Graphs of Rational Functions

A rational function is one polynomial divided by another. This page focuses on the simplest interesting kind, a linear expression over a linear expression, like $f(x) = \dfrac{2x + 1}{x - 1}$. Once you can find its two asymptotes and its intercepts, you can sketch it in a minute. You’ll also see what happens when the top and bottom share a factor.

Key ideas

The form

$$f(x) = \frac{ax + b}{cx + d}, \qquad c \ne 0$$

Its graph looks like a transformed $y = \dfrac{1}{x}$: two branches separated by a vertical asymptote, both levelling off toward a horizontal asymptote.

Vertical asymptote

The function is undefined where the denominator is $0$. If the numerator is not also $0$ there, the graph has a vertical asymptote:

$$cx + d = 0 \quad\Rightarrow\quad x = -\frac{d}{c}$$

Near the asymptote, the numerator is close to a fixed number and the denominator is tiny, so $f(x)$ is huge, positive on one side and negative on the other. Test a value just to each side to see which.

Horizontal asymptote

For very large $x$ (positive or negative), the constants $b$ and $d$ hardly matter, so

$$f(x) = \frac{ax + b}{cx + d} \approx \frac{ax}{cx} = \frac{a}{c}$$

The horizontal asymptote is $y = \dfrac{a}{c}$, the ratio of the leading coefficients. For example, $f(1000) = \dfrac{2001}{999} \approx 2.003$ for $f(x) = \dfrac{2x + 1}{x - 1}$, very close to $\dfrac{2}{1} = 2$.

Intercepts, domain, and range

• $x$-intercept: the fraction is $0$ when the numerator is $0$: $ax + b = 0$.
• $y$-intercept: $f(0) = \dfrac{b}{d}$ (if $d \ne 0$).
• Domain: $\left\{x \in \mathbb{R} \mid x \ne -\frac{d}{c}\right\}$.
• Range: $\left\{y \in \mathbb{R} \mid y \ne \frac{a}{c}\right\}$ (as long as the top and bottom have no common factor; otherwise the graph is a horizontal line with a hole).

Increasing or decreasing

On each branch, a graph of this type is either always increasing or always decreasing, and both branches do the same. Once you have the asymptotes and one or two points, the sketch shows you which.

Holes

If the numerator and denominator share a factor, cancel it, but remember the restriction. Where the cancelled factor is $0$, the graph has a hole (a single missing point), not an asymptote. For example,

$$g(x) = \frac{x^2 - 9}{x^2 + x - 6} = \frac{(x - 3)(x + 3)}{(x - 2)(x + 3)} = \frac{x - 3}{x - 2}, \quad x \ne -3$$

has a vertical asymptote $x = 2$ but only a hole at $x = -3$ (Example 3).

Oblique asymptotes (a preview)

If the numerator’s degree is one more than the denominator’s, use polynomial division to split off a linear part:

$$\frac{x^2 + 2x - 1}{x - 1} = x + 3 + \frac{2}{x - 1}$$

For large $x$, the fraction $\dfrac{2}{x - 1}$ is almost $0$, so the graph gets closer and closer to the slanted line $y = x + 3$. That line is an oblique asymptote. There’s no horizontal asymptote in this case.

Steps for sketching

1. Factor and cancel any common factors (note the holes).
2. Draw the vertical and horizontal asymptotes as dashed lines.
3. Plot the intercepts.
4. Test a point just to each side of the vertical asymptote, and draw each branch approaching the asymptotes.

Worked examples

Example 1: Sketching from the key features

Sketch $f(x) = \dfrac{2x + 1}{x - 1}$. State the domain, range, intercepts, positive and negative intervals, and intervals of increase or decrease.

Solution.

