1. **State the problem:** Write the quadratic equation $y = -x^2 - 4x + 5$ in vertex form, identify the axis of symmetry, extrema, and zeros. 2. **Recall vertex form:** The vertex form of a quadratic is $$y = a(x - h)^2 + k$$ where $(h, k)$ is the vertex. 3. **Complete the square:** Start with $$y = -x^2 - 4x + 5$$ Factor out the coefficient of $x^2$ from the first two terms: $$y = -\left(x^2 + 4x\right) + 5$$ 4. To complete the square inside the parentheses, take half of 4, which is 2, and square it: $2^2 = 4$. Add and subtract 4 inside the parentheses: $$y = -\left(x^2 + 4x + 4 - 4\right) + 5$$ Rewrite: $$y = -\left((x + 2)^2 - 4\right) + 5$$ Distribute the negative sign: $$y = - (x + 2)^2 + 4 + 5$$ Simplify: $$y = - (x + 2)^2 + 9$$ 5. **Vertex form:** $$y = -1(x - (-2))^2 + 9$$ 6. **Axis of symmetry:** The axis of symmetry is the vertical line through the vertex $x = -2$. 7. **Extrema:** Since $a = -1 < 0$, the parabola opens downward, so the vertex is a maximum point at $(-2, 9)$. 8. **Find zeros:** Set $y=0$: $$0 = - (x + 2)^2 + 9$$ Rearranged: $$(x + 2)^2 = 9$$ Take square root: $$x + 2 = \pm 3$$ Solve for $x$: $$x = -2 + 3 = 1$$ $$x = -2 - 3 = -5$$ Zeros are at $(-5, 0)$ and $(1, 0)$, listed from least to greatest. **Final answers:** - Vertex form: $$y = -1(x - (-2))^2 + 9$$ - Axis of symmetry: $$x = -2$$ - Extrema: maximum at $$(-2, 9)$$ - Zeros: $$(-5, 0)$$ and $$(1, 0)$$