1. **State the problem:** We need to graph the quadratic function $$y = 3x^2$$ and understand its properties. 2. **Formula and rules:** The general form of a quadratic function is $$y = ax^2 + bx + c$$. Here, $$a = 3$$, $$b = 0$$, and $$c = 0$$. - Since $$a > 0$$, the parabola opens upwards. - The vertex is at the origin $$(0,0)$$ because $$b = 0$$ and $$c = 0$$. 3. **Intermediate work:** - Calculate some points to plot: - When $$x = -2$$, $$y = 3(-2)^2 = 3 \times 4 = 12$$. - When $$x = -1$$, $$y = 3(-1)^2 = 3 \times 1 = 3$$. - When $$x = 0$$, $$y = 0$$. - When $$x = 1$$, $$y = 3(1)^2 = 3$$. - When $$x = 2$$, $$y = 3(2)^2 = 12$$. 4. **Explanation:** The parabola is symmetric about the y-axis. It is narrower than the basic parabola $$y = x^2$$ because the coefficient 3 stretches it vertically. 5. **Final answer:** The graph of $$y = 3x^2$$ is a parabola opening upwards with vertex at the origin and points as calculated above.