1. **State the problem:** We need to identify the graph of the quadratic function $$y = 12x^2 - 80x + 270$$. 2. **Recall the properties of quadratic functions:** The graph of a quadratic function $$y = ax^2 + bx + c$$ is a parabola. - If $$a > 0$$, the parabola opens upward. - If $$a < 0$$, it opens downward. - The vertex (turning point) is at $$x = -\frac{b}{2a}$$. 3. **Identify the direction of the parabola:** Here, $$a = 12 > 0$$, so the parabola opens upward. 4. **Find the vertex:** $$x = -\frac{b}{2a} = -\frac{-80}{2 \times 12} = \frac{80}{24} = \frac{10}{3} \approx 3.33$$ 5. **Calculate the y-coordinate of the vertex:** $$y = 12\left(\frac{10}{3}\right)^2 - 80\left(\frac{10}{3}\right) + 270$$ $$= 12 \times \frac{100}{9} - \frac{800}{3} + 270$$ $$= \frac{1200}{9} - \frac{2400}{9} + \frac{2430}{9} = \frac{1230}{9} = 136.67$$ 6. **Interpretation:** The vertex is at approximately $$(3.33, 136.67)$$, and the parabola opens upward. 7. **Conclusion:** Among the given graphs, the correct one is the parabola opening upward with a vertex near the right side (around $$x=3.33$$). This matches **Graph C**. **Final answer:** The graph of $$y = 12x^2 - 80x + 270$$ is **Graph C**.