Topic 1.8

Rational Functions and Zeros

11 MIN READ9 IDEAS23 PROBLEMS14 flashcards

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01
Odd multiplicity at $x = a$ → the factor flips sign → the whole expression changes sign as you cross $a$ .

even multiplicity at $x = a$ → the factor does not flip sign → the expression keeps the same sign on both sides.
02
1. Move everything to one side so the other side is $0$ .
2. Combine into a single fraction over a common denominator.
Only then factor, find boundaries, and build the sign chart.

Practice 23 problems Next: Chapter 1.9 Rational Functions and Vertical Asymptotes

Four Key Features to Find, Every Time

For any rational function $f(x) = N(x)/D(x)$ , there are four things worth finding right away: domain, zeros, holes, and vertical asymptotes. This note walks through each one, then puts them all together to solve inequalities.

f(x) = \frac{N(x)}{D(x)}

Zeros

CONCEPT

Zeros

Write the function as $f(x) = N(x)/D(x)$ with common factors already canceled.

Then $f(x) = 0$ exactly at the inputs where the numerator is zero, $N(x) = 0$ — as long as that input is in the domain, so the denominator is not also zero there.

Worked Example

Worked example

Find the domain and zero(s) of $p(x) = (x^{3} - 4x) / (x^{2} + 3)$ .

p(x) = \frac{x^3 - 4x}{x^2 + 3}

01
Check the denominator for real zeros:
$x^2 + 3 = 0 \;\Rightarrow\; x^2 = -3 \quad (\text{no real solution})$
02
Since $x^{2} = -3$ has no real solution, the denominator is NEVER zero. Domain: all real numbers.
03
Factor the numerator to find zeros — start by pulling out the GCF, then factor the difference of squares:
$x^3 - 4x = x(x^2 - 4) = x(x - 2)(x + 2)$
04
Zeros: $x = 0$ , $x = 2$ , and $x = -2$ (a cubic numerator can give up to three zeros, and all three are in the domain here).

The graph confirms all three zeros — and no vertical asymptotes, since the denominator is never zero.

Holes — A First Look

Sometimes a factor cancels completely between the numerator and denominator. When that happens, that x-value creates a hole instead of a vertical asymptote — a single missing point in an otherwise continuous graph.

Worked Example

Worked example

Find the domain, zero(s), and hole(s) of $q(t) = (t^{2} - 5t + 6) / (t - 2)$ .

q(t) = \frac{t^2 - 5t + 6}{t - 2}

01
Factor the numerator:
$t^2 - 5t + 6 = (t - 2)(t - 3)$
02
The factor $(t-2)$ appears in both numerator and denominator — it cancels:
$q(t) = \frac{(t - 2)(t - 3)}{t - 2} = t - 3 \quad (t \neq 2)$
03
Domain: all reals except $t=2$ (excluded because the ORIGINAL denominator is zero there).
04
Zero: from the simplified form $t-3$ , the zero is $t=3$ (safely in the domain).
05
Hole: plug $t=2$ into the SIMPLIFIED form to get the y-coordinate:
$q(2)_{\text{simplified}} = 2 - 3 = -1$
06
The hole is located at $(2, -1)$ — notice the y-coordinate can be negative, there's nothing special required about its sign.

A straight line with one point missing — the hole at $(2, -1)$ .

COMMON MISTAKE

A factor that cancels creates a HOLE, not a vertical asymptote — always check for common factors first.

To find a hole's y-coordinate, plug the x-value into the SIMPLIFIED function, never the original (unsimplified) one.

Domain restrictions come from the ORIGINAL denominator, even after simplifying — don't forget the excluded value just because it canceled algebraically.

Quick check

$k(x) = \frac{(x - 5)(x + 1)}{(x - 5)(x + 5)}$ . What happens at $x = 5$ , and what is the $y$ -coordinate there?

Vertical Asymptotes — A First Look

A vertical asymptote happens at an x-value that makes the denominator zero — as long as that factor doesn't cancel with the numerator (the distinction you just practiced in Section 3).

Worked Example

Worked example

Find the domain, zero(s), and vertical asymptote(s) of $r(k) = (2k - 6) / (k^{2} + 3k - 4)$ .

r(k) = \frac{2k - 6}{k^2 + 3k - 4}

01
Factor the denominator:
$k^2 + 3k - 4 = (k + 4)(k - 1)$
02
The denominator is zero at $k = -4$ and $k = 1$ — these are excluded from the domain. Domain: all reals except $k=-4$ , 1.
03
Factor the numerator and check it doesn't share these factors:
$2k - 6 = 2(k - 3)$
04
No common factors — so $k=-4$ and $k=1$ both become vertical asymptotes, and the numerator's zero is $k=3$ .

