User:
Asymptotes

# Vertical asymptote

A line $\fs2x=a$ is a vertical asymptote of a function $f$ when $\lim_{\fs1x\to\;a^-}f(x)=\pm\infty$ or $\lim_{\fs1x\to\;a^+}f(x)=\pm\infty$

An asymptote is a line that is not part of the graph, but one that the graph approaches closely. When the graph gets close to the vertical asymptote, it curves either upward or downward very steeply so that it looks almost vertical itself. Remember that the graph can get very close to the asymptote but can't touch it.

The following figure illustrates some of the ways in wich the graph of a function may approach its vertical asymptote:

As an example, let's see the graph of $f(x)=\;\frac{1}{x-1}$:

 On this graph, you will notice that the curves keep getting closer and closer to x=1. This line is called vertical asymptote.

In the rational functions, the vertical asymptotes are the points outside the domain of the function:

Theorem on Vertical Asymptotes of Rational Functions
If the real number a is a zero of the demoninator Q(x) of a rational function, then the graph of f(x)=P(x)/Q(x), where P(x) and Q(x) have no common factors, has the vertical asymptote x=a.

Algorithm for finding the vertical asymptotes for the graph of the quotient of two polynomials with no common factors.

 Step 1. Set the denominator equal to zero. Step 2. Factor and solve for x, if possible. Step 3. Substitute the value(s) for x into the numerator of the quotient. Those values for which the numerator is not zero are vertical asymptotes. Note: The values where the numerator is zero are points of discontinuity

Find the vertical asymptotes for $p(x)=\;\frac{2x-2}{x^2-4}$

 Step 1. x2-4=0 Step 2. (x-2)(x+2)=0; x=-2 and x=2 are candidates for vertical asymptotes. Step 3. The numerator is 2x-2 with $2(-2)-2=-6\;\neq\;0$ and $2(2)-2=2\;\neq\;0$. Therefore the lines x=-2 and x=2 are both vertical asymptotes.

Graphically:

Find the vertical asymptotes for $f(x)=\frac{x+1}{x^2-5x+6}$

 Step 1. The vertical asymptotes are the points outside the domain of the function: x2-5x+6=0 Step 2. $\fs2x^2-5x+6=0\;\Leftrightarrow\;(x-2)(x-3)=0\;\Leftrightarrow\;x=2\;o\;x=3$; x=2 and x=3 are candidates for vertical asymptotes. Step 3. The numerator is x+1 with $2+1=3\;\neq\;0$ and $3+1=4\;\neq\;0$. Therefore the lines x=2 and x=3 are both vertical asymptotes.

Remember that a function f admits vertical asymptotes in points where at least one lateral limit is infinite

$\lim_{\fs1x\to\;2^-}\frac{x+1}{(x-2)(x-3)}=+\infty$ $\lim_{\fs1x\to\;2^+}\frac{x+1}{(x-2)(x-3)}=-\infty\;\;\fs2\text{therefore\;x\;=\;2\;is\;a\;vertical\;asymptote\;for}\;f$
$\lim_{\fs1x\to\;3^-}\frac{x+1}{(x-2)(x-3)}=-\infty$ $\lim_{\fs1x\to\;3^+}\frac{x+1}{(x-2)(x-3)}=+\infty\;\;\fs2\text{therefore\;x\;=\;3\;is\;a\;vertical\;asymptote\;for}\;f$

Therefore the lines x=2 and x=3 are both vertical asymptotes.

Graphically:

After studying rational functions may seem that all functions work the same. We make the mistake of thinking that the vertical asymptotes are found only in the points outside the domain. To avoid falling into this error, we see that we can find functions with vertical asymptotes at points belonging to the domain of the function. Consider the following piecewise function that has a vertical asymptote at x =2 and this point is part of its domain:

Find the vertical asymptotes of $f(x)=\left\{\frac{1}{x-2}\;\;\;si\;x<2\\x-1\;\;\;si\;x\ge2$:

$f(x)=\left\{\frac{1}{x-2}\;\;\;si\;x<2\\x-1\;\;\;si\;x\ge2$ has a vertical asymptote at x=2, and x=2 is a point of its domain.

Graphically:

Find the vertical asymptote(s) of the function $f(x)=\frac{-x-5}{-2x^3+2x}$
There are vertical asymptotes.