DP Maths AA SL - Number and Algebra - Polynomials and Rational Functions

DP Maths AA SL - Number and Algebra - Polynomials and Rational Functions - Full notes

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DP Maths AA SL | Polynomials and Rational Functions

Polynomials and Rational Functions

This topic contains 2 notes:

2.6 Quadratic Functions
2.8 Rational, Exponential and Log Functions

DP MATHEMATICS AA · TOPIC 2: FUNCTIONS

2.6

Quadratic Functions and Equations

Full notes for SL · Syllabus SL 2.6, 2.7

Contents

1	Forms of a quadratic function
2	Completing the square and the vertex
3	Solving quadratic equations by factorizing
4	The quadratic formula
5	The discriminant
6	Lines and curves: tangency and intersections
7	Quadratic inequalities
8	Finding a quadratic from its graph
9	Quadratic models
10	Completing the square: more practice
11	Solving quadratic equations: choosing a method
12	The discriminant and the number of roots
13	Intersections of lines and parabolas
14	Quadratic inequalities
15	Finding a quadratic from information
16	Optimization with quadratics
17	Sum and product of roots
18	Mixed review: worked problems
19	Additional exam-style practice
20	Quadratic graphs in transformations
21	Exam-style questions with markscheme
22	Common misconceptions and links
23	Check your understanding
24	Summary and topic map
25	Answers to practice exercises

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and work through every worked example before attempting the Quick check.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
use the general, factorized and vertex forms of a quadratic and identify key features from each
complete the square to find the vertex and axis of symmetry
solve quadratic equations by factorizing, completing the square, the formula and technology
derive the quadratic formula
use the discriminant to determine the nature of roots
find conditions for a line to cut, touch or miss a curve
solve quadratic inequalities
find the equation of a quadratic from its graph and model real situations

This booklet covers SL 2.6, 2.7, which is common to SL and HL.

Guiding questions

What do the different forms of a quadratic reveal about its graph? How does the discriminant tell us about roots without solving?

How this topic is examined

Quadratics appear on almost every paper. Discriminant questions ('find k for equal roots', 'find the values of m for which the line is tangent') are very common on Paper 1, and quadratic models (projectiles, areas, profit) on Paper 2.

Write the condition on the discriminant explicitly (Δ > 0, Δ = 0 or Δ < 0) before solving. For inequalities, always sketch.

Command term	What the examiner wants
Write down	an answer with little or no working; it should be obtainable by inspection
Find / Calculate / Determine	obtain the answer showing relevant working
Solve	find the values of the unknowns, using algebra or technology as allowed
Show that	obtain the given result; every step must be shown and the result must not be assumed
Hence / Hence or otherwise	use the previous result (or another valid method)
Sketch	a diagram showing the general shape and key features, labelled
Prove	a formal, rigorous argument from accepted facts
Justify / Explain	give valid reasons or evidence

How to use these notes

Each section teaches one part of the syllabus. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section
Why it works	a derivation or proof of a key result, so you understand where formulas come from
Key point	a result or idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
GDC tip	how to use your graphic display calculator efficiently on Paper 2
In the real world	where the mathematics is used outside the classroom
Worked example	a full solution with IB-style marks (M1 method, A1 answer, R1 reasoning, AG answer given); examples are graded basic, standard and exam level
Practice exercise	graded questions for each section; full answers are at the end of the booklet
Quick check	short questions to check the key ideas

1 Forms of a quadratic function

Three forms, three sets of information.

Think about it

A ball is thrown upwards and its height follows a parabola. Which form of the equation would tell you its maximum height most directly? Which would tell you when it lands?

A quadratic function has the form f(x) = ax2 + bx + c, where a ≠ 0. Its graph is a parabola. If a > 0 it opens upward and has a minimum vertex; if a < 0 it opens downward and has a maximum vertex. The larger |a|, the narrower the parabola.

Every parabola is symmetric about a vertical line, the axis of symmetry, which passes through the vertex.

Form	Equation	What it shows directly
General	y = ax2 + bx + c	y-intercept c; axis x = −b/(2a)
Factorized (intercept)	y = a(x − p)(x − q)	x-intercepts p and q; axis x = (p + q)/2
Vertex	y = a(x − h)2 + k	vertex (h, k); axis x = h

Figure 1. The same parabola in vertex and factorized forms; expanded, it is x2 − 2x − 3.

Why it works

The axis of symmetry x = −b/(2a) comes from completing the square: ax2 + bx + c = a(x + b/(2a))2 + c − b2/(4a). The squared term is smallest (zero) when x = −b/(2a). For the factorized form, the axis is midway between the roots by symmetry.

Practice exercise 1

1. State the vertex and axis of y = −2(x + 3)2 + 5.

2. State the x-intercepts and axis of y = 3(x − 2)(x + 4).

3. Find the axis of symmetry of y = 2x2 − 12x + 1.

4. Does y = −x2 + 4x − 1 have a maximum or a minimum? Find it.

2 Completing the square and the vertex

Rewrite in vertex form.

To write ax2 + bx + c in vertex form: factor a out of the x2 and x terms; inside the bracket use x2 + px = (x + p/2)2 − (p/2)2; then multiply back and simplify.

WORKED EXAMPLE 1 Completing the square, a = 1 (basic)

Write x2 + 8x + 3 in the form (x + p)2 + q.

Solution

(x + 4)2 − 16 + 3M1

= (x + 4)2 − 13A1

WORKED EXAMPLE 2 Completing the square, a ≠ 1 (standard)

Write f(x) = 2x2 − 12x + 7 in the form a(x − h)2 + k and state the vertex.

Solution

= 2(x2 − 6x) + 7M1

= 2[(x − 3)2 − 9] + 7 = 2(x − 3)2 − 11A1

Vertex (3, −11), a minimumA1

WORKED EXAMPLE 3 Negative leading coefficient (exam level)

Write −3x2 + 6x + 1 in vertex form and state the range of the function.

Solution

= −3(x2 − 2x) + 1 = −3[(x − 1)2 − 1] + 1M1

= −3(x − 1)2 + 4A1

Maximum 4: range f(x) ≤ 4A1

Common mistake

Forgetting to multiply the subtracted square by a: 2[(x − 3)2 − 9] gives −18, not −9.

Practice exercise 2

1. Write x2 − 10x + 4 in vertex form.

2. Write 3x2 + 12x − 5 in vertex form.

3. Write 5 + 4x − x2 in vertex form and state the maximum value.

4. Find the range of f(x) = 2x2 + 8x + 11.

3 Solving quadratic equations by factorizing

Use the zero product property.

If AB = 0 then A = 0 or B = 0. So if a quadratic factorizes, set each factor to zero. To factorize ax2 + bx + c, find two numbers that multiply to ac and add to b, then split the middle term.

Watch for special forms: difference of two squares a2 − b2 = (a − b)(a + b), and perfect squares a2 ± 2ab + b2 = (a ± b)2. Also watch for equations that are quadratic in disguise, such as x4 − 5x2 + 4 = 0 (let u = x2) or x − 5√x + 6 = 0 (let u = √x).

