3x-1/x^2-10x+26

Question

Find all (real) zeros of the function
Find all (real) poles of the function
The function has...
A horizontal asymptote at 0?
A non-zero horizontal asymptote?
A slant asymptote?
None of the above

Answer 1

I see no = sign in your function, functions are equations.

I see no brackets, but I will assume you mean

f(x) = (3x-1)/(x^2 - 10x + 26)

solving for x^2-10x+26=0 gives no real roots, so there are no vertical asymptotes
if f(x) = 0, then x=1/3
so the x-intercept is 1/3
as x ---> ∞ , f(x) ---> 0
so y=0 or the x-axis is a horizontal asymptote
(from above for positive x's , from below for negative x/s
there is no slant asymptote

I don't know what you mean by "poles"

Here are two graphs of your function by Wolfram
http://www.wolframalpha.com/input/?i=%283x-1%29%2F%28x%5E2+-+10x+%2B+26%29

Answer 2

odd to be using the term "pole" which is usually reserved for functions of a complex variable. Here it is apparently used to indicate a vertical asymptote, since for a complex function

f(z), a pole at z=c means

lim f(z) = ∞
z→c