1. The problem is to analyze the quadratic function $$y = ax^2 + bx + c$$ and understand how changing the coefficients $a$, $b$, and $c$ affects the graph and the discriminant $$D = b^2 - 4ac$$. 2. The discriminant $$D$$ determines the nature of the roots of the quadratic equation: - If $$D > 0$$, there are two distinct real roots. - If $$D = 0$$, there is exactly one real root (a repeated root). - If $$D < 0$$, there are no real roots (the roots are complex). 3. Given $$a=1$$, $$b=4$$, and $$c=1$$, calculate the discriminant: $$D = b^2 - 4ac = 4^2 - 4 \times 1 \times 1 = 16 - 4 = 12$$ 4. Since $$D = 12 > 0$$, the quadratic has two distinct real roots. 5. The roots can be found using the quadratic formula: $$x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-4 \pm \sqrt{12}}{2}$$ 6. Simplify the roots: $$\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$$ $$x = \frac{-4 \pm 2\sqrt{3}}{2} = \frac{\cancel{-4} \pm \cancel{2}\sqrt{3}}{\cancel{2}} = -2 \pm \sqrt{3}$$ 7. Approximate the roots: $$-2 + \sqrt{3} \approx -2 + 1.732 = -0.268$$ $$-2 - \sqrt{3} \approx -2 - 1.732 = -3.732$$ 8. The vertex of the parabola is at: $$x = -\frac{b}{2a} = -\frac{4}{2} = -2$$ 9. Calculate the vertex's y-coordinate: $$y = 1 \times (-2)^2 + 4 \times (-2) + 1 = 4 - 8 + 1 = -3$$ 10. The parabola opens upward because $$a = 1 > 0$$. Summary: Changing $$a$$ affects the parabola's width and direction (up/down). Changing $$b$$ shifts the vertex horizontally and affects the roots. Changing $$c$$ shifts the parabola vertically. Final answer: The quadratic $$y = x^2 + 4x + 1$$ has two distinct real roots approximately at $$-3.732$$ and $$-0.268$$, with vertex at $$(-2, -3)$$, and opens upward.