WORKED EXAMPLE 4 Factorizing (basic)

Solve 2x2 + 5x − 3 = 0.

Solution

ac = −6: numbers 6 and −1; 2x2 + 6x − x − 3 = 0M1

(2x − 1)(x + 3) = 0A1

x = ½ or x = −3A1

WORKED EXAMPLE 5 Quadratic in disguise (exam level)

Solve x − 5√x + 6 = 0.

Solution

Let u = √x: u2 − 5u + 6 = 0M1

u = 2 or 3A1

x = 4 or x = 9A1

Practice exercise 3

1. Solve x2 − x − 12 = 0.

2. Solve 6x2 − x − 2 = 0.

3. Solve x4 − 10x2 + 9 = 0.

4. Solve 9x2 = 25.

5. Solve 22x − 6 × 2x + 8 = 0.

4 The quadratic formula

A method that always works.

Why it works

Complete the square on ax2 + bx + c = 0: divide by a, x2 + (b/a)x = −c/a; add (b/(2a))2: (x + b/(2a))2 = (b2 − 4ac)/(4a2); take square roots: x + b/(2a) = ±√(b2 − 4ac)/(2a); rearrange to obtain the formula.

WORKED EXAMPLE 6 Exact answers (standard)

Solve 3x2 − 5x − 1 = 0, giving exact answers.

Solution

x = (5 ± √(25 + 12)) ÷ 6M1

x = (5 ± √37)/6A1

WORKED EXAMPLE 7 Sum and product of roots (exam level)

The roots of 2x2 − 6x + 1 = 0 are α and β. Find α + β and αβ, and hence α2 + β2.

Solution

α + β = −b/a = 3; αβ = c/a = ½A1

α2 + β2 = (α + β)2 − 2αβM1

= 9 − 1 = 8A1

GDC tip

On Paper 2, use the polynomial root finder. Enter the coefficients in order (a, b, c) and record roots to 3 s.f.

Practice exercise 4

1. Solve x2 + 4x − 1 = 0 exactly.

2. Solve 5x2 − 2x − 4 = 0 to 3 s.f.

3. Solve 2x2 = 6x + 3 exactly.

4. If the roots of x2 − 7x + 3 = 0 are α and β, find 1/α + 1/β.

5 The discriminant

Δ = b2 − 4ac tells the nature of the roots.

The expression under the square root in the formula, Δ = b2 − 4ac, is the discriminant. Its sign determines how many real roots there are:

Discriminant	Roots	Graph of y = ax2 + bx + c
Δ > 0	two distinct real roots	crosses the x-axis twice
Δ = 0	two equal (repeated) real roots	touches the x-axis at the vertex
Δ < 0	no real roots	does not meet the x-axis

If Δ is a perfect square (and a, b, c are rational), the roots are rational. If a > 0 and Δ < 0, the quadratic is positive definite: positive for every x.

WORKED EXAMPLE 8 Equal roots (standard)

Find k such that kx2 + 4x + (k − 3) = 0 has equal roots.

Solution

Δ = 16 − 4k(k − 3) = 0M1

k2 − 3k − 4 = 0, (k − 4)(k + 1) = 0A1

k = 4 or k = −1A1

WORKED EXAMPLE 9 Two distinct roots (exam level)

Find the values of m for which x2 + mx + 9 = 0 has two distinct real roots.

Solution

m2 − 36 > 0M1

m < −6 or m > 6A1

WORKED EXAMPLE 10 Positive for all x (exam level)

Find the values of p for which px2 + 4x + p is positive for all x.

Solution

Need p > 0 and Δ < 0: 16 − 4p2 < 0M1

p2 > 4, so p > 2 or p < −2A1

With p > 0: p > 2R1 A1

Practice exercise 5

1. Find k if x2 − 6x + k = 0 has equal roots.

2. Find the range of k for which 2x2 + 3x + k = 0 has no real roots.

3. Find the range of k for which x2 + kx + k + 3 = 0 has no real roots.

4. Show that x2 + (k − 2)x − 2k = 0 has real roots for all k.

6 Lines and curves: tangency and intersections

Substitute, then use the discriminant.

To find where a line meets a parabola, substitute the line into the curve to get a quadratic. Its discriminant tells you how they meet: Δ > 0 the line cuts the curve at two points; Δ = 0 the line is a tangent (one point of contact); Δ < 0 the line misses the curve.

WORKED EXAMPLE 11 Tangent line (standard)

Find c such that y = 2x + c is tangent to y = x2 − 4x + 6, and find the point of contact.

Solution

x2 − 4x + 6 = 2x + c, so x2 − 6x + (6 − c) = 0M1

Tangent: Δ = 36 − 4(6 − c) = 0, c = −3M1 A1

Repeated root x = 3, y = 3: (3, 3)A1

WORKED EXAMPLE 12 Tangent with unknown gradient (exam level)

Find the values of m for which y = mx − 3 is tangent to y = x2 − 4x + 1.

Solution

x2 − (4 + m)x + 4 = 0M1

(4 + m)2 − 16 = 0M1

4 + m = ±4: m = 0 or m = −8A1

Practice exercise 6

1. Find the points where y = x + 1 meets y = x2 − 1.

2. Show that y = 2x − 5 does not meet y = x2 + 1.

3. Find k such that y = kx is tangent to y = x2 + 4.

7 Quadratic inequalities

Use the graph.

To solve ax2 + bx + c > 0 (or < 0, ≥, ≤): rearrange so one side is zero; find the roots (critical values); sketch the parabola; read off where the graph is above (> 0) or below (< 0) the x-axis. For an upward parabola, f(x) < 0 between the roots and f(x) > 0 outside them.

WORKED EXAMPLE 13 An inequality (standard)

Solve x2 − 2x − 8 ≤ 0.

Solution

(x − 4)(x + 2) ≤ 0: roots −2 and 4M1

Upward parabola is below the axis between the rootsR1

−2 ≤ x ≤ 4A1

WORKED EXAMPLE 14 Rearranging first (exam level)

Solve 2x2 > 3x + 5.

Solution

2x2 − 3x − 5 > 0: (2x − 5)(x + 1) > 0M1

Roots −1 and 2.5A1

x < −1 or x > 2.5A1

Common mistake

Taking square roots of an inequality: x2 > 9 does not mean x > 3 only; it means x < −3 or x > 3.

Practice exercise 7

1. Solve x2 < 16.

2. Solve x2 + x − 6 ≥ 0.

3. Solve 3 + 2x − x2 > 0.

4. Solve x(x + 5) < 14.

8 Finding a quadratic from its graph

Choose the most useful form.

If the graph shows the x-intercepts p and q, use y = a(x − p)(x − q) and find a from another point. If it shows the vertex (h, k), use y = a(x − h)2 + k. If three general points are given, set up three equations in a, b, c (1.16).

WORKED EXAMPLE 15 From intercepts (standard)

A parabola crosses the x-axis at −2 and 5 and passes through (0, −20). Find its equation.

