1. **State the problem:** We have two functions: $$f(x) = -\frac{1}{4}x^2$$ $$g(x) = -\frac{2}{3}x$$ Points A and B are the intersections of these graphs. Point C is on the x-axis and is symmetric to B with respect to the y-axis. We need to: - Find the coordinates of point B. - Calculate the area of triangle ABC, rounding the result to one decimal place. 2. **Find the intersection points (A and B):** Set $f(x) = g(x)$: $$-\frac{1}{4}x^2 = -\frac{2}{3}x$$ Multiply both sides by 12 (common denominator) to clear fractions: $$12 \times \left(-\frac{1}{4}x^2\right) = 12 \times \left(-\frac{2}{3}x\right)$$ $$-3x^2 = -8x$$ Bring all terms to one side: $$-3x^2 + 8x = 0$$ Factor out $x$: $$x(-3x + 8) = 0$$ So, $$x = 0 \quad \text{or} \quad -3x + 8 = 0$$ Solve for $x$ in second equation: $$-3x + 8 = 0 \Rightarrow -3x = -8 \Rightarrow x = \frac{8}{3}$$ 3. **Find corresponding y-values:** For $x=0$: $$f(0) = -\frac{1}{4} \times 0^2 = 0$$ $$g(0) = -\frac{2}{3} \times 0 = 0$$ So point A is $(0,0)$. For $x=\frac{8}{3}$: $$f\left(\frac{8}{3}\right) = -\frac{1}{4} \times \left(\frac{8}{3}\right)^2 = -\frac{1}{4} \times \frac{64}{9} = -\frac{64}{36} = -\frac{16}{9}$$ $$g\left(\frac{8}{3}\right) = -\frac{2}{3} \times \frac{8}{3} = -\frac{16}{9}$$ So point B is $\left(\frac{8}{3}, -\frac{16}{9}\right)$. 4. **Find point C:** Point C is symmetric to B with respect to the y-axis, so its x-coordinate is the negative of B's x-coordinate, and its y-coordinate is 0 (since it lies on the x-axis): $$C = \left(-\frac{8}{3}, 0\right)$$ 5. **Calculate the area of triangle ABC:** Vertices: $$A = (0,0), B = \left(\frac{8}{3}, -\frac{16}{9}\right), C = \left(-\frac{8}{3}, 0\right)$$ Use the formula for area of triangle given coordinates: $$\text{Area} = \frac{1}{2} \left| x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B) \right|$$ Substitute values: $$= \frac{1}{2} \left| 0 \times \left(-\frac{16}{9} - 0\right) + \frac{8}{3} \times (0 - 0) + \left(-\frac{8}{3}\right) \times (0 - \left(-\frac{16}{9}\right)) \right|$$ Simplify terms: $$= \frac{1}{2} \left| 0 + 0 + \left(-\frac{8}{3}\right) \times \frac{16}{9} \right| = \frac{1}{2} \left| -\frac{128}{27} \right| = \frac{1}{2} \times \frac{128}{27} = \frac{64}{27}$$ 6. **Convert to decimal and round:** $$\frac{64}{27} \approx 2.370370...$$ Rounded to one decimal place: $$2.4$$ **Final answers:** - Coordinates of point B: $\left(\frac{8}{3}, -\frac{16}{9}\right)$ - Area of triangle ABC: $2.4$