1. **Problem:** Graph the quadratic function $f(x) = 3x^2 + 6x + 1$ and determine its domain and range. 2. **Formula and rules:** The domain of any quadratic function is all real numbers, $\mathbb{R}$, because you can plug any $x$ value into a polynomial. The range depends on the vertex and whether the parabola opens up or down. Since the coefficient of $x^2$ is positive ($3 > 0$), the parabola opens upwards, so the range is $[y_{\text{min}}, \infty)$ where $y_{\text{min}}$ is the vertex's $y$-value. 3. **Find the vertex:** The vertex $x$-coordinate is given by $$x = -\frac{b}{2a} = -\frac{6}{2 \times 3} = -\frac{6}{6} = -1.$$ 4. **Find the vertex $y$-value:** Substitute $x = -1$ into $f(x)$: $$f(-1) = 3(-1)^2 + 6(-1) + 1 = 3(1) - 6 + 1 = 3 - 6 + 1 = -2.$$ 5. **Domain:** All real numbers, so $$\text{Domain} = (-\infty, \infty).$$ 6. **Range:** Since parabola opens up and vertex is minimum point, $$\text{Range} = [-2, \infty).$$ --- **Summary:** - Domain: $(-\infty, \infty)$ - Range: $[-2, \infty)$