Solution

y = a(x + 2)(x − 5)M1

−20 = a(2)(−5), a = 2A1

y = 2(x + 2)(x − 5) = 2x2 − 6x − 20A1

WORKED EXAMPLE 16 From the vertex (standard)

A parabola has vertex (3, 8) and passes through (1, 0). Find its equation.

Solution

y = a(x − 3)2 + 8M1

0 = 4a + 8, a = −2A1

y = −2(x − 3)2 + 8A1

Practice exercise 8

1. Find the parabola with vertex (−1, −5) through (1, 3).

2. Find the parabola through (1, 0), (4, 0) and (0, 8).

9 Quadratic models

Projectiles, areas and profits.

Quadratics model situations with a single maximum or minimum: the height of a projectile, the area of a rectangle with a fixed perimeter, the profit of a business. The vertex gives the optimum; the roots give when the quantity is zero. Always interpret answers in context and reject values outside the domain (such as negative time).

WORKED EXAMPLE 17 Projectile (standard)

A ball's height is h(t) = −4.9t2 + 19.6t + 1.5 metres. Find the maximum height and when the ball hits the ground.

Solution

Vertex at t = −19.6 ÷ (2 × −4.9) = 2 sM1

h(2) = 21.1 mA1

h(t) = 0: t = 4.08 s (the positive root)A1

WORKED EXAMPLE 18 Maximum area (exam level)

A farmer has 60 m of fencing to make a rectangular pen against a wall (fence on three sides). Find the maximum area.

Solution

Sides x, x and 60 − 2x: A = x(60 − 2x)M1

A = −2x2 + 60x, vertex at x = 15M1

A = 15 × 30 = 450 m2A1

Practice exercise 9

1. A rectangle has perimeter 40 cm. Find its maximum area.

2. Profit P = −2x2 + 120x − 1000 for x items. Find the number of items for maximum profit and the maximum profit.

3. A stone is thrown from a 20 m cliff with h = 20 + 15t − 5t2. When does it hit the sea?

10 Completing the square: more practice

Vertex form from standard form.

To write ax2 + bx + c in the form a(x − h)2 + k: factor a from the x terms, halve the coefficient of x, square it, and adjust the constant. The vertex is (h, k).

WORKED EXAMPLE 19 Leading coefficient not 1 (standard)

Write 3x2 − 12x + 5 in vertex form.

Solution

3(x2 − 4x) + 5 = 3((x − 2)2 − 4) + 5M1

= 3(x − 2)2 − 7A1

WORKED EXAMPLE 20 Negative leading coefficient (standard)

Write −2x2 + 8x + 1 in vertex form and state the maximum.

Solution

−2(x2 − 4x) + 1 = −2(x − 2)2 + 9M1 A1

Maximum 9 at x = 2A1

Practice exercise 10

1. Write x2 + 10x + 7 in vertex form.

2. Write 2x2 + 6x − 1 in vertex form.

3. Find the minimum value of 5x2 − 20x + 23.

11 Solving quadratic equations: choosing a method

Factorize, formula or complete the square.

Factorize when integer roots are likely; use the quadratic formula otherwise; complete the square when the vertex form is also needed. On Paper 2, a GDC solver is acceptable. Always check the discriminant if the question asks about the number of roots.

WORKED EXAMPLE 21 Formula (standard)

Solve 2x2 − 5x − 4 = 0 exactly.

Solution

x = (5 ± √(25 + 32))/4M1

x = (5 ± √57)/4A1

WORKED EXAMPLE 22 Disguised quadratic (exam level)

Solve x4 − 5x2 + 4 = 0.

Solution

(x2 − 1)(x2 − 4) = 0M1

x = ±1, ±2A1 A1

Practice exercise 11

1. Solve 3x2 + 2x − 8 = 0.

2. Solve x − 5√x + 6 = 0.

3. Solve 22x − 6(2x) + 8 = 0.

12 The discriminant and the number of roots

Δ = b2 − 4ac.

Δ > 0: two distinct real roots (the graph crosses the x-axis twice); Δ = 0: one repeated root (the graph touches the axis); Δ < 0: no real roots (the graph lies entirely above or below the axis).

WORKED EXAMPLE 23 Range of k (exam level)

Find k such that x2 + (k − 2)x + 4 = 0 has two distinct real roots.

Solution

(k − 2)2 − 16 > 0M1

|k − 2| > 4A1

k < −2 or k > 6A1

WORKED EXAMPLE 24 Always positive (exam level)

Find k such that kx2 + 4x + k > 0 for all x.

Solution

Need k > 0 and 16 − 4k2 < 0M1

k > 2A1

Practice exercise 12

1. Find k if 2x2 + kx + 8 = 0 has a repeated root.

2. Show that x2 + px + p2 = 0 has no real roots for p ≠ 0.

13 Intersections of lines and parabolas

Substitute and use Δ.

Substituting a line into a quadratic gives a quadratic equation. Two solutions: the line cuts the curve; one: it is a tangent; none: it misses. This gives a quick way to find tangent lines.

WORKED EXAMPLE 25 Tangent condition (exam level)

Find m such that y = mx − 1 is tangent to y = x2 + 3.

Solution

x2 − mx + 4 = 0M1

Δ = m2 − 16 = 0A1

m = ±4A1

Practice exercise 13

1. Find the intersections of y = x2 − 2x and y = x + 4.

2. Find c such that y = 2x + c is tangent to y = x2.

14 Quadratic inequalities

Use a sketch.

Solve ax2 + bx + c > 0 by finding the roots and sketching the parabola: for a > 0, the expression is positive outside the roots and negative between them.

WORKED EXAMPLE 26 Inequality (standard)

Solve x2 − x − 12 ≤ 0.

Solution

(x − 4)(x + 3) ≤ 0M1

−3 ≤ x ≤ 4A1

Practice exercise 14

1. Solve 2x2 + 5x − 3 > 0.

2. Solve x2 < 6x.

15 Finding a quadratic from information

Choose the best form.

Use the factorized form if the roots are known, the vertex form if the vertex is known, and solve simultaneous equations from three points if nothing else is given.

WORKED EXAMPLE 27 From vertex and a point (standard)

A parabola has vertex (3, −2) and passes through (5, 6). Find its equation.

Solution

y = a(x − 3)2 − 2: 6 = 4a − 2M1

y = 2(x − 3)2 − 2A1

WORKED EXAMPLE 28 From roots and a point (standard)

A parabola cuts the x-axis at −1 and 4 and passes through (0, −8). Find it.

Solution

y = a(x + 1)(x − 4): −8 = −4aM1

y = 2(x + 1)(x − 4)A1

Practice exercise 15

1. Find the quadratic through (0, 3), (1, 4) and (2, 9).

2. Find the quadratic with axis x = 1, y-intercept 5, through (3, 11).

16 Optimization with quadratics

The vertex gives the maximum or minimum.

Many problems lead to a quadratic in one variable whose maximum or minimum is at the vertex x = −b/(2a). Typical contexts: maximum area for a given perimeter, maximum revenue, maximum height of a projectile.

