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 $b$?
-12
-3
$\frac{1}{2}$
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 axis of symmetry value

In the function \( f(x)=\frac{1}{4}x^2 + bx + 10 \), \( a=\frac{1}{4} \) and the axis of symmetry \( x = 6 \).

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

Substitute \( a=\frac{1}{4} \) and \( x = 6 \) into \( x=-\frac{b}{2a} \):
\[
6=-\frac{b}{2\times\frac{1}{4}}
\]
Simplify the denominator: \( 2\times\frac{1}{4}=\frac{1}{2} \), so the equation becomes:
\[
6 = -\frac{b}{\frac{1}{2}}
\]
Which is equivalent to \( 6=-2b \) (since dividing by \( \frac{1}{2} \) is multiplying by 2).
Solve for \( b \): \( b = 6\div(-2)=-3 \)

Answer:

-3