we have
[tex]f(x)=-(x+8)(x-14)[/tex]
Convert to vertex form
[tex]f(x)=-(x+8)(x-14)\\f(x)=-( x^{2}-14x+8x-112) \\f(x)=-( x^{2}-6x-112)\\f(x)=-x^{2}+6x+112[/tex]
we know that
the equation of a vertical parabola in vertex form is equal to
[tex]y=a(x-h)^{2} +k[/tex]
where
(h,k) is the vertex
[tex]f(x)=-x^{2}+6x+112[/tex]
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex]f(x)-112=-(x^{2}-6x)[/tex]
Complete the square. Remember to balance the equation by adding the same constants to each side
[tex]f(x)-112-9=-(x^{2}-6x+9)[/tex]
[tex]f(x)-121=-(x^{2}-6x+9)[/tex]
Rewrite as perfect squares
[tex]f(x)-121=-(x-3)^{2}[/tex]
[tex]f(x)=-(x-3)^{2}+121[/tex]
the vertex is the point [tex](3,121)[/tex]
therefore
the answer is
The y-value of the vertex is [tex]121[/tex]