• Vertical asymptote: $x - 1 = 0$, so $x = 1$. (The numerator is $3 \ne 0$ there.)
• Horizontal asymptote: $y = \dfrac{2}{1} = 2$.
• $x$-intercept: $2x + 1 = 0$, so $x = -\dfrac{1}{2}$.
• $y$-intercept: $f(0) = \dfrac{1}{-1} = -1$.
• Near $x = 1$: $f(1.1) = \dfrac{3.2}{0.1} = 32$ and $f(0.9) = \dfrac{2.8}{-0.1} = -28$. So the graph goes up on the right of the asymptote and down on the left.

$y = \dfrac{2x + 1}{x - 1}$ has asymptotes $x = 1$ and $y = 2$.

Domain: $\{x \in \mathbb{R} \mid x \ne 1\}$. Range: $\{y \in \mathbb{R} \mid y \ne 2\}$.

The sign changes only at the $x$-intercept and the asymptote. From the graph (or test points), $f(x) \gt 0$ for $x \lt -\frac{1}{2}$ or $x \gt 1$, and $f(x) \lt 0$ for $-\frac{1}{2} \lt x \lt 1$.

The graph is decreasing on both branches: for $x \lt 1$ and for $x \gt 1$.

Example 2: Different leading coefficients

Find the key features of $h(x) = \dfrac{x - 4}{2x + 6}$ and describe its graph.

Solution.

• Vertical asymptote: $2x + 6 = 0$, so $x = -3$.
• Horizontal asymptote: $y = \dfrac{1}{2}$.
• $x$-intercept: $x = 4$. $y$-intercept: $h(0) = \dfrac{-4}{6} = -\dfrac{2}{3}$.
• Near the asymptote: $h(-2.9) = \dfrac{-6.9}{0.2} = -34.5$, and $h(-3.1) = \dfrac{-7.1}{-0.2} = 35.5$.

For very large $x$, $h(100) = \dfrac{96}{206} \approx 0.47$ and $h(-100) = \dfrac{-104}{-194} \approx 0.54$.

Putting this together: the left branch starts just above $y = \frac{1}{2}$ on the far left and rises toward the top beside $x = -3$. The right branch starts very low beside $x = -3$, rises through $\left(0, -\frac{2}{3}\right)$ and $(4, 0)$, and levels off just below $y = \frac{1}{2}$.

Domain: $\{x \in \mathbb{R} \mid x \ne -3\}$. Range: $\left\{y \in \mathbb{R} \mid y \ne \frac{1}{2}\right\}$. The graph is increasing on both branches. It is positive for $x \lt -3$ or $x \gt 4$ and negative for $-3 \lt x \lt 4$.

Example 3: A graph with a hole

Sketch $g(x) = \dfrac{x^2 - 9}{x^2 + x - 6}$.

Solution. Factor and cancel:

$$g(x) = \frac{(x - 3)(x + 3)}{(x - 2)(x + 3)} = \frac{x - 3}{x - 2}, \qquad x \ne -3,\ 2$$

• The factor $x + 3$ cancelled, so there’s a hole at $x = -3$. Its height comes from the simplified form: $\dfrac{-3 - 3}{-3 - 2} = \dfrac{6}{5} = 1.2$. The hole is at $(-3, 1.2)$.
• Vertical asymptote: $x = 2$. Horizontal asymptote: $y = \dfrac{1}{1} = 1$.
• $x$-intercept: $x = 3$. $y$-intercept: $g(0) = \dfrac{-3}{-2} = 1.5$.

The cancelled factor leaves a hole at $(-3, 1.2)$; the remaining factor gives the asymptote $x = 2$.

Domain: $\{x \in \mathbb{R} \mid x \ne -3, 2\}$. Range: $\{y \in \mathbb{R} \mid y \ne 1,\ y \ne 1.2\}$. (The graph reaches the height $1.2$ only at $x = -3$, where the hole is.)

Example 4: Writing an equation from features

Find an equation of the form $f(x) = \dfrac{ax + b}{cx + d}$ with vertical asymptote $x = 3$, horizontal asymptote $y = -2$, and $x$-intercept $1$.

