Module 0d — Polynomial and Rational Functions

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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

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. $\checkmark$

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: $x = 3, x = 1/2, x = -2$

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).


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 $q(x) = 0$.

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 $x = 2$

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. $\checkmark$

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

  1. Describe the end behaviour (as $x\to\infty$ and $x\to-\infty$) of $p(x) = 4x^6 - 3x^2 + 1$.

  2. Describe the end behaviour of $p(x) = -x^7 + 2x^4 - x$.

  3. Factor $p(x) = x^3 - 7x + 6$ completely (find one root by inspection, then synthetic divide).

  4. 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.

  5. Factor $x^3 + 3x^2 - 4x - 12$ by grouping.

  6. 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)?

  7. Find the domain, and classify each excluded point as a vertical asymptote or a hole, for $r(x) = (x^2 - 4)/(x^2 - x - 6)$.

  8. Find the domain and vertical asymptotes of $r(x) = (x + 5)/(x^2 + 3x - 10)$.

  9. Find the horizontal asymptote of $r(x) = (5x^3 - x)/(2x^3 + 4)$.

  10. Find the horizontal asymptote of $r(x) = (2x + 7)/(x^2 - 9)$.

  11. 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?).

  12. For $r(x) = (x - 1)/(x^2 - 1)$, identify the hole (with its $y$-coordinate) and the vertical asymptote, then state the horizontal asymptote.

  13. 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.

  14. (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.

  15. (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.