Question

the axis of symmetry for the graph of the function $f(x) = \frac{1}{4}x^2 + bx + 10$ is $x = 6$. what is the value of $b$?\
\
$\circ$ $-12$\
$\circ$ $-3$\
$\circ$ $\frac{1}{2}$\
$\circ$ $3$

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 \( a \) and substitute \( x \)

In the function \( f(x) = \frac{1}{4}x^2 + bx + 10 \), \( a = \frac{1}{4} \) and the axis of symmetry \( x = 6 \). Substitute into the formula: \( 6 = -\frac{b}{2\times\frac{1}{4}} \).

Step3: Simplify and solve for \( b \)

Simplify the denominator: \( 2\times\frac{1}{4} = \frac{1}{2} \), so the equation becomes \( 6 = -\frac{b}{\frac{1}{2}} \), which is \( 6 = -2b \). Solve for \( b \): \( b = 6\div(-2) = -3 \).

Answer:

-3