WORKED EXAMPLE 29 Revenue (exam level)

A theatre sells 500 tickets at $40. Each $2 increase in price loses 20 sales. Find the price for maximum revenue.

Solution

Price 40 + 2n, sales 500 − 20n: R = (40 + 2n)(500 − 20n)M1 A1

R = −40n2 + 200n + 20 000: maximum at n = 2.5M1

Price $45, revenue $20 250A1

Practice exercise 16

1. A rectangular field against a river uses 200 m of fence on three sides. Find the maximum area.

2. A ball's height is h = 1.5 + 18t − 5t2. Find the maximum height.

17 Sum and product of roots

Relationships between roots and coefficients.

If α and β are the roots of ax2 + bx + c = 0, then α + β = −b/a and αβ = c/a. These give values of symmetric expressions such as α2 + β2 = (α + β)2 − 2αβ without solving.

WORKED EXAMPLE 30 Symmetric expression (exam level)

The roots of 2x2 − 6x + 1 = 0 are α, β. Find α2 + β2 and 1/α + 1/β.

Solution

α + β = 3, αβ = 1/2A1

α2 + β2 = 9 − 1 = 8A1

1/α + 1/β = 3/(1/2) = 6A1

Practice exercise 17

1. Find a quadratic with roots 2 + √3 and 2 − √3.

2. The roots of x2 + px + 12 = 0 differ by 1. Find p.

18 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 31 Mixed problem (exam level)

f(x) = x2 − 4x + k. (a) Find k if the minimum value is 3. (b) For this k, find the range of f on 0 ≤ x ≤ 5.

Solution

(a) f(2) = k − 4 = 3: k = 7M1 A1

(b) f(0) = 7, f(5) = 12: 3 ≤ f(x) ≤ 12A1

Practice exercise 18

1. Find the range of values of x for which x2 − 4x − 5 > 0 and x > 0.

2. Find the vertex of y = (x − 1)(x − 7).

19 Additional exam-style practice

Paper 1 and Paper 2.

WORKED EXAMPLE 32 Paper 1 question (exam level)

The graph of y = ax2 + bx + c has vertex (2, −1) and passes through (0, 7). (a) Find a, b, c. (b) Find the x-intercepts.

Solution

(a) y = a(x − 2)2 − 1; 7 = 4a − 1: a = 2; y = 2x2 − 8x + 7M1 A1 A1

(b) x = 2 ± 1/√2A1

Practice exercise 19

1. Find k such that the line y = k does not meet y = x2 − 6x + 10.

2. Solve (x2 − 3)2 = 4x2.

20 Quadratic graphs in transformations

y = x2 transformed.

The vertex form a(x − h)2 + k shows that every parabola is y = x2 stretched vertically by factor |a| (reflected if a < 0) and translated by (h, k). The axis of symmetry x = h is the image of the y-axis.

Figure 2. Transformations of y = x2.

WORKED EXAMPLE 33 Describing (standard)

Describe y = −3(x + 1)2 + 5 as transformations of y = x2.

Solution

Vertical stretch factor 3 and reflection in the x-axisA1

Translation by (−1, 5)A1

Practice exercise 20

1. Write the equation of y = x2 after a translation by (4, −1) and a vertical stretch factor 0.5 (stretch first).

21 Exam-style questions with markscheme

Attempt each question under timed conditions before reading the markscheme.

Question 1 (Paper 1 style) (about 15 minutes)

f(x) = x2 − 4x + 1.

(a) Write f(x) in the form (x − h)2 + k. [2]

(b) State the vertex and axis of symmetry. [2]

(d) Find the values of m for which y = mx − 3 is tangent to y = f(x). [4]

Markscheme

Part	Answer	Marks
(a)	(x − 2)2 − 3	M1 A1
(b)	vertex (2, −3)	A1
	x = 2	A1
(c)	p < −3	A2
(d)	x2 − (4 + m)x + 4 = 0	M1
	(4 + m)2 − 16 = 0	M1
	4 + m = ±4	A1
	m = 0 or m = −8	A1

Examiner's comments. In (d), forming the equation correctly is the key method mark.

Question 2 (Paper 2 style) (about 12 minutes)

A company's daily profit, in dollars, from selling x items is P(x) = −0.5x2 + 40x − 350.

(a) Find the number of items for maximum profit and the maximum profit. [3]

(b) Find the range of x for which the company makes a profit. [3]

Markscheme

Part	Answer	Marks
(a)	x = −40/(2 × −0.5) = 40	M1 A1
	P(40) = $450	A1
(b)	−0.5x2 + 40x − 350 > 0	M1
	roots 10 and 70	A1
	10 < x < 70	A1
(c)	profit becomes very negative; demand and costs would change	R1

Examiner's comments. Answers in context (number of items, dollars) were expected.

22 Common misconceptions and links

Students often think…	Actually…
the vertex of a(x − h)2 + k is (−h, k)	it is (h, k)
Δ = 0 means no roots	one repeated root
x2 > 9 means x > 3 only	x < −3 or x > 3
a quadratic model is valid for all x	check the domain in context

How 2.6 connects to the rest of the course

Topic	Connection
2.1 Functions	key features
2.11 Transformations	vertex form as a translation
2.12 Polynomials (HL)	sum and product of roots
5.8 Optimization	maxima and minima by calculus

23 Check your understanding

Section A: multiple choice

1. The vertex of y = 3(x + 2)2 − 5 is

A. (−2, −5) B. (2, −5) C. (−2, 5) D. (3, −5)

2. Δ for x2 + 4x + 4 is

A. 16 B. 32 C. 0 D. −4

3. x2 − 9 < 0 when

A. x > 3 B. −3 < x < 3 C. x < −3 or x > 3 D. x < 3

4. The axis of symmetry of y = 2x2 − 8x + 1 is

A. x = −2 B. x = 4 C. x = 8 D. x = 2

5. If Δ < 0 and a > 0, the parabola is

A. entirely below the axis B. entirely above the x-axis C. tangent to the axis D. crosses twice

Section B: structured

6. Find the range of k for which x2 + kx + k + 3 = 0 has no real roots. [4]

7. Write −2x2 + 8x + 3 in vertex form. [3]

8. Find c if y = 3x + c is tangent to y = x2 + x + 5. [3]

Answers

1. A 2. C 3. B 4. D 5. B

6. k2 − 4k − 12 < 0: −2 < k < 6.

7. −2(x − 2)2 + 11.

8. x2 − 2x + (5 − c) = 0; Δ = 4 − 4(5 − c) = 0: c = 4.

24 Summary and topic map

Idea	Remember
Forms	general, factorized, vertex
Vertex	complete the square; x = −b/(2a)
Solving	factorize, formula, complete the square, GDC
Discriminant	Δ > 0, = 0, < 0
Tangency	substitute; Δ = 0
Inequalities	sketch and read off
Models	vertex gives the optimum

Before the exam, make sure you…

•state the discriminant condition
•sketch for inequalities
•check signs in vertex form
•give exact answers on Paper 1
•interpret in context

Quadratic Functions on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 3. Topic map for Quadratic Functions.

