Answer
The vertex is labeled with its coordinates, and the green line is the axis of symmetry.
Work Step by Step
$f(x)=(x-4)^2-2$
$f(x)=(x-4)(x-4)-2$
$f(x)=x^2-8x+16-2$
$f(x)=x^2-8x+14$
$a=1$, $b=-8$, $c=14$
The axis of symmetry is at $x=-b/2a$
$x=-(-8)/2*1$
$x=8/2*1$
$x=8/2$
$x=4$
The vertex is at $x=4$
$f(x)=(x-4)^2-2$
$f(4)=(4-4)^2-2$
$f(4)=0^2-2$
$f(4)=0-2$
$f(4)=-2$
$(4, -2)$ is the vertex.
