Minimum And Zeros Ad7000

1. **State the problem:** We are given the function $$f(x) = 3(x - 1)^2 + 1$$ and a graph of a parabola with vertex at $$(-4, 0)$$. We need to compare the minimum values of the equation and the graph, and determine which has more zeros. 2. **Find the minimum value of the equation:** The function is in vertex form $$f(x) = a(x - h)^2 + k$$ where $$a = 3 > 0$$, so the parabola opens upward and the vertex is the minimum point. Here, the vertex is at $$(h, k) = (1, 1)$$. Therefore, the minimum value of the equation is $$f(1) = 1$$. 3. **Find the minimum value of the graph:** The graph's vertex is given at $$(-4, 0)$$. Since the parabola opens upward, the minimum value is the $$y$$-coordinate of the vertex, which is $$0$$. 4. **Compare minimum values:** Minimum value of equation: $$1$$ Minimum value of graph: $$0$$ Since $$0 < 1$$, the graph has the smaller minimum value, so the equation has the larger minimum value. 5. **Find the zeros of the equation:** Set $$f(x) = 0$$: $$0 = 3(x - 1)^2 + 1$$ $$3(x - 1)^2 = -1$$ Since the left side is always $$\geq 0$$ and the right side is negative, there are no real solutions. Therefore, the equation has **0 zeros**. 6. **Find the zeros of the graph:** The graph's vertex is at $$(-4, 0)$$ and it passes through points $$(-5, 1)$$ and $$(-3, 1)$$. Since the vertex is on the $$x$$-axis, the graph has exactly **1 zero** at $$x = -4$$. 7. **Compare zeros:** Equation zeros: 0 Graph zeros: 1 Therefore, the graph has more zeros. **Final answers:** - Larger minimum value: the equation - More zeros: the graph