Let $$A = \{(x, y) : y^2 \leq 4x, y - 2x \geq -4\}$$. The area of the region $$A$$ in square units is:

The region $$A$$ is defined by the inequalities $$y^2 \leq 4x$$ and $$y - 2x \geq -4$$. To find the area, we first analyze these inequalities.

The inequality $$y^2 \leq 4x$$ can be rewritten as $$x \geq \frac{y^2}{4}$$, which represents the region to the right of the parabola $$x = \frac{y^2}{4}$$, including the parabola itself. The vertex of this parabola is at $$(0, 0)$$, and it opens to the right.

The second inequality $$y - 2x \geq -4$$ can be rearranged to $$y \geq 2x - 4$$, which represents the region above the line $$y = 2x - 4$$, including the line. This line has a slope of 2 and a y-intercept of $$-4$$.

To determine the bounded region, we find the points of intersection between the parabola $$x = \frac{y^2}{4}$$ and the line $$y = 2x - 4$$. Substitute $$x = \frac{y^2}{4}$$ into the line equation:

$$y = 2 \left( \frac{y^2}{4} \right) - 4$$

Simplify:

$$y = \frac{y^2}{2} - 4$$

Bring all terms to one side:

$$\frac{y^2}{2} - y - 4 = 0$$

Multiply both sides by 2 to clear the denominator:

$$y^2 - 2y - 8 = 0$$

Solve the quadratic equation using the discriminant $$D = b^2 - 4ac = (-2)^2 - 4(1)(-8) = 4 + 32 = 36$$:

$$y = \frac{2 \pm \sqrt{36}}{2} = \frac{2 \pm 6}{2}$$

So, the solutions are:

$$y = \frac{2 + 6}{2} = \frac{8}{2} = 4 \quad \text{and} \quad y = \frac{2 - 6}{2} = \frac{-4}{2} = -2$$

Find the corresponding $$x$$ values using $$x = \frac{y^2}{4}$$:

For $$y = 4$$:

$$x = \frac{4^2}{4} = \frac{16}{4} = 4$$

For $$y = -2$$:

$$x = \frac{(-2)^2}{4} = \frac{4}{4} = 1$$

Thus, the points of intersection are $$(4, 4)$$ and $$(1, -2)$$.

The region $$A$$ is bounded between $$y = -2$$ and $$y = 4$$. For each $$y$$ in this interval, the $$x$$ values range from the parabola $$x = \frac{y^2}{4}$$ (left boundary) to the line $$x = \frac{y + 4}{2}$$ (right boundary), derived from $$y \geq 2x - 4$$ which gives $$x \leq \frac{y + 4}{2}$$.

Verify that $$\frac{y^2}{4} \leq \frac{y + 4}{2}$$ for $$y \in [-2, 4]$$:

• At $$y = -2$$: $$\frac{(-2)^2}{4} = 1$$ and $$\frac{-2 + 4}{2} = 1$$, so $$1 \leq 1$$.
• At $$y = 0$$: $$\frac{0}{4} = 0$$ and $$\frac{0 + 4}{2} = 2$$, so $$0 \leq 2$$.
• At $$y = 4$$: $$\frac{4^2}{4} = 4$$ and $$\frac{4 + 4}{2} = 4$$, so $$4 \leq 4$$.

The inequality holds, so the area is given by the integral:

$$\text{Area} = \int_{y=-2}^{4} \left( \frac{y + 4}{2} - \frac{y^2}{4} \right) dy$$

Simplify the integrand:

$$\frac{y + 4}{2} - \frac{y^2}{4} = \frac{1}{2}y + 2 - \frac{1}{4}y^2 = -\frac{1}{4}y^2 + \frac{1}{2}y + 2$$

Now integrate:

$$\int_{-2}^{4} \left( -\frac{1}{4}y^2 + \frac{1}{2}y + 2 \right) dy$$

Find the antiderivative:

$$\int \left( -\frac{1}{4}y^2 + \frac{1}{2}y + 2 \right) dy = -\frac{1}{4} \cdot \frac{y^3}{3} + \frac{1}{2} \cdot \frac{y^2}{2} + 2y + C = -\frac{1}{12}y^3 + \frac{1}{4}y^2 + 2y + C$$

Evaluate from $$-2$$ to $$4$$:

At $$y = 4$$:

$$F(4) = -\frac{1}{12}(4)^3 + \frac{1}{4}(4)^2 + 2(4) = -\frac{1}{12}(64) + \frac{1}{4}(16) + 8 = -\frac{64}{12} + 4 + 8 = -\frac{16}{3} + 12 = -\frac{16}{3} + \frac{36}{3} = \frac{20}{3}$$

At $$y = -2$$:

$$F(-2) = -\frac{1}{12}(-2)^3 + \frac{1}{4}(-2)^2 + 2(-2) = -\frac{1}{12}(-8) + \frac{1}{4}(4) - 4 = \frac{8}{12} + 1 - 4 = \frac{2}{3} - 3 = \frac{2}{3} - \frac{9}{3} = -\frac{7}{3}$$

Subtract:

$$\text{Area} = F(4) - F(-2) = \frac{20}{3} - \left( -\frac{7}{3} \right) = \frac{20}{3} + \frac{7}{3} = \frac{27}{3} = 9$$

Thus, the area of region $$A$$ is 9 square units.

Hence, the correct answer is Option C.