1. **State the problem:** Rewrite the quadratic expression $x^2 - 6x + 1$ in the form $(x + a)^2 + b$ where $a$ and $b$ are integers. 2. **Formula and rule:** To complete the square for $x^2 + bx + c$, use the formula: $$x^2 + bx + c = (x + \frac{b}{2})^2 - \left(\frac{b}{2}\right)^2 + c$$ This helps rewrite the quadratic in vertex form. 3. **Apply to the given expression:** Here, $b = -6$ and $c = 1$. Calculate $\frac{b}{2} = \frac{-6}{2} = -3$. 4. **Rewrite:** $$x^2 - 6x + 1 = (x - 3)^2 - ( -3 )^2 + 1 = (x - 3)^2 - 9 + 1$$ 5. **Simplify:** $$= (x - 3)^2 - 8$$ 6. **Express in the form $(x + a)^2 + b$:** Note that $(x - 3)^2 = (x + (-3))^2$, so $a = -3$ and $b = -8$. 7. **Turning point coordinates:** The vertex form $y = (x + a)^2 + b$ shows the turning point at $(-a, b)$. Here, turning point is at $(3, -8)$. **Final answers:** (a) $x^2 - 6x + 1 = (x - 3)^2 - 8$ (b) Turning point coordinates: $(3, -8)$