Posted by **Kenzie** on Monday, November 5, 2012 at 11:52pm.

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

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

- College Algebra -
**Reiny**, Tuesday, November 6, 2012 at 8:30am
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

- College Algebra -
**Steve**, Tuesday, November 6, 2012 at 10:58am
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

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