1. **State the problem:** We are given the quadratic function $y = -x^{2} + 6x - 5$ and want to analyze it. 2. **Formula and rules:** This is a quadratic function in standard form $y = ax^{2} + bx + c$ where $a = -1$, $b = 6$, and $c = -5$. 3. **Find the vertex:** The vertex $x$-coordinate is given by $x = -\frac{b}{2a} = -\frac{6}{2 \times -1} = -\frac{6}{-2} = 3$. 4. **Calculate the vertex $y$-coordinate:** Substitute $x=3$ into the function: $$y = -(3)^{2} + 6(3) - 5 = -9 + 18 - 5 = 4$$ 5. **Vertex:** The vertex is at $(3, 4)$. 6. **Find the $x$-intercepts:** Set $y=0$ and solve: $$0 = -x^{2} + 6x - 5$$ Multiply both sides by $-1$ to simplify: $$0 = \cancel{-}x^{2} + \cancel{-}6x + \cancel{-}5 \Rightarrow 0 = x^{2} - 6x + 5$$ 7. **Factor the quadratic:** $$x^{2} - 6x + 5 = (x - 5)(x - 1)$$ 8. **Solve for $x$:** $$x - 5 = 0 \Rightarrow x = 5$$ $$x - 1 = 0 \Rightarrow x = 1$$ 9. **$x$-intercepts:** $(1, 0)$ and $(5, 0)$. 10. **Find the $y$-intercept:** Set $x=0$: $$y = -0^{2} + 6(0) - 5 = -5$$ 11. **$y$-intercept:** $(0, -5)$. 12. **Summary:** The parabola opens downward (since $a = -1 < 0$), has vertex at $(3,4)$, $x$-intercepts at $(1,0)$ and $(5,0)$, and $y$-intercept at $(0,-5)$. **Final answer:** Vertex $(3,4)$, $x$-intercepts $(1,0)$ and $(5,0)$, $y$-intercept $(0,-5)$.