25 Answers to practice exercises

Check your working against these answers. Full methods are expected in the exam.

Practice exercise 1

1. (−3, 5); x = −3

2. x = 2, x = −4; x = −1

3. x = 3

4. maximum, at (2, 3)

Practice exercise 2

1. (x − 5)2 − 21

2. 3(x + 2)2 − 17

3. 9 − (x − 2)2; 9

4. f(x) ≥ 3

Practice exercise 3

1. x = 4 or −3

2. x = 2/3 or −½

3. x = ±1, ±3

4. x = ±5/3

5. x = 1 or 2

Practice exercise 4

1. −2 ± √5

2. 1.12, −0.717

3. (3 ± √15)/2

4. 7/3

Practice exercise 5

1. 9

2. k > 9/8

3. −2 < k < 6

4. Δ = (k − 2)2 + 8k = (k + 2)2 ≥ 0

Practice exercise 6

1. (−1, 0) and (2, 3)

2. x2 − 2x + 6 = 0 has Δ = −20 < 0

3. k = ±4

Practice exercise 7

1. −4 < x < 4

2. x ≤ −3 or x ≥ 2

3. −1 < x < 3

4. −7 < x < 2

Practice exercise 8

1. y = 2(x + 1)2 − 5

2. y = 2(x − 1)(x − 4)

Practice exercise 9

1. 100 cm2

2. 30 items; 800

3. t = 4 s

Practice exercise 10

1. (x + 5)2 − 18

2. 2(x + 1.5)2 − 5.5

3. 3

Practice exercise 11

1. 4/3, −2

2. √x = 2 or 3: x = 4 or 9

3. 2x = 2 or 4: x = 1 or 2

Practice exercise 12

1. k = ±8

2. Δ = −3p2 < 0

Practice exercise 13

1. (4, 8), (−1, 3)

2. c = −1

Practice exercise 14

1. x < −3 or x > 1/2

2. 0 < x < 6

Practice exercise 15

1. y = 2x2 − x + 3

2. y = a(x − 1)2 + k: a + k = 5, 4a + k = 11: a = 2, k = 3

Practice exercise 16

1. 5000 m2

2. 17.7 m

Practice exercise 17

1. x2 − 4x + 1 = 0

2. roots 3, 4 or −3, −4: p = ±7

Practice exercise 18

1. x > 5

2. (4, −9)

Practice exercise 19

1. k < 1

2. x2 − 3 = ±2x: x = 3, −1, −3, 1

Practice exercise 20

1. y = 0.5(x − 4)2 − 1

DP MATHEMATICS AA · TOPIC 2: FUNCTIONS

2.8

Rational, Exponential and Logarithmic Functions

Full notes for SL · Syllabus SL 2.8, 2.9, 2.10

Contents

1	The reciprocal function
2	Rational functions of the form (ax + b)/(cx + d)
3	Exponential functions
4	The number e and natural growth
5	Logarithmic functions
6	Transformed exponential and logarithmic graphs
7	Solving equations analytically and graphically
8	Modelling with exponential functions
9	The reciprocal function and its transformations
10	Rational functions (ax + b)/(cx + d)
11	Exponential functions and their features
12	Logarithmic functions and their features
13	Modelling with exponential functions
14	Equations with exponentials and logarithms
15	Comparing growth rates
16	Mixed review: worked problems
17	Additional exam-style practice
18	Graphs of y = f(x) combined with exponentials
19	Exam-style questions with markscheme
20	Common misconceptions and links
21	Check your understanding
22	Summary and topic map
23	Answers to practice exercises

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and work through every worked example before attempting the Quick check.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
sketch y = 1/x and recognize it as self-inverse
find asymptotes and intercepts of (ax + b)/(cx + d) and sketch its graph
sketch exponential functions ax and ex, identifying intercepts and asymptotes
explain the special role of e and write ax as ex ln a
sketch logarithmic functions as inverses of exponentials
find asymptotes, domains and ranges of transformed exponential and log functions
solve equations analytically and graphically, including with technology
model growth and decay, including half-life and Newton's law of cooling

This booklet covers SL 2.8, 2.9, 2.10, which is common to SL and HL.

Guiding questions

How do asymptotes shape the graphs of rational, exponential and logarithmic functions? Why is e the natural base for growth?

How this topic is examined

These functions appear on both papers. Paper 1 asks for sketches with asymptotes and intercepts, domains and ranges, and exact solutions; Paper 2 asks for modelling questions and graphical solutions.

Always write asymptotes as equations (x = 2, y = 3), not just numbers, and label them on sketches with dashed lines.

Command term	What the examiner wants
Write down	an answer with little or no working; it should be obtainable by inspection
Find / Calculate / Determine	obtain the answer showing relevant working
Solve	find the values of the unknowns, using algebra or technology as allowed
Show that	obtain the given result; every step must be shown and the result must not be assumed
Hence / Hence or otherwise	use the previous result (or another valid method)
Sketch	a diagram showing the general shape and key features, labelled
Prove	a formal, rigorous argument from accepted facts
Justify / Explain	give valid reasons or evidence

How to use these notes

Each section teaches one part of the syllabus. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section
Why it works	a derivation or proof of a key result, so you understand where formulas come from
Key point	a result or idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
GDC tip	how to use your graphic display calculator efficiently on Paper 2
In the real world	where the mathematics is used outside the classroom
Worked example	a full solution with IB-style marks (M1 method, A1 answer, R1 reasoning, AG answer given); examples are graded basic, standard and exam level
Practice exercise	graded questions for each section; full answers are at the end of the booklet
Quick check	short questions to check the key ideas

1 The reciprocal function

y = 1/x.

Think about it

If you double the speed of a journey, the time halves. If you triple it, the time is a third. What shape does a graph of time against speed have, and what happens at very low speeds?

The reciprocal function f(x) = 1/x, x ≠ 0, describes quantities that are inversely proportional. Its graph is a rectangular hyperbola in the first and third quadrants. As x → 0, |f(x)| grows without bound: the vertical asymptote is x = 0. As x → ±∞, f(x) → 0: the horizontal asymptote is y = 0.

1/x is self-inverse: f−1(x) = 1/x, so its graph is symmetric about y = x (and also about y = −x). It is an odd function: f(−x) = −f(x).

Figure 1. Reciprocal functions: y = k/x is a vertical stretch of y = 1/x.

2 Rational functions of the form (ax + b)/(cx + d)

Hyperbolas with shifted asymptotes.

For f(x) = (ax + b)/(cx + d), c ≠ 0:

Vertical asymptote: where the denominator is zero, x = −d/c (provided the numerator is not also zero there).

Horizontal asymptote: y = a/c. For large |x|, the constants b and d become negligible, so f(x) ≈ ax/(cx) = a/c.

Intercepts: x-intercept where the numerator is zero, x = −b/a; y-intercept f(0) = b/d.

Writing f(x) in the form a/c + k/(cx + d) by division shows it is a transformation of 1/x (2.11).

