The curve $xy =1$ has a slope that is negative everywhere in the first quadrant. This is apparent both visually and algebraically ($dy/dx = -y/x$).
Its reflection in the line $x=4$ is the curve $y(8-x) = 1$ (Reflection of rectangular hyperbola in vertical line)
When I plot the curve using R it seems as if the slope of the reflected curve is positive everywhere in the first quadrant. But algebraically the slope $dy/dx$ is given by $y/(8-x)$ which indicates that when x is greater than 8 the slope is negative.
I can't believe that my eyes are deceiving me. Is the equation for the slope correct?
When $x>8$, I agree that $8-x<0$ but also notice that since $y=\dfrac{1}{8-x}$ , $y<0 ; ~\text{when}~ x>8$,
Hence the slope = $\dfrac{y}{8-x}= \dfrac{\text{negative}}{\text{negative}}=\boxed{\text{positive}}$ as expected by you.