Quadratic Minimum Zeros 5753Fe

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. **Find the vertex of the parabola (minimum point):** The vertex form of a quadratic is $$y = a(x-h)^2 + k$$ where $(h,k)$ is the vertex. The vertex $x$-coordinate is given by $$x = -\frac{b}{2a}$$ where $a=3$ and $b=6$. Calculate: $$x = -\frac{6}{2 \times 3} = -\frac{6}{6} = -1$$ 3. **Find the minimum value by substituting $x=-1$ into the equation:** $$y = 3(-1)^2 + 6(-1) + 3 = 3(1) - 6 + 3 = 3 - 6 + 3 = 0$$ So the minimum value of the equation is $$0$$ at $$x = -1$$. 4. **Compare with the graph's vertex:** The graph's vertex is approximately at $(-2, 1)$, which means the minimum value on the graph is about $$1$$. 5. **Answer the first question:** The equation's minimum value is $$0$$, the graph's minimum value is about $$1$$, so the equation has the smaller minimum value. 6. **Find zeros of the equation:** Solve $$3x^2 + 6x + 3 = 0$$. Divide both sides by 3: $$\cancel{3}x^2 + \cancel{3}2x + \cancel{3}1 = 0 \Rightarrow x^2 + 2x + 1 = 0$$ Factor: $$(x+1)^2 = 0$$ So the equation has one zero at $$x = -1$$. 7. **Check zeros on the graph:** The graph's vertex is at $(-2,1)$ and opens upward, so it does not touch or cross the x-axis, meaning it has no zeros. 8. **Answer the second question:** The equation has zeros, the graph has no zeros. **Final answers:** - Smaller minimum value: the equation. - Has no zeros: the graph.