Mathematics for Academic Studies 4 Student Book sample

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N O I T A C U D E Y R A D EC O N

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Building SUMA

Blocks piezaS

THIS IS

YOUR BOOK F EACH UNIT

Reading and listening

THE OPENING PAGES O

5

We read and listen a brief historical introduction of the contents you are going to learn in the unit.

ons

functi a certain road and has to brake, actual ers a hazard on the the When a driver encount between their decision to brake and e, the 3/4 of a second. Therefor ore, amount of time passes (reaction time), about in that time. Furtherm an moment they brake the further it will travel it travels faster the car is going, when the brake is applied; ically automat stop the car does not due to inertia. (braking distance) moment the to additional distance hazard the aware of greater the driver becomes (d ) that increases the From the moment an additional distance the car stops, it travels following values: experiment gives the the velocity (v). An

Evidently useful

ns Basic Functio g

Reading and listenin

cy exactly what were debates about th these 20th centuries there . Over the course of During the 19 and in defining a function Riemann and was and was not essential notably Dirichlet, s mathematicians, a function. discussions, numerou formulate a more precise definition of to which is very similar 1923, in Weierstrass, helped offered g definition was Finally, the followin today. use nds to a to the one we value of x correspo function of x if each the equation y = f (x).

One says that y is a ndence is shown in value of y. This correspo

Seeking accura

speed of the car (km/h)

(based on reaction time in m)

distance

50

The equations are: = 0.21v d reaction

95

74

21

100

79

60

19

90

64

47

17

80

50.5

36

14.5

70

39.5

27

12.5

60

ing to Poincaré

s were of bizarre function the for accuracy, a series unhappy with the direction However, in this search Poincaré, who was in 1899: invented that irritated had taken. This irritation led him to say s s which definition of function a mass of bizarre function seen have we functions century ‘For half a as possible honest to resemble as little new function was a appear to be forced when , purpose. Formerly end. Today, they are which serve some view of some practical ’s reasoning was at invented, it was in to show that our ancestor invented on purpose

7

29.5

19

10.5

dbraking = 0.0074v

2

3

20.5

12

8.5

40

13.5

7

6.5

30

total stopping distance (m)

1 3

4

20

accord Honest functions,

distance

(braking distance in m)

2

10

2 0.21v dtotal = 0.0074v +

of them g words and use each for each of the followin , span. irritate, mass, resemble in a sentence: bizarre, be slower if a reaction time may to explain why the attention.

2 Use modal verbs and not fully paying tion conversa a person driving is having

1 Write a synonym

can be fault.’ honest functions, which . going to focus on these ting or disproving concepts In this unit we are more than just construc used for something

Solve

for of the formulas above in to check the validity Use your calculator in each row, keeping (only check a few values them. the values in the table and graph approximate values) mind that they are

1 12) is one of Henri Poincaré (1854–19 ticians in history. the greatest mathema of spanned all fields His contributions mathematics.

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Solve. You can do these motivating activities to activate your previous knowledge.

LANG

108

The audios of each unit’s content are available at www.anayaeducacion.es

CONTENT DEVELOPMENT AND ACTIVITIES Unit

1

Science is full of functions in which variations in the cause have a proportional impact on variations in the effect. All such functions are called linear functions and are graphed using straight lines. Let’s look at an example:

y = 30 + 15x (y: length in cm; x: weight in kg) Based on the presumption that the spring becomes deformed when weights of over 6 kg are hung from it, the domain of definition of this function is [0, 6].

F ocus on English

This is a very useful formula. You should learn how to use it!

Y

y = mx

X

• If we give x the value x0 8 y = y0 + m (x0 – x0) = y0 + m · 0 = y0. x = x0, then

(x2, y2)

y = 30 + 15x

y = y0. In other words, it passes through (x0, y0).

y2 – y1 m = –––––– x –x 2

➜ finding straight lines using two points

In order to work out the equation of a line that passes through two points, we do the following:

1

(x1, y1) x – x 2 1

• We can find the slope using the two points.

• We can find the equation based on the slope and one of the points.

weight (kg)

0

1

2

3

4

5

6

Problem solved

anayaeducacion.es Review the point-slope equation.

Find the equation of each of the following straight lines:

a) Equation: y = 7 – 3 (x + 5). This is the equation of the line.

5

Functions of proportionality are graphed with straight lines that pass through the origin. They describe a ratio between the values of both variables.

a) It passes through (–5, 7) and its slope is –3 .

5

The slope of the straight line is the constant of proportionality, m.

b) It passes through (–2, 7) and (4, 5).

We can simplify it: y = 7 – 3 x – 3 · 5 8 y = 4 – 3 x 5 5 5 b) We begin by finding its slope: m = 5 – 7 = –2 = – 1 3 6 4 – (–2)

For example, the space covered at a constant velocity, v, as a function of time is: s = v · t, where v is the slope of the line that relates s to t. anayaeducacion.es GeoGebra. Graphic representation of a linear function.

The icons included with some activities indicate the keys to the project.

ã Constant function: y = n Y n

It is represented with a straight line parallel to the X axis.

y=n

X

The line y = 0 coincides with the X axis.

For example, the distance of an artificial satellite from Earth is constant. It does not depend on time, t. The equation for this would be d = 36 000, d: distance, in km; t: time, which does not appear in the equation.

200 100 32

F

ã General expression of linear functions: y = mx + n This is represented on a graph as a straight line with slope m that intersects the Y axis at point (0, n).

The number n is called the ordinate at the origin.

Y F = 32 + 1.8C C 50 100

y = mx + n n X

For example, the line F = 32 + 1.8C, represented in the margin, allows us to change from degrees Celsius, C, to degrees Fahrenheit, F.

anayaeducacion.es GeoGebra. Calculating the slope of a line.

1 Graph the following functions:

b) y = 2 x 3

c) y = – 1 x d) y = – 7 x 3 4

2 Graph the following:

a) y = 3

b) y = –2

Examples and solved problems.

To put into practice the most important methods.

Equation of the line that passes through (–2, 7) and has a slope of – 1 : 3 y = 7 – 1 (x + 2) 8 y = 19 – 1 x 3 3 3

Think and practise

a) y = 2x

Its slope is 0.

y=0

Equation: y = y0 + m(x – x0)

• The coefficient of x is m. Therefore, its slope is m.

lenght (cm)

50

Slope: m

• y = y0 + m (x – x0) is a first-degree expression. Therefore, it is a straight line.

100

ã Functions of proportionality: y = mx

point out: to mention some information you think is important.

Point: P (x0, y0) explanation

If different weights are hung from a spring, they will stretch to different lengths.

In other words, the length of the spring is a function of the weight hanging from it. It is worth pointing out* that this is a linear function.

More specifically, let’s suppose that the spring is 30 cm long before it is stretched and that is stretches 15 cm per kilogram added to it. The relationship is:

We often need to write down the equation of a straight line of which we only know one point and the slope. It is written as follows:

Remember

ã Linear functions in our daily lives

stretching a spring

5

ã Point-slope equation of a line

LINEAR FUNCTIONS

y2 – y1

Each unit is divided into epigraphs and subepigraphs. The most important contents are in bold.

Focus on English. Do you think Mathematics and English haveanything in common?

