1. The problem is to analyze the quadratic function $s = x^2 - 5x + 4$ and find its vertex, which represents the maximum or minimum point of the parabola. 2. The formula for the vertex $x$-coordinate of a quadratic function $ax^2 + bx + c$ is given by: $$x_v = \frac{-b}{2a}$$ 3. For the function $s = x^2 - 5x + 4$, we identify $a = 1$, $b = -5$, and $c = 4$. 4. Calculate the $x$-coordinate of the vertex: $$x_v = \frac{-(-5)}{2 \cdot 1} = \frac{5}{2}$$ 5. Next, find the $y$-coordinate of the vertex by substituting $x_v$ back into the function: $$y_v = \left(\frac{5}{2}\right)^2 - 5 \cdot \frac{5}{2} + 4$$ 6. Simplify step-by-step: $$y_v = \frac{25}{4} - \frac{25}{2} + 4$$ 7. Convert all terms to have a common denominator of 4: $$y_v = \frac{25}{4} - \frac{50}{4} + \frac{16}{4}$$ 8. Combine the fractions: $$y_v = \frac{25 - 50 + 16}{4} = \frac{-9}{4}$$ 9. Therefore, the vertex of the parabola is at: $$\left(\frac{5}{2}, -\frac{9}{4}\right)$$ 10. Since $a = 1 > 0$, the parabola opens upward, so this vertex is a minimum point. Final answer: The vertex is at $\left(\frac{5}{2}, -\frac{9}{4}\right)$ and the parabola opens upward, indicating a minimum at this point.