1. **State the problem:** We are given the function $f(x) = (x - 1)^2 + 2$ and need to find its domain and range. 2. **Domain:** The domain of a function is the set of all possible input values ($x$) for which the function is defined. Since $f(x)$ is a quadratic function (a polynomial), it is defined for all real numbers. \[ \text{Domain} = (-\infty, \infty) \] 3. **Range:** The function $f(x) = (x - 1)^2 + 2$ is a parabola opening upwards because the coefficient of the squared term is positive. - The vertex form of a parabola is $f(x) = a(x - h)^2 + k$, where $(h, k)$ is the vertex. - Here, $a = 1$, $h = 1$, and $k = 2$, so the vertex is at $(1, 2)$. - Since the parabola opens upwards, the minimum value of $f(x)$ is at the vertex, which is $2$. - Therefore, the range is all $y$ values greater than or equal to $2$. \[ \text{Range} = [2, \infty) \] 4. **Graph description:** The graph is a parabola with vertex at $(1, 2)$ opening upwards. The $x$-axis and $y$-axis intersect at the origin $(0,0)$. 5. **Summary:** - Domain: $(-\infty, \infty)$ - Range: $[2, \infty)$ This completes the solution.