1. The problem asks which functions have graphs with symmetry to the left of the vertex and axis of symmetry. 2. The given function is $f(x) = (x - 1)^2 + 1$. This is a parabola with vertex at $(1,1)$ and axis of symmetry $x=1$. 3. The axis of symmetry is the vertical line through the vertex. For $f(x)$, it is $x=1$. 4. To determine if the symmetry is to the left of the vertex, we check the axis of symmetry of each function. 5. Analyze each option: - A. $g(x) = 2(x - 1)^2 - 1$ has vertex at $(1,-1)$, axis $x=1$ (same as $f$), so symmetry is at $x=1$ (not left). - B. $g(x) = - (x + 1)^2 - 2$ has vertex at $(-1,-2)$, axis $x=-1$ which is to the left of $x=1$. - C. $g(x) = 2(x - 1)^2 - 2$ has vertex at $(1,-2)$, axis $x=1$ (not left). - D. $g(x) = - (x - 1)^2 + 1$ has vertex at $(1,1)$, axis $x=1$ (not left). - E. $g(x) = - (x + 2)^2 + 2$ has vertex at $(-2,2)$, axis $x=-2$ which is to the left of $x=1$. 6. Therefore, the functions with symmetry to the left of the vertex of $f(x)$ are options B and E. Final answer: B and E.