1. **State the problem:** Find the maximum or minimum value of the function $y = x^2 - 8x$ and the corresponding value of $x$. 2. **Formula and rules:** For a quadratic function $y = ax^2 + bx + c$, the vertex (which gives the maximum or minimum) occurs at $x = -\frac{b}{2a}$. If $a > 0$, the parabola opens upwards and the vertex is a minimum. If $a < 0$, it opens downwards and the vertex is a maximum. 3. **Identify coefficients:** Here, $a = 1$, $b = -8$, and $c = 0$. 4. **Calculate vertex $x$-value:** $$x = -\frac{b}{2a} = -\frac{-8}{2 \times 1} = \frac{8}{2} = 4$$ 5. **Calculate corresponding $y$-value:** $$y = (4)^2 - 8 \times 4 = 16 - 32 = -16$$ 6. **Interpretation:** Since $a = 1 > 0$, the parabola opens upwards, so the vertex at $(4, -16)$ is a minimum point. **Final answer:** The minimum value of $y$ is $-16$ at $x = 4$.