Minimum And Zeros Ba99A8

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 the vertex is at $$(h, k)$$. Here, $$a = 3 > 0$$, so the parabola opens upward and the vertex is a minimum point. The vertex is at $$(1, 1)$$, so the minimum value of the equation is $$f(1) = 1$$. 3. **Analyze the graph's minimum value:** The graph's vertex is at $$(-4, 0)$$, so its minimum value is $$0$$. 4. **Compare minimum values:** The equation's minimum value is $$1$$. The graph's minimum value is $$0$$. Since $$1 > 0$$, the graph has the smaller minimum value, so the equation has the larger minimum value. 5. **Find zeros of the equation:** Set $$f(x) = 0$$: $$0 = 3(x - 1)^2 + 1$$ $$3(x - 1)^2 = -1$$ Since $$3(x - 1)^2 \geq 0$$ for all $$x$$, and the right side is negative, there are no real solutions. So, the equation has **0 zeros**. 6. **Find zeros of the graph:** The graph's vertex is at $$(-4, 0)$$ and it crosses the x-axis at $$x = -4$$ (the vertex), so it has at least one zero. From the description, the parabola passes through $$(-5, 1)$$ and $$(-3, 1)$$, so it only touches the x-axis at $$x = -4$$. Therefore, the graph has **1 zero**. **Final answers:** - The equation has the larger minimum value. - The graph has more zeros.