The curve crosses the x-axis at x = 1. Which part of the formula explains the crossing?

Both the top and the bottom are 0 at x = 1

Which inputs are candidates for vertical asymptotes of p(x) = \dfrac{x^2 + 1}{x^2 - 6x + 8}?

None, because x^2 + 1 is never 0

Find candidate vertical asymptotes

Builds on recognizing a break that one value cannot repair. 6 minutes.

A bottom that shrinks to zero

The curve shows \(f(x) = \dfrac{2x + 1}{x^2 - x - 6}\). The bottom is 0 at \(x = 3\) and at \(x = -2\), so the formula has no value at those two inputs.

Drag the slider to move the orange point toward 3. At \(x = 3.1\) the top is 7.2 and the bottom is 0.51, so the height is 14.12. At \(x = 3.01\) the bottom is 0.0501 and the height is 140.12.

The point leaves the figure, so the readouts give its height. The top stays close to 7 while the bottom shrinks toward 0, so the height grows without bound. Left of 3 the bottom is negative, so the curve falls instead.

If the bottom approaches 0 and the top does not, the curve climbs or falls beside a vertical line, which is called a vertical asymptote. Press the toggle to mark the two inputs where this happens.

Zeros of the top and zeros of the bottom

The curve is a fraction. Its top is 0 at the input \(a\), and its bottom is 0 at the input \(c\).

Move the orange point toward \(a\). The top becomes small, so the whole fraction becomes small, and the curve crosses the x-axis at \(a\).

Move the point toward \(c\). The bottom becomes small and the top is not 0, so the size of the fraction becomes large. The curve climbs on one side of the dashed vertical line and falls on the other.

A candidate is an input where a vertical asymptote might be. The zeros of the bottom are candidates, and the zeros of the top are not.

Notation

Write the function as \(\dfrac{N(x)}{D(x)}\). Set the bottom \(D(x)\) equal to 0 and solve. Each solution is a candidate.
\[D(x) = 0\]
Substitute each candidate \(c\) into the top. If the top is not 0 there, the size of the fraction grows without bound near \(c\), so the line \(x = c\) is a vertical asymptote.
\[\begin{gathered} D(c) = 0 \ \text{ and } \ N(c) \neq 0 \\ \Rightarrow\ \cy{x = c} \text{ is a vertical asymptote} \end{gathered}\]
A zero of the top is an x-intercept, not an input where the curve rises or falls without bound.
\[\begin{gathered} N(a) = 0 \\ \Longrightarrow\ \text{x-intercept at } x = a \end{gathered}\]
A function that is not a fraction has candidates where its formula has no value. For \(\ln x\), the candidate is \(x = 0\). For \(\tan x = \dfrac{\sin x}{\cos x}\), the candidates are the inputs where \(\cos x = 0\).

When it fails

If the top is also 0 at the candidate, the test above does not decide. A common factor can cancel, and the break can be a hole instead of an asymptote.

In that case, cancel the common factor first. Then test the simplified fraction again.

Example 1: factor the bottom

Find the candidates for vertical asymptotes of \(g(x) = \dfrac{x + 4}{x^2 - 2x - 8}\), and decide which ones are vertical asymptotes.

Set the bottom equal to 0 and factor it.
\[x^2 - 2x - 8 = (x - 4)(x + 2) = 0\]
The candidates are \(x = 4\) and \(x = -2\).
Substitute each candidate into the top. At \(x = 4\) the top is 8, and at \(x = -2\) it is 2. Neither is 0, so the fraction is large near both inputs.
\[\begin{gathered} g(x) \text{ is large near } \cy{x = 4} \\ \text{and near } \cy{x = -2} \end{gathered}\]
Both lines are vertical asymptotes. The zero of the top, at \(x = -4\), is only an x-intercept.

Example 2: top or bottom

Find the vertical asymptotes of \(h(x) = \dfrac{x - 5}{x^2 - 9}\).

A tempting wrong move: Set the top equal to 0. The top is 0 at \(x = 5\), so \(x = 5\) is a candidate.
\[x - 5 = 0\]
At \(x = 5\) the height is \(\tfrac{0}{16} = 0\), and the orange dot shows the curve crossing the x-axis there. The curve does not climb or fall along the red line, so 5 is not a candidate.
A vertical asymptote comes from the bottom. Set the bottom equal to 0 and factor it.
\[x^2 - 9 = (x - 3)(x + 3) = 0\]
The candidates are \(x = 3\) and \(x = -3\). The top is \(-2\) at 3 and \(-8\) at \(-3\). Neither is 0, so both lines are vertical asymptotes. The input 5 is only an x-intercept.

Practice

Find the candidates for vertical asymptotes of \(y = \tan 2x\) for \(0 < x < \pi\).

Write \(\tan 2x\) as a fraction. The curve can leave the figure only where the bottom is 0.

Solution

Write the function as a fraction. The bottom is 0 where \(\cos 2x = 0\).
\[\tan 2x = \frac{\sin 2x}{\cos 2x}\]
For \(0 < x < \pi\), the angle \(2x\) is between 0 and \(2\pi\). In this range \(\cos 2x = 0\) at \(2x = \tfrac{\pi}{2}\) and at \(2x = \tfrac{3\pi}{2}\).
\[x = \cy{\tfrac{\pi}{4}} \quad \text{and} \quad x = \cy{\tfrac{3\pi}{4}}\]
At these inputs the top, \(\sin 2x\), is 1 and \(-1\). It is not 0, so both lines are vertical asymptotes.

Questions

The graph shows \(r(x) = \dfrac{x - 1}{(x + 1)(x - 2)}\). Dashed vertical lines mark the inputs where the curve has no point.

At which inputs does the graph have vertical asymptotes?
- \(x = -1\) and \(x = 2\)
- \(x = 1\) only
- \(x = -1\), \(x = 1\) and \(x = 2\)
- \(x = 1\) and \(x = -2\)
Answer
\(x = -1\) and \(x = 2\)
The curve crosses the x-axis at \(x = 1\). Which part of the formula explains the crossing?
- The top is 0 at \(x = 1\)
- The bottom is 0 at \(x = 1\)
- Both the top and the bottom are 0 at \(x = 1\)
Answer
The top is 0 at \(x = 1\)
Which inputs are candidates for vertical asymptotes of \(p(x) = \dfrac{x^2 + 1}{x^2 - 6x + 8}\)?
- \(x = 2\) and \(x = 4\)
- \(x = -2\) and \(x = -4\)
- \(x = 6\) and \(x = 8\)
- None, because \(x^2 + 1\) is never 0
Answer
\(x = 2\) and \(x = 4\)

Summary

You solve for the inputs where the bottom is 0, or where the function has no value, and list them as candidates.

A candidate with a nonzero top is a vertical asymptote, because a nonzero number divided by a number near 0 is large.

The list shows where to look for vertical asymptotes before you study the behavior on each side.

This is one of four skills in topic 1.14 of Which & Why: AP Calculus.