Unit 1 · Topic 1.8 · about 30 minutes · 7 parts · 11 scored checks, 3 of them multiple choice

Rational Functions and Zeros

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

Graph of r(x) = (x + 3)(x - 2) / ((x - 1)(x - 6)). It crosses the x-axis at its zeros, x = -3 and x = 2, and has dashed vertical asymptotes at x = 1 and x = 6. Plus and minus signs above the graph show its sign in each interval: positive, negative, positive, negative, positive.

Part 1 Think first

A zero that is not there

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

Predict first

Let $r (x) = \frac{x^{2} - 4}{x - 2}$ . The numerator is zero at $x = 2$ and at $x = - 2$ . Where is $r (x) = 0$ ?

In short. The real zeros of a rational function are the real zeros of its numerator that are also in its domain.

A value that makes the denominator zero is not in the domain, so it cannot be a zero, even if it also makes the numerator zero.
The real zeros of the numerator and of the denominator are the only places the sign of a rational function can change.
To solve $r (x) \geq 0$ or $r (x) \leq 0$ , test the sign between those values; include the zeros of $r$ and never the zeros of the denominator.

Part 2 Learn it

Three short ideas, each checked as you go

2a. Zeros come from the numerator, inside the domain

A fraction is zero only when its top is zero and its bottom is not. So the real zeros of a rational function $r (x) = \frac{p (x)}{q (x)}$ are the real zeros of the numerator $p$ that are in the domain of $r$ .

To find them, factor both polynomials. List the zeros of the numerator, then cross out any that also make the denominator zero.

For $r (x) = \frac{(x + 1) (x - 4)}{x (x - 4)}$ , the numerator is zero at $- 1$ and 4. But 4 also makes the denominator zero, so it is not in the domain. The only zero of $r$ is $x = - 1$ .

Quick check · 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

2b. Where the sign can change

A rational function can change sign only where its numerator or its denominator is zero. Between those values, every factor keeps its sign, so the whole quotient does too.

That is how to solve a rational inequality such as $r (x) \geq 0$ . The real zeros of both polynomials split the number line into intervals. Each zero of the numerator is a possible endpoint of a solution interval, and each zero of the denominator is a possible asymptote at the edge of one.

Pick one test value in each interval, find the sign of $r$ there, and keep the intervals with the sign you want. The diagram at the top shows the pattern: the sign of $r$ changes only at its two zeros and its two vertical asymptotes.

Quick check · Put it in order

No calculator. Tap the steps for solving $r (x) \geq 0$ in the order you do them.

2c. Which endpoints belong

With $\geq$ or $\leq$ , the solution includes the places where $r (x) = 0$ , so the zeros of $r$ are included endpoints.

The zeros of the denominator are never included, because $r$ is not defined there. With the sign chart for $\frac{x - 2}{x + 4} \geq 0$ :

left of $- 4$ : $\frac{-}{-}$ , positive
between $- 4$ and 2: $\frac{-}{+}$ , negative
right of 2: $\frac{+}{+}$ , positive

The solution is $x < - 4$ or $x \geq 2$ . The 2 is in, because $r (2) = 0$ . The $- 4$ is out, because $r (- 4)$ does not exist.

Quick check · Multiple choice

No calculator. Which set of values solves $\frac{x + 1}{x - 3} \leq 0$ ?

Part 3 Watch it done

Solving a rational inequality

Watch each step, then do the last one yourself.

Solve $\frac{(x + 1) (x - 3)}{x - 5} \geq 0$ .

Step 1 · The values that split the line
The numerator is zero at $x = - 1$ and $x = 3$ . The denominator is zero at $x = 5$ . These split the number line into four intervals.
Step 2 · Test each interval
At $x = - 2$ : $\frac{(-) (-)}{(-)}$ , negative. At $x = 0$ : $\frac{(+) (-)}{(-)}$ , positive. At $x = 4$ : $\frac{(+) (+)}{(-)}$ , negative. At $x = 6$ : $\frac{(+) (+)}{(+)}$ , positive.
Why this step?
Why do the endpoints $- 1$ and 3 belong to the solution, but 5 does not?
At $- 1$ and 3 the quotient equals 0, and $\geq 0$ includes 0. At 5 the denominator is zero, so the quotient is not defined there; 5 is an asymptote at the edge of a solution interval, never part of it.
Your step · Multiple choice
No calculator. Finish it. Which is the solution of $\frac{(x + 1) (x - 3)}{x - 5} \geq 0$ ?

