Answer
a. $f(x)=(x-4)^{2}-8$
b. see image below
c. Minimum value $f(4)=-8$
Work Step by Step
a. Complete the square:
$x^{2}-8x=x^{2}-2\cdot(x)(4)+4^{2}-4^{2}$
$f(x)=(x-4)^{2}-16+8$
$f(x)=(x-4)^{2}-8$
b.
To sketch, begin with the parent function $f_{1}(x)=x^{2},$
(blue, dashed line in the image)
and, since $f(x)=f_{1}(x-4)-8,$
shift the graph to the right by $4$ units,
and down $8$ units
(solid red line, see image)
c.
The graph of $f(x)=a(x-h)^{2}+k$
is a parabola, and,
if $a > 0$, then the quadratic function $f$ opens upward and
has the minimum value $k$ at $x=h=-\displaystyle \frac{b}{2a}$.
Minimum value $f(4)=-8$
