← All notesMaths AA · Polynomials, rational functions and inequalities⬇ Download PDF
IB DP Maths AA · Topic 2 Functions · SL 2.8 · AHL 2.12, 2.13, 2.15 · ~6 min read

Polynomials, rational functions and inequalities

Reciprocal and rational functions with their asymptotes, and for HL: the factor and remainder theorems, polynomial graphs, sum and product of roots, oblique asymptotes, and solving polynomial and rational inequalities.

Reciprocal functionRational functions & asymptotesFactor theorem (HL)Remainder theorem (HL)Sum & product of roots (HL)Oblique asymptotes (HL)Inequalities (HL)
RSby Riyaz Sir
mentorriyaz.in
No tag = SL and HLHL ONLY HL students onlyFORMULA BOOKLET given in the examM1A1 how marks are awardedhighlight = comes up again and againTest yourself at the end

In 30 seconds

  • A rational function y = (ax + b)/(cx + d) has a vertical asymptote where the denominator is zero (x = −d/c) and a horizontal asymptote y = a/c.
  • HL factor theorem: (x − a) is a factor of p(x) ⇔ p(a) = 0. Remainder theorem: dividing p(x) by (x − a) leaves remainder p(a).
  • HL: for a polynomial anxn + … + a0, the roots add to −an−1/an and multiply to (−1)na0/an.
  • HL inequalities: find the critical values (roots, asymptotes), then test the sign between them; a sketch or sign chart beats algebra tricks.

By the end you can

  • Sketch y = 1/x and y = (ax + b)/(cx + d), with intercepts and asymptotes.
  • Use the factor and remainder theorems to factorise polynomials and find unknowns. HL
  • Sketch polynomial graphs from their factors, including repeated roots. HL
  • Use the sum and product of roots. HL
  • Sketch rational functions with oblique asymptotes. HL
  • Solve polynomial and rational inequalities, analytically and graphically. HL

1Reciprocal and rational functions

xyx = 3y = 2y = (2x + 1)/(x − 3)vertical: x = 3horizontal: y = 2
Two asymptotes split the curve into two branches.
  • y = 1/x is self-inverse, with asymptotes x = 0 and y = 0. y = k/x is a stretch of it.
  • For y = ax + bcx + d: vertical asymptote x = −d/c; horizontal asymptote y = a/c; x-intercept where ax + b = 0; y-intercept b/d.
Worked example

Sketch y = 2x + 1x − 3: give asymptotes and intercepts.

  1. vertical x = 3, horizontal y = 2 A1A1
  2. x-intercept (−0.5, 0), y-intercept (0, −⅓) A1
Common mistakeThinking the curve can never cross a horizontal asymptote. Horizontal asymptotes describe the behaviour as x → ±∞; for more complicated rational functions the curve can cross them in the middle.

2Polynomials HL

p(a) = 0 ⇔ (x − a) is a factor
factor theorem
remainder when p(x) ÷ (x − a) = p(a)
remainder theorem; for (ax − b) the remainder is p(b/a)
Worked example

Factorise p(x) = x3 − 4x2 + x + 6 completely.

  1. p(−1) = −1 − 4 − 1 + 6 = 0, so (x + 1) is a factor M1
  2. divide: x3 − 4x2 + x + 6 = (x + 1)(x2 − 5x + 6) A1
  3. = (x + 1)(x − 2)(x − 3) A1
Worked example

When 2x3 + ax2 − 5 is divided by (x − 2) the remainder is 15. Find a.

  1. p(2) = 16 + 4a − 5 = 15 M1
  2. a = 1 A1
xy-3-2-112345-8-4480single root: crossesdouble root: touchesy = (x + 1)(x − 2)²: positive leading coefficient, so it rises to the right.
A single root crosses the axis; a double root touches it; the leading coefficient fixes the ends.
sum of roots = −an−1an, product of roots = (−1)na0an FORMULA BOOKLET
for anxn + an−1xn−1 + … + a0 = 0
Worked example

Find the sum and product of the roots of 2x3 − 3x2 + 4x − 5 = 0.

