Respuesta :

gmany
[tex]f(x)=ax^2+bx+c\\\\h=\dfrac{-b}{2a};\ k=f(h)[/tex]

We have:
[tex]h(x)=x^2-8x+14\\\\a=1;\ b=-8;\ c=14[/tex]

Substitute:
[tex]h=\dfrac{-(-8)}{2\cdot1}=\dfrac{8}{2}=4[/tex]

Completing the squqre:

h(x) = x^2 - 8x + 14  =  x^2 - 8x + 16 - 16 + 14  =  (x-4)^2 - 2

Here h=4 and k = -2.