1. The problem is to find the y-coordinate of the vertex of a parabola given by a quadratic function. 2. The general form of a quadratic function is $$y = ax^2 + bx + c$$ where $a$, $b$, and $c$ are constants. 3. The vertex of the parabola is the point where the function reaches its maximum or minimum value. The x-coordinate of the vertex is given by the formula $$x = -\frac{b}{2a}$$. 4. To find the y-coordinate of the vertex, substitute this x-value back into the quadratic function: $$y = a\left(-\frac{b}{2a}\right)^2 + b\left(-\frac{b}{2a}\right) + c$$ 5. Simplify the expression: $$y = a\frac{b^2}{4a^2} - \frac{b^2}{2a} + c = \frac{b^2}{4a} - \frac{b^2}{2a} + c$$ 6. Combine the terms: $$y = -\frac{b^2}{4a} + c$$ 7. Therefore, the y-coordinate of the vertex is $$y = c - \frac{b^2}{4a}$$. This formula allows you to find the y-coordinate of the vertex directly from the coefficients of the quadratic function.