1. Let's analyze the function $y = 4x - x^2$. 2. This is a quadratic function, which is a parabola opening downwards since the coefficient of $x^2$ is negative (-1). 3. To find its vertex, use the vertex formula $x = -\frac{b}{2a}$, where $a = -1$ and $b = 4$: $$x = -\frac{4}{2 \times -1} = 2$$ 4. Compute the y-coordinate of the vertex by substituting $x=2$: $$y = 4(2) - (2)^2 = 8 - 4 = 4$$ 5. The vertex is at $(2,4)$, which is the maximum point of the curve. 6. The x-intercepts (roots) are found by solving $4x - x^2 = 0$: $$x(4 - x) = 0$$ $$x = 0 \text{ or } x = 4$$ 7. Intercept points are at $(0,0)$ and $(4,0)$. 8. For the shaded area in the top-left quadrant: the area under the curve between $x=0$ and $x=4$ is given by the definite integral: $$\int_0^4 (4x - x^2) \, dx$$ 9. Compute the integral: $$\int_0^4 4x \, dx - \int_0^4 x^2 \, dx = \left[2x^2\right]_0^4 - \left[\frac{x^3}{3}\right]_0^4 = (2 \times 16) - \frac{64}{3} = 32 - \frac{64}{3} = \frac{96 - 64}{3} = \frac{32}{3}$$ 10. So, the shaded area under the curve $y = 4x - x^2$ from $x=0$ to $x=4$ is $\frac{32}{3}$. Similarly, for the other functions and their shaded areas: --- 1. For $y = e^x$ between $x = -1$ and $x = 3$: $$\int_{-1}^3 e^x \, dx = \left[e^x\right]_{-1}^3 = e^3 - e^{-1}$$ --- 2. For $y = x^3 + 2$ between $x = -1$ and $x = 1$: $$\int_{-1}^1 (x^3 + 2) \, dx = \left[\frac{x^4}{4} + 2x\right]_{-1}^1 = \left(\frac{1}{4} + 2\right) - \left(\frac{1}{4} - 2\right) = \left(\frac{1}{4} + 2\right) - \left(\frac{1}{4} - 2\right) = 4$$ --- 3. For $y = x + 3$ between $x = 0$ and $x = 4$: $$\int_0^4 (x + 3) \, dx = \left[\frac{x^2}{2} + 3x\right]_0^4 = \left(\frac{16}{2} + 12\right) - 0 = 8 + 12 = 20$$