1. **State the problem:** We are given a quadratic function $f(x) = x^2 + bx + 1$ and asked to find its minimum value. 2. **Recall the formula for the vertex of a quadratic:** For a quadratic function $f(x) = ax^2 + bx + c$, the vertex occurs at $x = -\frac{b}{2a}$. Since $a=1$ here, the vertex is at $x = -\frac{b}{2}$. 3. **Calculate the minimum value:** Substitute $x = -\frac{b}{2}$ into $f(x)$: $$ f\left(-\frac{b}{2}\right) = \left(-\frac{b}{2}\right)^2 + b\left(-\frac{b}{2}\right) + 1 = \frac{b^2}{4} - \frac{b^2}{2} + 1 $$ 4. **Simplify the expression:** $$ \frac{b^2}{4} - \frac{b^2}{2} + 1 = \frac{b^2}{4} - \frac{2b^2}{4} + 1 = 1 - \frac{b^2}{4} $$ 5. **Interpretation:** The minimum value of $f(x)$ is $1 - \frac{b^2}{4}$. **Final answer:** $\boxed{1 - \frac{b^2}{4}}$