Two vertical asymptotes, at $k=-4$ and $k=1$ — notice the U-shaped middle branch, different from a typical S-curve.

Even Multiplicity: When the Sign Does NOT Change

A sign chart works by asking "does this factor flip sign as I cross its root?" A repeated factor answers that question differently depending on whether the power is odd or even.

KEY RULE

ODD multiplicity at $x = a$ → the factor FLIPS sign → the whole expression changes sign as you cross $a$ .

EVEN multiplicity at $x = a$ → the factor does NOT flip sign → the expression keeps the SAME sign on both sides.

CONCEPT

Why Even Powers Never Flip

$(x-3)^2$ is a square, so it is positive on both sides of $3$ — it is only $0$ at $3$ itself.

$(x-3)^3$ keeps the sign of $(x-3)$ : negative on the left, positive on the right.

The rule applies to factors in the DENOMINATOR too — an even power under the bar holds its sign the same way.

Worked example

Where is $h(x) \ge 0$ , for $h(x) = \dfrac{(x-1)^2 (x+4)}{x - 2}$ ?

01
Boundaries: zeros at $x = 1$ (multiplicity $2$ ) and $x = -4$ ; vertical asymptote at $x = 2$ .
02
Test a point in the leftmost interval, say $x = -5$ :
$\frac{(-6)^2(-1)}{-7} = \frac{36 \cdot (-1)}{-7} > 0$
03
Now walk right, flipping only at ODD-multiplicity boundaries:

cross $x = -4$ (multiplicity $1$ , odd) → FLIP → negative

cross $x = 1$ (multiplicity $2$ , even) → NO flip → still negative

cross $x = 2$ (asymptote, multiplicity $1$ ) → FLIP → positive
04
So $h > 0$ on $(-\infty, -4)$ and on $(2, \infty)$ , and $h < 0$ on $(-4, 1)$ and $(1, 2)$ .
05
The inequality allows equality, so include the zeros $x = -4$ and $x = 1$ . The asymptote $x = 2$ is never included.
$(-\infty, -4] \cup \{1\} \cup (2, \infty)$

COMMON MISTAKE

Flipping the sign at every boundary out of habit. At an even-multiplicity zero the sign stays put — the graph touches the axis and turns back.

Dropping the isolated point. With $\ge$ , the even-multiplicity zero $x = 1$ still SATISFIES the inequality even though the intervals around it do not, so it goes in the answer on its own.

With a STRICT inequality ( $>$ or $<$ ) that same point is excluded — $h(1) = 0$ is not $> 0$ .

Getting a Rational Inequality Ready to Solve

Everything above assumes the inequality is already "expression $\gtrless 0$ ". When it is not, there are two steps to do first — and one very tempting move that is wrong.

KEY RULE

Move everything to ONE side so the other side is $0$ .
Combine into a SINGLE fraction over a common denominator.

Only then factor, find boundaries, and build the sign chart.

COMMON MISTAKE

Multiplying both sides by the denominator to "clear" it.

You do not know the SIGN of the denominator — it changes across the vertical asymptote. Multiplying by a negative reverses the inequality, so one rule cannot be right for the whole line. Doing it silently gives the wrong intervals.

The same move is fine for an EQUATION, which is why the habit sneaks in.

Worked example

Solve $\dfrac{x}{x - 3} \ge 2$ .

01
Move the $2$ across — do NOT multiply by $x - 3$ :
$\frac{x}{x - 3} - 2 \ge 0$
02
Combine over the common denominator $x - 3$ :
$\frac{x - 2(x - 3)}{x - 3} \ge 0 \;\Longrightarrow\; \frac{-x + 6}{x - 3} \ge 0$
03
Boundaries: zero at $x = 6$ , vertical asymptote at $x = 3$ .
04
Test $x = 4$ : $\dfrac{2}{1} > 0$ . Both boundaries have multiplicity $1$ , so the sign flips at each.
05
Positive on $(3, 6)$ . Include $x = 6$ (the inequality allows equality); exclude $x = 3$ (not in the domain):
$(3, 6]$

CONCEPT

What Clearing the Denominator Would Have Given

Multiplying both sides by $x - 3$ gives $x \ge 2x - 6$ , so $x \le 6$ — which wrongly sweeps in everything to the left of $3$ , where the function is actually negative.

That is the whole reason for the one-fraction step.

