Answer:
\(b=-12\)
Step-by-step explanation:
we have
\(f(x)=x^{2}+bx+10\)
we know that
The equation of a vertical parabola into vertex form is equal to
\(y=a(x-h)^{2} +k\)
where
(h,k) is the vertex of the parabola
and the axis of symmetry is equal to \(x=h\)
In this problem we have the axis of symmetry \(x=6\)
so
the x-coordinate of the vertex is equal to \(6\)
therefore
For \(x=6+1=7\) -----> one unit to the right of the vertex
Find the value of \(f(7)\)
\(f(7)=7^{2}+b(7)+10\)
\(f(7)=59+7b\)
For \(x=6-1=5\) -----> one unit to the left of the vertex
Find the value of \(f(5)\)
\(f(5)=5^{2}+b(5)+10\)
\(f(5)=35+5b\)
Remember that
\(f(5)=f(7)\) ------> the x-coordinates are at the same distance from the axis of symmetry
so
\(59+7b=35+5b\) ------> solve for b
\(7b-5b=35-59\)
\(2b=-24\)
\(b=-12\)
