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

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

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

To find the zeros of the function, we need to set the function equal to zero and solve for x. In this case, the function is:

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

Setting f(x) = 0, we have:

3x - 1 / (x^2 - 10x + 26) = 0

To simplify the equation, we can multiply both sides by (x^2 - 10x + 26):

3x - 1 = 0

Now, we solve for x:

3x = 1

x = 1/3

Therefore, the only real zero of the function is x = 1/3.

To find the poles of the function, we need to find the values of x that make the denominator equal to zero. In this case, we have:

x^2 - 10x + 26 = 0

We can solve this quadratic equation by factoring or using the quadratic formula. However, upon solving this equation, we find that it has no real roots.

Therefore, the function has no real poles.

Moving on to the asymptotes:

To determine if the function has a horizontal asymptote at 0, we need to check the behavior of the function as x approaches positive or negative infinity. By finding the limit of the function as x goes to infinity and negative infinity, we can determine if the function approaches a certain value or tends to zero.

Taking the limit as x approaches infinity:

lim(x->∞) [(3x - 1) / (x^2 - 10x + 26)]

As the degree of the numerator (3x) is lower than the degree of the denominator (x^2), the limit as x approaches infinity will be 0. Therefore, the function has a horizontal asymptote at 0.

Since the function has a horizontal asymptote at 0, it does not have a non-zero horizontal asymptote.

Lastly, to see if the function has a slant asymptote, we can use long division or synthetic division to divide the numerator by the denominator. If the result is a polynomial of degree 1, then the function has a slant asymptote.

Performing long division or synthetic division, we find that the quotient is a polynomial of degree 1. Therefore, the function has a slant asymptote.

In summary:
- The real zero of the function is x = 1/3.
- The function has no real poles.
- The function has a horizontal asymptote at 0.
- The function does not have a non-zero horizontal asymptote.
- The function has a slant asymptote.