Solution. The $x$-intercept $1$ means the numerator has the factor $(x - 1)$. The asymptote $x = 3$ means the denominator has the factor $(x - 3)$. So try

$$f(x) = \frac{k(x - 1)}{x - 3}$$

The horizontal asymptote is the ratio of leading coefficients, $\dfrac{k}{1}$, so $k = -2$:

$$f(x) = \frac{-2(x - 1)}{x - 3} = \frac{-2x + 2}{x - 3}$$

Check: numerator $0$ at $x = 1$ ✓; denominator $0$ at $x = 3$ (numerator $-4 \ne 0$) ✓; leading coefficients give $\frac{-2}{1} = -2$ ✓.

Common mistakes

Setting the numerator to zero to find the vertical asymptote. The numerator gives the $x$-intercept. The denominator gives the vertical asymptote.

Using the constants for the horizontal asymptote. For $\dfrac{x - 4}{2x + 6}$, the asymptote is $y = \frac{1}{2}$ (leading coefficients), not $y = \frac{-4}{6}$. That second number is the $y$-intercept.

Drawing an asymptote where a factor cancels. If a factor cancels, it leaves a hole. Factor first, every time.

Forgetting the hole in the domain and range. The hole’s $x$-value is still excluded from the domain, and its $y$-value is missing from the range (unless the graph reaches that height somewhere else).

Guessing which way each branch goes. Don’t rely on memory: test one value just left and one just right of the vertical asymptote.

Thinking a graph can never cross its horizontal asymptote. For the functions on this page it doesn’t, but for other rational functions it can, especially for small $x$. A horizontal asymptote describes the far-left and far-right behaviour only.

Practice

1. (Warm-up) State the asymptotes, domain, and range of $y = \dfrac{3x - 2}{x + 4}$.

Solution

Vertical asymptote $x = -4$; horizontal asymptote $y = \frac{3}{1} = 3$.

Domain: $\{x \in \mathbb{R} \mid x \ne -4\}$. Range: $\{y \in \mathbb{R} \mid y \ne 3\}$.

2. (Warm-up) Find the intercepts of $y = \dfrac{2x - 6}{x + 3}$.

Solution

$x$-intercept: $2x - 6 = 0$, so $x = 3$.

$y$-intercept: $\dfrac{-6}{3} = -2$.

3. (Core) For $f(x) = \dfrac{x + 2}{x - 3}$, find the asymptotes, intercepts, positive and negative intervals, and intervals of increase or decrease. Sketch the graph.

Solution

• Vertical asymptote $x = 3$; horizontal asymptote $y = 1$.
• $x$-intercept $-2$; $y$-intercept $-\frac{2}{3}$.
• $f(3.1) = \frac{5.1}{0.1} = 51$ and $f(2.9) = \frac{4.9}{-0.1} = -49$: up on the right of $x = 3$, down on the left.

Positive for $x \lt -2$ or $x \gt 3$; negative for $-2 \lt x \lt 3$.

The left branch comes in just below $y = 1$, passes through $(-2, 0)$ and $\left(0, -\frac{2}{3}\right)$, and drops beside $x = 3$. The right branch comes down from the top and levels off above $y = 1$. It’s decreasing on both branches.

4. (Core) Sketch $y = \dfrac{-x + 5}{2x + 4}$, and state where it is increasing or decreasing.

Solution

• Vertical asymptote: $2x + 4 = 0$, so $x = -2$.
• Horizontal asymptote: $y = \frac{-1}{2} = -\frac{1}{2}$.
• $x$-intercept $5$; $y$-intercept $\frac{5}{4}$.
• Test: at $x = -1.9$, $y = \frac{6.9}{0.2} = 34.5$; at $x = -2.1$, $y = \frac{7.1}{-0.2} = -35.5$.
• Far right: at $x = 100$, $y = \frac{-95}{204} \approx -0.47$, just above $-\frac{1}{2}$. Far left: at $x = -100$, $y = \frac{105}{-196} \approx -0.54$, just below $-\frac{1}{2}$.

