Question

the axis of symmetry for the graph of the function $f(x)=3x^2+bx+4$ is $x = \frac{3}{2}$. what is the value of $b$?\
\bigcirc -18\
\bigcirc -9\
\bigcirc 9\
\bigcirc 18

Explanation:

Step1: Recall axis of symmetry formula

For a quadratic function \( f(x) = ax^2 + bx + c \), the axis of symmetry is given by \( x = -\frac{b}{2a} \).

Step2: Identify values of \( a \) and axis of symmetry

In the function \( f(x) = 3x^2 + bx + 4 \), \( a = 3 \) and the axis of symmetry is \( x = \frac{3}{2} \).

Step3: Substitute into the formula and solve for \( b \)

Substitute \( a = 3 \) and \( x = \frac{3}{2} \) into \( x = -\frac{b}{2a} \):
\[
\frac{3}{2} = -\frac{b}{2 \times 3}
\]
Simplify the right - hand side: \( \frac{3}{2}=-\frac{b}{6} \)
Multiply both sides by 6 to get rid of the denominator: \( 6\times\frac{3}{2}=-b \)
Calculate \( 6\times\frac{3}{2} = 9 \), so \( 9=-b \)
Multiply both sides by - 1: \( b=-9 \)

Answer:

\( b=-9 \), so the correct option is -9 (the option with -9, i.e., the second option: -9)