1. The problem is to analyze and graph the quadratic function $f(x) = 3x^2 + 3x - 6$. 2. The general form of a quadratic function is $f(x) = ax^2 + bx + c$, where $a$, $b$, and $c$ are constants. 3. For this function, $a=3$, $b=3$, and $c=-6$. 4. To find the vertex, use the formula for the x-coordinate of the vertex: $$x = -\frac{b}{2a} = -\frac{3}{2 \times 3} = -\frac{3}{6} = -\frac{1}{2}.$$ 5. Substitute $x = -\frac{1}{2}$ into $f(x)$ to find the y-coordinate of the vertex: $$f\left(-\frac{1}{2}\right) = 3\left(-\frac{1}{2}\right)^2 + 3\left(-\frac{1}{2}\right) - 6 = 3\times \frac{1}{4} - \frac{3}{2} - 6 = \frac{3}{4} - \frac{3}{2} - 6.$$ 6. Simplify the y-coordinate: $$\frac{3}{4} - \frac{3}{2} - 6 = \frac{3}{4} - \frac{6}{4} - \frac{24}{4} = \frac{3 - 6 - 24}{4} = \frac{-27}{4} = -6.75.$$ 7. The vertex is at $$\left(-\frac{1}{2}, -6.75\right).$$ 8. The parabola opens upwards because $a=3 > 0$. 9. To find the y-intercept, set $x=0$: $$f(0) = -6.$$ 10. To find the x-intercepts, solve $3x^2 + 3x - 6 = 0$: Divide both sides by 3: $$\cancel{3}x^2 + \cancel{3}x - \cancel{6} = 0 \Rightarrow x^2 + x - 2 = 0.$$ 11. Factor the quadratic: $$(x + 2)(x - 1) = 0.$$ 12. So, the solutions are: $$x = -2 \quad \text{or} \quad x = 1.$$ 13. The x-intercepts are at $(-2, 0)$ and $(1, 0)$. 14. Summary: - Vertex: $\left(-\frac{1}{2}, -6.75\right)$ - Opens upward - Y-intercept: $(0, -6)$ - X-intercepts: $(-2, 0)$ and $(1, 0)$ This information can be used to sketch the graph of the function.