Unit 1 · Topic 1.9 · about 25 minutes · 7 parts · 11 scored checks, 2 of them multiple choice

Rational Functions and Vertical Asymptotes

Built from College Board's course framework · Every answer key independently checked

Graph of C(p) = 120p / (100 - p), the cost in thousands of dollars of removing p percent of a pollutant. The curve rises slowly, then shoots upward without bound as p approaches the dashed vertical asymptote at p = 100. The point (90, 1080) is marked.

Part 1 Think first

The last bit of pollution

Guess before you learn. A wrong guess still makes the answer stick.

Predict first

A factory's cost, in thousands of dollars, to remove $p$ percent of a pollutant from its smokestack is modeled by $C (p) = \frac{120 p}{100 - p}$ . Removing 90% costs 1,080 thousand dollars. What does the model say about removing 100%?

In short. A rational function has a vertical asymptote at $x = a$ when $a$ is a real zero of the denominator that the numerator does not cancel.

If $a$ is a zero of the denominator and not of the numerator, there is a vertical asymptote at $x = a$ .
If $a$ is a zero of both, there is still a vertical asymptote when its multiplicity in the denominator is greater than in the numerator.
Near the asymptote the denominator is close to zero, so the outputs increase or decrease without bound, written with one-sided limits such as $\lim_{x \to a^{+}} r (x) = \infty$ .

Part 2 Learn it

Three short ideas, each checked as you go

2a. Where a vertical asymptote comes from

If $a$ is a real zero of the denominator of a rational function $r$ , and not a zero of the numerator, then the graph of $r$ has a vertical asymptote at $x = a$ .

The reason is the size of the quotient. Near $x = a$ the denominator is very close to zero while the numerator is not, and dividing a number that is not close to zero by a number that is gives an output of huge size.

For $r (x) = \frac{x + 1}{x^{2} - 4 x} = \frac{x + 1}{x (x - 4)}$ , the denominator is zero at 0 and 4, and the numerator is zero at neither, so there are vertical asymptotes at $x = 0$ and $x = 4$ . Factor the denominator first; its zeros are your candidates.

Quick check · Calculate

No calculator. The graph of $r (x) = \frac{x + 4}{x^{2} + x - 12}$ has exactly one vertical asymptote, $x = k$ . What is $k$ ?

Your answer

2b. When the numerator shares the zero

Sometimes $a$ makes both the numerator and the denominator zero. Then count multiplicities, the number of times each factor $(x - a)$ appears.

If the multiplicity of $a$ in the denominator is greater than its multiplicity in the numerator, there is still a vertical asymptote at $x = a$ . After cancelling, a factor $(x - a)$ is left on the bottom.

For $r (x) = \frac{x - 2}{(x - 2)^{2}}$ , the zero 2 appears once on top and twice on the bottom. One factor cancels, leaving $\frac{1}{x - 2}$ , so there is a vertical asymptote at $x = 2$ .

If the numerator has at least as many copies, the factor cancels completely and leaves a hole instead. That is 1.10.

Quick check · Decide and justify

No calculator. Let $r (x) = \frac{(x - 1)^{2} (x + 3)}{(x - 1)^{3}}$ . Does the graph of $r$ have a vertical asymptote at $x = 1$ ?

Which reason supports your answer?

2c. Approaching from each side

Near a vertical asymptote $x = a$ , the denominator gets arbitrarily close to zero, so the outputs increase or decrease without bound. The two sides can go different ways, so each gets its own one-sided limit:

$\lim_{x \to a^{+}} r (x)$ is for inputs near $a$ and greater than $a$ .
$\lim_{x \to a^{-}} r (x)$ is for inputs near $a$ and less than $a$ .

To decide the sign, check each factor just to one side. For $r (x) = \frac{x + 1}{x - 2}$ with $x$ slightly more than 2, the top is near 3 and the bottom is a tiny positive number, so $\lim_{x \to 2^{+}} r (x) = \infty$ . Just left of 2 the bottom is a tiny negative number, so that side goes to $- \infty$ .

Quick check · Multiple choice

No calculator. Let $r (x) = \frac{x + 5}{x - 3}$ . Which statement is true?

Part 3 Watch it done

How expensive is the last percent?

Watch each step, then do the last one yourself.

Use $C (p) = \frac{120 p}{100 - p}$ , the cost in thousands of dollars to remove $p$ percent of the pollutant, for $0 \leq p < 100$ . Describe what happens near $p = 100$ , and find the cost of removing 99%.

Step 1 · Find the asymptote
The denominator $100 - p$ is zero at $p = 100$ , and the numerator there is $12,000$ , not zero. So $p = 100$ is a vertical asymptote.
Step 2 · Which side, which direction
Only $p < 100$ makes sense here. Just left of 100, the top is near 12,000 and the bottom is a tiny positive number, so $\lim_{p \to 100^{-}} C (p) = \infty$ .
Why this step?
Why does the cost grow so fast at the end and not at the start?
The cost divides by $100 - p$ , the percent still left to remove. Going from 90% to 99% shrinks that from 10 to 1, which divides by a number ten times smaller. Near the asymptote, tiny changes in the denominator make huge changes in the output.
Step 3 · One value to compare
At 90%: $C (90) = \frac{120 \cdot 90}{10} = 1080$ thousand dollars.
Your step · Calculate

No calculator. Finish it. What does removing 99% cost, in thousands of dollars?

