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