1. **Problem Statement:** Show that the function $y = x^2 - 6x + 4$ is increasing at $x = 8$, decreasing at $x = 1$, and stationary at $x = 3$. 2. **Formula and Rules:** To determine whether a function is increasing, decreasing, or stationary at a point, we use the first derivative $y' = \frac{dy}{dx}$. - If $y'(x) > 0$, the function is increasing at $x$. - If $y'(x) < 0$, the function is decreasing at $x$. - If $y'(x) = 0$, the function is stationary at $x$. 3. **Find the derivative:** $$y = x^2 - 6x + 4$$ $$y' = \frac{d}{dx}(x^2) - \frac{d}{dx}(6x) + \frac{d}{dx}(4) = 2x - 6 + 0 = 2x - 6$$ 4. **Evaluate the derivative at the given points:** - At $x=8$: $$y'(8) = 2(8) - 6 = 16 - 6 = 10 > 0$$ So, the function is increasing at $x=8$. - At $x=1$: $$y'(1) = 2(1) - 6 = 2 - 6 = -4 < 0$$ So, the function is decreasing at $x=1$. - At $x=3$: $$y'(3) = 2(3) - 6 = 6 - 6 = 0$$ So, the function is stationary at $x=3$. 5. **Conclusion:** The function $y = x^2 - 6x + 4$ is increasing at $x=8$, decreasing at $x=1$, and stationary at $x=3$ as required.