Rational Functions

A rational function is a function that can be written as a quotient of two polynomial functions $P(x)$ and $Q(x)$ which means $f(x) = \frac{P(x)}{Q(x)} = \frac{q_px^p + a_{p - 1}x^{p - 1} + ... + a_1x + a_0}{b_qx^q + b_{q - 1}x^{q - 1} + ... + b_1x + b_0}$ where $Q(x) \neq 0$. These types of functions are useful as many problems involving rates and concentrations often involve rational functions.

Domain

The domain of a rational function includes all real numbers except those that cause the denominator to equal zero.

$\underline{Example}$

Find the domain of $f(x) = \frac{4x}{5(x - 1)(x - 5)}$.

If $5(x - 1)(x - 5) = 0$ then $x = 1, 5$.

The domain of $f(x)$ is all real numbers except for $x = 1, 5$.

Intercepts

A rational function will have a $y$-intercept at $f(0)$ only if the function is defined at zero. If the domain does not include $x = 0$ then the function has no $y$-intercept.

A rational function will have $x$-intercepts at the inputs that cause $f(x) = 0$ which can only happen if the numerator is equal to $0$. This means to find the zeros of a rational function, we need to find the zeros of the numerator. After this, we just need to exclude all zeros not within the domain to get the zeros of $f(x)$.

$\underline{Example}$

Find the intercepts of $f(x) = \frac{(x - 2)(x + 3)}{(x - 1)(x + 2)(x - 5)}$.

We find the zeros of the numerator. In this case we solve for $(x - 2)(x + 3) = 0$ which is true if $x = 2, -3$.

We exclude any of those zeros if the domain does not include it. The domain can be found by finding the zeros of the denominator. In this case we solve for $(x - 1)(x + 2)(x - 5) = 0$ which is true if $x = 1, -2, 5$. This means we exclude $x = 1, -2, 5$ from the domain.

The zeros of $f(x)$ are $x = 2, -3$.

Vertical Asymptote

A vertical asymptote of a graph is a vertical line $x = a$ where the graph tends toward positive or negative infinity as the input approaches $a$ from either the left or the right. We write as $x \to a^-, f(x) \to \pm \infty$ or $x \to a^+, f(x) \to \pm \infty$.

Given a rational function, we can find any vertical asymptotes of its graph by...

1. Factor the numerator and denominator.
2. Reduce the function by cancelling common factors in the numerator and denominator.
3. All the values that cause the denominator to be zero in this simplified version are where the vertical asymptotes occur.

$\underline{Example}$

Find the vertical asymptotes of the graph of $f(x) = \frac{x - 2}{x^2 + 2x - 8}$.

First we factor the function and then reduce it if possible...

$f(x) = \frac{x - 2}{x^2 + 2x - 8} = \frac{(x - 2)}{(x - 2)(x + 4)} = \frac{\bcancel{(x - 2)}}{\bcancel{(x - 2)}{(x + 4)}} = \frac{0}{(x + 4)}$

The denominator for the reduced form is $x + 4$ which is zero only if $x = -4$.

The only vertical asymptote of the graph is $x = -4$.

Multiplicity

The multiplicity of a vertical asymptote is the number of times the factor is repeated in the denominator. This is important because it determines the behavior of the graph at a vertical asymptote.

1. If the multiplicity is even, the graph will tend towards the same direction on both sides of the asymptote.
2. If the multiplicity is odd, the graph will tend towards infinity on one side and negative infinity on the other side of the asymptote.

Fig. 1 - Vertical Asymptote Multiplicity

Removable Discontinuity

A removable discontinuity occurs in the graph of a rational function at $x = a$. These are single points where the graph is not defined. We often refer to these discountinuity as holes as indicate them by a open circle in the graph.

We can find these discontinuities through the cancelled out factors during reduction. Given a rational function, the steps are...

1. Factor the numerator and denominator.
2. Reduce the function by cancelling common factors in the numerator and denominator.
3. All the values that cause the denominator to be zero in this simplified version are where the vertical asymtotes occur.
4. All the values that cause the non-simplified version to be zero but are not vertical asymptotes are where the removable discontinuities occur.

$\underline{Example}$

Find the removable discontinuities of the graph of $f(x) = \frac{x - 2}{x^2 + 2x - 8}$.

First we factor the function and then reduce it if possible...

$f(x) = \frac{x - 2}{x^2 + 2x - 8} = \frac{(x - 2)}{(x - 2)(x + 4)} = \frac{\bcancel{(x - 2)}}{\bcancel{(x - 2)}{(x + 4)}} = \frac{0}{(x + 4)}$

The only factor that was cancelled out was $(x - 2)$ and $x - 2 = 0$ when $x = 2$.

The only removable discontinuity of the graph is at $x = 2$.

Fig. 2 - Vertical Asymptote and Removable Discontinuity of f(x)

note

A graph will continually approach a given asymptote or discontinuity but will never meet it at any finite distance.

Horizontal Asymptote

A horizontal asymptote of a graph is a horizontal line $y = b$ where the graph approaches the line as the inputs increase or decreases without bound. We write as $x \to \infty$ or $x \to -\infty$, $f(x) \to b$.

The horizontal asymptote of a rational function can determined through the degrees of the numerator and denominator...

1. If the degree of the numerator is less than the degree of the denominator, then there is a horizontal asymptote at $y = 0$.
2. If the degree of the numerator is greater than the degree of the denominator by one, then there is no horizontal asymptote.
3. If the degree of the numerator is equal to the degree of the denominator, then the horizontal asymptote is at the ratio of leading coefficients.

