Write both curves in convenient form

The parabola is $y^2=4(x-2)$ so $x=2+\frac{y^2}{4}.$

The line is $y=2x-8$ so $x=\frac{y+8}{2}=4+\frac y2.$

Find the points of intersection

Substitute the line expression for $x$ into the parabola equation: $y^2=4\left(\frac{y+8}{2}-2\right).$

Simplify: $y^2=4\left(\frac{y+4}{2}\right)=2y+8.$

Hence, $y^2-2y-8=0$ $(y-4)(y+2)=0.$

So the intersection values are $y=4 \quad \text{and} \quad y=-2.$

Corresponding $x$-coordinates:

• For $y=4$: $x=4+\frac{4}{2}=6$
• For $y=-2$: $x=4-1=3$

Thus intersection points are $(6,4) \quad \text{and} \quad (3,-2).$

Determine which curve lies to the right

For a fixed $y$ between $-2$ and $4$:

• Line: $x=4+\frac y2$
• Parabola: $x=2+\frac{y^2}{4}$

Check at $y=0$: $x_{\text{line}}=4, \qquad x_{\text{parabola}}=2.$

So the line is to the right and the parabola is to the left.

Set up the area integral

Area bounded by the curves is $A=\int_{-2}^{4}\left[\left(4+\frac y2\right)-\left(2+\frac{y^2}{4}\right)\right]dy.$

Simplify the integrand: $A=\int_{-2}^{4}\left(2+\frac y2-\frac{y^2}{4}\right)dy.$

Evaluate the integral

$A=\left[2y+\frac{y^2}{4}-\frac{y^3}{12}\right]_{-2}^{4}.$

At $y=4$: $2(4)+\frac{16}{4}-\frac{64}{12}=8+4-\frac{16}{3}=12-\frac{16}{3}=\frac{20}{3}.$

At $y=-2$: $2(-2)+\frac{4}{4}-\frac{-8}{12}=-4+1+\frac{2}{3}=-3+\frac{2}{3}=-\frac{7}{3}.$

Therefore, $A=\frac{20}{3}-\left(-\frac{7}{3}\right)=\frac{27}{3}=9.$

Match with the options

$\boxed{9}$

So the correct option is C.