Understanding the Problem

The question asks for the area of a specific region in the Cartesian plane. This region is defined by the boundaries formed by the line $y = 3 - x$, the parabola $y = x^2 - 9$, and the constraints $x \ge -4$ and $y \ge 0$. We need to find the finite area enclosed by these conditions.

Analyzing the Curves and Constraints

1. Line: $y = 3 - x$
   • This is a straight line with a slope of -1 and y-intercept of 3.
   • The constraint $y \ge 0$ implies $3 - x \ge 0$, which means $x \le 3$.
   • Combined with $x \ge -4$, the relevant part of the line is for $x$ in the interval $[-4, 3]$.
2. Parabola: $y = x^2 - 9$
   • This is an upward-opening parabola with its vertex at $(0, -9)$ and x-intercepts at $x = \pm 3$.
   • The constraint $y \ge 0$ implies $x^2 - 9 \ge 0$, which means $x^2 \ge 9$. This holds true for $x \le -3$ or $x \ge 3$.
   • Combined with $x \ge -4$, the relevant parts of the parabola where $y \ge 0$ are for $x$ in the interval $[-4, -3]$ and for $x \ge 3$.
3. Constraints: $x \ge -4$ and $y \ge 0$. These restrict the region of interest to the right of the vertical line $x = -4$ and above the x-axis ($y = 0$).

Finding Intersection Points

To find where the line and the parabola intersect, we set their equations equal: $3 - x = x^2 - 9$ Rearranging the terms gives a quadratic equation: $x^2 + x - 12 = 0$ Factoring the quadratic equation: $(x + 4)(x - 3) = 0$ The intersection points occur at $x = -4$ and $x = 3$.

• At $x = -4$, $y = 3 - (-4) = 7$. The point is $(-4, 7)$.
• At $x = 3$, $y = 3 - 3 = 0$. The point is $(3, 0)$.

Defining the Area Boundaries

We need to find the area of the region that satisfies all conditions. Let's consider the intervals defined by the intersection points and constraints:

• Interval 1: $x \in [-4, -3]$
  • The line $y = 3 - x$ ranges from $y=7$ down to $y=6$. It is above the x-axis ($y \ge 0$).
  • The parabola $y = x^2 - 9$ ranges from $y=7$ down to $y=0$. It is above or on the x-axis ($y \ge 0$).
  • In this interval, the line $y = 3 - x$ is above the parabola $y = x^2 - 9$.
  • The area in this interval ($A_1$) is the integral of the difference between the upper curve (line) and the lower curve (parabola).
• Interval 2: $x \in [-3, 3]$
  • The line $y = 3 - x$ ranges from $y=6$ down to $y=0$. It is above or on the x-axis ($y \ge 0$).
  • The parabola $y = x^2 - 9$ ranges from $y=0$ down to $y=-9$ and back to $y=0$. It is below the x-axis for $x \in (-3, 3)$.
  • The constraint $y \ge 0$ means we are only interested in the region above the x-axis. Therefore, in this interval, the region is bounded above by the line $y = 3 - x$ and below by the x-axis ($y = 0$).
  • The area in this interval ($A_2$) is the integral of the line equation from $x=-3$ to $x=3$.
• Interval 3: $x > 3$
  • The line $y = 3-x$ is negative.
  • The parabola $y = x^2-9$ is positive.
  • The constraint $y \ge 0$ applies. If we consider the region *between* the curves bounded by $y \ge 0$, this interval does not form a finite closed region relevant to the question as typically interpreted.

The total area is the sum of the areas calculated in the relevant intervals: $A = A_1 + A_2$.

Calculating the Area

Calculating Area $A_1$ (for $x \in [-4, -3]$): The integrand is $(3 - x) - (x^2 - 9) = -x^2 - x + 12$. $A_1 = \int_{-4}^{-3} (-x^2 - x + 12) dx$ $A_1 = \left[ -\frac{x^3}{3} - \frac{x^2}{2} + 12x \right]_{-4}^{-3}$ $A_1 = \left( -\frac{(-3)^3}{3} - \frac{(-3)^2}{2} + 12(-3) \right) - \left( -\frac{(-4)^3}{3} - \frac{(-4)^2}{2} + 12(-4) \right)$ $A_1 = \left( -\frac{-27}{3} - \frac{9}{2} - 36 \right) - \left( -\frac{-64}{3} - \frac{16}{2} - 48 \right)$ $A_1 = \left( 9 - \frac{9}{2} - 36 \right) - \left( \frac{64}{3} - 8 - 48 \right)$ $A_1 = \left( -27 - \frac{9}{2} \right) - \left( \frac{64}{3} - 56 \right)$ $A_1 = \left( -\frac{54}{2} - \frac{9}{2} \right) - \left( \frac{64}{3} - \frac{168}{3} \right)$ $A_1 = -\frac{63}{2} - \left( -\frac{104}{3} \right)$ $A_1 = -\frac{63}{2} + \frac{104}{3} = \frac{-189 + 208}{6} = \frac{19}{6}$

Calculating Area $A_2$ (for $x \in [-3, 3]$): The integrand is $(3 - x) - 0 = 3 - x$. $A_2 = \int_{-3}^{3} (3 - x) dx$ $A_2 = \left[ 3x - \frac{x^2}{2} \right]_{-3}^{3}$ $A_2 = \left( 3(3) - \frac{3^2}{2} \right) - \left( 3(-3) - \frac{(-3)^2}{2} \right)$ $A_2 = \left( 9 - \frac{9}{2} \right) - \left( -9 - \frac{9}{2} \right)$ $A_2 = \frac{9}{2} - \left( -\frac{27}{2} \right)$ $A_2 = \frac{9}{2} + \frac{27}{2} = \frac{36}{2} = 18$

Total Area Calculation

The total area is the sum of $A_1$ and $A_2$. Total Area $= A_1 + A_2 = \frac{19}{6} + 18$ Total Area $= \frac{19}{6} + \frac{18 \times 6}{6} = \frac{19 + 108}{6} = \frac{127}{6}$

Conclusion

The calculated area is $\frac{127}{6}$. Comparing this value to the given options:

• Option 1: $\frac{7}{2} = \frac{21}{6}$
• Option 2: $\frac{11}{2} = \frac{33}{6}$
• Option 3: $\frac{9}{2} = \frac{27}{6}$

Since $\frac{127}{6}$ does not match any of the options 1, 2, or 3, the correct answer is 'None of these'.