1. **Problem:** For each quadratic function, find the axis of symmetry and the turning point. 2. **Formula:** The axis of symmetry for a quadratic function $y = ax^2 + bx + c$ is given by: $$x = -\frac{b}{2a}$$ The turning point (vertex) coordinates are: $$\left(-\frac{b}{2a}, f\left(-\frac{b}{2a}\right)\right)$$ 3. **(a) For** $y = x^2 - 8x + 8$: - Here, $a=1$, $b=-8$, $c=8$ - Axis of symmetry: $$x = -\frac{-8}{2 \times 1} = \frac{8}{2} = 4$$ - Turning point: $$y = (4)^2 - 8(4) + 8 = 16 - 32 + 8 = -8$$ - Coordinates: $(4, -8)$ 4. **(b) For** $y = 3x^2 + 12x + 27$: - Here, $a=3$, $b=12$, $c=27$ - Axis of symmetry: $$x = -\frac{12}{2 \times 3} = -\frac{12}{6} = -2$$ - Turning point: $$y = 3(-2)^2 + 12(-2) + 27 = 3(4) - 24 + 27 = 12 - 24 + 27 = 15$$ - Coordinates: $(-2, 15)$ 5. **(c) For** $y = -x^2 + 18x - 70$: - Here, $a=-1$, $b=18$, $c=-70$ - Axis of symmetry: $$x = -\frac{18}{2 \times (-1)} = -\frac{18}{-2} = 9$$ - Turning point: $$y = -(9)^2 + 18(9) - 70 = -81 + 162 - 70 = 11$$ - Coordinates: $(9, 11)$