1. **Problem:** Find the domain and range of the function $f(x) = 1 + 3x + x^2$. 2. The function is a quadratic polynomial which is defined for all real $x$, so the domain is all real numbers: $\text{Domain} = (-\infty, \infty)$. 3. To find the range, rewrite $f(x)$ in vertex form by completing the square: $$f(x) = x^2 + 3x + 1 = (x^2 + 3x + \left(\frac{3}{2}\right)^2) - \left(\frac{3}{2}\right)^2 + 1 = (x + \frac{3}{2})^2 - \frac{9}{4} + 1 = (x + \frac{3}{2})^2 - \frac{5}{4}$$ 4. Since $(x + \frac{3}{2})^2 \geq 0$ for all $x$, the minimum value of $f(x)$ is $-\frac{5}{4}$ when $x = -\frac{3}{2}$. 5. Therefore, the range is $[-\frac{5}{4}, \infty)$. **Final answer:** - Domain: $(-\infty, \infty)$ - Range: $[-\frac{5}{4}, \infty)$