### Graphing rational functions, asymptotes

This section shows another kind of function whose graphs we can
understand effectively by our methods. There is one new item here, the
idea of *asymptote* of the graph of a function.

A **vertical asymptote** of the graph of a function $f$ most
commonly occurs when $f$ is defined as a *ratio* $f(x)=g(x)/h(x)$
of functions $g,h$ continuous at a point $x_o$, but with the
denominator going to zero at that point while the numerator
doesn't. That is, $h(x_o)=0$ but $g(x_o) \ne 0$. Then we say that $f$
*blows up* at $x_o$, and that the line $x=x_o$ is a **vertical
asymptote** of the graph of $f$.

And as we take $x$ closer and closer to $x_o$, the graph of $f$ zooms
off (either up or down or both) *closely to the line* $x=x_o$.

A very simple example of this is $f(x)=1/(x-1)$, whose denominator is
$0$ at $x=1$, so causing a *blow-up* at that point, so that $x=1$
is a *vertical asymptote*. And as $x$ approaches $1$ from the
right, the values of the function zoom *up* to $+\infty$. When $x$
approaches $1$ from the *left*, the values zoom *down* to
$-\infty$.

A **horizontal asymptote** of the graph of a function $f$
occurs if either limit
$$\lim_{x\rightarrow +\infty}f(x)$$
or
$$\lim_{x\rightarrow -\infty}f(x)$$
exists. If $R=\lim_{x\rightarrow +\infty}f(x)$, then $y=R$ is a **horizontal asymptote** of the function, and if $L=\lim_{x\rightarrow
-\infty}f(x)$ exists then $y=L$ is a horizontal asymptote.

As $x$ goes off to $+\infty$ the graph of the function gets closer and
closer to the horizontal line $y=R$ if *that* limit exists. As $x$
goes of to $-\infty$ the graph of the function gets closer and
closer to the horizontal line $y=L$ if *that* limit exists.

So in rough terms *asymptotes* of a function are *straight lines* which the graph of the function approaches *at
infinity*. In the case of *vertical asymptotes*, it is the
$y$-coordinate that goes off to infinity, and in the case of *horizontal asymptotes* it is the $x$-coordinate which goes off to
infinity.

#### Example

Find asymptotes, critical points, intervals of increase and
decrease, inflection points, and intervals of concavity up and down of
$f(x)={x+3\over 2x-6}$: First, let's find the asymptotes. The
denominator is $0$ for $x=3$ (and this is *not* cancelled by the
numerator) so the line $x=3$ is a *vertical asymptote*. And as $x$
goes to $\pm\infty$, the function values go to $1/2$, so the line
$y=1/2$ is a horizontal asymptote.

The derivative is
$$f'(x)={1\cdot(2x-6)-(x+3)\cdot
2\over (2x-6)^2}={-12\over (2x-6)^2}$$
Since a ratio of polynomials can be zero only if the numerator is
zero, this $f'(x)$ can *never* be zero, so there are *no
critical points*. There is, however, the discontinuity at $x=3$ which
we must take into account. Choose auxiliary points $0$ and $4$ to the
left and right of the discontinuity. Plugging in to the derivative, we
have
$f'(0)=-12/(-6)^2
<0$, so the function is *decreasing* on the
interval $(-\infty,3)$. To the right, $f'(4)=-12/(8-6)^2
<0$, so the
function is also decreasing on $(3,+\infty)$.

The second derivative is $f''(x)=48/(2x-6)^3$. This is never zero, so
there are *no inflection points*. There is the discontinuity at
$x=3$, however. Again choosing auxiliary points $0,4$ to the left and
right of the discontinuity, we see $f''(0)=48/(-6)^3
<0$ so the curve is *concave downward* on the interval $(-\infty,3)$. And
$f''(4)=48/(8-6)^3>0$, so the curve is concave *upward* on
$(3,+\infty)$.

Plugging in just two or so values into the function then is enough to enable a person to make a fairly good qualitative sketch of the graph of the function.

#### Exercises

- Find all asymptotes of $f(x)={ x-1 \over x+2 }$.
- Find all asymptotes of $f(x)={ x+2 \over x-1 }$.
- Find all asymptotes of $f(x)={ x^2-1 \over x^2-4 }$.
- Find all asymptotes of $f(x)={ x^2-1 \over x^2+1 }$.