Question 1064351
The first one cannot have x=8, because the denominator would be 0.
The function is (x-8)(x-3)/(x-8), which is x-3 after simplification.  There would be a hole in the graph at x=8, where the function is 0/0.
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This denominator factors into (n-2)(n+1)
n cannot equal 2 or -1.
9(n+1)/(n-2)(n+1)=9/(n-2).  There would be a hole in the graph at n= -1 and a vertical asymptote at n=2.