1. **State the problem:** Convert the quadratic function $y = x^2 + 2x - 5$ into vertex form and identify the axis of symmetry, extrema, and zeros. 2. **Recall the 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 $$y = x^2 + 2x - 5$$ 4. Group the $x$ terms: $$y = (x^2 + 2x) - 5$$ 5. To complete the square, take half of the coefficient of $x$, which is $2$, half is $1$, and square it: $1^2 = 1$. 6. Add and subtract $1$ inside the parentheses: $$y = (x^2 + 2x + 1 - 1) - 5$$ 7. Rewrite as: $$y = ((x + 1)^2 - 1) - 5$$ 8. Simplify: $$y = (x + 1)^2 - 6$$ 9. **Vertex:** The vertex is at $(-1, -6)$. 10. **Axis of symmetry:** The vertical line through the vertex, $x = -1$. 11. **Extrema:** Since $a=1 > 0$, the parabola opens upward, so the vertex is a minimum point at $y = -6$. 12. **Zeros:** Solve $y=0$: $$0 = (x + 1)^2 - 6$$ $$ (x + 1)^2 = 6$$ $$x + 1 = \pm \sqrt{6}$$ $$x = -1 \pm \sqrt{6}$$ Approximations: $$x \approx -1 + 2.45 = 1.45$$ and $$x \approx -1 - 2.45 = -3.45$$ 13. **Y-intercept:** Set $x=0$: $$y = (0 + 1)^2 - 6 = 1 - 6 = -5$$ 14. **Summary:** - Vertex form: $$y = (x + 1)^2 - 6$$ - Axis of symmetry: $$x = -1$$ - Minimum at $$(-1, -6)$$ - Zeros at $$x = -1 \pm \sqrt{6}$$ - Y-intercept at $(0, -5)$ These key features help graph the parabola accurately.