WORKED EXAMPLE 1 Sketching a rational function (standard)

For f(x) = (2x − 1)/(x + 3), find the asymptotes and intercepts, and sketch the graph.

Solution

Vertical asymptote: x = −3A1

Horizontal asymptote: y = 2A1

x-intercept (½, 0); y-intercept (0, −1/3)A1

Figure 2. Asymptotes x = −3 and y = 2.

WORKED EXAMPLE 2 Domain, range and inverse (exam level)

f(x) = (3x + 1)/(x − 2). State the domain and range of f, and find f−1(x).

Solution

Domain x ≠ 2; range f(x) ≠ 3A1 A1

x(y − 2) = 3y + 1, so y(x − 3) = 2x + 1M1

f−1(x) = (2x + 1)/(x − 3), x ≠ 3A1

The asymptotes of the inverse are swapped: x = 3 and y = 2.

Common mistake

Writing the horizontal asymptote of (2x − 1)/(x + 3) as y = 0. It is the ratio of the x coefficients, y = 2.

Practice exercise 1

1. Write down the asymptotes of y = 5/(x − 4).

2. Find the asymptotes and intercepts of y = (4x + 2)/(2x − 3).

3. Find the range of f(x) = (x + 5)/(x + 1).

4. Show that (3x + 7)/(x + 2) = 3 + 1/(x + 2).

3 Exponential functions

Growth by a constant factor.

An exponential function has the form f(x) = ax, where a > 0 and a ≠ 1. The variable is in the exponent. Equal steps in x multiply the output by the same factor.

If a > 1 the function shows exponential growth; if 0 < a < 1, exponential decay. All graphs y = ax pass through (0, 1), have the horizontal asymptote y = 0, domain x ∈ ℝ and range y > 0. Note that (1/a)x = a−x, the reflection of ax in the y-axis.

Exponential growth eventually outgrows any polynomial: 2x > x100 for large enough x.

Figure 3. Exponential growth and decay all pass through (0, 1).

Practice exercise 2

1. Write down the y-intercept and asymptote of y = 3x.

2. Sketch y = 2−x, stating its range.

3. Which is larger for x = 30: 2x or x5?

4 The number e and natural growth

The base for continuous growth.

Suppose $1 is invested at 100% interest. Compounded n times a year, it grows to (1 + 1/n)n after one year. As n increases this approaches a limit, the number e ≈ 2.718 28. So e arises naturally from continuous growth.

The natural exponential function ex has a remarkable property: its gradient at every point equals its value (5.6). For this reason it is the standard base in calculus and science. Any exponential ax can be written in base e, since a = eln a:

n	(1 + 1/n)n
1	2
10	2.5937
100	2.7048
10 000	2.7181
1 000 000	2.7183

Practice exercise 3

1. Write 5x in the form ekx.

2. Write e0.3x in the form ax.

5 Logarithmic functions

Inverses of exponentials.

The function f(x) = loga x is the inverse of ax. Its graph is the reflection of y = ax in y = x: it passes through (1, 0), has the vertical asymptote x = 0, domain x > 0 and range ℝ. The natural logarithm ln x is the inverse of ex.

Logarithmic functions grow very slowly: ln x only reaches 10 when x ≈ 22 026.

Figure 4. ln x is the reflection of ex in y = x.

6 Transformed exponential and logarithmic graphs

Track the asymptote.

Transformations (2.11) move asymptotes. For y = kex + c, the horizontal asymptote is y = c, and the y-intercept is k + c. For y = ln(x − h) + c, the vertical asymptote is x = h, the domain is x > h, and the x-intercept solves ln(x − h) = −c.

WORKED EXAMPLE 3 Transformed exponential (standard)

For f(x) = 3 − 2e−x, find the asymptote, intercepts and range.

Solution

As x → ∞, e−x → 0: asymptote y = 3A1

y-intercept (0, 1); x-intercept: e−x = 3/2, x = −ln 1.5 = −0.405A1

Range f(x) < 3A1

WORKED EXAMPLE 4 Transformed logarithm (standard)

For g(x) = ln(2x − 4) + 1, find the domain, asymptote and x-intercept.

Solution

2x − 4 > 0: domain x > 2, asymptote x = 2A1

ln(2x − 4) = −1: 2x − 4 = e−1M1

x = 2 + 1/(2e) ≈ 2.18A1

Practice exercise 4

1. State the asymptote and range of y = 4ex − 1.

2. State the domain and asymptote of y = ln(x + 3) − 2, and find its x-intercept.

3. Sketch y = 5 − e2x, labelling the asymptote and intercepts.

7 Solving equations analytically and graphically

Algebra where possible; technology otherwise.

Analytically, use logarithm laws and algebra (1.5). Graphically, find the intersection points of the two sides with the GDC. Some equations mix types, such as ex = 3 − x or ln x = x − 2, and can only be solved with technology.

WORKED EXAMPLE 5 Graphical solution (standard)

Solve ln x = x − 2.

Solution

Graph y = ln x and y = x − 2M1

x = 0.159 or x = 3.15 (3 s.f.)A1 A1

WORKED EXAMPLE 6 Rational equation (standard)

Solve (2x − 1)/(x + 3) = x − 1.

Solution

2x − 1 = (x − 1)(x + 3) = x2 + 2x − 3M1

x2 = 2A1

x = ±√2A1

GDC tip

When graphs approach each other slowly (for example a logarithm and a horizontal line), a standard window may miss an intersection. Think about end behaviour and extend the window if needed.

Practice exercise 5

1. Solve ex = 3 − x.

2. Solve 2x = x2 for x > 0.

3. Solve 3/(x − 1) = x + 1.

4. Solve ex = 5 − x2.

8 Modelling with exponential functions

Growth and decay models.

Exponential models have the form N(t) = N0ekt or N(t) = N0at, where N0 is the initial value, k > 0 for growth and k < 0 for decay. The half-life (time to halve) is ln 2 ÷ |k|; the doubling time for growth is ln 2 ÷ k.

Newton's law of cooling gives T(t) = Troom + (T0 − Troom)e−kt: the temperature approaches room temperature, a horizontal asymptote. Similar 'limited growth' models, such as P = A − Be−kt, describe learning curves and drug absorption.

WORKED EXAMPLE 7 Exponential decay (standard)

A drug's concentration is C(t) = 40e−0.15t mg L−1 after t hours. (a) Find the concentration after 6 hours. (b) Find the half-life.

Solution

(a) 40e−0.9 = 16.3 mg L−1A1

(b) e−0.15t = ½, t = ln 2 ÷ 0.15M1

= 4.62 hoursA1

WORKED EXAMPLE 8 Finding a model (exam level)

A population is 500 at t = 0 and 800 at t = 4. Assuming P = 500ekt, find k and predict P at t = 10.

Solution

800 = 500e4k, k = ln 1.6 ÷ 4 = 0.1175M1 A1

P(10) = 500e1.175 = 1619A1

WORKED EXAMPLE 9 Newton's law of cooling (exam level)

Coffee at 90 °C cools in a 20 °C room: T = 20 + 70e−0.05t, t in minutes. When does it reach 50 °C?