Part 4 Try it

When does a phone case cost 14 dollars or less?

Try it · Calculate

No calculator. From 1.7, the average cost per phone case for $x$ cases is $C (x) = \frac{5000 + 12 x}{x}$ dollars. The inequality $C (x) \leq 14$ rearranges to $\frac{5000 - 2 x}{x} \leq 0$ . What is the fewest cases the business can make so that the average cost is at most 14 dollars?

Your answer cases

Practice

Practice until it is automatic

Zeros and horizontal asymptotes of rational functions in factored form. Check each zero of the numerator against the denominator.

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 $r (x) = \frac{x^{2} - 9}{x^{2} - 3 x}$ . A classmate says the zeros of $r$ are $x = 3$ and $x = - 3$ , because those are the zeros of the numerator. Explain whether the classmate is correct, and give every real zero of $r$ .

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

Model answer

Factored, $r (x) = \frac{(x - 3) (x + 3)}{x (x - 3)}$ . The numerator is zero at 3 and $- 3$ , but the denominator is zero at 0 and 3, so 3 is not in the domain of $r$ and cannot be a zero. The classmate is not correct. The only real zero is $x = - 3$ , where the numerator is zero and the denominator is $(- 3) (- 6) = 18$ , not zero.

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

Factors the numerator and the denominator.
Says that 3 makes the denominator zero, so it is not in the domain and is not a zero.
Concludes that the only real zero is $x = - 3$ , and checks that the denominator is not zero there.

Part 6 Check yourself

Reading a sign chart

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

Let $r (x) = \frac{(x + 3) (x - 2)}{(x - 1) (x - 6)}$ , the function in the diagram at the top. A student built this sign chart for it.

Sign chart for $r$
Interval	Test value	Sign of $r$
$x < - 3$	$- 4$	positive
$- 3 < x < 1$	0	negative
$1 < x < 2$	1.5	positive
$2 < x < 6$	4	negative
$x > 6$	7	positive

Case question 1 · Calculate

No calculator. The student found the sign for $- 3 < x < 1$ by evaluating $r (0)$ . What is $r (0)$ ?

Your answer

Case question 2 · Multiple choice

No calculator. Which set of values solves $r (x) \geq 0$ ?

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

Check yourself · Spot the error

No calculator. A student solves $\frac{x}{x + 2} \geq 0$ . Tap the first line with a mistake.

Check yourself · Fill the statement

No calculator. Let $r (x) = \frac{x (x + 6)}{(x + 6) (x - 2)}$ . Tap a blank, then tap an answer.

Of the two zeros of the numerator, also makes the denominator zero, so the only real zero of $r$ is .

Part 7 Keep it

Bring back earlier ideas

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

From 1.7 Rational Functions and End Behavior · Sort it

No calculator. Tap a function, then its end behavior.

Horizontal asymptote y = 0

Horizontal asymptote at another constant

No horizontal asymptote

A small company models its monthly profit, in thousands of dollars, from making $x$ hundred units of a product: $P (x) = - 0.5 x (x - 2) (x - 9)$ for $0 < x \leq 10$ . Below 200 units it cannot cover its costs, and above 900 it has to pay for overtime.

Testing the sign of $P$ between its zeros
Interval	Test value	Value of $P$
$0 < x < 2$	1	−4, negative
$2 < x < 9$	5	30, positive
$9 < x \leq 10$	10	−40, negative

From 1.5 Polynomial Functions and Complex Zeros · Calculate

No calculator. What is the greatest production level, in hundreds of units, at which the company at least breaks even, $P (x) \geq 0$ ?

Your answer

In three lines

The real zeros of a rational function are the real zeros of its numerator that are in its domain.
A rational function can change sign only at a zero of its numerator or its denominator.
For $r (x) \geq 0$ or $r (x) \leq 0$ , test each interval, include the zeros of $r$ , and never include a zero of the denominator.

Vocabulary

Zero of a rational function: A real zero of the numerator that is in the domain, so the denominator is not zero there.
Domain of a rational function: Every real number except the real zeros of the denominator.
Rational inequality: An inequality such as $r (x) \geq 0$ or $r (x) \leq 0$ with a rational function $r$ .
Sign chart: A list of the intervals between the real zeros of the numerator and denominator, with the sign of the function in each.

Course alignment, for teachers

AP Precalculus topic 1.8, Unit 1: Polynomial and Rational Functions.

Skill 1.A: Solve equations and inequalities represented analytically, 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.