Question 536099
<pre>
f(x) = {{{(x^2+6x-8)/(x-5)}}}

Since the denominator is not a factor of the numerator,
we find all vertical asymptotes by setting the denominator = 0

x - 5 = 0
    x = 5

So the vertical asymptote is the line whose equation is x = 5,
which is a vertical line through 5 on the x-axis, which I will
draw in green:

{{{drawing(400,400,-15,15,-100,100, graph(400,400,-15,15,-100,100),green(line(5,100,5,-100))  )}}}

It has no horizontal asymptote because the numerator has a greater
degree than the denominator.  However since the degree of the
numerator is exactly 1 more than the degree of the denominator,
it does have a oblique (slanted) anymptote.  We find that by dividing
the denominator into the numerator:

      <u>      x + 11</u>
x - 5)x² + 6x -  8
      <u>x² - 5x</u>
          11x -  8  
          <u>11x - 55</u>
                47

The oblique asymptote is the line which has the equation

y = x + 11

So we draw that oblique asymptote:

{{{drawing(400,400,-15,15,-100,100, graph(400,400,-15,15,-100,100,200,(x+11)*sqrt(sin(10x))/sqrt(sin(10x))),green(line(5,100,5,-100))  )}}}

We get a few points and sketch in the graph:

{{{drawing(400,400,-15,15,-100,100, graph(400,400,-15,15,-100,100,200,(x+11)*sqrt(sin(10x))/sqrt(sin(10x))),green(line(5,100,5,-100)),
 graph(400,400,-15,15,-100,100,(x^2+6x-8)/(x-5))  

)}}}

The domain is all real numbers except where there is a vertical asymptote,
which is at 5, so the domain in interval notation is: 

{{{(matrix(1,3,-infinity,",",5))}}} U {{{(matrix(1,3,5,",",infinity))}}}

Edwin</pre>