Graph of $\log(3-x)$ So if $y=\log(3-x) = \log(-x+3)$ then you reflect $\log(x)$ in the $y$ axis to get $\log(-x)$.
Then because it is $+3$ inside brackets you then shift to the left by $3$ giving an asymptote of $x=-3$ and the graph crossing the $x$ axis at $(-4,0)$. 
However this does not work. The answer shows the $+3$ in the bracket shifting the curve to the right by $3$ giving an asymptote of $x=3$ and the curve crossing the $x$ axis at $(2,0)$. 
Why does it do this? Can anyone please explain?

Start with $y=\log(x)$. To shift this left three units, replace "$x$" with "$x+3$". Now you have $y=\log(x+3)$.
Now reflect over the $y$-axis. To do this, replace "$x$" with "$-x$". Now you have $y=\log(-x+3)$.
The order that the horizontal graph transformations happen is opposite from what you might think by the order of operations. If you first do "$x\mapsto-x$ and then do $x\mapsto x+3$, you get $\log(x)\mapsto\log(-x)\mapsto\log(-(x+3))$ which is not what you set out with.