Solution

70e−0.05t = 30M1

t = ln(7/3) ÷ 0.05M1

= 16.9 minutesA1

Practice exercise 6

1. A sample of 200 g decays by 8% per year. Find the mass after 10 years and the half-life.

2. Bacteria double every 3 hours. Find k in N = N0ekt.

3. A car's value is V = 30 000e−0.18t. When is it worth $10 000?

4. T = 25 + 75e−kt. If T = 70 when t = 10, find k.

9 The reciprocal function and its transformations

y = a/(x − h) + k.

y = 1/x has asymptotes x = 0 and y = 0 and is self-inverse. Transformations give y = a/(x − h) + k with asymptotes x = h and y = k; a stretches the curve and, if negative, reflects it.

Figure 5. Transformations of y = 1/x.

WORKED EXAMPLE 10 Features (standard)

For y = 3/(x − 2) + 1, find the asymptotes and intercepts.

Solution

x = 2, y = 1A1

y-intercept (0, −0.5); x-intercept (−1, 0)A1 A1

Practice exercise 7

1. Find the asymptotes and intercepts of y = −2/(x + 1) + 4.

2. Write y = (2x + 5)/(x + 1) in the form a/(x + 1) + k.

10 Rational functions (ax + b)/(cx + d)

Asymptotes from coefficients.

For f(x) = (ax + b)/(cx + d): vertical asymptote x = −d/c; horizontal asymptote y = a/c; x-intercept −b/a; y-intercept b/d. Its inverse is (−dx + b)/(cx − a), with the asymptotes swapped.

WORKED EXAMPLE 11 Inverse (exam level)

Find the inverse of f(x) = (3x − 1)/(x + 2) and its asymptotes.

Solution

f−1(x) = (−2x − 1)/(x − 3)M1 A1

Asymptotes of f−1: x = 3, y = −2A1

Practice exercise 8

1. Find the range of f(x) = (4x + 1)/(2x − 3).

2. Find the intercepts of y = (x − 6)/(2x + 3).

11 Exponential functions and their features

Growth, decay and asymptotes.

f(x) = kax + c (a > 0, a ≠ 1) has horizontal asymptote y = c and y-intercept k + c. The base e gives f(x) = kerx + c, with r > 0 for growth and r < 0 for decay.

Figure 6. Exponential growth and decay with asymptotes.

WORKED EXAMPLE 12 Features (standard)

For f(x) = 5 − 2e0.3x, find the asymptote, intercepts and range.

Solution

Asymptote y = 5 (as x → −∞)A1

f(0) = 3; f(x) = 0 when e0.3x = 2.5: x = 3.05A1 A1

Range f(x) < 5A1

Practice exercise 9

1. Find the asymptote and y-intercept of y = 2(0.5)x − 3.

2. Find the x-intercept of y = 4 − ex.

12 Logarithmic functions and their features

Inverses of exponentials.

f(x) = loga x has vertical asymptote x = 0 and passes through (1, 0) and (a, 1). y = ln(x − h) + k has asymptote x = h. The domain is where the argument is positive.

WORKED EXAMPLE 13 Features (standard)

For y = 2 ln(x + 3) − 1, find the domain, asymptote and intercepts.

Solution

Domain x > −3; asymptote x = −3A1

x-intercept: ln(x + 3) = 0.5, x = e0.5 − 3 = −1.35A1

y-intercept 2 ln 3 − 1 = 1.20A1

Practice exercise 10

1. Find the inverse of y = ln(2x − 1).

2. State the domain of y = log(5 − x).

13 Modelling with exponential functions

From two points to a model.

To find a model y = Aekt from two data points, substitute both and divide to eliminate A. For y = Aekt + c, the asymptote c is often given by the context (for example, room temperature).

WORKED EXAMPLE 14 Model from data (exam level)

A population is 800 at t = 0 and 1250 at t = 5. Find P = Aekt and predict t = 12.

Solution

A = 800; e5k = 1.5625, k = 0.0893M1 A1

P(12) = 800e1.071 = 2335A1

WORKED EXAMPLE 15 Cooling (exam level)

T = 20 + Ae−kt. T(0) = 85, T(10) = 50. Find A, k and T(25).

Solution

A = 65; 30 = 65e−10k, k = 0.0773A1 A1

T(25) = 20 + 65e−1.933 = 29.4 °CA1

Practice exercise 11

1. A substance decays from 120 g to 90 g in 4 hours. Find its half-life.

2. A car's value V = 30 000e−kt halves in 5 years. Find k.

14 Equations with exponentials and logarithms

Isolate, then take logs.

Isolate the exponential term before taking logarithms. For equations mixing exponentials and polynomials, use technology.

WORKED EXAMPLE 16 Exact solution (standard)

Solve 4e2x − 3 = 17 exactly.

Solution

e2x = 5M1

x = ½ ln 5A1

WORKED EXAMPLE 17 Technology (standard)

Solve ex = x2 + 2.

Solution

Graph both sidesM1

x = 1.32A1

Practice exercise 12

1. Solve 3(2x) = 7x−1 exactly.

2. Solve ln(x + 1) = 2 − x.

15 Comparing growth rates

Exponential beats polynomial beats logarithm.

For large x, exponential functions (with base > 1) grow faster than any polynomial, which in turn grows faster than any logarithm. This explains why exponential models eventually predict huge values and must be used with care.

Figure 7. Logarithmic, polynomial and exponential growth.

WORKED EXAMPLE 18 Overtaking (standard)

Use technology to find where 1.2x = x2, and describe which function is larger for large x.

Solution

Intersections x = −0.920, 1.11 and 40.6M1 A1

For x > 40.6, 1.2x > x2 and the gap grows rapidlyR1

Practice exercise 13

1. Which is larger at x = 100: x3 or 1.1x?

16 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 19 Mixed problem (exam level)

f(x) = 2 + 3/(x − 1), g(x) = ex. (a) State the asymptotes of f. (b) Find (f ∘ g)(x) and its asymptotes.

Solution

(a) x = 1, y = 2A1

(b) 2 + 3/(ex − 1)A1

Vertical asymptote x = 0; horizontal y = 2 (x → ∞) and y = −1 (x → −∞)A1 A1

Practice exercise 14

1. Find the inverse of f(x) = 3 + e2x and state its domain.

2. Find the asymptotes of y = (ex + 2)/(ex − 1).

17 Additional exam-style practice

Paper 1 and Paper 2.

WORKED EXAMPLE 20 Paper 2 question (exam level)

The concentration of a drug is C(t) = 10te−0.4t. (a) Find the maximum concentration. (b) Find the time during which C > 5.

Solution

(a) GDC maximum at t = 2.5: C = 9.20A1 A1

(b) C = 5 at t = 0.648 and 6.36: 5.71 hoursM1 A1

Practice exercise 15

1. Find the range of f(x) = 2 − e−x for x ≥ 0.

2. Find where y = ln x and y = 1/x intersect.

18 Graphs of y = f(x) combined with exponentials

Products and quotients.

