In the graph [(3x+12)(x-2)]/[(x-1)(x+5)], the graph at one point crosses the horizontal asypmtote. How come?

Question

I did the graph on my calculator and saw that it crosses the horizontal asypmtote right in middle. But, how is that possible?

(I don't know how to insert graphs just yet ... )

Answer 1

A horizontal asymptote of a function $f$ is not a horizontal line that the graph of $f$ cannot cross, but rather a horizontal line that the graph of $f$ approaches as $x$ approaches $-\infty$ and as $x$ approaches $\infty$. Said another way, $y = c$ is a horizontal asymptote of $f$ if $\lim\limits_{x\to -\infty}f(x) = c$ and $\lim\limits_{x\to\infty}f(x) = c$.

Added later: A vertical asymptote of a function $f$ is a vertical line such that the graph of $f$ approaches infinity or negative infinity as $x$ gets closer to the line (from at least one direction). Said another way, $x = a$ is a vertical asymptote of $f$ if $\lim\limits_{x\to a^-}f(x) = \pm\infty$ or $\lim\limits_{x\to a^+}f(x) = \pm\infty$.

Note, $\lim\limits_{x\to a^-}f(x) = \pm\infty$ means that either $\lim\limits_{x\to a^-}f(x) = \infty$ or $\lim\limits_{x\to a^-}f(x) = -\infty$.

Some sources may only call $x = a$ a vertical asymptote of $f$ if $\lim\limits_{x\to a^-}f(x) = \pm\infty$ and $\lim\limits_{x\to a^+}f(x) = \pm\infty$. In this case, provided that $f$ is continuous, $f$ is not defined at $x = a$, so the graph of $f$ does not cross the line $x = a$.