Module 0d — Polynomial and Rational Functions

Course: MAT1CJ101 — Differential Calculus (prerequisite refresher)
Status: Prerequisite background material — not part of the official 60 taught hours, not examined.

0. Why This Module Exists

Module IV of the main notes (Graphing and Asymptotes) uses calculus — derivatives — to locate turning points and confirm asymptotic behaviour precisely. But you should already be able to sketch a rough shape of a polynomial or rational function from algebra alone: degree, leading coefficient, roots, and simple limits at infinity. This module builds that algebraic groundwork so Module IV can focus on what calculus adds, not on re-teaching factoring.

1. Polynomial Functions

1.1 Definition and Terminology

A polynomial function has the form $p(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0,$ where $n$ is a non-negative integer and $a_n \neq 0$.

Degree $= n$ (the highest power of $x$ with a nonzero coefficient)
Leading coefficient $= a_n$
Leading term $= a_n x^n$
Constant term $= a_0$

Degree	Name	General shape
0	Constant	Horizontal line
1	Linear	Straight line
2	Quadratic	Parabola
3	Cubic	S-shaped curve
$n$	Degree-$n$ polynomial	Up to $n - 1$ turning points, up to $n$ real roots

Example 1 (Polynomial function): Is $f(x) = 4x^3 - 2x + 1$ a polynomial function?

Terms: $4x^3$ (power 3), $-2x$ (power 1), $1$ (power 0). Every exponent is a non-negative integer, and coefficients are real numbers. $\checkmark$ $f$ is a polynomial, degree 3.

Non-Example 1 (Polynomial function): Is $g(x) = 3x^2 + 2/x - 5$ a polynomial function?

The term $2/x$ can be written $2x^{-1}$ — exponent $-1$ is NOT a non-negative integer. $\times$ $g$ is NOT a polynomial (it’s a rational function instead — see Section 2). Also note $h(x) = \sqrt{x} = x^{1/2}$ fails the same way, for the same reason: exponent $1/2$ isn’t a non-negative integer.

1.2 End Behaviour

The leading term alone determines what happens as $x \to \pm\infty$ — every other term becomes negligible in comparison for large $\lvert x \rvert$.

Degree $n$	Leading coeff $a_n$	As $x \to \infty$	As $x \to -\infty$
Even	positive	$\to +\infty$	$\to +\infty$
Even	negative	$\to -\infty$	$\to -\infty$
Odd	positive	$\to +\infty$	$\to -\infty$
Odd	negative	$\to -\infty$	$\to +\infty$

Worked Example 1: Describe the end behaviour of $p(x) = -2x^5 + 3x^3 - x + 7$.

Leading term: $-2x^5$. Degree 5 is odd, leading coefficient $-2$ is negative.

From the table (odd, negative): $x \to \infty$ gives $p(x) \to -\infty$; $x \to -\infty$ gives $p(x) \to +\infty$.

Sanity check by direct reasoning: for very large $x$, $-2x^5$ dominates all other terms. $x = 100: \ -2(100)^5 = -2 \times 10^{10}, \text{ overwhelmingly negative — matches "} \to -\infty \text{ as } x\to\infty\text{."}$

1.3 Factoring Techniques

Technique	When to use	Example
Common factor	Every term shares a factor	$3x^3-6x^2 = 3x^2(x-2)$
Difference of squares	$a^2-b^2$	$x^2-9 = (x-3)(x+3)$
Difference/sum of cubes	$a^3 \pm b^3$	$x^3-8 = (x-2)(x^2+2x+4)$
Trinomial factoring	$ax^2+bx+c$	$x^2-x-6 = (x-3)(x+2)$
Grouping	Four-term polynomials	$x^3+2x^2-3x-6 = x^2(x+2)-3(x+2) = (x+2)(x^2-3)$
Rational Root Theorem + synthetic division	Higher-degree polynomials	see Example 3

Worked Example 2: Factor $p(x) = x^3 - 2x^2 - 5x + 6$ by grouping after finding one root by inspection.

Try $x = 1$: $p(1) = 1 - 2 - 5 + 6 = 0$. So $(x - 1)$ is a factor.

Divide $p(x)$ by $(x - 1)$ using synthetic division (coefficients $1, -2, -5, 6$):

  1 | 1  -2  -5   6
    |     1  -1  -6
    | 1  -1  -6   0     ← remainder 0, confirms (x-1) is a factor

Quotient: $x^2 - x - 6 = (x-3)(x+2)$

\[p(x) = (x - 1)(x - 3)(x + 2)\]

Roots: $x = 1, 3, -2$. Check degree: 3 roots for a degree-3 polynomial. $\checkmark$

Example 2 (Root/factor of a polynomial): Is $x = 2$ a root of $p(x) = x^3 - 2x^2 - 5x + 6$ (from Worked Example 2, factored as $(x-1)(x-3)(x+2)$)?

