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

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

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DP Maths AA HL | Polynomials and Rational Functions

Polynomials and Rational Functions

This topic contains 5 notes:

2.6 Quadratic Functions
2.12 Polynomial Functions
2.13 Rational Functions and Properties
1.11 Partial Fractions
1.12 Complex Numbers

DP MATHEMATICS AA · TOPIC 2: FUNCTIONS

2.6

Quadratic Functions and Equations

Full notes for HL · 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. HL students extend this topic in booklet 2.12.

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