Question 547530
f(x) = {{{(2x^2-x-10)/(2x-5)}}}

f(x) = {{{((x+2)(2x-5))/(2x-5)}}}

2x-5 must not be 0 since denominators are never 0.

So

2x - 5 ≠ 0
    2x ≠ 5
     x ≠ {{{5/2}}}

We can cancel those as long as we require x ≠ {{{5/2}}}

f(x) = {{{((x+2)(cross(2x-5)))/(cross(2x-5))}}}, where x ≠ {{{5/2}}}

f(x) = x+2,  x ≠ {{{5/2}}}

Since x ≠ {{{5/2}}}, f(x) ≠ {{{5/2+2}}} or {{{9/2}}}

So the graph is this line with a hole:

{{{drawing(400,400,-5,5,-3,7,graph(400,400,-5,5,-3,7),
circle(2.5,4.5,.1), line(-11,-9,2.4,4.4), line(2.6,4.6,11,13),
locate(2.5,4.5,H(5/2,9/2))

 )}}}

with a hole at the point H({{{5/2}}},{{{9/2}}})

There are no asymptotes.

Edwin</pre>