Question 865322
You have the correct domain and range. Also, you have the correct vertical asymptote. Nice work on getting those right. 



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To find the x-intercept, you plug in y = 0 or f(x) = 0 and solve for x.



{{{f(x) = log(3,(9-3x))}}}



{{{0 = log(3,(9-3x))}}} Plug in f(x) = 0. Remember that f(x) and y are interchangeable.



{{{3^0 = 9-3x}}}



{{{1 = 9-3x}}}



{{{1-9 = -3x}}}



{{{-8 = -3x}}}



{{{-8/(-3)=x}}}



{{{8/3=x}}}



{{{x=8/3}}}



The x-intercept is {{{8/3}}} which is the point (8/3, 0)



Note: since {{{8/3=2.6666667}}} approximately, this means that the x-intercept is approximately the point (2.6666667,0)



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To find the y-intercept, you plug in x = 0 and solve for y (or f(x)).



{{{f(x) = log(3,(9-3x))}}}



{{{f(0) = log(3,(9-3(0)))}}} Plug in x = 0.



{{{f(0) = log(3,(9-0))}}}



{{{f(0) = log(3,(9))}}}



{{{f(0) = log(3,(3^2))}}}



{{{f(0) = 2*log(3,(3))}}}



{{{f(0) = 2*1}}}



{{{f(0) = 2}}}



The y-intercept is {{{2}}} which is the point (0,2)



Here is the graph of {{{f(x) = log(3,(9-3x))}}} (the red curve). The blue vertical dashed line is the vertical asymptote. The red curve will never touch or cross the blue dashed line (even though it looks like it does). It will just get closer and closer to it.



<img src="http://i150.photobucket.com/albums/s91/jim_thompson5910/4-20-20144-13-36PM_zpsab3116ad.png">