Asymptotes

Let p(x) and q(x) define polynomials. Consider the rational function $LaTeX: f\left(x\right)=\frac{p\left(x\right)}{q\left(x\right)}$ $f\left(x\right)=\frac{p\left(x\right)}{q\left(x\right)}$

written in lowest terms, and real numbers a and b.

If $LaTeX: \mid f\left(x\right)\mid\longrightarrow\infty\:$ $\mid f\left(x\right)\mid\longrightarrow\infty\:$ as $LaTeX: x\longrightarrow a,\:$ $x\longrightarrow a,\:$ then the line $LaTeX: x=a\:$ $x=a\:$ is a vertical asymptote.
If $LaTeX: f\left(x\right)\longrightarrow b\:$ $f\left(x\right)\longrightarrow b\:$ as $LaTeX: \mid x\mid\longrightarrow\infty,\:$ $\mid x\mid\longrightarrow\infty,\:$ then the line $LaTeX: y=b\:$ $y=b\:$ is the horizontal asymptote.

Determining Asymptotes

To find the asymptotes of a rational function defined by a rational expression in lowest terms, use the following procedures.

Vertical Asymptotes
Find any vertical asymptotes by setting the denominator equal to 0 and solving for x. If a is a zero of the denominator, then the line x= a is a vertical asymptote.
Other Asymptotes
Determine any other asymptotes by considering three possibilities:
1. If the numerator has lesser degree than the denominator, then there is a horizontal asymptote y = 0 (the x-axis).
2. If the numerator and denominator have the same degree, and the function is of the form $LaTeX: f\left(x\right)=\frac{a_nx^n+...+a_0}{b_nx^n+...+b_0},\:$ $f\left(x\right)=\frac{a_nx^n+...+a_0}{b_nx^n+...+b_0},\:$ where $LaTeX: a_n,b_n\ne0,\:$ $a_n,b_n\ne0,\:$ then the horizontal asymptote has equation $LaTeX: y=\frac{a_n}{b_n}.$ $y=\frac{a_n}{b_n}.$
3. If the numerator is of degree exactly one more than the denominator, then there will be an oblique (slanted) asymptote. To find it, divide the numerator by the denominator and disregard the remainder. Set the rest of the quotient equal to y to obtain the equation of the asymptote.

Note

The graph of a rational function may have more than one vertical asymptote, or it may have none at all. The graph cannot intersect any vertical asymptote. There can be at most one other (nonvertical) asymptote, and the graph may intersect that asymptote.

Examples

Give the equations of any vertical, horizontal, or oblique asymptotes for the graph of each rational function.

$LaTeX: f\left(x\right)=\frac{x+1}{\left(2x-1\right)\left(x+3\right)}$ $f\left(x\right)=\frac{x+1}{\left(2x-1\right)\left(x+3\right)}$
$LaTeX: f\left(x\right)=\frac{2x+1}{x-3}$ $f\left(x\right)=\frac{2x+1}{x-3}$
$LaTeX: f\left(x\right)=\frac{x^2+1}{x-2}$ $f\left(x\right)=\frac{x^2+1}{x-2}$

Solutions

$LaTeX: f\left(x\right)=\frac{x+1}{\left(2x-1\right)\left(x+3\right)}$ $f\left(x\right)=\frac{x+1}{\left(2x-1\right)\left(x+3\right)}$

Solution To find the vertical asymptotes, set the denominator equal to 0 and solve.
$LaTeX: \left(2x-1\right)\left(x+3\right)=0$ $\left(2x-1\right)\left(x+3\right)=0$
Using the zero-factor property:
$LaTeX: 2x-1=0\:\:or\:\:x+3=0$ $2x-1=0\:\:or\:\:x+3=0$
Solve each equation.
$LaTeX: x=\frac{1}{2}\:\:or\:\:x=-3$ $x=\frac{1}{2}\:\:or\:\:x=-3$

The equations of the vertical asymptotes are $LaTeX: x=\frac{1}{2}\:\:$ $x=\frac{1}{2}\:\:$ and $LaTeX: \:x=-3.$ $\:x=-3.$

To find the equation of the horizontal asymptote, divide each term by the greatest power of x in the expression. First, multiply the factors in the denominator.
$LaTeX: f\left(x\right)=\frac{x+1}{\left(2x-1\right)\left(x+3\right)}=\frac{x+1}{2x^2+5x-3}$ $f\left(x\right)=\frac{x+1}{\left(2x-1\right)\left(x+3\right)}=\frac{x+1}{2x^2+5x-3}$

Now divide each term in the numerator and denominator by $x^2$ since 2 is the greatest power of x.

