Asymptotes

A vertical asymptote appears at an $x$ where the denominator is 0 and $y\to\pm\infty$. For $y=\frac{1}{(x-4)(x+2)}$ there are two: $x=4$ and $x=-2$.

A horizontal asymptote comes from the limit as $x\to\pm\infty$. Dividing the top and bottom of $y=\frac{4x+5}{x-4}$ by $x$ gives $\frac{4+5/x}{1-4/x}\to 4$, so the asymptote is $y=4$.

Examples

1. 01List all vertical asymptotes:$\displaystyle y=\frac{1}{(x-4)(x+2)}$AnswerHide

   Answer$\displaystyle x=4,\ -2$

   Steps

   1. A vertical asymptote sits where the denominator $(x-4)(x+2)$ is 0, and a product is 0 when one factor is
   2. $\displaystyle x-4 = 0$$\displaystyle \Rightarrow x = 4$$\displaystyle x+2 = 0$$\displaystyle \Rightarrow x = -2$
2. 02What is the horizontal asymptote?$\displaystyle y=\frac{4x+5}{x-4}$AnswerHide

   Answer$\displaystyle y=4$

   Steps

   1. Split the top into a multiple of the bottom, plus a remainder
   2. $\displaystyle 4x+5$$\displaystyle = 4(x-4)+21$
   3. So
   4. $\displaystyle \frac{4x+5}{x-4}$$\displaystyle = 4+\frac{21}{x-4}$
   5. As x grows the fraction on the right goes to 0, leaving $4$
3. 03What is the slant asymptote?$\displaystyle y=\frac{x^{2}+3}{x} = x+\frac{3}{x}$AnswerHide

   Answer$\displaystyle y=x$
4. 04What is the horizontal asymptote?$\displaystyle y=\frac{5x+1}{x-3}$AnswerHide

   Answer$\displaystyle y=5$

   Steps

   1. Split the top into a multiple of the bottom, plus a remainder
   2. $\displaystyle 5x+1$$\displaystyle = 5(x-3)+16$
   3. So
   4. $\displaystyle \frac{5x+1}{x-3}$$\displaystyle = 5+\frac{16}{x-3}$
   5. As x grows the fraction on the right goes to 0, leaving $5$
5. 05List all vertical asymptotes:$\displaystyle y=\frac{1}{(x-1)(x+4)}$AnswerHide

   Answer$\displaystyle x=1,\ -4$

   Steps

   1. A vertical asymptote sits where the denominator $(x-1)(x+4)$ is 0, and a product is 0 when one factor is
   2. $\displaystyle x-1 = 0$$\displaystyle \Rightarrow x = 1$$\displaystyle x+4 = 0$$\displaystyle \Rightarrow x = -4$
6. 06List all vertical asymptotes:$\displaystyle y=\frac{1}{(x-2)(x+3)}$AnswerHide

   Answer$\displaystyle x=2,\ -3$

   Steps

   1. A vertical asymptote sits where the denominator $(x-2)(x+3)$ is 0, and a product is 0 when one factor is
   2. $\displaystyle x-2 = 0$$\displaystyle \Rightarrow x = 2$$\displaystyle x+3 = 0$$\displaystyle \Rightarrow x = -3$

FAQ

Q1Is there always an asymptote where the denominator is 0?AnswerHide

A

No. $y=\frac{x^2-1}{x-1}$ equals $y=x+1$ for $x\neq 1$, so $x=1$ is only a missing point. If the numerator is also 0 there, cancel first and check.

Q2Is there a quick way to get the horizontal asymptote of $y=\frac{ax+b}{cx+d}$?AnswerHide

A

When $c\neq 0$ and $ad-bc\neq 0$, it is the ratio of the $x$ coefficients, $y=\frac{a}{c}$. For $y=\frac{5x+1}{x-3}$, it is $y=5$.