \):
[tex]\[
b = -9
\][/tex]
Thus, the value of \( b \) is:
[tex]\[
\boxed{-9}
\][/tex]
