#### Answer

Option (H)

#### Work Step by Step

Note that:
$x^2-2x = x^2-2x+(1-1)
\\x^2-2x=(x^2-2x+1)-1
\\x^2-2x=(x-1)^2-1$
Thus, the given function can be written as:
$f(x) =(x-1)^2-1$
RECALL:
(1) The vertex form of a quadratic function whose vertex is at $(h, k)$ is: $f(x) = a(x-h)^2+k$
(2) The graph of the quadratic function $ax^2+bx+c$ is a parabola that opens:
(i) upward when $a \gt 0$;
(ii) downward when $a\lt 0$.
Using the vertex form in (1) above, the given function has its vertex at $(1,-1)$.
Using the rule in (2) above, with $a=1$, the given function's graph is a parabola that opens upward.
The parabola that opens upward and whose vertex is at $(1, -1)$ is the one in Option (H).