Substitute directly: $p(2) = 2^3 - 2(2)^2 - 5(2) + 6 = 8 - 8 - 10 + 6 = -4$. Is $p(2) = 0$? No. $\times$ $x = 2$ is NOT a root, and correspondingly $(x - 2)$ is NOT a factor — 2 doesn’t appear in the factored form $(x-1)(x-3)(x+2)$.

Non-Example 2 (Root/factor of a polynomial): Is $(x - 4)$ a factor of $p(x) = x^3 - 2x^2 - 5x + 6$?

Test: if $(x-4)$ were a factor, then $x = 4$ would need to be a root, i.e. $p(4) = 0$. $p(4) = 4^3 - 2(4)^2 - 5(4) + 6 = 64 - 32 - 20 + 6 = 18$. Is $p(4) = 0$? No, it’s 18. $\times$ $(x - 4)$ is NOT a factor of $p(x)$ — a candidate factor $(x - r)$ only works when substituting $x = r$ gives exactly 0, not just a small or “plausible-looking” value.

1.4 The Rational Root Theorem

For $p(x) = a_n x^n + \cdots + a_0$ with integer coefficients, every rational root $p/q$ (in lowest terms) must have $p$ dividing $a_0$ and $q$ dividing $a_n$. This narrows an infinite search down to a short finite list of candidates.

Worked Example 3: Find all rational roots of $p(x) = 2x^3 - 3x^2 - 11x + 6$.

$a_0 = 6$, divisors: $\pm1, \pm2, \pm3, \pm6$
$a_n = 2$, divisors: $\pm1, \pm2$

Candidates $p/q$: $\pm1, \pm2, \pm3, \pm6, \pm1/2, \pm3/2$

Test $x = 3$: $2(27) - 3(9) - 11(3) + 6 = 54 - 27 - 33 + 6 = 0$. Root found: $x = 3$.

Synthetic division by $(x - 3)$, coefficients $2, -3, -11, 6$:

  3 | 2  -3  -11   6
    |     6   9  -6
    | 2   3   -2   0

Quotient: $2x^2 + 3x - 2 = (2x - 1)(x + 2)$

Full factorization: $p(x) = (x - 3)(2x - 1)(x + 2)$
Roots: $x = 3, x = 1/2, x = -2$

Example 3 (Valid Rational Root Theorem candidate): For $p(x) = 2x^3 - 3x^2 - 11x + 6$ ($a_0 = 6$, $a_n = 2$), is $3/2$ a valid candidate to test?

Need: numerator 3 divides $a_0 = 6$? Yes, $6/3 = 2$. $\checkmark$ Need: denominator 2 divides $a_n = 2$? Yes, $2/2 = 1$. $\checkmark$ Both conditions hold. $\checkmark$ $3/2$ is a legitimate candidate (it’s in the list $\pm1, \pm2, \pm3, \pm6, \pm1/2, \pm3/2$ from Worked Example 3) — though being a candidate does NOT guarantee it’s an actual root; it still must be tested.

Non-Example 3 (Invalid Rational Root Theorem candidate): For the same $p(x) = 2x^3 - 3x^2 - 11x + 6$, is $4/3$ a valid candidate?

Need: numerator 4 divides $a_0 = 6$? $6/4 = 1.5$, not an integer. $\times$ Already fails — 4 is not a divisor of 6, so $4/3$ cannot appear in the candidate list, no matter what the denominator does. $4/3$ is NOT a valid candidate and testing it would be a wasted step; the theorem guarantees the true rational roots are among $\pm1, \pm2, \pm3, \pm6, \pm1/2, \pm3/2$ only.

1.5 Multiplicity of Roots and Graph Behaviour

If $(x - r)^k$ is a factor of $p(x)$ with $k$ the largest such power, $r$ is a root of multiplicity $k$.

Multiplicity	Graph behaviour at $x = r$
Odd ($1, 3, \ldots$)	Graph crosses the $x$-axis
Even ($2, 4, \ldots$)	Graph touches the $x$-axis and turns back (doesn’t cross)

Worked Example 4: Describe the roots and crossing behaviour of $p(x) = (x-1)^2(x+2)^3$.