Your answer thousand dollars

Part 4 Try it

Asymptotes, holes and zeros together

Try it · Sort it

No calculator. Each function has a denominator that is zero at $x = 2$ . Tap a function, then what its graph has at $x = 2$ .

Vertical asymptote at x = 2

No vertical asymptote at x = 2

Practice

Practice until it is automatic

Vertical asymptotes, zeros and horizontal asymptotes of rational functions in factored form. One shared factor makes a hole, not an asymptote.

Asymptotes, zeros and holes practice page

Part 5 Explain it

Write it the way the exam asks

Nothing you type leaves this page.

Free response is 37.5% of the exam score. This one is shaped like Question 1, Function Concepts.

Free-response practice · Question 1 style

Graphing calculator required. Let $r (x) = \frac{x - 3}{x^{2} - 9}$ and $s (x) = \frac{x - 3}{x^{2} - 6 x + 9}$ . Both denominators are zero at $x = 3$ . Explain why the graph of $s$ has a vertical asymptote at $x = 3$ but the graph of $r$ does not.

Your answer stays on this page. It is never sent or saved.

Model answer

Factored, $r (x) = \frac{x - 3}{(x - 3) (x + 3)}$ . The zero 3 has multiplicity 1 in the numerator and 1 in the denominator, so the factor cancels completely and $r$ has no vertical asymptote at $x = 3$ ; its only one is $x = - 3$ . For $s (x) = \frac{x - 3}{(x - 3)^{2}}$ , the zero 3 has multiplicity 2 in the denominator, greater than 1 in the numerator, so a factor $(x - 3)$ is left on the bottom and $s$ has a vertical asymptote at $x = 3$ .

Score your own answer. This is a self-check, not a grade. Tick each point your answer makes.

Factors both denominators: $(x - 3) (x + 3)$ and $(x - 3)^{2}$ .
Compares the multiplicity of 3 in each numerator and denominator.
Concludes $s$ has an asymptote at 3 because the denominator's multiplicity is greater, and $r$ does not because the factor cancels completely.

Part 6 Check yourself

Close to an asymptote

The two case questions use the same data. Answer them from the table.

The table gives values of $r (x) = \frac{2 x}{x - 4}$ for inputs close to 4.

Values of $r$ near $x = 4$
$x$	$r (x)$
3.9	−78
3.99	−798
4.01	802
4.1	82

Case question 1 · Multiple choice

No calculator. Which pair of statements matches the table?

Case question 2 · Calculate

No calculator. Continue the table: what is $r (4.001)$ ?

Your answer

The rest stand on their own and do not use the case.

Check yourself · Fill the statement

No calculator. Tap a blank, then tap a word.

In the notation $\lim_{x \to 4^{+}} r (x)$ , the inputs are close to 4 and than 4. Near a vertical asymptote the is close to zero, so the outputs grow without bound.

Check yourself · Spot the error

No calculator. A student finds the vertical asymptotes of $r (x) = \frac{x^{2} - 1}{x^{2} - 2 x - 3}$ . Tap the line with the mistake.

Part 7 Keep it

Bring back earlier ideas

Pulling an older idea back up is what makes it last.

From 1.8 Rational Functions and Zeros · Calculate

No calculator. The function $r (x) = \frac{(x - 5) (x + 3)}{(x - 5) (x - 1)}$ has exactly one real zero. What is it?

Your answer

From 1.6 Polynomial Functions and End Behavior · Match

No calculator. Tap a statement in limit notation, then the words that say the same thing.

In three lines

A real zero of the denominator that is not a zero of the numerator gives a vertical asymptote.
A shared zero still gives one when its multiplicity in the denominator is greater than in the numerator.
Near $x = a$ the outputs increase or decrease without bound, and $\lim_{x \to a^{+}}$ and $\lim_{x \to a^{-}}$ describe each side.

Vocabulary

Vertical asymptote: A line $x = a$ near which the outputs of a function increase or decrease without bound.
One-sided limit: What the outputs do as inputs approach $a$ from one side: $x \to a^{+}$ from the right, $x \to a^{-}$ from the left.
Multiplicity: The number of times a factor $(x - a)$ appears; compared top and bottom, it decides between an asymptote and a hole.

Course alignment, for teachers

AP Precalculus topic 1.9, Unit 1: Polynomial and Rational Functions.

Skill 2.A: Identify information from graphical, numerical, analytical, and verbal representations to answer a question or construct a model, with and without technology.

This lesson maps all 2 essential knowledge statements of the topic to the part that teaches each one: 2 are checked by a scored question. The map is in the lesson file.