Quadratic Minimum Zeros 2Df34B

1. **State the problem:** We are given the quadratic equation $$y = 3x^2 + 6x + 3$$ and a graph of this parabola. We need to determine which has the smaller minimum value: the equation or the graph, and which has no zeros. 2. **Recall the formula for the vertex of a parabola:** For a quadratic equation $$y = ax^2 + bx + c$$, the vertex's x-coordinate is given by $$x = -\frac{b}{2a}$$. 3. **Calculate the vertex x-coordinate:** Here, $$a = 3$$ and $$b = 6$$. $$x = -\frac{6}{2 \times 3} = -\frac{6}{6} = -1$$ 4. **Calculate the vertex y-coordinate (minimum value):** Substitute $$x = -1$$ into the equation: $$y = 3(-1)^2 + 6(-1) + 3 = 3(1) - 6 + 3 = 3 - 6 + 3 = 0$$ 5. **Compare with the graph's vertex:** The graph's vertex is given as $$(-2, 1)$$, which means the minimum value on the graph is $$1$$. 6. **Conclusion on minimum values:** The equation's minimum value is $$0$$ at $$x = -1$$, but the graph's vertex is at $$(-2, 1)$$ with minimum value $$1$$. Since the parabola opens upward, the minimum value of the function is at the vertex. The graph's vertex does not match the equation's vertex, so the graph shown does not represent the equation exactly. Therefore, the equation has the smaller minimum value. 7. **Check zeros of the equation:** Solve $$3x^2 + 6x + 3 = 0$$. Divide both sides by 3: $$\cancel{3}x^2 + \cancel{3} \times 2x + \cancel{3} = 0 \Rightarrow x^2 + 2x + 1 = 0$$ This factors as: $$(x + 1)^2 = 0$$ So, the equation has one zero at $$x = -1$$. 8. **Check zeros of the graph:** The graph has no x-intercepts (zeros) as stated. 9. **Conclusion on zeros:** The equation has one zero at $$x = -1$$, but the graph has no zeros. **Final answers:** - The equation has the smaller minimum value. - The graph has no zeros, but the equation has one zero. Hence: - Smaller minimum value: the equation. - No zeros: the graph.