An equation I have is $$F(x) = \frac{9x(x-9)}{3x^2-11x-4}.$$
Upon calculating using the rules taught in class,
There is an H.A. at $y = 3$ and a V.A. at $x = -\frac13$ and at $4.$
After graphing, V.A. seems to be correct, but if I plug in $23/11$ as the $x$ value, it still returns $3$.
Are there any exceptions to when to use the rule to find out H.A.?
On
$$y=\lim _{ x\rightarrow +\infty }{ \frac { 9x\left( x-9 \right) }{ 3{ x }^{ 2 }-11x-4 } = } \lim _{ x\rightarrow +\infty }{ \frac { { x }^{ 2 }\left( 9-\frac { 81 }{ x } \right) }{ { x }^{ 2 }\left( 3-\frac { 11 }{ x } -\frac { 4 }{ { x }^{ 2 } } \right) } = } 3$$ so $$y=3$$ is your horizontal asymptote
Remember that an horizontal asymptote can cross the curve (there are even examples when this happens infinitely many times). The horizontal asymptote only describes the curve's behaviour for $x\to\pm \infty$.