Root $x = 1$, multiplicity 2 (even) $\to$ graph touches $x$-axis at $x=1$, doesn’t cross
Root $x = -2$, multiplicity 3 (odd) $\to$ graph crosses $x$-axis at $x=-2$

Degree $= 2 + 3 = 5$ (odd), leading coefficient positive (both factors expand with $+1$ leading terms) $\to$ end behaviour: $x\to\infty$ gives $p\to+\infty$, $x\to-\infty$ gives $p\to-\infty$ (odd, positive row).

Example 4 (Odd multiplicity → crosses): For $p(x) = (x+2)^3(x-5)$, what happens at $x = -2$?

$(x+2)$ appears to power 3 — multiplicity 3, which is odd. By the table: odd multiplicity $\implies$ graph CROSSES the $x$-axis at $x = -2$. Sanity check with signs: just left of $-2$ (say $x=-2.1$), $(x+2)^3$ is a small negative cubed $\to$ negative; just right ($x=-1.9$), $(x+2)^3$ is small positive cubed $\to$ positive. The sign flips across $x=-2$ $\implies$ the graph does cross. $\checkmark$

Non-Example 4 (Even multiplicity does NOT cross): For the same $p(x) = (x+2)^3(x-5)$, does the graph cross at $x = 5$?

$(x-5)$ appears to power 1 (odd) here, not even — so this is actually still a crossing root. To see the “does not cross” case, consider $q(x) = (x+2)^3(x-5)^2$: at $x = 5$, $(x-5)$ has multiplicity 2 (even). By the table: even multiplicity $\implies$ graph TOUCHES the $x$-axis and turns back — it does NOT cross. Sanity check with signs: just left of 5 ($x=4.9$), $(x-5)^2$ is a small number squared $\to$ positive; just right ($x=5.1$), $(x-5)^2$ is also positive (any square is non-negative). The sign does NOT flip across $x=5$ $\implies$ the graph touches but does not cross. $\times$ (no sign change, unlike the odd-multiplicity case above)

2. Rational Functions

2.1 Definition

A rational function has the form $r(x) = p(x)/q(x)$, where $p$ and $q$ are polynomials and $q$ is not the zero polynomial. Its natural domain excludes every $x$ where $q(x) = 0$.

Example 5 (Rational function, x in the domain): For $r(x) = (x+1)/(x-3)$, is $x = 0$ in the domain?

Denominator at $x=0$: $0 - 3 = -3$. Is $-3 = 0$? No. $\times$ (the “bad” condition fails, which is good — it means $x=0$ is fine) Denominator is nonzero at $x=0$. $\checkmark$ $x = 0$ IS in the domain; $r(0) = 1/(-3) = -1/3$.

Non-Example 5 (Rational function, x excluded from the domain): For the same $r(x) = (x+1)/(x-3)$, is $x = 3$ in the domain?

Denominator at $x=3$: $3 - 3 = 0$. $\times$ Division by zero is undefined — no matter what the numerator does (even if it were also 0), $x = 3$ can never be plugged into $r(x)$. $x = 3$ is NOT in the domain.

2.2 Vertical Asymptotes and Holes

Find the zeros of the denominator. For each zero $x = a$:

If $(x - a)$ is not also a factor of the numerator: vertical asymptote at $x = a$.
If $(x - a)$ is also a factor of the numerator (and cancels): usually a hole (removable discontinuity) at $x = a$ instead of an asymptote — provided it doesn’t remain in the denominator with higher multiplicity than in the numerator.

Worked Example 5: Find the vertical asymptotes/holes of $r(x) = (x^2 - 1)/(x^2 - 3x + 2)$.

Factor numerator: $x^2 - 1 = (x-1)(x+1)$
Factor denominator: $x^2 - 3x + 2 = (x-1)(x-2)$

\[r(x) = \frac{(x-1)(x+1)}{(x-1)(x-2)}\]

Common factor $(x-1)$ cancels (for $x \neq 1$): $r(x) = \dfrac{x+1}{x-2}, \ x \neq 1$

$x = 1$: cancelled factor $\to$ HOLE at $x = 1$ (not an asymptote). Hole’s $y$-coordinate: plug $x=1$ into the simplified form $\dfrac{x+1}{x-2} = \dfrac{2}{-1} = -2$. Hole at $(1, -2)$

$x = 2$: remains in the denominator of the simplified form, not in the numerator $\to$ VERTICAL ASYMPTOTE at $x = 2$

Example 6 (Vertical asymptote — factor does NOT cancel): For $r(x) = (x+3)/(x^2-4)$, what happens at $x = 2$ (a zero of the denominator)?

