If the line $L$ in the $xy$-plane has half the slope and twice the $y$-intercept of the line $y = \frac{2}{3} x + 4$, then an equation for $L$ is:
$\textbf{(A)}\ y = \frac{1}{3} x + 8 \qquad \textbf{(B)}\ y = \frac{4}{3} x + 2 \qquad \textbf{(C)}\ y =\frac{1}{3}x+4\qquad\\  \textbf{(D)}\ y =\frac{4}{3}x+4\qquad \textbf{(E)}\ y =\frac{1}{3}x+2$
The original slope and $y$-intercept are $\frac{2}{3}$ and $4$, so the new ones are $\frac{1}{3}$ and $8$ respectively. Thus, using the slope-intercept form ($y = mx+c$), the new equation is $y=\frac{1}{3}x + 8$, which is $\boxed{A}$.