1. **State the problem:** Complete the square for the quadratic function $y = -x^2 + 6x - 5$. 2. **Recall the formula:** To complete the square for a quadratic $ax^2 + bx + c$, factor out $a$ from the $x^2$ and $x$ terms, then add and subtract $(\frac{b}{2a})^2$ inside the parentheses. 3. **Start with the given function:** $$y = -x^2 + 6x - 5$$ 4. **Factor out the coefficient of $x^2$ from the first two terms:** $$y = - (x^2 - 6x) - 5$$ 5. **Find the value to complete the square:** Calculate $\left(\frac{-6}{2}\right)^2 = (-3)^2 = 9$. 6. **Add and subtract 9 inside the parentheses:** $$y = - (x^2 - 6x + 9 - 9) - 5$$ 7. **Group the perfect square trinomial and simplify:** $$y = - \big((x - 3)^2 - 9\big) - 5$$ 8. **Distribute the negative sign:** $$y = - (x - 3)^2 + 9 - 5$$ 9. **Combine constants:** $$y = - (x - 3)^2 + 4$$ **Final answer:** $$y = - (x - 3)^2 + 4$$ This is the completed square form of the quadratic function, showing the vertex at $(3, 4)$ and that the parabola opens downward.