1. **State the problem:** Solve the inequality $$-x^2 + 6x \leq 5$$ and graph the solution on the number line. 2. **Rewrite the inequality:** Move all terms to one side to set the inequality to zero: $$-x^2 + 6x - 5 \leq 0$$ 3. **Multiply both sides by -1** to make the quadratic coefficient positive. Remember to reverse the inequality sign when multiplying by a negative number: $$\cancel{-1}(-x^2 + 6x - 5) \geq \cancel{-1}(0)$$ which simplifies to $$x^2 - 6x + 5 \geq 0$$ 4. **Factor the quadratic:** $$x^2 - 6x + 5 = (x - 1)(x - 5)$$ 5. **Analyze the inequality:** We want to find where $$(x - 1)(x - 5) \geq 0$$ 6. **Determine intervals:** The critical points are $x=1$ and $x=5$. Test intervals: - For $x < 1$, pick $x=0$: $(0-1)(0-5) = (-1)(-5) = 5 \geq 0$ (True) - For $1 < x < 5$, pick $x=3$: $(3-1)(3-5) = (2)(-2) = -4 \geq 0$ (False) - For $x > 5$, pick $x=6$: $(6-1)(6-5) = (5)(1) = 5 \geq 0$ (True) 7. **Write the solution set:** $$(-\infty, 1] \cup [5, \infty)$$ 8. **Interpretation:** The solution includes all $x$ values less than or equal to 1 and greater than or equal to 5. **Final answer:** $$\boxed{(-\infty, 1] \cup [5, \infty)}$$