1. **Problem statement:** We are given the quadratic function $$f(x) = -x^2 + 6x - 5$$ and need to analyze its graph and properties. 2. **Concavity:** The coefficient of $$x^2$$ is $$-1$$, which is negative, so the parabola opens downward (concave down). 3. **Vertex:** Use the vertex formula $$x = -\frac{b}{2a}$$ where $$a = -1$$ and $$b = 6$$. $$x = -\frac{6}{2 \times (-1)} = -\frac{6}{-2} = 3$$ Find $$f(3)$$: $$f(3) = -(3)^2 + 6(3) - 5 = -9 + 18 - 5 = 4$$ So the vertex is at $$(3, 4)$$. 4. **Axis of symmetry:** The vertical line through the vertex, $$x = 3$$. 5. **Y-intercept:** Set $$x=0$$: $$f(0) = -0 + 0 - 5 = -5$$ So the y-intercept is $$(0, -5)$$. 6. **X-intercepts:** Solve $$f(x) = 0$$: $$-x^2 + 6x - 5 = 0$$ Multiply both sides by $$-1$$ to simplify: $$\cancel{-}x^2 + 6x - 5 = 0 \Rightarrow x^2 - 6x + 5 = 0$$ Factor or use quadratic formula: $$x = \frac{6 \pm \sqrt{(-6)^2 - 4(1)(5)}}{2(1)} = \frac{6 \pm \sqrt{36 - 20}}{2} = \frac{6 \pm \sqrt{16}}{2}$$ $$x = \frac{6 \pm 4}{2}$$ So, $$x = \frac{6 + 4}{2} = 5$$ and $$x = \frac{6 - 4}{2} = 1$$ X-intercepts are $$(1, 0)$$ and $$(5, 0)$$. 7. **Domain:** Quadratic functions have domain all real numbers: $$\text{Domain} = (-\infty, \infty)$$ 8. **Range:** Since the parabola opens downward and vertex is maximum at $$y=4$$, $$\text{Range} = (-\infty, 4]$$ 9. **Increasing and decreasing intervals:** - The function increases on the interval to the left of the vertex: $$(-\infty, 3)$$ - The function decreases on the interval to the right of the vertex: $$(3, \infty)$$ 10. **Where $$f(x) > 0$$ and $$f(x) < 0$$:** - $$f(x) > 0$$ between the x-intercepts: $$(1, 5)$$ - $$f(x) < 0$$ outside the x-intercepts: $$(-\infty, 1) \cup (5, \infty)$$ **Summary:** - Concave down - Vertex at $$(3,4)$$ - Axis of symmetry $$x=3$$ - Y-intercept $$(0,-5)$$ - X-intercepts $$(1,0)$$ and $$(5,0)$$ - Domain $$(-\infty, \infty)$$ - Range $$(-\infty, 4]$$ - Increasing on $$(-\infty, 3)$$ - Decreasing on $$(3, \infty)$$ - $$f(x) > 0$$ on $$(1, 5)$$ - $$f(x) < 0$$ on $$(-\infty, 1) \cup (5, \infty)$$