Denominator zero: $x^2-4 = (x-2)(x+2) = 0$ at $x = 2$ (and $x = -2$). Does $(x-2)$ also divide the numerator $(x+3)$? Substitute $x=2$ into the numerator: $2+3 = 5 \neq 0$. $\times$ $(x-2)$ is NOT a common factor — nothing cancels. $\checkmark$ Genuine VERTICAL ASYMPTOTE at $x = 2$ (the function blows up there instead of having a removable gap).

Non-Example 6 (Looks like an asymptote, but is actually a hole): For $r(x) = (x-2)(x+5)/(x^2-4)$, what happens at $x = 2$?

Denominator zero at $x = 2$ (same as above: $x^2-4 = (x-2)(x+2)$). Does $(x-2)$ also divide the numerator? Numerator is already written as $(x-2)(x+5)$ — yes, $(x-2)$ is an explicit common factor. $\checkmark$ Cancel: $r(x) = (x+5)/(x+2)$ for $x \neq 2$. This is NOT a vertical asymptote at $x = 2$, even though $x = 2$ zeroes the original denominator — it’s a HOLE instead, at $(2, 7/4)$, since the simplified form is finite there: $(2+5)/(2+2) = 7/4$. The asymptote only survives at $x = -2$, which remains in the denominator after cancelling.

2.3 Horizontal Asymptotes (Algebraic Preview)

Compare the degree of the numerator ($\deg p = m$) to the degree of the denominator ($\deg q = n$):

Comparison	Horizontal asymptote
$m < n$	$y = 0$
$m = n$	$y = $ (leading coeff of $p$)/(leading coeff of $q$)
$m > n$	none (function grows without bound; may have a slant asymptote if $m = n+1$)

This table is an algebraic shortcut for limits at infinity — Module I proves it rigorously using limit laws; here we just state the pattern so you can sketch a rough shape early.

Worked Example 6: Find the horizontal asymptote of $r(x) = (3x^2 + x)/(2x^2 - 5)$.

$\deg(\text{numerator}) = 2 = \deg(\text{denominator})$. Equal degree case.

\[\text{Horizontal asymptote: } y = \frac{\text{leading coeff of numerator}}{\text{leading coeff of denominator}} = \frac{3}{2}\]

Sanity check with a large $x$, say $x = 1000$: numerator $\approx 3(1000)^2 = 3{,}000{,}000$, denominator $\approx 2(1000)^2 = 2{,}000{,}000$, ratio $\approx 1.5 = 3/2$ $\checkmark$ matches

Worked Example 7: Find the horizontal asymptote of $r(x) = (x + 4)/(x^2 - 1)$.

$\deg(\text{numerator}) = 1 < \deg(\text{denominator}) = 2$.

Horizontal asymptote: $y = 0$

Sanity check, $x = 1000$: $1004/999999 \approx 0.001$, very close to 0. $\checkmark$

Example 7 (Horizontal asymptote exists): Does $r(x) = (2x^2 + 1)/(x^2 - 5)$ have a horizontal asymptote?

$\deg(\text{numerator}) = 2$, $\deg(\text{denominator}) = 2$ — equal degrees. By the table, equal degrees $\implies$ a horizontal asymptote exists, at $y = $ (leading coeff of num)/(leading coeff of denom) $= 2/1 = 2$. Sanity check, $x = 1000$: $2{,}000{,}001/999{,}995 \approx 2.0000\ldots$ $\checkmark$

Non-Example 7 (No horizontal asymptote): Does $r(x) = (x^3 + 1)/(x^2 - 5)$ have a horizontal asymptote?

$\deg(\text{numerator}) = 3$, $\deg(\text{denominator}) = 2$. Is $\deg(\text{num}) \leq \deg(\text{denom})$? No, $3 > 2$. $\times$ By the table, $m > n$ means NO horizontal asymptote — the function grows without bound instead (here $\deg(\text{num}) = \deg(\text{denom})+1$, so it actually has a slant asymptote, found by polynomial division, but never a horizontal one). Sanity check, $x = 1000$: numerator $\approx 10^9$, denominator $\approx 10^6$, ratio $\approx 1000$ — this keeps growing as $x$ grows, confirming it never settles at a fixed height.

2.4 Putting It Together — Sketching a Rough Shape

Worked Example 8: Sketch the key features of $r(x) = (x - 3)/(x^2 - 4)$.

