To determine the value of [tex]\( b \)[/tex] for the function [tex]\( f(x) = \frac{1}{4}x^2 + bx + 10 \)[/tex] given that the axis of symmetry is [tex]\( x = 6 \)[/tex], follow these steps:
1. Recall that for a quadratic function [tex]\( ax^2 + bx + c \)[/tex], the axis of symmetry is given by the formula:
[tex]\[
x = -\frac{b}{2a}
\][/tex]
2. For the given function [tex]\( f(x) = \frac{1}{4}x^2 + bx + 10 \)[/tex], identify the coefficients:
[tex]\[
a = \frac{1}{4}, \quad b = b, \quad c = 10
\][/tex]
3. Substitute [tex]\( a = \frac{1}{4} \)[/tex] into the axis of symmetry formula:
[tex]\[
x = -\frac{b}{2 \cdot \frac{1}{4}} = -\frac{b}{\frac{1}{2}} = -2b
\][/tex]
4. We know the axis of symmetry is [tex]\( x = 6 \)[/tex]. Set this equal to our derived expression for the axis of symmetry:
[tex]\[
-2b = 6
\][/tex]
5. Solve for [tex]\( b \)[/tex]:
[tex]\[
-2b = 6 \\
b = -\frac{6}{2} \\
b = -3
\][/tex]
Therefore, the value of [tex]\( b \)[/tex] is [tex]\( \boxed{-3} \)[/tex].
