we have
\(f(x)=\frac{1}{4} x^{2} +bx+10\)
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 \(x=h\)
In this problem
\(x=6\)
so the x-coordinate of the vertex is \(h=6\)
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
\(f(x)-10=\frac{1}{4} x^{2} +bx\)
Factor the leading coefficient
\(f(x)-10=\frac{1}{4}(x^{2} +4bx)\)
Complete the square. Remember to balance the equation by adding the same constants to each side
\(f(x)-10+b^{2}=\frac{1}{4}(x^{2} +4bx+4b^{2})\)
Rewrite as perfect squares
\(f(x)+(b^{2}-10)=\frac{1}{4}(x+2b)^{2}\)
\(f(x)=\frac{1}{4}(x+2b)^{2}-(b^{2}-10)\)
remember that
\(h=6\)
so
\((x+2b)=(x-6)\)
\(2b=-6\)
\(b=-3\)
substitute in the equation
\(f(x)=\frac{1}{4}(x-6)^{2}-((-3)^{2}-10)\)
\(f(x)=\frac{1}{4}(x-6)^{2}+1\)
therefore
the answer is
\(b=-3\)