Discover how language and mathematics are linked so you can learn both: Mathematics and English.

c) y = 0

d) y = –5

3 Graph the following functions:

b) y = 2 x + 2 3 c) y = – 1 x + 5 d) y = –3x – 1 4 4 At an initial point in time, a moving object is 3 m from its origin and is travelling away from it at a velocity of 2 m/s. Find the equation for its distance from the origin as a function of time and graph it. a) y = 2x – 3

5

The price of potatoes at the market is €1/kg, and the price of tomatoes is €2/kg. a) Write the equation for the price of a bag of potatoes as a function of its weight. b) Write the equation for the price of a bag of tomatoes as a function of its weight. c) Graph the functions above.

6 Find the equation of these straight lines using the

given data:

a) It passes through (–3, –5) and its slope is 4 .

9 b) It passes through (0, –3) and its slope is 4. c) It passes through (3, –5) and (–4, 7).

111

110

Think and practice. These are exercises to apply the theory you have learnt.

KEYS

PROJECT

SDG SDG Commitment Discover the Sustainable Development Goals and be an active part of our commitment to make a more equal and liveable world.

Developing thinking Work on strategies for thinking: reflect on the content you are learning, generate ideas, organise them, debate them, explain them…

Cooperative learning Get involved in your learning and participate in the group’s learning; you will find that cooperating improves performance and harmony in the class.

Emotional education Get to know yourself; identify the situations that bring up complicated emotions and manage them with constructive, self-affirming experiences.

EXERCISES AND PROBLEMS EXERCISES AND

to choose Remember portfolio.

ED

PROBLEMS SOLV

Exercises and problems solved.

Methods, suggestions, tricks and thinking strategies that will come in handy to solve similar problems.

In order to express it as a piecewise-defined function, we first need to see where the function within the absolute value intersects with the X axis. In this way, we can work out in which intervals it is positive and in which it is negative.

y = 1 x2 – 5 x + 9 2 4 4

1 x 2 – 5 x + 9 = 0 8 x = 1, x = 9 2 4 4 As the branches of the parabola point upwards, since the coefficient of x 2 is positive, the function is only negative between the points of intersection with the X axis.

Z ] 1 x2 – 5 x + 9 2 ]4 4 ] y = [– 1 x 2 + 5 x – 9 2 4 ] 4 ]] 1 x 2 – 5 x + 9 2 4 \4

Your turn Express as a piecewisedefined function and graph: y = |x 2 – 2x – 8|

1

|2x – 4|

a) y = |x 2 – 4| – 2 b) y = |x + 3| + |x|

x<0

0≤x≤2

x>2

–2x + 4

–2x + 4

2x – 4

|x|

–x

x

x

|2x – 4| + |x|

–3x + 4

–x + 4

3x – 4

Therefore, the function is: Z –3x + 4 if x < 0 ]] y = [–x + 4 if 0 ≤ x ≤ 2 ]3x – 4 if x > 2 \

6 Functions related to the function of inverse proportionality

y = 2x + 3 x +1 Your turn Graph this function: y = 3x – 1 x+2

Your turn. Exercises and problems to practice the strategies you have just learnt.

5

b) P (2, –1), m = –2

c) A (–2, 1), m = 1 2

d) A (1, 3), m = – 5 3

C

6

A

D

4 2 –2

2

8

B D

Find the equation of the straight line that passes through points A and B in each case: a) A (3, 0), B (5, 0)

b) A (–2, – 4), B (2, –3)

c) A (0, –3), B (3, 0)

d) A (0, –5), B (–3, 1)

4

6

X

–2

Calculate the equation of these linear functions:

9

Find the equation in each case and graph it: a) Line that passes through (2, –3) and is parallel to the one passing through (1, –2) and (– 4, 3). b) Proportionality function that passes through (–4, 2).

10

c) Constant function that passes through (18; –1.5).

6

d) y = x 2 – 6x + 6 Y

B

a) P (0, 0), m = 1

A

4

Match each expression to its graph: a) y = x 2 b) y = (x – 3)2 c) y = x 2 – 3

Find the value of the unknown parameters so that the straight lines and points comply with these conditions. Graph them.

Graph the following functions by making a table of values like the one below for each of them: x

–4

–3

–2

–1

0

1

2

3

4

y

…

…

…

…

…

…

…

…

…

a) y = x 2 + 1

b) y = –x 2 + 4

c) y = –3x 2

d) y = 0.4x 2

Graph the following parabolas by finding their vertex, some points close to it and the points where they intersect with the axes: b) y = x 2 – 4x a) y = (x + 2)2 d) y = x 2 – 9 c) y = 1 x 2 + 2x + 1 2 Give the point (abscissa and ordinate) where the vertex of the following parabolas is found. In each case, state whether it is a maximum or a minimum.

Then, graph them. a) y = 8 – x 2 c) y = –x 2 – 2x + 4

a) The line that passes through points (4, 0) and (–2, a) has a slope of –1.

We start by dividing: 2x + 3 x + 1 This function can be expressed as: – 2x – 2 2 2x + 3 = 2 + 1 1 x +1 x +1 It is the function y = 1 shifted 1 unit to the left and x 2 units up.

Express the following function k + b. in the form y = x–a Then, graph it.

Using the given slope and point, calculate the equation of each line:

C

a) The parabola of y = x 2 – 1 intersects with the X axis at x = –1 and x = 1.

Since its branches point upwards, the piecewise-defined function is: Z Z if x < –1 ]]x 2 – 1 + 1 if x < –1 ]]x 2 y = [–x 2 + 1 + 1 if –1 ≤ x ≤ 1 8 y = [–x 2 + 2 if –1 ≤ x ≤ 1 ]x 2 – 1 + 1 if x > 1 ]x 2 if x > 1 \ \ b) We make a table to study how the expression varies within each absolute value depending on the section. The endpoints of each section are those that make each of the expressions equal to 0:

Your turn Express each function as a piecewise-defined function:

2

if x ≥ 9

5 Functions with absolute values

a) y = |x 2 – 1| + 1

Graph the following linear functions: a) y = 2x – 3 b) y = 4 x 7 d) y = 2.5 c) y = –3x + 10 5

X

9

7

1

3

if x ≤ 1 if 1 < x < 9

Exercises and problems. To put into practice all the knowledge acquired throughout the unit.

Unit 5

Quadratic functions

Linear functions

Therefore, we have three sections: (– ∞, 1] and [9, +∞) where the function is positive or zero, so there is no need to change the sign, and the interval (1, 9), in which the function is negative. In this Y section, we change the sign of the function.

Let’s see what it looks like:

b) y = |2x – 4| + |x|

unit for your

PROBLEMS

Practise

Express the following function without using the absolute value sign and graph it:

Express each of the following functions as a piecewise-defined function:

this resources from

EXERCISES AND

4 The absolute value of a function

e) y = 15 – 1 x 2 + 1 x 2 4 4

b) The line y = bx + 2 passes through point (–3, 4).

11

c) The lines of the equations y = 3x + c and y = cx + 3 intersect at ordinate 2. Which is the corresponding abscissa?

The exercises are divided into topics.

Each one is also marked with its degree of difficulty, from one to three.

Calculate the vertex, the axis of symmetry and the points of intersection with the axes (if any) of each of these parabolas: a) y = 2x2

d) The points (d, –2) and (4, e) belong to the line of the equation y = 1 x – 3 .

