Answer
Please see the graph.
Axis of symmetry: $x=4$
Vertex: $(4, 2)$
Work Step by Step
$f(x)=(x-4)^2-2$
$f(x)=(x-4)(x-4)-2$
$f(x)=x*x+x*(-4)+(-4)*x+(-4)(-4)-2$
$f(x)=x^2-4x-4x+16-2$
$f(x)=x^2-8x+14$
$a=1$, $b=-8$, $c=14$
$x=-b/2a$ is the vertex.
$x=-b/2a$
$x=-(-8)/2*1$
$x=8/2$
$x=4$
$f(x)=(x-4)^2-2$
$f(4)=(4-4)^2-2$
$f(4)=0^2-2$
$f(4)=0-2$
$f(4)=-2$
