1. **State the problem:** Find the points of intersection between the functions $f(x) = \frac{1}{3}x^2$ and $g(x) = -x$, and analyze the right triangle $ABC$ formed by points $A$, $B$, and $C$ where $A$ and $B$ are intersections of the graphs and $C$ lies on the x-axis. 2. **Find intersection points:** Set $f(x) = g(x)$ to find $x$ values where the graphs intersect. $$\frac{1}{3}x^2 = -x$$ 3. **Rewrite the equation:** Multiply both sides by 3 to clear the fraction: $$3 \times \frac{1}{3}x^2 = 3 \times (-x)$$ $$x^2 = -3x$$ 4. **Bring all terms to one side:** $$x^2 + 3x = 0$$ 5. **Factor the quadratic:** $$x(x + 3) = 0$$ 6. **Solve for $x$:** $$x = 0 \quad \text{or} \quad x = -3$$ 7. **Find corresponding $y$ values:** For $x=0$: $$y = f(0) = \frac{1}{3} \times 0^2 = 0$$ For $x=-3$: $$y = f(-3) = \frac{1}{3} \times (-3)^2 = \frac{1}{3} \times 9 = 3$$ 8. **Coordinates of intersection points:** $$A = (-3, 3), \quad B = (0, 0)$$ 9. **Point $C$ lies on the x-axis below $A$:** Since $C$ is on the x-axis, its $y$-coordinate is 0, and it shares the same $x$-coordinate as $A$: $$C = (-3, 0)$$ 10. **Calculate the area of triangle $ABC$:** Triangle $ABC$ is right-angled at $C$, with legs $AC$ (vertical) and $BC$ (horizontal). Length of $AC$: $$|y_A - y_C| = |3 - 0| = 3$$ Length of $BC$: $$|x_B - x_C| = |0 - (-3)| = 3$$ Area formula for right triangle: $$\text{Area} = \frac{1}{2} \times AC \times BC = \frac{1}{2} \times 3 \times 3 = \frac{9}{2} = 4.5$$ **Final answer:** - Intersection points: $A(-3, 3)$ and $B(0, 0)$ - Point $C(-3, 0)$ on x-axis - Area of triangle $ABC$ is $4.5$ square units.