I didn't quite understand how the horizontal asymptote of the function $r(x) = \frac{3x^2 - 2x - 1}{2x^2 + 3x - 2}$ can be $\frac{3}{2}$, since when you use desmos, for example, to graph it, part of the line crosses that horizontal asymptote.
I don't know what I'm missing! Is it called an asymptote just 'cause when $x \rightarrow\infty$ and when $x \rightarrow -\infty$ it truly just approaches $y = \frac{3}{2}$, but it can indeed even cross that y value at other parts of the graph?
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$\begingroup$ Your post is indeed a duplicate, and the answer to your question is just "Yes". $\endgroup$Anne Bauval– Anne Bauval2024-06-04 19:00:03 +00:00Commented Jun 4, 2024 at 19:00
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2 Answers
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The value which $f(x)$ approaches as "$x$" tends to $\pm\infty$ will give us the horizontal asymptote of our function. (See this source for additional information.) The value is independent of the behavior for $f(x)$ whenever $x\in(-\infty,\infty)$.
For instance, plug the function $f(x)=\frac{2x}{3x^2+1}$ into Desmos. The horizontal asymptote of this function will be $y=0$, even in spite of the fact that the function will intersect this line when $x\in(-\infty,\infty)$.
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Suppose we have a function
$y=f(x)$
Horizontal asymptote of this function is the value that $f(x)$ will tend to achieve as $x\to\infty$
In a similar reasoning, vertical asymptote of this function is the value of $x$ at which $f(x)$ will tend to go at $\infty$.
And yes, the graph can cross it's horizontal asymptote. There's nowhere written that it can't.
