How do I tell a hole from a vertical asymptote?

Factor the numerator and denominator and cancel common factors. If the factor is gone from the denominator, there's a hole. If it's still there, there's a vertical asymptote.

Can a graph cross its horizontal asymptote?

Yes. A horizontal asymptote only describes the ends of the graph, so the graph can cross it at a finite input. A graph never crosses a vertical asymptote, because the function is undefined there.

When does a rational function have a slant asymptote?

When the numerator's degree is exactly one more than the denominator's. Divide, and the quotient is the line the graph approaches.

Why can't I multiply both sides of a rational inequality by the denominator?

The denominator is positive for some inputs and negative for others, and multiplying by a negative flips the inequality. A sign chart handles both cases at once.

Rational functions for AP Precalculus: asymptotes, holes, and sign charts

Updated November 10, 2026

A rational function is one polynomial divided by another. Almost every question about one comes down to two moves: factor the top and bottom, and compare their degrees.

This guide shows how to tell a vertical asymptote from a hole, how to read end behavior from the degrees, where the zeros are, and how to solve a rational inequality with a sign chart. Each part has a worked example with new numbers.

Vertical asymptotes or holes: factor first

A rational function $r(x) = \frac{p(x)}{q(x)}$ is undefined wherever $q(x) = 0$ . At each of those inputs, the graph has either a vertical asymptote or a hole. You can't tell which until you factor both the numerator and the denominator.

Vertical asymptote: a factor $(x - a)$ is still in the denominator after you cancel. Near $x = a$ , the outputs grow without bound in size.
Hole: every copy of $(x - a)$ in the denominator cancels with a copy on top. The graph is an ordinary curve with one point missing.
Height of a hole: cancel the common factors, then plug $x = a$ into what's left. If a copy of $(x - a)$ is left on top, the hole sits on the $x$ -axis.
Zero: a factor of the numerator that is left after cancelling. The graph crosses or touches the $x$ -axis there.

To find which way the graph goes on each side of an asymptote, pick an input just to one side and check the sign of each factor. Limit notation records the result: $\lim_{x \to a^+} r(x) = \infty$ means the outputs increase without bound as $x$ approaches $a$ from the right.

Sorting out every feature

Let

f(x) = \frac{x^2 - 2x - 8}{x^2 - 7x + 12}

. Find every hole, vertical asymptote, and zero of

f

, and describe the behavior of

f

on each side of the vertical asymptote.

Factor both parts: $f(x) = \frac{(x - 4)(x + 2)}{(x - 4)(x - 3)}$ . The denominator is $0$ at $x = 4$ and $x = 3$ .

The factor $x - 4$ appears once on top and once on the bottom, so it cancels. That gives a hole at $x = 4$ . The simplified form is $\frac{x + 2}{x - 3}$ , which equals $\frac{6}{1} = 6$ at $x = 4$ . The hole is at $(4, 6)$ .

The factor $x - 3$ stays in the denominator, so $x = 3$ is a vertical asymptote. The factor $x + 2$ stays on top, so $x = -2$ is a zero.

Just right of $3$ , $x + 2$ is about $5$ and $x - 3$ is a tiny positive number, so $\lim_{x \to 3^+} f(x) = \infty$ . Just left of $3$ , $x - 3$ is a tiny negative number, so $\lim_{x \to 3^-} f(x) = -\infty$ .

y = f(x)

y = 1

The graph splits at the asymptote x = 3. The hole at (4, 6) is marked here, but Desmos draws the curve straight through it.

Answer: Hole at $(4, 6)$ , vertical asymptote $x = 3$ with $\lim_{x \to 3^-} f(x) = -\infty$ and $\lim_{x \to 3^+} f(x) = \infty$ , and a zero at $x = -2$

End behavior: compare the degrees

As $x \to \infty$ or $x \to -\infty$ , a rational function acts like the ratio of the leading terms of its numerator and denominator. Compare the degrees.

Top degree less than bottom: the horizontal asymptote is $y = 0$ .
Degrees equal: the horizontal asymptote is $y = \frac{a}{b}$ , the ratio of the leading coefficients.
Top degree one more than bottom: there's a slant asymptote. Divide to find it.
Top degree two or more higher: no horizontal or slant asymptote. The ends act like a polynomial, so $\frac{x^3 + 1}{x - 1}$ acts like $x^2$ and goes to $\infty$ at both ends.

In the first example, both degrees are $2$ and both leading coefficients are $1$ , so the horizontal asymptote is $y = 1$ . That means $\lim_{x \to \infty} f(x) = 1$ and $\lim_{x \to -\infty} f(x) = 1$ . A graph can cross a horizontal asymptote at some finite $x$ . It can never cross a vertical one.

A slant asymptote by long division

Let

g(x) = \frac{x^2 + 5x + 3}{x + 2}

. Find the slant asymptote of the graph of

g

, and decide whether the graph is above or below it for large positive

x

Start with $x^2 \div x = x$ . Subtract $x(x + 2) = x^2 + 2x$ to leave $3x + 3$ .

Next, $3x \div x = 3$ . Subtract $3(x + 2) = 3x + 6$ to leave $-3$ .