Quick check

To solve $\frac{x + 7}{x - 1} \ge 3$ , why can't you multiply both sides by $x - 1$ ?

Solving Rational Inequalities

Once you know a rational function's zeros AND vertical asymptotes, you have every boundary point you need to build a complete sign chart — just like with polynomials (1.5A), except now vertical asymptotes are boundaries too, not just zeros.

Worked example

Where is $s(x) < 0$ , for $s(x) = (x+2)(x-5) / (x^{2} - 1)$ ?

s(x) = \frac{(x + 2)(x - 5)}{x^2 - 1}

01
The numerator is already factored: $(x+2)(x-5)$ . Factor the denominator:
$x^2 - 1 = (x - 1)(x + 1)$
02
No common factors, so there are no holes. Zeros: $x=-2$ , $x=5$ . Vertical asymptotes: $x=-1$ , $x=1$ .
03
Domain: all reals except $x=-1$ and $x=1$ .
04
Find the y-intercept while we're at it:
$s(0) = \frac{(2)(-5)}{-1} = \frac{-10}{-1} = 10$
05
Order all four boundary points on a number line: −2, −1, 1, 5. Test one point in each of the 5 regions:
06
Since we want $s(x) < 0$ (strictly negative): answer = $(-2, -1)$ ∪ $(1, 5)$ .

COMMON MISTAKE

Vertical asymptotes are boundary points for a sign chart too — not just zeros. Missing one will scramble your intervals.

Since the inequality here is strict (<0, not ≤0), the zeros themselves are NOT included in the answer — use open intervals.

Vertical asymptotes are NEVER included in the answer (the function isn't even defined there), regardless of the inequality symbol.

Practice Problems

Worked example

Problem 1. Find the domain and zeros of $f(x) = (x-1) / (x^{2}-9)$ .

01
Denominator zero at x=±3 (excluded from domain). Domain: all reals except $x=3$ , −3.
02
Numerator zero at $x=1$ (safely in the domain). Zero: $x=1$ .

Worked example

Problem 2. Find any holes in $r(x) = (x-1) / x$ .

01
Numerator and denominator share no common factor — $(x-1)$ and $x$ are different factors.
02
No holes. The only domain restriction is x≠0, which is a vertical asymptote, not a hole.

Common slips

A factor that cancels creates a hole, not a vertical asymptote — always check for common factors first.

To find a hole's y-coordinate, plug the x-value into the simplified function, never the original (unsimplified) one.

Domain restrictions come from the original denominator, even after simplifying — don't forget the excluded value just because it canceled algebraically.
Flipping the sign at every boundary out of habit. At an even-multiplicity zero the sign stays put — the graph touches the axis and turns back.

Dropping the isolated point. With $\ge$ , the even-multiplicity zero $x = 1$ Still satisfies the inequality even though the intervals around it do not, so it goes in the answer on its own.

With a strict inequality ( $>$ or $<$ ) that same point is excluded — $h(1) = 0$ is not $> 0$ .
Multiplying both sides by the denominator to "clear" it.

You do not know the sign of the denominator — it changes across the vertical asymptote. Multiplying by a negative reverses the inequality, so one rule cannot be right for the whole line. Doing it silently gives the wrong intervals.

The same move is fine for an equation, which is why the habit sneaks in.
Vertical asymptotes are boundary points for a sign chart too — not just zeros. Missing one will scramble your intervals.

Since the inequality here is strict (<0, not ≤0), the zeros themselves are not included in the answer — use open intervals.

Vertical asymptotes are never included in the answer (the function isn't even defined there), regardless of the inequality symbol.

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14 cards · Rational functions, Hole or asymptote?

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Recap card

5 lines to re-read the night before.

01
Zeros come from the numerator: $N(x)=0$ , checked against the domain.
02
Holes come from factors that cancel — find the y-coordinate using the simplified function.
03
Vertical asymptotes come from denominator zeros that do not cancel with the numerator.
04
For inequalities, both zeros and vertical asymptotes become boundary points on your sign chart.
05
Strict inequalities (< or >) use open intervals and exclude the zeros; asymptotes are always excluded.

Practice 23 problems Next: Chapter 1.9 Rational Functions and Vertical Asymptotes

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Rational Functions and Zeros

Four Key Features to Find, Every Time

Zeros

Worked Example

Holes — A First Look

Worked Example

Vertical Asymptotes — A First Look

Worked Example

Even Multiplicity: When the Sign Does NOT Change

Getting a Rational Inequality Ready to Solve

Solving Rational Inequalities

Practice Problems

Recap card

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