Domain: $x^2 - 4 = 0 \implies x = \pm2$. Domain: $\mathbb{R} \setminus \{-2, 2\}$

Numerator zero: $x = 3$ $\to$ $x$-intercept at $(3, 0)$
$y$-intercept: $r(0) = (0-3)/(0-4) = -3/-4 = 3/4$ $\to$ $(0, 3/4)$

Vertical asymptotes: denominator zero at $x=\pm2$; numerator $(x-3)$ does NOT vanish at $\pm2$, so BOTH are genuine vertical asymptotes (no cancellation). $x = -2$ and $x = 2$

Horizontal asymptote: $\deg(\text{num})=1 < \deg(\text{denom})=2$ $\to$ $y = 0$

Sign chart (critical points $-2, 2, 3$ split the line into 4 intervals) to know which side of each asymptote the curve approaches from:

$x < -2: \ \text{test } x=-3: \ \frac{-6}{5} = -1.2 \ \to \text{ negative}$ $-2<x<2: \ \text{test } x=0: \ \frac{-3}{-4} = 0.75 \ \to \text{ positive}$ $2<x<3: \ \text{test } x=2.5: \ \frac{-0.5}{2.25} \approx -0.22 \ \to \text{ negative}$ $x > 3: \ \text{test } x=4: \ \frac{1}{12} \approx 0.083 \ \to \text{ positive}$

This sign pattern plus the two vertical asymptotes and one horizontal asymptote is enough to sketch the overall shape without calculus.

3. Summary

Concept	Key Idea	Worked Example
Degree & end behaviour	Leading term controls $x\to\pm\infty$	$-2x^5+\ldots \to -\infty$ as $x\to\infty$
Factoring by grouping	Find one root, synthetic-divide, factor quotient	$x^3-2x^2-5x+6 = (x-1)(x-3)(x+2)$
Rational Root Theorem	Candidates are $\pm$(factor of $a_0$)/(factor of $a_n$)	$2x^3-3x^2-11x+6$ has root $x=3$
Multiplicity	Odd $\to$ crosses axis; even $\to$ touches and turns	$(x-1)^2(x+2)^3$
Vertical asymptote vs. hole	Uncancelled factor $\to$ asymptote; cancelled $\to$ hole	$(x^2-1)/(x^2-3x+2)$: hole at 1, asymptote at 2
Horizontal asymptote	Compare degrees of numerator/denominator	deg equal $\to$ ratio of leading coeffs

4. Practice Problems

Describe the end behaviour (as $x\to\infty$ and $x\to-\infty$) of $p(x) = 4x^6 - 3x^2 + 1$.
Describe the end behaviour of $p(x) = -x^7 + 2x^4 - x$.
Factor $p(x) = x^3 - 7x + 6$ completely (find one root by inspection, then synthetic divide).
Use the Rational Root Theorem to list all candidate rational roots of $p(x) = 3x^3 + 2x^2 - 7x + 2$, then find all actual rational roots and factor completely.
Factor $x^3 + 3x^2 - 4x - 12$ by grouping.
For $p(x) = (x+1)^3(x-4)^2$, state each root’s multiplicity and whether the graph crosses or touches the $x$-axis there. What is the degree of $p$, and what is its end behaviour (assume positive leading coefficient)?
Find the domain, and classify each excluded point as a vertical asymptote or a hole, for $r(x) = (x^2 - 4)/(x^2 - x - 6)$.
Find the domain and vertical asymptotes of $r(x) = (x + 5)/(x^2 + 3x - 10)$.
Find the horizontal asymptote of $r(x) = (5x^3 - x)/(2x^3 + 4)$.
Find the horizontal asymptote of $r(x) = (2x + 7)/(x^2 - 9)$.
Find the horizontal asymptote of $r(x) = (x^3 - 1)/(x + 2)$, or explain why there is none (what happens instead, given the degree comparison?).
For $r(x) = (x - 1)/(x^2 - 1)$, identify the hole (with its $y$-coordinate) and the vertical asymptote, then state the horizontal asymptote.
Sketch the key features (domain, intercepts, vertical/horizontal asymptotes, and sign on each interval) of $r(x) = (2x)/(x^2 - 1)$, following the method of Worked Example 8.
(Harder) A degree-4 polynomial has roots at $x = -1$ (multiplicity 1), $x = 2$ (multiplicity 3), a positive leading coefficient, and no other real roots. Sketch its end behaviour and crossing/touching pattern, and write one possible formula for it.
(Harder) Explain, using the degree-comparison table, why a rational function can never have both a horizontal asymptote and a slant asymptote at the same time.

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