The right branch comes down from the top beside $x = -2$, passes through $\left(0, \frac{5}{4}\right)$ and $(5, 0)$, and levels off just above $y = -\frac{1}{2}$. The left branch starts just below $y = -\frac{1}{2}$ on the far left and drops toward the bottom beside $x = -2$.

So the graph is decreasing on both branches: for $x \lt -2$ and for $x \gt -2$.

5. (Core) Find any holes and asymptotes of $y = \dfrac{x^2 - 1}{x^2 + 4x + 3}$, and state the domain.

Solution

$$\frac{x^2 - 1}{x^2 + 4x + 3} = \frac{(x - 1)(x + 1)}{(x + 1)(x + 3)} = \frac{x - 1}{x + 3}, \qquad x \ne -1,\ -3$$

Hole at $x = -1$: $\frac{-1 - 1}{-1 + 3} = -1$, so the hole is at $(-1, -1)$.

Vertical asymptote $x = -3$; horizontal asymptote $y = 1$.

Domain: $\{x \in \mathbb{R} \mid x \ne -3, -1\}$.

6. (Core) Let $f(x) = \dfrac{3x}{x + 2}$. Find $f(-2.1)$, $f(-2.01)$, $f(-1.99)$, $f(-1.9)$, and $f(1000)$. What do these values tell you about the graph?

Solution

$$f(-2.1) = \frac{-6.3}{-0.1} = 63, \quad f(-2.01) = \frac{-6.03}{-0.01} = 603$$$$f(-1.99) = \frac{-5.97}{0.01} = -597, \quad f(-1.9) = \frac{-5.7}{0.1} = -57$$$$f(1000) = \frac{3000}{1002} \approx 2.994$$

As $x$ approaches $-2$ from the left, $f(x)$ grows without bound; from the right, it drops without bound. So $x = -2$ is a vertical asymptote, with the left branch going up and the right branch going down. For large $x$, $f(x)$ is close to $3$, which is the horizontal asymptote $y = \frac{3}{1}$.

7. (Core) Write an equation of a function of the form $y = \dfrac{ax + b}{cx + d}$ with vertical asymptote $x = -1$ and horizontal asymptote $y = 3$, whose graph passes through the origin.

Solution

Through the origin means an $x$-intercept at $0$, so the numerator is $kx$. The asymptote $x = -1$ gives the denominator $x + 1$. The horizontal asymptote $\frac{k}{1} = 3$ gives $k = 3$:

$$y = \frac{3x}{x + 1}$$

Check: at $x = 0$, $y = 0$ ✓; denominator $0$ at $x = -1$ ✓; ratio of leading coefficients $3$ ✓.

8. (Challenge) Use long division to write $f(x) = \dfrac{2x^2 - x + 3}{x + 1}$ as a linear function plus a fraction. What happens to the graph of $f$ for very large positive and negative $x$?

Solution

Divide $2x^2 - x + 3$ by $x + 1$: the quotient is $2x - 3$ and the remainder is $6$. Check: $(x + 1)(2x - 3) + 6 = 2x^2 - x - 3 + 6 = 2x^2 - x + 3$ ✓.

$$f(x) = 2x - 3 + \frac{6}{x + 1}$$

For very large $|x|$, $\dfrac{6}{x + 1}$ is close to $0$, so the graph gets closer and closer to the line $y = 2x - 3$, an oblique asymptote. (For large positive $x$ the fraction is positive, so the graph is just above the line; for large negative $x$ it is just below.) There’s also a vertical asymptote at $x = -1$.

9. (Challenge) For which value of $k$ does $f(x) = \dfrac{2x + k}{x - 3}$ have a hole instead of a vertical asymptote? Describe the graph for that value of $k$.

Solution

A hole needs the numerator to be $0$ at $x = 3$ too: $2(3) + k = 0$, so $k = -6$.

Then

$$f(x) = \frac{2x - 6}{x - 3} = \frac{2(x - 3)}{x - 3} = 2, \qquad x \ne 3$$

The graph is the horizontal line $y = 2$ with a hole at $(3, 2)$.