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{{ printedBook.courseTrack.name }} {{ printedBook.name }} A rational function is a function that contains a rational expression. That is, any function that can be written in the form $f(x) = \dfrac{p(x)}{q(x)},$ where $p$ and $q$ are polynomial functions. For any values of $x$ where $q(x) = 0,$ the rational function is undefined. One example of a simple rational function is

$f(x) = \dfrac{1}{x}.$An asymptote of a function is a line that the function's graph approaches as the distance to the origin approaches infinity. It is very common for rational functions to have at least one asymptote. These can arise when $x$ extends to the right or left infinitely, and around $x$-values making the denominator equal to zero. The rational function
$f(x) = \dfrac{1}{x}$
has two asymptotes, the $x$-axis and the $y$-axis. The shape of this function's graph is called a *hyperbola*, which means that it consists of two mirror images with bow-like appearances.

Rational functions of the form $f(x) = \dfrac{a}{x - h} + k$ can be called simple rational functions, as their parent function is $g(x) = \dfrac{1}{x}.$

When $x$ is equal to $h,$ the denominator becomes $0.$ Thus, these functions have a vertical asymptote at $x = h.$ As $x \to \infty$ and $x \to \text{-} \infty,$ the fraction tends toward $0.$ This gives the horizontal asymptote $y = k.$
The identifiable asymptotes of simple rational functions can be used to simplify the process of graphing them. As an example, consider the function
$f(x) = \dfrac{\text{-} 1}{x - 3} + 2.$
### 1

Comparing the function rule to the general form of a simple rational function makes it possible to identify the asymptotes $x = h$ and $y = k.$ For $f,$ these are $x = 3$ and $y = 2.$ These can be drawn in a coordinate plane.

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Draw the asymptotes

Plot points around the vertical asymptote

Plot some points both to the left and the right of the vertical asymptote. For the example, it would be appropriate to use, for instance, the $x$-values $1,$ $2,$ $2.5,$ $3.5,$ $4,$ and $5.$

$x$ | $\dfrac{\text{-} 1}{x - 3} + 2$ | $f(x)$ |
---|---|---|

${\color{#0000FF}{1}}$ | $\dfrac{\text{-} 1}{{\color{#0000FF}{1}} - 3} + 2$ | $2.5$ |

${\color{#0000FF}{2}}$ | $\dfrac{\text{-} 1}{{\color{#0000FF}{2}} - 3} + 2$ | $3$ |

${\color{#0000FF}{2.5}}$ | $\dfrac{\text{-} 1}{{\color{#0000FF}{2.5}} - 3} + 2$ | $4$ |

${\color{#0000FF}{3.5}}$ | $\dfrac{\text{-} 1}{{\color{#0000FF}{3.5}} - 3} + 2$ | $0$ |

${\color{#0000FF}{4}}$ | $\dfrac{\text{-} 1}{{\color{#0000FF}{4}} - 3} + 2$ | $1$ |

${\color{#0000FF}{5}}$ | $\dfrac{\text{-} 1}{{\color{#0000FF}{5}} - 3} + 2$ | $1.5$ |

The points $(x, f(x))$ can be added to the coordinate plane to begin to see the behavior of $f.$

Draw the graph

The graph can now be drawn by connecting the points with a smooth curve. It must approach both the horizontal and vertical asymptotes.

Another common form of rational functions is $f(x) = \dfrac{ax + b}{cx + d},$ where $p(x) = ax + b$ and $q(x) = cx + d$ are linear functions with different $x$-intercepts. Just like simple rational functions, these are hyperbolas and have horizontal and vertical asymptotes. The horizontal asymptote is the line $y = \dfrac{a}{c}.$ That is, the quotient of the leading coefficients of $p$ and $q.$ The vertical asymptote is the line $x = \text{-} \dfrac{d}{c},$

which is the $x$-value that makes the denominator $0.$Identify the asymptotes of the given rational function. Then, rewrite it in the form, $f(x) = \frac{a}{x - h} + k$ and draw its graph. $f(x) = \dfrac{4x+2}{x + 1}$

Show Solution

The function is currently written in the form
$f(x) = \dfrac{ax + b}{cx + d}.$
The horizontal asymptote can be identified as the quotient of the leading coefficients.
$y = \dfrac{a}{c} = \dfrac{4}{1} = 4$
The vertical asymptote is located at the $x$-value for which the denominator equals $0.$
$x = \text{-} \dfrac{d}{c} = \text{-} \dfrac{1}{1} = \text{-} 1.$
There is more than one viable strategy to rewrite the expression. Two of which are long division and synthetic division. However, here, we'll use *inspection.* Notice that, if the numerator was $4x + 4$ the quotient would be $4.$ Let's manipulate the expression to make that happen.
As a last step in writing the function in the form $f(x)=\frac{a}{x-h}+k,$ we need to rewrite the denominator.
$f(x) = \dfrac{\text{-} 2}{x + 1} + 4 \quad \Leftrightarrow \quad f(x) = \dfrac{\text{-} 2}{x - (\text{-} 1)} + 4$
We have now written the function in the desired form. The asymptotes $y = 4$ and $x = \text{-} 1$ can be confirmed here as well. To graph this function, we start by drawing the asymptotes in a coordinate plane.

