1. **Problem 16:** Write $y = x^2 - 8x + 12$ in vertex form, identify axis of symmetry, extrema, and zeros. 2. **Formula:** Vertex form is $y = a(x-h)^2 + k$ where $(h,k)$ is the vertex. 3. **Complete the square:** $$y = x^2 - 8x + 12 = (x^2 - 8x + 16) - 16 + 12 = (x - 4)^2 - 4$$ 4. **Vertex:** $(4, -4)$, axis of symmetry: $x = 4$. 5. **Extrema:** Since $a=1 > 0$, parabola opens upward, vertex is minimum point at $y = -4$. 6. **Zeros:** Solve $0 = (x - 4)^2 - 4$: $$ (x - 4)^2 = 4 $$ $$ x - 4 = \pm 2 $$ $$ x = 4 \pm 2 $$ Zeros: $x=2$ and $x=6$. --- 7. **Problem 17:** Write $y = -x^2 - 2x + 3$ in vertex form, identify axis of symmetry, extrema, and zeros. 8. **Complete the square:** $$y = -(x^2 + 2x) + 3 = -\left(x^2 + 2x + 1\right) + 3 + 1 = -(x + 1)^2 + 4$$ 9. **Vertex:** $(-1, 4)$, axis of symmetry: $x = -1$. 10. **Extrema:** Since $a = -1 < 0$, parabola opens downward, vertex is maximum point at $y = 4$. 11. **Zeros:** Solve $0 = -(x + 1)^2 + 4$: $$ (x + 1)^2 = 4 $$ $$ x + 1 = \pm 2 $$ $$ x = -1 \pm 2 $$ Zeros: $x = -3$ and $x = 1$. --- **Summary:** - Equation 16 vertex form: $y = (x - 4)^2 - 4$ - Axis of symmetry: $x=4$ - Minimum at $(4, -4)$ - Zeros at $x=2, 6$ - Equation 17 vertex form: $y = -(x + 1)^2 + 4$ - Axis of symmetry: $x=-1$ - Maximum at $(-1, 4)$ - Zeros at $x=-3, 1$