2

b) y = 4 + (3 – x)2 d) y = 3x – 1 x 2 + 1 2 f ) y = 1 x 2 + 2x + 3 3

b) y = 2(x – 5)2

c) y = 2(x – 5)2 + 2

d) y = –x2 + 1

e) y = –(x + 1)2 + 1

f ) y = –3x + 2x2 127

126

MATHS WORKSHOP Unit 5

P

MATHS WORKSHO LEARN

Here, you will find readings, activities, advice, information...

PRACTICE MAKES PERFECT!

Radioactive decay

• A candle lasts an hour. With the leftover wax of 10 candles you can make a new one.

Radioactive substances decay by emitting radiation and transforming into other substances. This process takes place over time and its rate varies greatly from one substance to another.

Uranium 92 radioactive substance 8 radiation + different substance 2 500 million The rate at which a radioactive substance decays is measured by its half-life, which is the time it takes for half of its original mass to decay. On the right, you can see the half-lives of some radioactive substances.

glossary

U

Ac

227.028

t

t/28

where M is the amount of actinium remaining after t years.

1

226.025

81

Tl

204.383

algebraic fraction alternate exterior angle alternate interior angle analytic expression of a function angle bisector apothem

Glossary. We learn the relevant terms that are underlined in the units with a clear definition.

t 57

= 100 · >c 1 m

1 57

t

H ≈ 100 · 0.988t 8 P = 100 · 0.988t, t function in centuries discontinuous direct motion

area

The size of a surface.

arithmetic

The branch of mathematics that studies numbers and the 2 operations that2we can do with them.

arithmetic progression

A sequence in which the next term is determined by1.2 adding number means that % willthe havesame decayed. (positive or negative), called the difference, to the previous term. 14

axial symmetry axis of symmetry

P = 100 · c 1 m

In other words, in 1 century there will be 100 · 0.988 = 98.8 % of the C14 left, which discrete quantitative variable

• What percentage of C will a 33 000-year-old fossil have compared to a living plant?

dispersion parameters A symmetry in which the points of one figure coincide withathe points another • How old must fossil be if itofonly has 10 % of the C14 of a living plant? figure when reflected across a line. domain of a function

Substitute P inthing the formula for 10equal and isolate t taking logarithms.

An imaginary line that divides a figure, a shape, or any other into two ellipse and symmetric parts.

bar chart

A graphic representation formed of thin 132 bars, where the height of each bar is proportional to the frequency of that value. They are used to represent tables of discrete quantitative or qualitative variables.

biquadratic equation

Fourth-degree equations that have no terms with odd degrees.

box-and-whisker plot

A graphic representation that visually describes the position parameters (median and quartiles).

case

How many eggs did she have at the beginning?

equalisation method

2

3

4

time in half lives

SELF-ASSESSMENT

Trust in your skills and knowledge, develop creativity, adapt to changing situations and have a proactive and responsible attitude.

anayaeducacion.es Answer key and interactive self-assessment.

1 Graph the piecewise function that has the following

equation:

Z ]] 2x + 6 if x < –2 y = [ x/2 + 3 if –2 ≤ x < 2 ] –x + 6 if x ≥ 2 An amount that can be t multiplied by itself three times to result in\the number.

Is it continuous? up with a function with the 1 m 5 700 of a value with the frequencies The sum the of all theCome previous values, P = of 100 · cfrequency same sections that is not continuous. when the values 2 are ordered from smallest to largest.

The result of a non-exact quotient. It has an integer and of a decimal 2 Findportion the vertex each of portion the following parabolas separated by a decimal point. and graph them: 2 independent variable x A function where the dependent variable y decreases asxthe –2 b) y = x 2 + 4x – 5 a) y = increases.

2

c) y =y variable.

(5 – x)(x + 1) The variable represented on the vertical axis. It is the 2

3 Express as piecewise-defined functions and graph

A variable that can only take certain numerical values. them:

a) y = |2x + 1|

b) y = 1 – x 4 d) y = |9 – (x – 2)2|

Parameters that tell us how far away from the centre the values in a distribution are.

The set of x values that have y values.

c) y = |–x 2 + 4x – 3|

a) y =

1 x +5

d) y = x + 2

b) y = 3 – 2 x

c) y =

e) y = 2 x – 1

f) y = – x – 3

An equality of algebraic expressions that is only true for certain values of the letters.

equation

An equality statement that includes a letter called an unknown.

equivalent fractions

Two fractions are equivalent when they have the same numerical value.

Every possible result of a random experiment.

equivalent systems

Systems of equations that have the same solution.

central angle in a circle

An angle with its vertex at the centre of the circumference.

exact root

The square root of a perfect square.

chance

A combination of circumstances that supposedly cause an unpredictable event.

factorising polynomials

circumference

A closed two-dimensional curve in which all points are equidistant from the centre.

A method for decomposing a polynomial into the product of other polynomials with the lowest possible degree.

class midpoint

The central value of each interval.

final value

The product of an original value and a variation index.

compatible determinate system

A system that has one unique solution. Its graphic representation is two lines that intersect at one point.

fraction

A given division. It is the result of dividing an integer into equal parts.

frequency polygon

compatible indeterminate system

A system that has infinite solutions. Its graphic representation is two coinciding lines that share all their points.

A graphic representation formed by connecting the ends of the bars in a bar graph or the midpoints of the rectangles in a histogram. It is used to represent quantitative variables.

compound experiment

An experiment that includes more than one random experiment.

frieze

Longitudinal decoration that contains a pattern that repeats in translations.

compound interest

The profit generated over a certain time by adding interest to the original capital.

function

A relationship between two variables that we usually call x and y.

conic section

The curves resulting from different intersections between a cone and a plane.

general term

An expression that represents any term in a sequence.

continuous function

A function without any discontinuities. Its graph can be drawn without lifting the pencil from the paper.

geographic coordinates

A reference system for specifying the location of any point on the Earth. There are two: latitude and longitude.

continuous quantitative statistical variable

A variable that can take any numerical value in an interval.

geometric progression

A sequence in which the next term is determined by multiplying the previous term by a fixed number, called the ratio.

coordinates

A system of values, references... used to determine the position of a point on a plane.

geometric transformation

corresponding angles

Angles that share one side that intersects two parallel lines formed by their other sides.

A transformation in which every point of one figure corresponds to another point of another figure.

geometry

The study of the properties and measurements of figures in the plane or in space.

3 +1 x –1

5 Graph these pairs of functions:

a) y = 1.2x; y = log1.2 x

b) y = 2.5x; y = log2.5 x

With respect to which line are the two functions of each pair symmetrical?

b) Which expression gives us the surface area, S, for any base, b ? Graph it. c) For which value of the base do we obtain the maximum surface area? What is the value of that surface area?

A closed geometric curve with two unequal perpendicular axes, resulting from video for target 6.a. Think of something you can do to contribute cutting the surface of a cone with a plane that is not perpendicularWatch to itsthe axis.

Commitment to achieve that goal. Make a commitment to put your idea into practice.

A procedure for solving a system of equations that involves solving for the same unknown in both equations and then equalising the resulting expressions.

equation

Practice makes perfect! In this section you will have to solve many different types of problems.

domains of definition:

make a picture frame. a) If the base of the frame is 0.5 m long, how tall is it? What is its surface area?

