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).