Question 914202
This is most of the way to an answer.


Instead of something as {{{f(x)=a*log((x))}}}, you have something like {{{f(x)=a*log((x+9))+k}}}.  That is based on the given asymptote.


This is what happens with the x-intercept:
{{{a*log((-8+9))+k=0}}}
{{{a*log((1))+k=0}}}
{{{a*0+k=0}}}
{{{highlight_green(k=0)}}}


This happens using the y-intercept:
{{{a*log((9))+k=2}}}
and you just found k=0, so 
{{{a*log((9))=2}}}
{{{highlight_green(a=2/log((9)))}}}


You could pick the base you want or need for the logarithm, but at least you know you have {{{highlight(f(x)=(2/log((9)))log((x-9)))}}}.