We are asked to find the area of the region bounded by the graphs of $x = \pm \sqrt{y-2}$ and $x = y-4$.

1. Problem Description

We are asked to find the area of the region bounded by the graphs of $x = \pm \sqrt{y-2}$ and $x = y-4$.

2. Solution Steps

First, let's rewrite the equation $x = \pm \sqrt{y-2}$ as $x^2 = y-2$, so $y = x^2 + 2$. The equation $x = y-4$ can be rewritten as $y = x+4$.

We need to find the points of intersection of the two curves $y = x^2 + 2$ and $y = x+4$.

Setting the two equations equal to each other, we have

$x^2 + 2 = x + 4$

$x^2 - x - 2 = 0$

$(x-2)(x+1) = 0$

So $x=2$ or $x=-1$.

If $x=2$, then $y = x+4 = 2+4 = 6$.

If $x=-1$, then $y = x+4 = -1+4 = 3$.

Therefore, the points of intersection are $(2,6)$ and $(-1,3)$.

Now, we can find the area of the region by integrating with respect to $x$. The area is given by

$$A = \int_{-1}^{2} (x+4 - (x^2+2)) dx = \int_{-1}^{2} (x+4-x^2-2) dx = \int_{-1}^{2} (-x^2 + x + 2) dx$$

$$A = \left[ -\frac{1}{3}x^3 + \frac{1}{2}x^2 + 2x \right]_{-1}^{2}$$

$$A = \left( -\frac{1}{3}(2)^3 + \frac{1}{2}(2)^2 + 2(2) \right) - \left( -\frac{1}{3}(-1)^3 + \frac{1}{2}(-1)^2 + 2(-1) \right)$$

$$A = \left( -\frac{8}{3} + \frac{4}{2} + 4 \right) - \left( \frac{1}{3} + \frac{1}{2} - 2 \right)$$

$$A = \left( -\frac{8}{3} + 2 + 4 \right) - \left( \frac{1}{3} + \frac{1}{2} - 2 \right)$$

$$A = -\frac{8}{3} + 6 - \frac{1}{3} - \frac{1}{2} + 2$$

$$A = 8 - \frac{9}{3} - \frac{1}{2} = 8 - 3 - \frac{1}{2} = 5 - \frac{1}{2} = \frac{10}{2} - \frac{1}{2} = \frac{9}{2} = 4.5$$

The area of the region is 4.5 square units.

3. Final Answer

C. 4.5 sq. units