1. **State the problem:** We need to graph the quadratic function $f(x) = x^2 + 4x + 1$ and find the coordinates of its minimum point. 2. **Recall the formula for the vertex of a parabola:** For a quadratic function $f(x) = ax^2 + bx + c$, the vertex (minimum or maximum) occurs at $x = -\frac{b}{2a}$. 3. **Identify coefficients:** Here, $a = 1$, $b = 4$, and $c = 1$. 4. **Calculate the x-coordinate of the vertex:** $$x = -\frac{4}{2 \times 1} = -\frac{4}{2} = -2$$ 5. **Calculate the y-coordinate of the vertex by substituting $x = -2$ into $f(x)$:** $$f(-2) = (-2)^2 + 4(-2) + 1 = 4 - 8 + 1 = -3$$ 6. **Therefore, the vertex (minimum point) is at $(-2, -3)$. 7. **Graphing notes:** The parabola opens upwards since $a = 1 > 0$, so the vertex is a minimum. **Final answer:** The minimum point of the graph of $f(x) = x^2 + 4x + 1$ is at $\boxed{(-2, -3)}$.