It will have a horizontal asymptote at y = 3
The range will be ( - infinity, 3) U ( 3 , infinity )
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)=3x+1x+8=3+−23x+8Now−23x+8cannot equal 0 so f(x)≠3Alsoy=−23x+8+3is of the from y=ax−h+kso 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. :)
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 )