1. **Problem Statement:** Find the axis of symmetry and vertex for the quadratic equation $y = x^2 + 6x + 4$. 2. **Formula for axis of symmetry:** The axis of symmetry for a quadratic equation $y = ax^2 + bx + c$ is given by $$x = -\frac{b}{2a}$$ 3. **Calculate axis of symmetry:** Here, $a = 1$, $b = 6$, so $$x = -\frac{6}{2 \times 1} = -\frac{6}{2} = -3$$ 4. **Find vertex:** Substitute $x = -3$ into the equation to find $y$: $$y = (-3)^2 + 6(-3) + 4 = 9 - 18 + 4 = -5$$ 5. **Vertex coordinates:** The vertex is at $(-3, -5)$. --- 1. **Problem Statement:** Find the axis of symmetry and vertex for the quadratic equation $y = -2x^2 + 8x + 5$. 2. **Formula for axis of symmetry:** Same as above, $$x = -\frac{b}{2a}$$ 3. **Calculate axis of symmetry:** Here, $a = -2$, $b = 8$, so $$x = -\frac{8}{2 \times (-2)} = -\frac{8}{-4} = 2$$ 4. **Find vertex:** Substitute $x = 2$ into the equation: $$y = -2(2)^2 + 8(2) + 5 = -2(4) + 16 + 5 = -8 + 16 + 5 = 13$$ 5. **Vertex coordinates:** The vertex is at $(2, 13)$. --- **Summary:** - For $y = x^2 + 6x + 4$, axis of symmetry is $x = -3$, vertex is $(-3, -5)$. - For $y = -2x^2 + 8x + 5$, axis of symmetry is $x = 2$, vertex is $(2, 13)$. This completes the first problem fully.