2

g) yin= 2(x – 1)(x + 3) gapsh)in y =the (x + 2)2 – 2x 2 A function where the independent variable moves jumps, creating graph.

4 Graph the following functions and find their

6 Using a 3-metre-long strip of wood, we want to

d) y = –(x – 3)2 – 1

e) y = 2x also + 4xcalled a shift. f ) y = 9 – (x – 1) A type of motion that maintains the direction of rotation,

306

Enterprising culture

• A farmer shared a flock of sheep among his children.

— He gave the eldest one sheep plus 1/7 of those remaining.

— He gave the second child two sheep plus 1/7 of those remaining.

— He gave the third child three sheep plus 1/7 of those remaining.

— And so on until he got to the youngest.

In this way, all of the children received the same inheritance and no sheep were left over.

How many children does the farmer have?

How many sheep were there in the flock?

1 –– 4

There are three naturally occurring isotopes of carbon: C12, C13 and C14. The first two are stable but the third one is radioactive and has a half-life of 5 700 years. This means that over this period of time the amount of C14 reduces by half. like value the other isotopes of carbon, C14 is present in the atmosphere The absolute value of the difference betweenJust a real and a rounded value. cube (CO root2) and is absorbed by plants (photosynthesis) which incorporate it in a specific proportion. C14 is then incorporated A given quotient of two polynomials. cumulative frequency in the same proportion, via these plants, into other living things.

When a living thing dies and Non-adjacent angle that is outside the parallel lines. fossilised, its C14 continues to decay according to the function on the right. becomes decimal number Non-adjacent angle that is inside the parallelEvery lines. 5 700 years, the amount of C14 left reduces by half. Therefore, by finding out the 14 proportion An equation that algebraically relates the two variables.of C in a fossil and checking it against the initial proportion (that in a living plant), the equation above can be used to find the fossil’s geological age,function in other words, decreasing the time when it was formed.

The locus of the points equidistant from the sides of an angle.

If we express the time, t, in centuries, the equation above can be expressed like this: dependent variable A perpendicular segment from the centre of a triangle to its side.

b) How many candles are needed for 1 000 hours of light?

• A farmer went to the market to sell a basket of eggs. The first customer bought half her eggs plus half an egg. The second customer bought half of the remaining eggs plus half an egg, and the third one did the same. The seller then had no more eggs left.

1 –– 2

1

absolute error

Thallium 3 minutes

mass (g)

Carbon-14 dating

Glossary

Radon 1 620 years

Ra

Actinium 28 years

M = c 1 m grams 2 where t is the time elapsed taking one half-life as a unit. If the substance was actinium and we wanted to express the time in years, the formula would be: M = c1m 2

88

years

238.029

89

If we have an initial mass of 1 g, the amount of mass of this substance that will be left after a given time is:

a) How many hours of light will 442 candles give you?

133

ment. By Self-assess s, se activitie doing the r u o heck y you can c ding and n ta unders you have how much learnt.

307

Academic and professional

ICT

orientation

Assessment

Linguistic Plan

Learn how to obtain information, select it and apply it; to plan, manage and work on projects; to collaborate online in an ethical and safe manner.

Evaluate your personal skills, discover and awaken your calling, train yourself to make decisions and learn to choose between different options.

Discover different strategies to analyse what you have learnt and how you learnt it; train yourself to take responsibility or overcome difficulties.

Use your communication skills in the different types of text that you will see.

Language is always present, communicate!

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BANK

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DIGITAL BOOK RESOURCE BANK

A digital version of your book to be used online or offline. It offers access to your digital resources which are grouped by type or linked to unit content.

A space with resources, techniques and activities, designed to strengthen your knowledge.

More about the keys

Resources related to THE PROJECT KEYS

SDG

SDG Commitment with short videos that will help you understand the targets for reaching the Sustainable Development Goals worked on in this project.

Linguistic Plan with infographics that will give you models to work with the four linguistic skills, using different text types (descriptive, narrative, explanatory, etc.).

Cooperative learning Preparing for the task In small groups, of four or five members: 1 All the members of the team will review how the assigned task can be accomplished.

2 To do this, the steps can be shared out to each team member who then in turn will explain how each part of the process can be done to the others. The others listen and participate if they think they can contribute something.

Authorship / adaptation : Variant of the Educational Innovation Laboratory of the colegio Ártica - David and Roger Johnson.

Developing thinking with explanations are included on how to apply the different thinking techniques proposed in the project.

3 Once everyone is in agreement on how to do each part, you will all complete the tasks and, finally, verify, among everyone, that you have solved it correctly.

Cooperative learning which includes the descriptions of the cooperative learning techniques proposed in the project.

Thinking techniques

Emotional education with resources to help you overcome any worries that may arise in different situations at school (beginning of the school year, taking a test, etc.).

Logic Wheel This thinking technique will help you to establish phases when analysing specific content that you have to study.

Identify What is it? What is it like? Are there different types?

By following a logical sequence (the logic wheel), and by asking yourself a series of questions in each phase, you can:

1

• Identify content by asking yourself: What is it? What is it like?

Are there different types?

• Compare the content by formulating questions such as: In what way is it similar to ...? In what way is it different from ...?

• Establish cause-effect relationships by asking yourself questions such as: Why? What impact does it have ...?

• Argue, assess and ask yourself questions such as: What conclusions can be drawn after the analysis? What can be assessed or scored about it?

Doing a data dump of these questions into a graphic organiser will help you.

Compare

Argue, assess What can we conclude?

4

Logic wheel

2

In what way is it similar to ...?

In what way is it different from ...?

3 Establish cause-effect relationships Why? What impact does it have ...?

Authorship: Hernández, P., and García, L. A.; adapted by Escamilla, A.

ICT resources to help you use information and communication technology in a healthy, correct and safe way.

Academic and professional orientation with information on different professions linked to the subject content.

0:40/1:33

Assessment which includes resources for your portfolio, as well as rubrics and targets that will help with your self-assessment.

resources SUBJECT KEY CONCEPTS Tutorials Activities with GeoGebra Glossary Learn by playing Self-assessments Language Bank

Resources classified by unit

All the resources are classified by unit so that you can find them more easily.

1

course contents REAL NUMBERS

Page 10

1. Irrational numbers...............................................................

2. Real numbers: the real number line..............................

3. Sections of the real number line: intervals and half-lines.........................................................................

4. Roots and radicals...............................................................

5. Approximate numbers. Errors.........................................

6. Numbers in scientific notation. Error control............

7. Logarithms.............................................................................

Exercises and problems solved..........................................

Exercises and problems.........................................................

Maths workshop........................................................................

Self-assessment........................................................................

2

POLYNOMIALS AND ALGEBRAIC FRACTIONS

3

16 18 22 24 26 29 30 34 35

4

FUNCTIONS. CHARACTERISTICS

Page 88

1. Basic concepts...................................................................... 90 2. How functions are presented.......................................... 91 3. Domain of definition........................................................... 94 4. Continuous functions. Discontinuities......................... 95 5. Increase, maximums and minimums............................ 96 6. Tendency and periodicity................................................. 98 Exercises and problems solved........................................... 100 Exercises and problems.......................................................... 102 Maths workshop......................................................................... 106 Self-assessment........................................................................ 107

Page 36

1. Polynomials. Operations...................................................

2. Ruffini’s rule...........................................................................

3. Root of a polynomial. Finding roots.............................

4. Factorising polynomials....................................................

5. Divisibility of polynomials.................................................

6. Algebraic fractions..............................................................

7. Decomposition of algebraic fractions into partial fractions...................................................................................

