1. **State the problem:** We need to find which quadratic equation best matches the given graph. 2. **Analyze the graph:** The parabola opens downward, indicating the coefficient of the squared term is negative. 3. **Identify the vertex:** The vertex is at approximately $(4, 1)$. 4. **Recall the vertex form of a parabola:** $$y = a(x - h)^2 + k$$ where $(h, k)$ is the vertex and $a$ determines the direction and width. 5. **Apply the vertex:** Since vertex is $(4, 1)$, the equation is: $$y = a(x - 4)^2 + 1$$ 6. **Determine the sign of $a$:** The parabola opens downward, so $a < 0$. 7. **Check the options:** - $y = -3(x - 4)^2 + 1$ (opens downward, vertex at $(4,1)$) - $y = -3(x + 4)^2 + 1$ (vertex at $(-4,1)$) - $y = 3(x + 4)^2 + 1$ (opens upward, vertex at $(-4,1)$) - $y = 3(x - 4)^2 + 1$ (opens upward, vertex at $(4,1)$) 8. **Conclusion:** The correct equation is: $$y = -3(x - 4)^2 + 1$$