Graphs of rational functions

The graph of $y=\frac{k}{x}$ $(k\neq 0)$ is a hyperbola with the $x$- and $y$-axes as asymptotes. Shifting it gives $y=\frac{k}{x-p}+q$, whose asymptotes are $x=p$ and $y=q$. The point $(p,\ q)$ where they cross is the center of the hyperbola.

For the form $y=\frac{ax+b}{x-p}$, divide the numerator by the denominator to get $\frac{k}{x-p}+q$.

Here the center is $(1,\ 5)$ and the asymptotes are $x=1$ and $y=5$.

Examples

1. 01What is the center of this hyperbola?$\displaystyle y=\frac{3}{x-2}-5$AnswerHide

   Answer$\displaystyle (2,\ -5)$

   Steps

   1. The centre's x is where x−p becomes 0
   2. $\displaystyle x-2 = 0$$\displaystyle \Rightarrow x = 2$
   3. The added -5 is the centre's y
2. 02What is x for the vertical asymptote?$\displaystyle y=\frac{4}{x-3}$AnswerHide

   Answer$\displaystyle 3$

   Steps

   1. The x that makes the denominator 0
   2. $\displaystyle x-3 = 0$$\displaystyle \Rightarrow x = 3$
3. 03Which one has x=0 as a vertical asymptote?$\displaystyle$A$\displaystyle y=\frac{1}{x+4}$B$\displaystyle y=x$C$\displaystyle y=\frac{1}{x}$D$\displaystyle y=\frac{x}{x-3}$AnswerHide

   Answer$\displaystyle y=\frac{1}{x}$

   Steps

   1. Look for the x that makes the denominator 0
   2. $\displaystyle x = 0$$\displaystyle \Rightarrow \frac{1}{0}$
   3. Meanwhile $\frac{1}{x-3}$ blows up at $x=3$
4. 04What goes in □?$\displaystyle y=\frac{5x+2}{x-1} = \frac{k}{x-1}+\square$AnswerHide

   Answer$\displaystyle 5$

   Steps

   1. Split the top into a multiple of the bottom, plus a remainder
   2. $\displaystyle 5x+2$$\displaystyle = 5(x-1)+7$
   3. So
   4. $\displaystyle \frac{5x+2}{x-1}$$\displaystyle = 5 + \frac{7}{x-1}$
5. 05What is the center of this hyperbola?$\displaystyle y=\frac{3}{x-5}-1$AnswerHide

   Answer$\displaystyle (5,\ -1)$

   Steps

   1. The centre's x is where x−p becomes 0
   2. $\displaystyle x-5 = 0$$\displaystyle \Rightarrow x = 5$
   3. The added -1 is the centre's y
6. 06What is the center of this hyperbola?$\displaystyle y=\frac{1}{x-4}-4$AnswerHide

   Answer$\displaystyle (4,\ -4)$

   Steps

   1. The centre's x is where x−p becomes 0
   2. $\displaystyle x-4 = 0$$\displaystyle \Rightarrow x = 4$
   3. The added -4 is the centre's y

FAQ

Q1Is the center of $y=\frac{1}{x+4}+4$ at $(4,\ 4)$?AnswerHide

A

No. $x+4=x-(-4)$, so $p=-4$ and the center is $(-4,\ 4)$. Thinking of the vertical asymptote as the $x$ that makes the denominator $0$ helps avoid sign errors.

Q2How do I find the horizontal asymptote of $y=\frac{5x+2}{x-1}$?AnswerHide

A

Divide to rewrite it as $\frac{7}{x-1}+5$. The constant $5$ tells you that $y=5$ is the horizontal asymptote.