1. The problem is to find the turning points of the function $y = x^2 - 6x + 6$. 2. To find turning points, we first find the derivative $y' = \frac{dy}{dx}$. 3. For $y = x^2 - 6x + 6$, the derivative is $y' = 2x - 6$. 4. Set the derivative equal to zero to find critical points: $2x - 6 = 0$. 5. Solve for $x$: $2x = 6 \Rightarrow x = 3$. 6. Find the $y$-coordinate by substituting $x=3$ into the original function: $$y = 3^2 - 6(3) + 6 = 9 - 18 + 6 = -3$$ 7. So the turning point is at $(3, -3)$. 8. Next, find the turning points of $y = 2x - 3x^2$. 9. Compute the derivative: $y' = 2 - 6x$. 10. Set derivative to zero: $2 - 6x = 0$. 11. Solve for $x$: $6x = 2 \Rightarrow x = \frac{1}{3}$. 12. Find $y$ at $x=\frac{1}{3}$: $$y = 2\left(\frac{1}{3}\right) - 3\left(\frac{1}{3}\right)^2 = \frac{2}{3} - 3\cdot \frac{1}{9} = \frac{2}{3} - \frac{1}{3} = \frac{1}{3}$$ 13. The turning point is at $\left(\frac{1}{3}, \frac{1}{3}\right)$. Final answers: - For $y = x^2 - 6x + 6$, turning point at $(3, -3)$. - For $y = 2x - 3x^2$, turning point at $\left(\frac{1}{3}, \frac{1}{3}\right)$.