  1. sum = −(−3)/2 = 32 A1
  2. product = (−1)3(−5)/2 = 52 A1
  • Complex roots count too: a polynomial of degree n has exactly n roots (some may be complex, some repeated).

3Rational functions with oblique asymptotes HL

  • Degree of top = degree of bottom + 1: divide to find an oblique asymptote. x2 + 3x + 5x + 1 = x + 2 + 3x + 1, so the asymptotes are x = −1 and y = x + 2.
  • Degree of top < degree of bottom (e.g. xx2 − 4): horizontal asymptote y = 0, vertical asymptotes at each root of the denominator.
Worked example

Find all asymptotes of y = 2x2 − x + 1x − 1.

  1. vertical x = 1 A1
  2. divide: 2x2 − x + 1 = (x − 1)(2x + 1) + 2 M1
  3. oblique y = 2x + 1 A1

4Inequalities HL

-214−+−+Sign of (x + 2)(x − 1)(x − 4)> 0 where the sign is +: −2 < x < 1 or x > 4
The sign can only change at a root (or an asymptote), so test one value in each interval.
Worked example

Solve (x + 2)(x − 1)(x − 4) > 0.

  1. critical values −2, 1, 4 A1
  2. sign chart (or sketch of the cubic) M1
  3. −2 < x < 1 or x > 4 A1
Worked example

Solve x + 1x − 2 ≤ 2.

  1. x + 1x − 2 − 2 ≤ 0 ⇒ 5 − xx − 2 ≤ 0 M1
  2. critical values x = 2 (excluded) and x = 5 A1
  3. x < 2 or x ≥ 5 A1
Common mistakeMultiplying both sides of an inequality by (x − 2). Its sign is unknown, so the inequality might flip. Move everything to one side instead, or multiply by (x − 2)2, which is positive.
GDC (calculator) methodPaper 2: graph both sides and read the x-values where one is above the other. State the critical values to 3 s.f. and use the correct strict or non-strict signs.
Linking questionPartial fractions (Proof, counting and partial fractions) undo the addition of rational functions, and the factor theorem connects to complex conjugate roots (Complex numbers).
Idea for your IA / explorationInvestigate how the shape of a rational function (asymptotes, turning points) changes with its parameters, or model cost per item, y = (fixed + variable × x)/x, with a rational function.

Test yourself — cover the answers first

Quick check

  1. State the asymptotes of y = 3/(x + 1) − 2.
  2. Find the horizontal asymptote of y = (4x − 1)/(2x + 5).
  3. HL Is (x − 2) a factor of x3 − 8?
  4. HL Find the remainder when x3 + 2x − 1 is divided by (x + 1).
  5. HL What does a double root look like on a graph?
  6. HL Solve x2(x − 3) ≥ 0.

Exam-style

  1. HL x = 2 is a root of x3 + kx2 − x − 2 = 0. Find k and the other roots.
  2. HL Solve 3x − 1 < 2x.
Show answers
  1. x = −1 and y = −2
  2. y = 2
  3. yes: p(2) = 0
  4. p(−1) = −4
  5. the curve touches the x-axis and turns back
  6. x = 0 or x ≥ 3
  7. 8 + 4k − 2 − 2 = 0 ⇒ k = −1
    x3 − x2 − x − 2 = (x − 2)(x2 + x + 1)
    other roots x = (−1 ± √3 i)/2 (complex)
  8. 3/(x − 1) − 2/x < 0 ⇒ x + 2x(x − 1) < 0
    critical values −2, 0, 1
    x < −2 or 0 < x < 1
Stuck on a question?

Send it to me on WhatsApp. I read every message.

Want regular help? Send this to your parent →
Share with a friendAsk Riyaz Sir

Independent notes written by Riyaz Sir. Not produced, endorsed or approved by the International Baccalaureate Organization. Syllabus codes refer to the IB DP subject guides.