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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.?

Jean Marie
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hs2345
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    It is alright for $F(x)$ to cross its horizontal asymptote. – peterwhy Jul 26 '15 at 21:27
  • @peterwhy I thought the definition of a horizontal asymptote at y = a is that limit as x approaching positive or negative infinity, a is never met... When I graph this, I'm quite not sure where I'd have to actually look at to see to make sure that y=3 is the horizontal asymptote... – hs2345 Jul 26 '15 at 22:18

2 Answers2

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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 $x\to\pm \infty$.

Dr. Heinz Doofenshmirtz
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  • I guess I will have to google for what other aspects of horizontal asymptote there are. Thank you! – hs2345 Jul 26 '15 at 22:17
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$$y=\lim _{ x\rightarrow +\infty }{ \frac { 9x\left( x-9 \right) }{ 3{ x }^{ 2 }-11x-4 } = } \lim _{ x\rightarrow +\infty }{ \frac { { x }^{ 2 }\left( 9-\frac { 81 }{ x } \right) }{ { x }^{ 2 }\left( 3-\frac { 11 }{ x } -\frac { 4 }{ { x }^{ 2 } } \right) } = } 3$$ so $$y=3$$ is your horizontal asymptote

haqnatural
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