1. The problem is to understand and apply the formula for the minimum value of a quadratic function. 2. The quadratic function is generally written as $f(x) = ax^2 + bx + c$ where $a$, $b$, and $c$ are constants and $a \neq 0$. 3. The formula for the vertex (which gives the minimum or maximum value) of the quadratic function is $x = -\frac{b}{2a}$. 4. This $x$ value is where the function reaches its minimum if $a > 0$ (the parabola opens upwards) or maximum if $a < 0$ (the parabola opens downwards). 5. To find the minimum value, substitute $x = -\frac{b}{2a}$ back into the function: $$f\left(-\frac{b}{2a}\right) = a\left(-\frac{b}{2a}\right)^2 + b\left(-\frac{b}{2a}\right) + c$$ 6. Simplify the expression step-by-step: $$= a \cdot \frac{b^2}{4a^2} - \frac{b^2}{2a} + c$$ $$= \frac{b^2}{4a} - \frac{b^2}{2a} + c$$ 7. Use a common denominator to combine the first two terms: $$= \frac{b^2}{4a} - \frac{2b^2}{4a} + c = -\frac{b^2}{4a} + c$$ 8. Therefore, the minimum value of the quadratic function is: $$f_{min} = c - \frac{b^2}{4a}$$ This formula gives the minimum value of the quadratic function when $a > 0$. In summary, the vertex formula $x = -\frac{b}{2a}$ gives the $x$-coordinate of the minimum point, and substituting it back into the function yields the minimum value $f_{min} = c - \frac{b^2}{4a}$.