Question 1126016
<br>
{{{f(x) = (x-5)/(x^2-36) = (x-5)/((x-6)(x+6))}}}<br>
f(x) is 0 at x = 5 only.<br>
f(x) is undefined (so there are vertical asymptotes) at x=-6 and x=6.<br>
The zeros of the numerator and denominator divide the domain into four parts: (-infinity,-6), (-6,5], [5,6), and (6, infinity).<br><pre>
                  |  x+6  x-5  x-6  |  f(x)
  --------------------------------------------
  (-infinity, -6) |   -    -    -   |   -
           (-6,5] |   +    -    -   |   +
            [5,6) |   +    +    -   |   -
    (6, infinity) |   +    +    +   |   +</pre><br>
The function is negative on (-infinity,-6) and on (5,6); it is 0 at x=5; the value goes to positive or negative infinity close to the vertical asymptotes, at x=-6 and x=6.<br>
The graph...<br>
{{{graph(400,400,-10,10,-1,1,(x-5)/(x^2-36))}}}