Answer
$\bf{Option \ (F)}$
Work Step by Step
1) The graph of the quadratic function $f(x) = ax^2+bx+c$ is a parabola that opens:
(a) Upward when $a \gt 0$ and its vertex has a minimum value.
(b) Downward when $a \lt 0$ and its vertex has a maximum value.
On comparing $f(x)=x^2-2x+1$ with $f(x) = ax^2+bx+c$, we get: $a=1, b=-2,c=1; a \gt 0$, we can see that the given function shows a graph of a parabola that opens upward.
2) The coordinates of the vertex of a quadratic function $f(x) = ax^2+bx+c$ are given by: $\displaystyle(\frac{-b}{2a}, f(-\frac{2}{a}))$
Therefore, the coordinates of the function's vertex can be expressed as: $\displaystyle(\frac{-(-2)}{2(1)}, f(-\frac{-2}{2(1)})) $
The minimum value is: $f(1)=1^2-(2)(1)+1=0$
Thus, the parabola that opens downward and whose vertex is at $(1,0)$ matches with $\bf{Option \ (F)}$.