Exercises and problems solved..........................................

Exercises and problems.........................................................

Maths workshop........................................................................

Self-assessment........................................................................

EQUATIONS, INEQUATIONS AND SYSTEMS

12 14

38 40 42 44 46 48 50 52 54 60 61

5

1. Equations................................................................................

2. Systems of equations.........................................................

3. Inequations with one unknown......................................

Exercises and problems solved..........................................

Exercises and problems.........................................................

Maths workshop........................................................................

Self-assessment........................................................................

64 70 74 78 79 84 85

Page 108

1. Linear functions.................................................................... 110 2. The parabola: a very interesting curve........................ 113 3. Quadratic functions............................................................ 114 4. Absolute value functions.................................................. 117 5. Radical functions.................................................................. 118 6. Inversely proportional functions.................................... 120 7. Exponential functions........................................................ 122 8. Logarithmic functions........................................................ 124 Exercises and problems solved........................................... 125 Exercises and problems......................................................... 127 Maths workshop......................................................................... 132 Self-assessment......................................................................... 133

6 Page 62

BASIC FUNCTIONS

SIMILARITY. APPLICATIONS

Page 136

1. Similarity................................................................................. 138 2. Homothety............................................................................. 140 3. Rectangles with interesting dimensions..................... 142 4. Similarity of triangles......................................................... 144 5. Similarity of right-angled triangles............................... 146 6. Similarity of triangles in geometric solids.................. 148 Exercises and problems solved........................................... 150 Exercises and problems.......................................................... 151 Maths workshop........................................................................ 156 Self-assessment........................................................................ 157

7

TRIGONOMETRY

Page 158

1. Trigonometric ratios of an acute angle....................... 160 2. Basic trigonometric identities......................................... 162 3. Using a calculator in trigonometry............................... 164 4. Solving right-angled triangles........................................ 166 5. Solving non-right-angled triangles............................... 167 6. Trigonometric ratios of 0° to 360°............................... 168 7. Angles of any measurement. Trigonometric ratios.. 171 8. Trigonometric functions. The radian............................ 172 Exercises and problems solved.......................................... 174 Exercises and problems......................................................... 175 Maths workshop......................................................................... 180 Self-assessment......................................................................... 181

8

ANALYTIC GEOMETRY

Page 182

1. Vectors in the plane.......................................................... 184 2. Operating with vectors.................................................... 185 3. Vectors which represent points................................... 187 4. Midpoint of a segment..................................................... 188 5. Aligned points..................................................................... 189 6. Equations of a straight line............................................ 190 7. Straight lines. Parallelism and perpendicularity..... 192 8. Straight lines parallel to the coordinate axes.......... 194 9. Relative positions of two straight lines..................... 195 10. Distance between two points....................................... 196 11. Equation of a circumference......................................... 197 Exercises and problems solved.......................................... 198 Exercises and problems......................................................... 199 Maths workshop........................................................................ 204 Self-assessment........................................................................ 205

9

STATISTICS

Page 208

1. Statistics and statistical methods.................................. 210 2. Frequency tables................................................................. 212 3. Statistical parameters: x and q....................................... 214 4. Dispersion parameters for isolated data.................... 216 5. Dispersion parameters for grouped data................... 218 6. Box-and-whisker plots....................................................... 220 7. Statistical inference............................................................. 222 8. Statistics in the media........................................................ 225 Exercises and problems solved........................................... 227 Exercises and problems......................................................... . 229 Maths workshop......................................................................... 234 Self-assessment ....................................................................... 235

10

TWO-DIMENSIONAL DISTRIBUTIONS

Page 236

1. Two-dimensional distributions........................................ 238 2. Correlation value.................................................................. 242 3. Using the line of best fit to make estimations.......... 244 4. Analyse: does correlation mean cause-effect?........ 246 5. Two-dimensional distributions using a calculator... 247 Exercises and problems solved.......................................... 248 Exercises and problems......................................................... 249 Maths workshop......................................................................... 252 Self-assessment......................................................................... 253

11

COMBINATORICS

Page 254

1. Strategies based on the product................................... 256 2. Variations and permutations (the order matters)... 262 3. When the order does not matter. Combinators....... 264 4. An interesting arithmetical triangle.............................. 266 Exercises and problems solved........................................... 268 Exercises and problems......................................................... 269 Maths workshop......................................................................... 272 Self-assessment......................................................................... 273

12

CALCULATING PROBABILITY

Page 274

1. Random events..................................................................... 276 2. Probability of events........................................................... 278 3. Probability in single experiments.................................. 280 4. Probability in compound experiments........................ 282 5. Creating individual experiments.................................... 283 6. Creating dependent experiments.................................. 284 7. Contingency tales................................................................ 286 Exercises and problems solved........................................... 288 Exercises and problems......................................................... 290 Maths workshop......................................................................... 294 Self-assessment......................................................................... 295

Annex • Glossary............................................................................................ 296

5

Basic Functions Reading and listening

Seeking accuracy During the 19th and 20th centuries there were debates about exactly what was and was not essential in defining a function. Over the course of these discussions, numerous mathematicians, notably Dirichlet, Riemann and Weierstrass, helped to formulate a more precise definition of a function.

Finally, the following definition was offered in 1923, which is very similar to the one we use today.

One says that y is a function of x if each value of x corresponds to a value of y. This correspondence is shown in the equation y = f (x).

Honest functions, according to Poincaré However, in this search for accuracy, a series of bizarre functions were invented that irritated Poincaré, who was unhappy with the direction the definition of functions had taken. This irritation led him to say in 1899: ‘For half a century we have seen a mass of bizarre functions which appear to be forced to resemble as little as possible honest functions which serve some purpose. Formerly, when a new function was invented, it was in view of some practical end. Today, they are invented on purpose to show that our ancestor’s reasoning was at fault.’

In this unit we are going to focus on these honest functions, which can be used for something more than just constructing or disproving concepts.

Henri Poincaré (1854–1912) is one of the greatest mathematicians in history.

His contributions spanned all fields of mathematics.

108

Evidently useful functions When a driver encounters a hazard on the road and has to brake, a certain amount of time passes between their decision to brake and the actual moment they brake (reaction time), about 3/4 of a second. Therefore, the faster the car is going, the further it will travel in that time. Furthermore, the car does not stop automatically when the brake is applied; it travels an additional distance (braking distance) due to inertia.

From the moment the driver becomes aware of the hazard to the moment the car stops, it travels an additional distance (d ) that increases the greater the velocity (v). An experiment gives the following values: speed of the car (km/h)

(based on reaction time in m)

distance

distance

(braking distance in m)

total stopping distance (m)

10

2

1

3

20

4

3

7

30

6.5

7

13.5

40

8.5

12

20.5

50

10.5

19

29.5

60

12.5

27

39.5

70

14.5

36

50.5

80

17

47

64

90

19

60

79

100

21

74

95

The equations are: dreaction = 0.21v

dbraking = 0.0074v 2

dtotal = 0.0074v 2 + 0.21v

1 Write a synonym for each of the following words and use each of them

in a sentence: bizarre, irritate, mass, resemble, span.

