1. **Problem:** For the quadratic function $y = x^2 - 8x + 8$, find (a) the axis of symmetry and (b) the turning point. 2. **Formula:** The axis of symmetry for a quadratic $y = ax^2 + bx + c$ is given by the formula: $$x = -\frac{b}{2a}$$ The turning point (vertex) coordinates are: $$\left(-\frac{b}{2a}, y\left(-\frac{b}{2a}\right)\right)$$ 3. **Calculate axis of symmetry:** Here, $a=1$, $b=-8$. $$x = -\frac{-8}{2 \times 1} = \frac{8}{2} = 4$$ 4. **Calculate turning point:** Substitute $x=4$ into the function: $$y = (4)^2 - 8(4) + 8 = 16 - 32 + 8 = -8$$ Turning point is at $(4, -8)$. **Answer:** - Axis of symmetry: $x=4$ - Turning point: $(4, -8)$