$f(x) = \dfrac{4x + 2}{x + 1}$

AddDiffZero$a = a+{\color{#0000FF}{2}}-{\color{#0000FF}{2}}$

$f(x) = \dfrac{4x + 2 + {\color{#0000FF}{2}} - {\color{#0000FF}{2}}}{x + 1}$

AddTermsAdd terms

$f(x) = \dfrac{4x + 4 - 2}{x + 1}$

The fraction can now be split into two, where one numerator is $4x + 4$ and the other is $\text{-} 2.$

$f(x) = \dfrac{4x + 4 - 2}{x + 1}$

WriteSumFracWrite as a sum of fractions

$f(x) = \dfrac{4x + 4}{x + 1} + \dfrac{\text{-} 2}{x + 1}$

FactorOutFactor out $4$

$f(x) = \dfrac{4(x + 1)}{x + 1} + \dfrac{\text{-} 2}{x + 1}$

CalcQuotCalculate quotient

$f(x) = 4 + \dfrac{\text{-} 2}{x + 1}$

CommutativePropAddCommutative Property of Addition

$f(x) = \dfrac{\text{-} 2}{x + 1} + 4$

Next, we can make a table to find points around the vertical asymptote.

$x$ | $\dfrac{\text{-} 2}{x + 1} + 4$ | $f(x)$ |
---|---|---|

$\text{-} 3$ | $\dfrac{\text{-} 2}{{\color{#0000FF}{\text{-} 3}} + 1} + 4$ | $5$ |

$\text{-} 2$ | $\dfrac{\text{-} 2}{{\color{#0000FF}{\text{-} 2}} + 1} + 4$ | $6$ |

$0$ | $\dfrac{\text{-} 2}{{\color{#0000FF}{0}} + 1} + 4$ | $2$ |

$1$ | $\dfrac{\text{-} 2}{{\color{#0000FF}{1}} + 1} + 4$ | $3$ |

Plotting the points gives the graph.

We can connect the points with two curves that tend toward the asymptote as $x$ and $y$ extend in both directions.

Below, two rational functions are graphed.

Match the graphs with their corresponding function rule. $\begin{aligned} f(x) &= \dfrac{4x + 1.5}{\text{-} 2x + 1} & g(x) &= \dfrac{2}{x - 1} - 1\\[0.8em] h(x) &= \dfrac{2}{x + 1} - 1 & t(x) &= \dfrac{4x + 1.5}{2x + 1} \end{aligned}$

Show Solution

We match the graphs with their corresponding rule by identifying the asymptotes. Observing the first graph, it looks as though the asymptotes are $\begin{aligned} \text{I:}\ x = 1 \quad \text{and} \quad y = \text{-} 1. \end{aligned}$ For the second graph, $\begin{aligned} \text{II:}\ x = \text{-} 0.5 \quad \text{and} \quad y = 2. \end{aligned}$ Let's now identify the asymptotes of the given rules. For rational functions in the form $\begin{aligned} f(x) = \dfrac{ax + b}{cx + d}, \end{aligned}$ the vertical asymptote is $x = \text{-} \frac d c$ and the horizontal asymptote is $y = \frac a c.$ For the ones written as $\begin{aligned} f(x) = \dfrac{a}{x - h} + k, \end{aligned}$ their vertical asymptote is $x = h$ and the horizontal asymptote is $y = k.$

Function | Vertical asymptote | Horizontal asymptote |
---|---|---|

$f$ | $x = \text{-} \dfrac{1}{\text{-} 2} = 0.5$ | $y = \dfrac{4}{\text{-} 2} = \text{-} 2$ |

$g$ | $x = 1$ | $y = \text{-} 1$ |

$h$ | $x = \text{-} 1$ | $y = \text{-} 1$ |

$t$ | $x = \text{-} \dfrac{1}{2} = \text{-} 0.5$ | $y = \dfrac{4}{2} = 2$ |

The function $g$ has the same asymptotes as Graph I, and $t$ has the same asymptotes as Graph II. Thus, the graphs' corresponding function rules are $\begin{aligned} \text{I:}\ g(x) \quad \text{and} \quad \text{II:}\ t(x). \end{aligned}$

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