2 Use modal verbs to explain why the reaction time may be slower if a

person driving is having a conversation and not fully paying attention.

Solve 1

Use your calculator to check the validity of the formulas above for the values in the table (only check a few values in each row, keeping in mind that they are approximate values) and graph them.

ANK ANK B E G A U LANG LANGUAGE B ANK ANK GE BANK B B E E G G A A U U GUA K LANG ANG N L A L AN GE BANK 109 ANK GE BANK B B E E G G A A U U G A LAN LANG LANGUA LANGU

1

LINEAR FUNCTIONS ãã Linear functions in our daily lives Science is full of functions in which variations in the cause have a proportional impact on variations in the effect. All such functions are called linear functions and are graphed using straight lines. Let’s look at an example:

stretching a spring

If different weights are hung from a spring, they will stretch to different lengths.

In other words, the length of the spring is a function of the weight hanging from it. It is worth pointing out* that this is a linear function.

More specifically, let’s suppose that the spring is 30 cm long before it is stretched and that is stretches 15 cm per kilogram added to it. The relationship is:

100

y = 30 + 15x (y: length in cm; x: weight in kg)

50

Based on the presumption that the spring becomes deformed when weights of over 6 kg are hung from it, the domain of definition of this function is [0, 6].

F ocu s on Eng lish

lenght (cm)

y = 30 + 15x

weight (kg)

0

1

2

3

4

5

6

ãã Functions of proportionality: y = mx Y

point out: to mention some information you think is important.

Functions of proportionality are graphed with straight lines that pass through the origin. They describe a ratio between the values of both variables.

y = mx

X

The slope of the straight line is the constant of proportionality, m.

For example, the space covered at a constant velocity, v, as a function of time is: s = v · t, where v is the slope of the line that relates s to t. anayaeducacion.es GeoGebra. Graphic representation of a linear function.

ãã Constant function: y = n Y n

It is represented with a straight line parallel to the X axis.

y=n

Its slope is 0.

y=0

X

The line y = 0 coincides with the X axis.

For example, the distance of an artificial satellite from Earth is constant. It does not depend on time, t. The equation for this would be d = 36 000, d: distance, in km; t: time, which does not appear in the equation.

200 100 32

F

ãã General expression of linear functions: y = mx + n This is represented on a graph as a straight line with slope m that intersects the Y axis at point (0, n).

The number n is called the ordinate at the origin.

Y F = 32 + 1.8C C 50 100

y = mx + n n X

110

For example, the line F = 32 + 1.8C, represented in the margin, allows us to change from degrees Celsius, C, to degrees Fahrenheit, F.

Unit

5

ãã Point-slope equation of a line We often need to write down the equation of a straight line of which we only know one point and the slope. It is written as follows:

Remember This is a very useful formula. You should learn how to use it!

Point: P(x0, y0)

Slope: m

Equation: y = y0 + m(x – x0)

explanation

• y = y0 + m (x – x0) is a first-degree expression. Therefore, it is a straight line.

• The coefficient of x is m. Therefore, its slope is m.

• If we give x the value x0 8 y = y0 + m (x0 – x0) = y0 + m · 0 = y0. x = x0, then

(x2, y2)

y = y0. In other words, it passes through (x0, y0).

y2 – y1

➜➜ finding straight lines using two points

y2 – y1 m = –––––– x –x 2

In order to work out the equation of a line that passes through two points, we do the following:

1

(x1, y1) x – x 2 1

• We can find the slope using the two points.

• We can find the equation based on the slope and one of the points.

Problem solved

anayaeducacion.es Review the point-slope equation.

Find the equation of each of the following straight lines:

a) Equation: y = 7 – 3 (x + 5). This is the equation of the line.

5

a) I t passes through (–5, 7) and its slope is –3 .

5

We can simplify it: y = 7 – 3 x – 3 · 5 8 y = 4 – 3 x 5 5 5 b) We begin by finding its slope: m = 5 – 7 = –2 = – 1 3 6 4 – (–2)

b) I t passes through (–2, 7) and (4, 5).

Equation of the line that passes through (–2, 7) and has a slope of – 1 : 3 y = 7 – 1 (x + 2) 8 y = 19 – 1 x 3 3 3

Think and practise

anayaeducacion.es GeoGebra. Calculating the slope of a line.

1 Graph the following functions:

a) y = 2x b) y = 2 x c) y = – 1 x d) y = – 7 x 3 3 4

2 Graph the following:

a) y = 3

b) y = –2

c) y = 0

d) y = –5

3 Graph the following functions:

a) y = 2x – 3

b) y = 2 x + 2 3 d) y = –3x – 1

c) y = – 1 x + 5 4 4 At an initial point in time, a moving object is 3 m from its origin and is travelling away from it at a velocity of 2 m/s. Find the equation for its distance from the origin as a function of time and graph it.

5

The price of potatoes at the market is €1/kg, and the price of tomatoes is €2/kg. a) Write the equation for the price of a bag of potatoes as a function of its weight. b) Write the equation for the price of a bag of tomatoes as a function of its weight. c) Graph the functions above.

6 Find the equation of these straight lines using the

given data:

a) It passes through (–3, –5) and its slope is 4 .

9 b) It passes through (0, –3) and its slope is 4. c) It passes through (3, –5) and (–4, 7).

111

1 LINEAR FUNCTIONS

y=x+4

y = –2x + 13

y=1

(3, 7)

ãã Piecewise linear functions It is common to see functions whose graphs are made up of different pieces of straight lines. Look at the function represented in the graph on the margin: • The first piece, x ≤ 3, is a line with a slope of 1 and its ordinate at the origin is

5

4. Its equation is: y = x + 4.

• The second piece, 3 < x < 6, is a line that passes through (3, 7) with a slope

of –2. Therefore, in the point-slope form its equation is y = 7 – 2(x – 3) 8 y = –2x + 13.

5 x≤3

3<x<6

x≥6

• The third piece, x ≥ 6, is the constant function y = 1.

Z if x ≤ 3 ]]x + 4 The function is described as follows: f (x) = [–2x + 13 if 3 < x < 6 ]1 if x ≥ 6 \ anayaeducacion.es Exercises to consolidate functions defined by two or three pieces.

The analytical description of a piecewise graph made up of straight lines gives the equation of the different pieces in order from left to right. It indicates the x values for which the function is defined on each of the pieces.

Problem solved

Graph the function that has the following analytical expression: Z if x < –1 ]4 ]]0.5 (x – 1) + 5 if –1 ≤ x ≤ 3 y =[ if 3 < x < 5 ]–2x + 12 ]–1 if x ≥ 5 \

We find the endpoints of each of the pieces, starting, for example, at x = –5 and ending at x = 11(we could also take –4 and 8): 1st 2nd 3rd 4th (–5, 4) 0.5(–1 – 1) + 5 = 4 8 (–1, 4) –2 · 3 + 12 = 6 8 (3, 6) (5, –1) (–1, 4) 0.5(3 – 1) + 5 = 6 8 (3, 6)

We plot the four pieces (taking into account if the endpoints are included in the mentioned piece or not) to obtain the graph of the function.

(3, 6)

We can see that the function is continuous at x = –1 because at that point the two sections have the same value, 4. This is also true at x = 3, but not at x = 5, where the function is discontinuous.

