1. **Problem:** Understand where the function $f(x) = -x^2 + 6x - 5$ is increasing and decreasing (part c), and where $f(x)$ is positive or negative (part d). 2. **Recall vertex:** The vertex is at $x=3$ with $f(3)=4$. 3. **Increasing and decreasing (part c):** - Since the parabola opens downward (coefficient of $x^2$ is negative), the function increases before the vertex and decreases after. - So, $f$ is increasing on $(-\infty, 3)$ and decreasing on $(3, \infty)$. 4. **Where $f(x) > 0$ and $f(x) < 0$ (part d):** - Find x-intercepts by solving $f(x)=0$: $$-x^2 + 6x - 5 = 0 \Rightarrow x^2 - 6x + 5 = 0$$ - Factor or use quadratic formula: $$x = \frac{6 \pm \sqrt{36 - 20}}{2} = \frac{6 \pm 4}{2}$$ - So, $x=1$ and $x=5$. - Since the parabola opens downward, $f(x) > 0$ between the roots: $(1, 5)$. - And $f(x) < 0$ outside the roots: $(-\infty, 1) \cup (5, \infty)$. This means the function is positive between $x=1$ and $x=5$, and negative elsewhere.