we have
[tex]f(x)=\frac{1}{4} x^{2} +bx+10[/tex]
This is a vertical parabola open upward
The axis of symmetry is equal to the x-coordinate of the vertex
The vertex is the point (h,k)
the equation of the axis of symmetry is [tex]x=h[/tex]
In this problem
[tex]x=6[/tex]
so the x-coordinate of the vertex is [tex]h=6[/tex]
Convert the quadratic equation in vertex form
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex]f(x)-10=\frac{1}{4} x^{2} +bx[/tex]
Factor the leading coefficient
[tex]f(x)-10=\frac{1}{4}(x^{2} +4bx)[/tex]
Complete the square. Remember to balance the equation by adding the same constants to each side
[tex]f(x)-10+b^{2}=\frac{1}{4}(x^{2} +4bx+4b^{2})[/tex]
Rewrite as perfect squares
[tex]f(x)+(b^{2}-10)=\frac{1}{4}(x+2b)^{2}[/tex]
[tex]f(x)=\frac{1}{4}(x+2b)^{2}-(b^{2}-10)[/tex]
remember that
[tex]h=6[/tex]
so
[tex](x+2b)=(x-6)[/tex]
[tex]2b=-6[/tex]
[tex]b=-3[/tex]
substitute in the equation
[tex]f(x)=\frac{1}{4}(x-6)^{2}-((-3)^{2}-10)[/tex]
[tex]f(x)=\frac{1}{4}(x-6)^{2}+1[/tex]
therefore
the answer is
[tex]b=-3[/tex]