$LaTeX: f\left(x\right)=\frac{\frac{x}{x^2}+\frac{1}{x^2}}{\frac{2x^2}{x^2}+\frac{5x}{x^2}-\frac{3}{x^2}}=\frac{\frac{1}{x}+\frac{1}{x^2}}{2+\frac{5}{x}-\frac{3}{x^2}}\:\:Stop\:here.\:Leave\:the\:expression\:in\:complex\:form.$ $f\left(x\right)=\frac{\frac{x}{x^2}+\frac{1}{x^2}}{\frac{2x^2}{x^2}+\frac{5x}{x^2}-\frac{3}{x^2}}=\frac{\frac{1}{x}+\frac{1}{x^2}}{2+\frac{5}{x}-\frac{3}{x^2}}\:\:Stop\:here.\:Leave\:the\:expression\:in\:complex\:form.$

As $LaTeX: \mid x\mid\:$ $\mid x\mid\:$ increases without bound, the quotients $LaTeX: \frac{1}{x},\:\frac{1}{x^2},\:\frac{5}{x},\:$ $\frac{1}{x},\:\frac{1}{x^2},\:\frac{5}{x},\:$ and $LaTeX: \frac{3}{x^{2\:}}$ $\frac{3}{x^{2\:}}$ all approach 0, and the value of f(x) approaches $LaTeX: \frac{0+0}{2+0-0}=\frac{0}{2}=0.$ $\frac{0+0}{2+0-0}=\frac{0}{2}=0.$
The line y = 0 (that is, the x-axis) is therefore the horizontal asymptote.
$LaTeX: f\left(x\right)=\frac{2x+1}{x-3}$ $f\left(x\right)=\frac{2x+1}{x-3}$

Set the denominator x − 3 equal to 0 to find that the vertical asymptote has equation x = 3. To find the horizontal asymptote, divide each term in the rational expression by x since the greatest power of x in the expression is 1. $LaTeX: f\left(x\right)=\frac{2x+1}{x-3}=\frac{\frac{2x}{x}+\frac{1}{x}}{\frac{x}{x}-\frac{3}{x}}=\frac{2+\frac{1}{x}}{1-\frac{3}{x}}$ $f\left(x\right)=\frac{2x+1}{x-3}=\frac{\frac{2x}{x}+\frac{1}{x}}{\frac{x}{x}-\frac{3}{x}}=\frac{2+\frac{1}{x}}{1-\frac{3}{x}}$
As $LaTeX: \mid x\mid\:$ $\mid x\mid\:$ increases without bound, both $LaTeX: \frac{1}{x}$ $\frac{1}{x}$ and $LaTeX: \frac{3}{x}\:$ $\frac{3}{x}\:$ approach 0, and f(x) approaches $LaTeX: \frac{2+0}{1-0}=\frac{2}{1}=2,\:$ $\frac{2+0}{1-0}=\frac{2}{1}=2,\:$ so the line y = 2 is the horizontal asymptote.
$LaTeX: f\left(x\right)=\frac{x^2+1}{x-2}$ $f\left(x\right)=\frac{x^2+1}{x-2}$

Setting the denominator x – 2 equal to 0 shows that the vertical asymptote has equation x = 2. If we divide by the greatest power of x as before ( $LaTeX: x^2\:$ $x^2\:$ in this case), we see there is no horizontal asymptote because $LaTeX: f\left(x\right)=\frac{x^2+1}{x-2}=\frac{\frac{x^2}{x^2}+\frac{1}{x^2}}{\frac{x}{x^2}-\frac{2}{x^2}}=\frac{1+\frac{1}{x^2}}{\frac{1}{x}-\frac{2}{x^2}}$ $f\left(x\right)=\frac{x^2+1}{x-2}=\frac{\frac{x^2}{x^2}+\frac{1}{x^2}}{\frac{x}{x^2}-\frac{2}{x^2}}=\frac{1+\frac{1}{x^2}}{\frac{1}{x}-\frac{2}{x^2}}$ does not approach any real number as $LaTeX: \mid x\mid\longrightarrow\infty,\:$ $\mid x\mid\longrightarrow\infty,\:$ since $LaTeX: \frac{1}{0}\:$ $\frac{1}{0}\:$ is undefined. This happens whenever the degree of the numerator is greater than the degree of the denominator. In such cases, divide the denominator into the numerator to write the expression in another form. We use synthetic division, as shown.

The result enables us to write the function as follows.
$LaTeX: f\left(x\right)=x+2+\frac{5}{x-2}$ $f\left(x\right)=x+2+\frac{5}{x-2}$
For very large values of $LaTeX: \mid x\mid,\:\frac{5}{x-2}\:$ $\mid x\mid,\:\frac{5}{x-2}\:$ is close to 0, and the graph approaches the line y = x + 2. This line is an oblique asymptote (slanted, neither vertical nor horizontal) for the graph of the function.