1. **State the problem:** Given the function $f(x) = -x^2 + 2x + 3$, find points $A$, $B$, and $C$ where the graph cuts the axes, then find the area of triangle $\Delta ABC$, and finally find the maximum possible area of $\Delta ABD$ where $D$ lies on the curve above the x-axis. 2. **Find points $A$ and $B$ (x-intercepts):** Solve $f(x) = 0$: $$-x^2 + 2x + 3 = 0$$ Multiply both sides by $-1$: $$x^2 - 2x - 3 = 0$$ Factor: $$(x - 3)(x + 1) = 0$$ So, $x = 3$ or $x = -1$. Since $A$ has negative x-coordinate, $A = (-1, 0)$ and $B = (3, 0)$. 3. **Find point $C$ (y-intercept):** Evaluate $f(0)$: $$f(0) = -0 + 0 + 3 = 3$$ So, $C = (0, 3)$. 4. **Find area of $\Delta ABC$:** Points are $A(-1,0)$, $B(3,0)$, $C(0,3)$. Base $AB$ length: $$|3 - (-1)| = 4$$ Height is the y-coordinate of $C$ since $AB$ lies on x-axis: $$3$$ Area formula: $$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 3 = 6$$ 5. **Find maximum area of $\Delta ABD$ where $D$ lies on $y = f(x)$ above x-axis:** Let $D = (x, f(x))$ with $f(x) > 0$. Points $A(-1,0)$ and $B(3,0)$ lie on x-axis. Base $AB$ length is 4. Height of $\Delta ABD$ is the vertical distance from $D$ to $AB$ (x-axis), which is $f(x)$. Area of $\Delta ABD$: $$A(x) = \frac{1}{2} \times 4 \times f(x) = 2 f(x)$$ Maximize $A(x)$ by maximizing $f(x)$. 6. **Find vertex of parabola $f(x)$:** Vertex $x$-coordinate: $$x = -\frac{b}{2a} = -\frac{2}{2 \times (-1)} = 1$$ Evaluate $f(1)$: $$f(1) = -(1)^2 + 2(1) + 3 = -1 + 2 + 3 = 4$$ 7. **Maximum area of $\Delta ABD$:** $$A_{max} = 2 \times 4 = 8$$ **Final answers:** - $A = (-1, 0)$ - $B = (3, 0)$ - $C = (0, 3)$ - Area of $\Delta ABC = 6$ - Maximum area of $\Delta ABD = 8$