1. The problem describes a parabola with the axis of symmetry $x=2$, vertex at $(2,1)$, minimum value $1$, no maximum, domain all real numbers, and range $y \geq 1$. 2. The general form of a parabola with vertex $(h,k)$ is given by: $$y = a(x - h)^2 + k$$ where $a$ determines the direction and width of the parabola. 3. Since the parabola opens upwards (minimum at vertex), $a > 0$. 4. Using the vertex $(2,1)$, substitute $h=2$ and $k=1$: $$y = a(x - 2)^2 + 1$$ 5. The minimum value is $1$ at $x=2$, confirming the vertex form. 6. The domain is all real numbers, so $x \in (-\infty, \infty)$. 7. The range is $y \geq 1$ because the parabola opens upwards and the vertex is the minimum point. 8. The end behavior as $x \to \pm \infty$ is $y \to \infty$. 9. Without additional points, $a$ cannot be numerically determined, but the function form is: $$y = a(x - 2)^2 + 1, \quad a > 0$$