So $g(x) = x + 3 - \frac{3}{x + 2}$ . Check by multiplying back: $(x + 3)(x + 2) - 3 = x^2 + 5x + 3$ .

As $x$ grows in size, $\frac{3}{x + 2}$ approaches $0$ , so the graph approaches the line $y = x + 3$ . For large positive $x$ , $-\frac{3}{x + 2}$ is negative, so the graph is just below the line.

Answer: The slant asymptote is $y = x + 3$ , and for large positive $x$ the graph is just below it

Which one levels off at 3

Which function's graph has the horizontal asymptote

y = 3

A. $\frac{3x^2 + 1}{x^2 - 4}$
B. $\frac{3x + 1}{x^2 - 4}$
C. $\frac{3x^3}{x^2 - 4}$
D. $\frac{x^2 + 3}{3x^2 - 4}$

In the first choice the degrees are equal, and the leading coefficients are $3$ and $1$ , so the asymptote is $y = 3$ . The second has a lower top degree, so its asymptote is $y = 0$ . The third has a higher top degree, so it has no horizontal asymptote. The fourth has equal degrees but leading coefficients $1$ and $3$ , so its asymptote is $y = \frac{1}{3}$ .

Answer: A

Solving a rational inequality with a sign chart

A rational function can only change sign at a zero or at an input where it is undefined. Between those inputs, its sign stays the same. That makes a sign chart the fastest way to solve an inequality like $r(x) \ge 0$ .

1Get $0$ on one side, then factor the numerator and denominator completely.
2Mark every zero of the numerator and of the denominator on a number line. Note any factor that cancels, because that input is still excluded.
3Test one input in each interval, or count the negative factors there. An even count means positive and an odd count means negative.
4Choose the intervals that match the inequality. Include the zeros for $\le$ or $\ge$ . Never include an input where the function is undefined.

Three factors, four intervals

Solve

\frac{x^2 - 3x - 4}{x - 2} \ge 0

Factor: $\frac{(x + 1)(x - 4)}{x - 2} \ge 0$ . The key inputs are $-1$ and $4$ , where the expression is $0$ , and $2$ , where it is undefined.

At $x = -2$ : $\frac{(-1)(-6)}{-4} = -\frac{3}{2}$ , negative.
At $x = 0$ : $\frac{(1)(-4)}{-2} = 2$ , positive.
At $x = 3$ : $\frac{(4)(-1)}{1} = -4$ , negative.
At $x = 5$ : $\frac{(6)(1)}{3} = 2$ , positive.

The expression is positive on $(-1, 2)$ and $(4, \infty)$ . Add the zeros $x = -1$ and $x = 4$ , since the inequality allows $0$ . Leave out $x = 2$ .

Answer: $[-1, 2) \cup [4, \infty)$

A sign chart with a hole in it

Solve

\frac{x^2 + x - 2}{x^2 - 4x + 3} \le 0

Factor: $\frac{(x + 2)(x - 1)}{(x - 1)(x - 3)}$ . For $x \ne 1$ this equals $\frac{x + 2}{x - 3}$ , so the sign only changes at $-2$ and $3$ .

$\frac{x + 2}{x - 3}$ is negative between $-2$ and $3$ , and $0$ at $x = -2$ . So the solution would be $[-2, 3)$ , except the original expression is undefined at $x = 1$ . Remove it.

Answer: $[-2, 1) \cup (1, 3)$

Checking your work in Desmos

On the calculator parts of the exam, Desmos is built into Bluebook. It's a fast check on everything above, as long as you know its one blind spot.

Holes don't show. Desmos draws $\frac{(x - 4)(x + 2)}{(x - 4)(x - 3)}$ as an unbroken curve through $x = 4$ . Find holes from the factors, then evaluate the function very close to them. In the first example, $f(4.0001) \approx 5.9995$ , which confirms the height $6$ .
Zeros: click where the graph crosses the $x$ -axis to see the coordinates of each zero.
End behavior: zoom out, or evaluate at a large input. Here $f(1000) \approx 1.005$ , close to the asymptote $y = 1$ .
Inequalities: graph $y = f(x)$ and read off where it is above or below the $x$ -axis. Then go back to the factors to decide which endpoints to include.

For more calculator moves, see how to use Desmos on the AP Precalculus exam. Rational functions can also appear on free response, where you explain your reasoning in words.

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Questions

How do I tell a hole from a vertical asymptote?: Factor the numerator and denominator and cancel common factors. If the factor is gone from the denominator, there's a hole. If it's still there, there's a vertical asymptote.
Can a graph cross its horizontal asymptote?: Yes. A horizontal asymptote only describes the ends of the graph, so the graph can cross it at a finite input. A graph never crosses a vertical asymptote, because the function is undefined there.
When does a rational function have a slant asymptote?: When the numerator's degree is exactly one more than the denominator's. Divide, and the quotient is the line the graph approaches.
Why can't I multiply both sides of a rational inequality by the denominator?: The denominator is positive for some inputs and negative for others, and multiplying by a negative flips the inequality. A sign chart handles both cases at once.