1. **State the problem:** Solve the quadratic equation $x^2 + x - 12 = 0$ and analyze its graph. 2. **Recall the quadratic formula:** For $ax^2 + bx + c = 0$, the solutions are given by $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=1$, and $c=-12$. 3. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = 1^2 - 4 \times 1 \times (-12) = 1 + 48 = 49$$ 4. **Find the roots:** $$x = \frac{-1 \pm \sqrt{49}}{2 \times 1} = \frac{-1 \pm 7}{2}$$ 5. **Evaluate each root:** - For $+$ sign: $$x_1 = \frac{-1 + 7}{2} = \frac{6}{2} = 3$$ - For $-$ sign: $$x_2 = \frac{-1 - 7}{2} = \frac{-8}{2} = -4$$ 6. **Find the vertex:** The vertex $x$-coordinate is given by $$x_v = -\frac{b}{2a} = -\frac{1}{2 \times 1} = -0.5$$ 7. **Calculate the vertex $y$-coordinate:** $$y_v = (x_v)^2 + x_v - 12 = (-0.5)^2 + (-0.5) - 12 = 0.25 - 0.5 - 12 = -12.25$$ 8. **Axis of symmetry:** The vertical line through the vertex is $$x = -0.5$$ 9. **Summary:** - Roots: $x_1 = 3$, $x_2 = -4$ - Vertex: $(-0.5, -12.25)$ - Axis of symmetry: $x = -0.5$ - The parabola opens upwards because $a=1 > 0$. Final answer: The solutions to the equation are $x=3$ and $x=-4$, with vertex at $(-0.5, -12.25)$ and axis of symmetry $x=-0.5$.