+129852 It will have a horizontal asymptote at y = 3
The range will be ( - infinity, 3) U ( 3 , infinity )
+118677 What is the range of the function (3x+1)/(x+8)
I am not sure how CPhill looked at this question. This is what I'd do.
Chris may well have an easier way to look at it 
x cannot equal -8 because the denominator cannot be equal to zero.
If I do the algebraic division here then
\(f(x)=\frac{3x+1}{x+8}=3+\frac{-23}{x+8}\\ Now\;\;\frac{-23}{x+8}\;\;\text{cannot equal 0 so } f(x)\ne 3\\ Also\\ y=\frac{-23}{x+8}+3\\ \text{is of the from }y=\frac{a}{x-h}+k \\ \text{so it must be a hyperbola with asymptotes at x=-8 and y=3.}\)
the range Chris gave is certainly correct, I just wanted to discuss the question more. :)
+129852 I should have expanded my answer a tad bit more
We will have a vertical asymptote ar x = -8
As the graph approaches x = -8 from the left , the graph will approach infinity
As the graph approaches x = -8 from the right , the graph will approach -infinity
And there is a horizontal asymptote at y = 3 which is not crossed by the graph
So.....the range is ( - infinity, 3) U ( 3 , infinity )