Functions such as xe−x, x2ex or ln x/x combine features: zeros of each factor, asymptotes from the exponential or logarithm, and a single maximum or minimum. Sketch them with technology and identify the key points.

Figure 8. Combined functions.

WORKED EXAMPLE 21 Features (standard)

Find the maximum of y = (ln x)/x and its asymptotes.

Solution

Maximum (e, 1/e)A1

Vertical asymptote x = 0; horizontal y = 0A1 A1

Practice exercise 16

1. Find the minimum of y = x2ex.

19 Exam-style questions with markscheme

Attempt each question under timed conditions before reading the markscheme.

Question 1 (Paper 1 style) (about 15 minutes)

f(x) = (3x + 2)/(x − 1), x ≠ 1; g(x) = ln(x + 2), x > −2.

(a) Write down the equations of the asymptotes of f. [2]

(b) Find the intercepts of f. [2]

(d) Find the x-intercept of g and state its range. [2]

Markscheme

Part	Answer	Marks
(a)	x = 1	A1
	y = 3	A1
(b)	(−2/3, 0)	A1
	(0, −2)	A1
(c)	x(y − 1) = 3y + 2	M1
	y(x − 3) = x + 2	A1
	f−1(x) = (x + 2)/(x − 3), x ≠ 3	A1
(d)	(−1, 0)	A1
	g(x) ∈ ℝ	A1

Examiner's comments. Asymptotes must be given as equations of lines.

Question 2 (Paper 2 style) (about 12 minutes)

A cup of tea is made at 95 °C in a room at 22 °C. Its temperature after t minutes is T(t) = 22 + 73e−kt. After 5 minutes it is 75 °C.

(a) Find k. [3]

(b) Find the temperature after 15 minutes. [2]

(d) State the long-term temperature, with a reason. [1]

Markscheme

Part	Answer	Marks
(a)	75 = 22 + 73e−5k	M1
	e−5k = 53/73	A1
	k = 0.0640	A1
(b)	T(15) = 22 + 73e−0.960	M1
	49.9 °C	A1
(c)	e−kt = 18/73	M1
	t = 21.9 minutes	A1
(d)	22 °C, since e−kt → 0	R1

Examiner's comments. Keeping k unrounded in later parts avoids accuracy errors.

20 Common misconceptions and links

Students often think…	Actually…
the horizontal asymptote of (ax + b)/(cx + d) is y = 0	y = a/c
ln x is defined for x ≤ 0	only x > 0
exponential graphs cross the x-axis	they approach their asymptote
e is just a large number	e is the base of continuous growth

How 2.8 connects to the rest of the course

Topic	Connection
1.5 Logarithms	laws and equations
2.5 Inverse functions	log as inverse of exp
2.13 Rational functions (HL)	oblique asymptotes
5.6 Differentiation	derivatives of ex and ln x

21 Check your understanding

Section A: multiple choice

1. The vertical asymptote of 5/(x − 4) is

A. x = 4 B. y = 4 C. x = 5 D. x = −4

2. The horizontal asymptote of (4x + 1)/(2x − 3) is

A. y = 4 B. y = 0 C. y = 2 D. y = −1/3

3. y = 3x passes through

A. (1, 0) B. (0, 1) C. (0, 3) D. (0, 0)

4. The domain of ln(x + 5) is

A. x > 5 B. x ≥ −5 C. x > 0 D. x > −5

5. The half-life for N = N0e−0.2t is

A. 0.2 B. ln 2/0.2 ≈ 3.47 C. 5 D. ln 0.2

Section B: structured

6. Sketch y = 2 + 3/(x + 1), labelling asymptotes and intercepts. [4]

7. Coffee cools: T = 20 + 70e−0.05t. When does it reach 50 °C? [3]

8. Explain why 1/x is self-inverse. [2]

Answers

1. A 2. C 3. B 4. D 5. B

6. Asymptotes x = −1, y = 2; intercepts (−2.5, 0), (0, 5).

7. 70e−0.05t = 30: t = 16.9 min.

8. Swapping x and y in y = 1/x gives y = 1/x; symmetric about y = x.

22 Summary and topic map

Idea	Remember
1/x	asymptotes x = 0, y = 0; self-inverse
(ax + b)/(cx + d)	x = −d/c, y = a/c
Exponential	through (0, 1); asymptote y = 0
e	limit of (1 + 1/n)n; ax = ex ln a
Logarithmic	inverse of exponential; asymptote x = 0
Transformations	kex + c: y = c; ln(x − h): x = h
Models	N0ekt; half-life ln 2/\|k\|; cooling

Before the exam, make sure you…

•write asymptotes as equations
•label intercepts
•use technology for non-algebraic equations
•interpret models in context

Rational, Exponential and Log Functions on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 9. Topic map for Rational, Exponential and Log Functions.

23 Answers to practice exercises

Check your working against these answers. Full methods are expected in the exam.

Practice exercise 1

1. x = 4, y = 0

2. x = 1.5, y = 2; (−½, 0), (0, −2/3)

3. f(x) ≠ 1

4. 3(x + 2) + 1 = 3x + 7

Practice exercise 2

1. (0, 1); y = 0

2. decay through (0, 1); range y > 0

3. 230 ≈ 1.07 × 109 > 305 ≈ 2.43 × 107

Practice exercise 3

1. ex ln 5, k = 1.61

2. 1.35x

Practice exercise 4

1. y = −1; y > −1

2. x > −3; x = −3; x = e2 − 3 ≈ 4.39

3. asymptote y = 5; (0, 4); (ln 5/2, 0) ≈ (0.805, 0)

Practice exercise 5

1. x = 0.792

2. x = 2 or x = 4

3. x = ±2

4. x = −2.21 or x = 1.24

Practice exercise 6

1. 86.9 g; 8.31 years

2. ln 2/3 = 0.231

3. 6.10 years

4. k = ln(75/45)/10 = 0.0511

Practice exercise 7

1. x = −1, y = 4; (0, 2) and (−0.5, 0)

2. 2 + 3/(x + 1)

Practice exercise 8

1. y ≠ 2

2. (6, 0) and (0, −2)

Practice exercise 9

1. y = −3; (0, −1)

2. ln 4 = 1.39

Practice exercise 10

1. (ex + 1)/2

2. x < 5

Practice exercise 11

1. 9.64 hours

2. 0.139

Practice exercise 12

1. x ln 2 + ln 3 = (x − 1) ln 7: x = ln 21/ln 3.5 = 2.43

2. 1.21

Practice exercise 13

1. x3 = 106; 1.1100 = 13 781: x3 is larger, but 1.1x overtakes later (x ≈ 160)

Practice exercise 14

1. ½ ln(x − 3), x > 3

2. x = 0; y = 1 and y = −2

Practice exercise 15

1. 1 ≤ f(x) < 2

2. x = 1.76

Practice exercise 16

1. (0, 0) (and a local maximum at x = −2)

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