(–1, 4) (5, 2) (5, –1)

Think and practise 7 Graph the function that has the following analytical

expression: Z if x < 0 ]]–3 y = [x – 3 if 0 ≤ x ≤ 5 ]2 if x > 5 \ Find the slope of each of the pieces that form the function. Is it a continuous function?

112

–2 · 5 + 12 = 2 8 (5, 2) (11, –1)

anayaeducacion.es GeoGebra. Graphing piecewise linear functions.

8 Write the equation that corresponds to the following

graph:

5

5

10

Unit

2

5

THE PARABOLA: A VERY INTERESTING CURVE The tennis player on the left takes advantage of the fact that his opponent is close to the net to beat him with a lob. The curve described by the ball is a parabola.

We encounter all sorts of parabolas in everyday life: jets of water, the flights of balls and other objects, sections of parabolic antennas and car headlights, etc. All of them are described using quadratic functions. ➜➜ quadratic functions

• The function y = x 2 gives us a typical parabola, which serves as a point of reference for the others.

• The equation of the parabola described by the tennis player’s ball in the image 2 above is y = 5 – x , where x is its horizontal path, in metres, and y is its 20 height, also in metres. The origin of coordinates is at the player standing at the net.

• The area of a square as a function of the length of a side (A = s 2), or of a circle as a function of its radius (A = πr 2).

• The height of a stone thrown up in the air, as a function of time (height = 20t – 5t 2). y = x2

x2 y = –– 4

➜➜ are all parabolas the same?

Yes, or to be more precise, they are all similar. Some parabolas may seem much wider than others, but that is because we are only looking at the ‘end’, the part near to the vertex. If we extended them, they would be the same shape as other seemingly narrower parabolas. ➜➜ do the parabolas of these functions fit their definitions as loci?

anayaeducacion.es • All parabolas are similar.

• Parabolas can be expressed using quadratic equations. They have a focus and a directrix.

Last year you learnt that a parabola is defined by a point, F (focus) and a line, d (directrix). The points P of the parabola are equidistant from F and d: 2 PF = dist(P, d). So, do the parabolas y = x 2 or y = x fit this definition? The 4 answer is, of course, yes. For y = x 2, the focus is F(0, 1/4), and the directrix 2 is d: y = –1 . For y = x , the focus is F(0, 1), and the directrix is d: y = –1.

4 4 You can check it yourself. ➜➜ what is special about parabolic antennas and car headlights?

A ray parallel to the axis of the parabola will be reflected by it so that it passes through the focus, which concentrates all of the rays at this one point.

Similarly, rays emitted from the focus are reflected in the parabola and are projected outwards as a beam parallel to the axis. anayaeducacion.es Property of parabolas which justify parabolic antennas, solar furnaces, car headlights, etc.

113

3

QUADRATIC FUNCTIONS. PARABOLAS ãã Quadratic functions Last year you learnt some things about parabolas. Let’s review what you know and learn a bit more.

Look at the following parabolas and their respective equations:

y y==x x2y 2= x 2

1— 2+ 21 x2x –+42x4 + 4 y1x=x2 — ––2x y y==— 22 2

1— 21 x 2 y1x=x2 — y y==— 22 2

y 2=2 3x 2 y y==3x3x 2 –+18x y 2=2–3x y y==3x3x –18x 18x +2424+ 24 2+ x6x –+66x6 + 6 y y==x x2y 2–=–6x

2– +–66x6 – 6 y 2=2+–x +6x6x y y==–x–x

y 2=2 –x 2 y y==–x–x

2 +–18x 2++18x y =2 –3x 18x –2424– 24 y y==–3x –3x

2 2 y =2 –3x y y==–3x –3x

12 x 2 1=— 1x–x2 — y y==––y— 22 2

12++2x 1=— 1x–x2 — 2– x2x +–42x4 – 4 y y==––y— 22 2

By analysing them, we can reach the following conclusions: The functions y = ax 2 + bx + c, with a ≠ 0 are called quadratic functions. They are all graphed using parabolas and are continuous in all Á. anayaeducacion.es Further theory and practice on the translation of parabolas.

eje axis

Each of these parabolas has an axis parallel to the Y axis.

Their shape depends on a, the coefficient of x 2, in the following way: • If two quadratic functions have the same coefficient of

x 2, their corresponding parabolas are identical, although they can be located in different positions.

• If a > 0, their branches point upwards, and if a < 0, they

point downwards.

• The larger the value of |a |, the narrower the parabola.

114

vértice vertex

Unit

ãã Graphing quadratic functions

Tables of values on a calculator

In order to graph a quadratic function given as an equation, we only need to find a few of its points. We begin by calculating the vertex of the parabola in order to then find some of the points around it. These are the steps you should follow: • Find the abscissa of the vertex of the parabola y = ax 2 + bx + c 8 p = – b 2a • Then, calculate the value of the function at a few values close to the vertex.

We can use a calculator to prepare a table of values quickly and efficiently corresponding to any function, for the interval we want and with the desired increments. Look at how we do this for the function y = –x 2 + 3x + 4.

• The points of intersection with the axes can be helpful for graphing it:

We select 9:Table from �. f (x) appears and we enter the expression.

Remember that to write x we have to press

-

x )

x )

.

x+3

x )

— With the X axis, we can solve the equation: ax 2 + bx + c = 0.

— With the Y axis, the (0, c ).

Problem solved

+4=

Graph the parabola for the equation y = –x 2 + 3x + 4.

Then, this screen appears, where we have to indicate the first and last values of x and the size of the steps:

First, we find the vertex: Abscissa: p = – 3 = 1.5 8 Ordinate: f (1.5) = 6.25 8 Vertex: (1.5; 6.25) –2 Then, we find points close to the vertex:

In our case, since the vertex is at x = 1.5, we give the values from –2 to 5 and choose an increment of 1 step.

x

–2

–1

0

1

2

3

4

5

y

– 6

0

4

6

6

4

0

– 6

Note how the points at the same distance from the vertex coincide in the value of their ordinate.

This is because the parabola is symmetric with regard to its axis. In other words, as the vertex is at x = 1.5, then f (1) = f (2); f (0) = f (3)…

To see the values that are not shown on the screen, you have to scroll down with the cursor.

We see that – x 2 + 3x + 4 = 0 has two solutions, x = –1 and x = 4; and that f (0) = 4. Nevertheless, these points of intersection with the axes already appear in the table.

Think and practise 1

5

anayaeducacion.es GeoGebra. Graphing quadratic functions.

Match each of the coefficients of x 2 to its corresponding parabola: A

• a = –1

C

D

• a = 2 • a = – 1 3 1 • a = 2

2 Graph the following parabolas:

a) y = x 2 – 2x + 2

b) y = –2x 2 – 2x – 3

c) y = 1 x 2 + x – 2 3 e) y = – 1 x 2 + 2 2

d) y = –x 2 + 4 f ) y = 3x 2 + 6x + 4

3 In your notebook, graph the following quadratic

functions:

E

• a = –3 B

a) y = (x – 1) · (x – 3)

b) y = 2(x – 2)2

c) y = 1 (x + 2) · (x – 2) 2

d) y = (x – 1)2 + 5

115

3 QUADRATIC FUNCTIONS. PARABOLAS

ãã Lines and parabolas To solve some problems we have to perform a combined study of a linear function and a quadratic function, or of two quadratic functions. To do this, we sometimes have to solve a second-degree system, whose solutions (two, one or none) will be the points at which the two graphs intersect.

