1. **State the problem:** Find the intercept form of the quadratic function $$y=2x^2 - 2x - 12$$ and find its vertex. 2. **Intercept form formula:** The intercept form of a quadratic is $$y = a(x - r_1)(x - r_2)$$ where $$r_1$$ and $$r_2$$ are the roots (x-intercepts). 3. **Find roots:** Solve $$2x^2 - 2x - 12 = 0$$ by dividing both sides by 2: $$x^2 - x - 6 = 0$$ Factor: $$(x - 3)(x + 2) = 0$$ So roots are $$x=3$$ and $$x=-2$$. 4. **Write intercept form:** Using roots, $$y = 2(x - 3)(x + 2)$$ which matches option A. 5. **Find vertex:** Vertex formula for $$y = ax^2 + bx + c$$ is $$x = -\frac{b}{2a}$$ Here, $$a=2$$ and $$b=-2$$, so $$x = -\frac{-2}{2 \times 2} = \frac{2}{4} = \frac{1}{2}$$. 6. **Find y-coordinate of vertex:** Substitute $$x=\frac{1}{2}$$ into original equation: $$y = 2\left(\frac{1}{2}\right)^2 - 2\left(\frac{1}{2}\right) - 12 = 2\times \frac{1}{4} - 1 - 12 = \frac{1}{2} - 1 - 12 = -\frac{25}{2}$$. 7. **Vertex is** $$\left(\frac{1}{2}, -\frac{25}{2}\right)$$ which matches option C. **Final answers:** - Intercept form: $$y=2(x-3)(x+2)$$ - Vertex: $$\left(\frac{1}{2}, -\frac{25}{2}\right)$$