1. **State the problem:** Find the range of values of $x$ for which $y = -(x-1)^2 + 2$ is less than $-1$. 2. **Write the inequality:** We want to solve $$-(x-1)^2 + 2 < -1$$ 3. **Isolate the squared term:** Subtract 2 from both sides: $$-(x-1)^2 < -3$$ Multiply both sides by $-1$ (remember to reverse the inequality sign when multiplying by a negative): $$(x-1)^2 > 3$$ 4. **Solve the inequality:** The square of a number is greater than 3 means $$x-1 > \sqrt{3} \quad \text{or} \quad x-1 < -\sqrt{3}$$ 5. **Find the intervals:** Add 1 to all parts: $$x > 1 + \sqrt{3} \quad \text{or} \quad x < 1 - \sqrt{3}$$ 6. **Interpretation:** The values of $x$ for which $y < -1$ are those less than $1 - \sqrt{3}$ or greater than $1 + \sqrt{3}$. 7. **Approximate values:** Since $\sqrt{3} \approx 1.732$, $$x < 1 - 1.732 = -0.732 \quad \text{or} \quad x > 1 + 1.732 = 2.732$$ **Final answer:** $$x < -0.732 \quad \text{or} \quad x > 2.732$$