1. **State the problem:** Find the area enclosed between the curves $y = |x|$ and $y = x^2 - 6$. 2. **Find the points of intersection:** Set $|x| = x^2 - 6$. Since $|x|$ is piecewise, consider two cases: - For $x \geq 0$, $|x| = x$: $$x = x^2 - 6 \implies x^2 - x - 6 = 0$$ - For $x < 0$, $|x| = -x$: $$-x = x^2 - 6 \implies x^2 + x - 6 = 0$$ 3. **Solve for $x \geq 0$:** $$x^2 - x - 6 = 0$$ Use quadratic formula: $$x = \frac{1 \pm \sqrt{1 + 24}}{2} = \frac{1 \pm 5}{2}$$ Possible roots: $x = 3$ or $x = -2$ (discard $-2$ since $x \geq 0$). 4. **Solve for $x < 0$:** $$x^2 + x - 6 = 0$$ Use quadratic formula: $$x = \frac{-1 \pm \sqrt{1 + 24}}{2} = \frac{-1 \pm 5}{2}$$ Possible roots: $x = 2$ or $x = -3$ (discard $2$ since $x < 0$). 5. **Intersection points are at $x = -3$ and $x = 3$.** 6. **Set up the integral for the area:** The area is symmetric about $x=0$, so calculate from 0 to 3 and double it. For $x \geq 0$, $y = |x| = x$. Area: $$\text{Area} = 2 \times \int_0^3 \left[(x^2 - 6) - x\right] dx = 2 \times \int_0^3 (x^2 - x - 6) dx$$ 7. **Calculate the integral:** $$\int_0^3 (x^2 - x - 6) dx = \left[\frac{x^3}{3} - \frac{x^2}{2} - 6x\right]_0^3$$ Evaluate at 3: $$\frac{3^3}{3} - \frac{3^2}{2} - 6 \times 3 = \frac{27}{3} - \frac{9}{2} - 18 = 9 - 4.5 - 18 = -13.5$$ Evaluate at 0: $$0$$ So the integral is $-13.5$. 8. **Area is:** $$2 \times (-13.5) = -27$$ Since area cannot be negative, take absolute value: $$\boxed{27}$$ **Final answer:** The area enclosed between the curves is $27$ square units.