Lesson 5.6 : Rational Functions

New Concept

Rational functions are fractions of polynomials, written as $f(x) = \frac{P(x)}{Q(x)}$. In this lesson, you'll find their domains, identify asymptotes, and graph them to model real-world problems like calculating concentrations or average costs.

What’s next

Next up, you'll work through interactive examples to find domains and asymptotes. A series of practice cards will then help you master graphing rational functions.

Property

To summarize, we use arrow notation to show that $x$ or $f(x)$ is approaching a particular value.

Examples

• For the function $f(x) = \frac{1}{x-3}$, as $x$ approaches 3 from the right, the function grows infinitely large. We write this as: As $x \to 3^+$, $f(x) \to \infty$.

Property

A rational function is a function that can be written as the quotient of two polynomial functions $P(x)$ and $Q(x)$.

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

To find the domain:

1. Set the denominator equal to zero.
2. Solve for the $x$-values.
3. The domain is all real numbers except the values found.

Examples

• Find the domain of $f(x) = \frac{x+5}{x^2-16}$. We set the denominator to zero: $x^2-16=0$, which gives $x = \pm 4$. The domain is all real numbers except $x=4$ and $x=-4$.

Property

A vertical asymptote of a graph is a vertical line $x = a$ where the graph tends toward positive or negative infinity as the inputs approach $a$. We write: As $x \to a$, $f(x) \to \pm\infty$.

A removable discontinuity (hole) occurs in the graph of a rational function at $x = a$ if $a$ is a zero for a factor in the denominator that is common with a factor in the numerator.

To find vertical asymptotes and holes:

1. Factor the numerator and denominator.
2. Zeros of common factors are locations of removable discontinuities (holes).
3. Zeros of the remaining factors in the denominator are locations of vertical asymptotes.

Property

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

The horizontal asymptote is determined by comparing the degree of the numerator ($p$) and the degree of the denominator ($q$):

1. If $p < q$: horizontal asymptote at $y = 0$.
2. If $p = q$: horizontal asymptote at $y = \frac{a_p}{b_q}$ (the ratio of the leading coefficients).
3. If $p > q$ by one: no horizontal asymptote; there is a slant asymptote, found by polynomial division.

Examples

• For $f(x) = \frac{8x+3}{2x^2+1}$, the degree of the numerator (1) is less than the denominator (2). The horizontal asymptote is $y=0$.

Property

A rational function will have a $y$-intercept at $f(0)$, if the function is defined at zero. A rational function will not have a $y$-intercept if the function is not defined at zero.

A rational function will have $x$-intercepts at the inputs that cause the output to be zero. Since a fraction is only equal to zero when the numerator is zero, $x$-intercepts can only occur when the numerator of the rational function is equal to zero.

Examples

• Find the intercepts of $f(x) = \frac{x-4}{x+2}$. The $y$-intercept is $f(0) = \frac{0-4}{0+2} = -2$, at $(0, -2)$. The $x$-intercept is when $x-4=0$, so $x=4$, at $(4, 0)$.

Property

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

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, and the stretch factor $a$ can be determined given a value of the function other than the $x$-intercept or by the horizontal asymptote if it is nonzero.

Examples

• Write a function with a vertical asymptote at $x=3$, an $x$-intercept at $x=-2$, and a horizontal asymptote at $y=0$. The simplest form is $f(x) = \frac{a(x+2)}{x-3}$. Since HA is $y=0$, this works. Let $a=1$, so $f(x) = \frac{x+2}{x-3}$.

• Write a function with vertical asymptotes at $x=\pm 1$, an $x$-intercept at $x=2$, and a horizontal asymptote at $y=4$. The form is $f(x) = a\frac{x-2}{(x-1)(x+1)}$. For HA $y=4$, degrees must match. So, $f(x) = a\frac{x^2-4}{(x-1)(x+1)}$ is not right. Instead, $f(x)=a\frac{x-2}{x^2-1}$ has HA at $y=0$. Let's adjust: $f(x) = a\frac{(x-2)(x+k)}{(x-1)(x+1)}$. For the HA to be $y=4$, the leading coefficients ratio must be 4. A simpler case is $f(x) = a\frac{(x-2)}{(x-1)}$. We need another VA. The form is $f(x)=a \frac{(x-x_1)...}{(x-v_1)...}$. For HA $y=4$, degrees must be equal. $f(x) = a\frac{(x-2)(x-c)}{(x-1)(x+1)}$. Let's use the given intercepts and asymptotes: $f(x)=a\frac{(x-2)}{(x-1)(x+1)}$ has HA at $y=0$. Let's try again with the rule: To get HA at $y=4$, degrees must match. $f(x) = a\frac{(x-2)(x-c)}{(x-1)(x+1)}$. Let's assume one x-intercept. $f(x) = a\frac{(x-2)^2}{(x-1)(x+1)}$ won't work. Let's try $f(x) = a\frac{x-2}{x-1}$. Let's follow the rules strictly. $x$-intercepts at $(2,0)$ and $x$-intercepts at $(5,0)$. Vertical asymptotes at $x=-4$ and $x=-1$. Horizontal asymptote at $y=7$. The function is $f(x)=a\frac{(x-2)(x-5)}{(x+4)(x+1)}$. The ratio of leading coefficients is $a$. So $a=7$. Final function: $f(x)=7\frac{(x-2)(x-5)}{(x+4)(x+1)}$.