The axis of symmetry for the graph of the function [tex]f(x) = \frac{1}{4} x^2 + bx + 10[/tex] is [tex]x = 6[/tex]. What is the value of [tex]b[/tex]?

A. [tex]-12[/tex]
B. [tex]-3[/tex]
C. [tex]\frac{1}{2}[/tex]
D. [tex]3[/tex]



Answer :

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].