$\underline{Example}$

Find the horizontal asymptote of $f(x) = \frac{6x^2 - 10x}{2x^2 + 5x}$.

The degree of the numerator is $2$ and the degree of the denominator is $2$. The degrees are equal and so the horizontal asymptote is at the ratio of leading coefficients which is $\frac{6}{2}$.

This means the horizontal asymptote is at $y = 3$.

Fig. 3 - Horizontal Asymptote of f(x)

Slant Asymptote

If the degree of the numerator is greater than the degree of the denominator by one, then there is no horizontal asymptote. Instead we have a slant asymptote which we can find by dividing the numerator by the denominator.

$\underline{Example}$

Find the slant asymptote of $f(x) = \frac{x^2 - 4x + 1}{x + 2}$.

The degree of the numerator is $2$ and the degree of the denominator is $1$. The degree of the numerator is one greater than the denominator which means we can find the slant asymptote by dividing the numerator by the denominator. In this case, we can use synthetic division as $x + 2$ is in the form of $x - k$.

$$\begin{array}{c|rrrr} -2 & 1 & -4 & 1 \\ \: & & -2 & 12 \\ \hline & 1 & -6 & 13 \\ \end{array}$$

The slant asymptote is at $y = x - 6$.

Fig. 4 - Slant Asymptote of f(x)

Arrow Notation

We have used arrow notation to write asymptotes because they help us show that $x$ or $f(x)$ is approaching a particular value. The components of arrow notation include...

Symbol	Definition
$x \to a^-$	$x$ approaches $a$ from the left ($x < a$ but close to $a$)
$x \to a^+$	$x$ approaches $a$ from the right ($x > a$ but close to $a$)
$x \to \infty$	$x$ approaches infinity ($x$ increases without bound)
$x \to -\infty$	$x$ approaches negative infinity ($x$ decreases without bound)
$f(x) \to \infty$	the output approaches infinity (the output increases without bound)
$f(x) \to -\infty$	the output approaches negative infinity (the output decreases without bound)
$f(x) \to a$	the output approaches $a$

$\underline{Example}$

Describe the basic function $f(x) = \frac{1}{x}$ using arrow notation.

Fig. 5 - Basic Rational Function

The basic function can be described as...

1. As $x \to 0^-$, $f(x) \to -\infty$.
2. As $x \to 0^+$, $f(x) \to \infty$.
3. As $x \to -\infty$, $f(x) \to 0$.
4. As $x \to \infty$, $f(x) \to 0$.

Graphing

Given a rational function, we can graph it by following these steps...

1. Evaluate the function at $x = 0$ to find the $y$-intercept.
2. Factor the numerator and denominator.
3. For factors in the numerator not common to the denominator, determine where each factor of the numerator is zero to find the $x$-intecepts.
4. Find the multiplicities of the $x$-intercepts to determine the behavior of the graph at those points.
5. For factors in the denominator that are not common to the numerator, find the vertical asymptotes by finding the zeros of those factors.
6. Note the multiplicity of the vertical asymptotes.
7. For factors in the denominator that are common to the numerator, find the removable discontinuities by finding the zeros of those factors.
8. Compare the degrees of the numerator and the denominator to determine the horizontal or slant asymptotes.
9. Sketch the graph using all the information found.

$\underline{Example}$

Sketch the graph of $f(x) = \frac{(x + 2)^2(x - 2)}{2(x - 1)^2(x - 3)}$.

The rational function is already factored, so lets find the $x$-intercepts and their multiplicities. The first $x$-intercept is at $x = -2$ with a multiplicity of $2$ because $(x + 2)^2 = 0$. The second $x$-intercept is at $x = 2$ with a multiplicity of $1$ because $(x - 2) = 0$.

The next step is to find the vertical asymptotes. The first vertical asymptote is at $x = 1$ because $(x - 1)^2 = 0$. The second vertical asymptote is at $x = 3$ because $(x - 3) = 0$.

The multiplicity of $x = 1$ is $2$ meaning the graph will tend towards the same direction on both sides of the asymptote. The multiplicity of $x = 3$ is $1$ meaning the graph will tend towards negative infinity on one side and positive infinity on the other side of the asymptote.

There are no removable discontinuities because all factors in the numerator are not in the denominator.

The final step before sketching the graph is to find the horizontal asymptote. The degree of the numerator is $3$ and the degree of the denominator is $3$. The degrees are equal which means the horizontal asymptote is at the ratio of leading coefficients which is $\frac{1}{2}$.

The graph of $f(x)$ is shown below...

Fig. 6 - Graph of f(x)

Writing Functions

If a rational function has $x$-intercepts at $x = x_1, x_2, ..., x_n$, vertical asymptotes at $x = v_1, v_2, ..., v_m$, and no $x_i =$ any $v_j$, then the function can be written in the form:

$f(x) = a\dfrac{(x - x_1)^{p_1}(x - x_2)^{p_2}...(x - x_n)^{p_n}}{(x - v_1)^{q_1}(x - v_2)^{q_2}...(x - v_m)^{q_m}}$

where the powers $p_i$ or $q_i$ on each factor can be determined by the behavior of the graph at the corresponding intercept or asymptote. Finally, the stretch factor $a$ can be determined given a value of the function other than the $x$-intercept or by the horizontal asymptote if its non-zero.