In some piecewise functions you will also find straight lines and parabolas together.

Problems solved 1 Solve

the following system analytically and graphically:

*

y = –2x 2 + 5x – 2 y=x – 2

Analytically: y = –2x 2 + 5x – 2 4 –2x 2 + 5x – 2 = x – 2 8 –2x 2 + 4x = 0 8 y=x –2 x = 0 8 y = –2 8 x (–2x + 4) = 0 x =2 8 y =0

There are two solutions. When graphed, we see that the parabola and the straight line intersect at points (0, –2) and (2, 0). y=x –2

Graphically: We graph the parabola y = –2x 2 + 5x – 2 and the straight line y = x – 2.

(2, 0)

(0, –2)

The points of intersection of the two curves are the solutions of the system: (0, –2) and (2, 0).

2 Graph this function:

y = –2x 2 + 5x – 2

• The first section of the function corresponds to a piece of parabola and is only

Z 2 ]]–x + 1 if x ≤ 1/2 [3/4 if 1/2 < x < 2 ]x – 5/4 if x ≥ 2 \

defined for x ≤ 1/2. The vertex is at point (0, 1), it intersects the X axis at (–1, 0) [we do not consider the point (1, 0), as it has to be x ≤ 1/2], and passes through points (1/2, 3/4) and (–2, –3).

2

• The second section is a piece of horizontal line that

1

begins at (1/2, 3/4) and ends at (2, 3/4).

• The last section is a piece of straight line that starts at

point (2, 3/4) and passes through (3, 7/4).

and graphically: a) *

x2

y = – 6x + 5

y=x –5

b) *

x2

y = – 2x y = –x + 2

Z 2 ]] y = x – 4x + 2 y = 2x + 2 2 c) * d) [ y = 5 – x2 ] y = –x + x + 6 2 \ 116

5 Graph the following functions:

Z 2 ]x + 2 ] a) y = [3 ]] –3x + 15 2 \

–1 –2

Think and practise 4 Solve the following systems of equations analytically

–1

if x ≤ 1 if 1 < x < 3 if x ≥ 3

4–x if x ≤ 1 b) y = )–x 2 + 4x if x > 1

1

2

3

Unit

4

5

ABSOLUTE VALUE FUNCTIONS Remember that the absolute value of a number a coincides with a if it is positive or zero, or with its opposite, –a, if it is negative: |a | = )

y = |x|

a if a ≥ 0 –a if a < 0

As a consequence, y = |x | is a piecewise function like the one below: |x | = )

–x if x < 0 x if x ≥ 0

This is called an absolute value function and it is graphed as shown on the left margin.

ãã The absolute value of a function: y = |f (x)| Look at how we find the absolute value of some functions:

y = |2x – 4|

intersects at x = 2

y = |2x – 4|

– (2x – 4) if x < 2 y = |2x – 4| = ) 2x – 4 if x ≥ 2 –2x + 4 if x < 2 In other words: y = ) 2x – 4 if x ≥ 2

y = 2x – 4

intersects at x = –1 x=4

if x ≤ –1 or x ≥ 4 x 2 – 3x – 4 y = |x 2 – 3x – 4| = * 2 – (x – 3x – 4) if –1 < x < 4

y = |x2 – 3x – 4|

y = |x 2 – 3x – 4|

y = x 2 – 3x – 4

Z 2 ]]x – 3x – 4 if x ≤ –1 In other words: y = [–x 2 + 3x + 4 if –1 < x < 4 ]x 2 – 3x – 4 if x ≥ 4 \

Graphing y = |f (x)| from y = f (x) is very simple: we just need to ‘make a negative positive’. In other words, the part of the graph that is under the X axis is replaced by its symmetrical version with respect to this axis.

The analytical expression is achieved by changing the sign of the function in the sections where f (x) takes negative values. To do this, we need to know the points where y = f (x) intersects with the X axis.

Think and practise 1 In your notebook,

show the absolute value of these functions:

2 Graph these functions and rewrite them in their A

B

analytical expression without the absolute value: a) y = |3 – 2x | b) y = 1 x + 1 2 c) y = |x 2 – 4x |

d) y = |–x 2 + 6x – 5|

117

5

RADICAL FUNCTIONS Let’s look at some examples of radical functions: • The length of the side of a square as a function of its area is: l= A • The radius of a circle as a function of its surface area is: S π • The period of a simple pendulum as a function of its length is: r=

T = 2 l ; l in m, T in s • The speed at which a ball that we drop from a height h reaches the ground is: v = 20 h ; v in m/s, h in m Now we are going to study them theoretically.

ãã Different types of radical functions Remember The function y = x is defined and is continuous in [0, +∞). It is also increasing, although it increases more and more slowly.

On the contrary, y = – x is defined and is continuous in [0, +∞) and decreasing.

F ocu s on Eng lish plot: to mark the position of a point using coordinates.

The functions y = x and y = – x can be plotted* point by point and result in the graphs below. They are halves of parabolas and together they describe a parabola that is identical to y = x2, but with the X axis as its axis of symmetry.

Y

Y y = √x X

X y = –√x

The domain of definition of these functions is [0, +∞).

The following curves belong to the same family: Y

— y = 2√x – 4

— y = 2√–x + 1

Y

X — y = –2√x – 5

X — y = –2√–x – 3

The functions y = a x + b and y = a –x + b are graphed using half parabolas.

Their domains of definition are, respectively: [–b, +@) and (– @, b] 118

Unit

5

Problems solved 1 Graph this function and

indicate its definition:

domain

of

y = –2 + x – 1

• The x in the radicand means that the curve goes to the right.

• The first value we give x is 1, because it cancels out the radicand.

• We give x the values 1, 2, 5, 10 and 17, which make the root exact. x

1

2

5

10

17

x–1

0

1

4

9

16

0 1 √x – 1 y = –2 + √x – 1 –2 –1

2

3

4

0

1

2

Y 5

10

15

X

Domain of definition: [1, +∞) 2 Graph this function:

y = –3 –x + 2

• The –x tells us that the curve goes to the –14 left.

–7

• The first value we give x is 2, because it cancels out the radicand.

12 (1, –3)

(–2, –6)

• We give x the values 2, 1, –2, –7 and –14, with which the root is exact.

• The coefficient –3 produces negative ordinates. Therefore, the curve is below the X axis.

–2

(–7, –9) (–14, –12)

Domain of definition: (–@, 2] 3 Graph

the following function and indicate its domain of definition: y= 3x

x 3

y = √x

–8

–1

0

1

8

27

–2

–1

0

1

2

3

Y

8

27 X

Domain of definition: (–∞, +∞) Think and practise

anayaeducacion.es GeoGebra. Graphing radical functions.

1 Graph the following functions and find the domain of definition of each one:

a) y = 2 x b) y = –2 x c) y = 2 x + 3 d) y = –2 x + 3 e) y = 4 –2 x + 3 f ) y = 2 –x g) y = –2 –x h) y = 2 –x + 3 i) y = –2 –x + 5 j) y = –3 –2 –x + 5

119

6

INVERSELY PROPORTIONAL FUNCTIONS • We have a rectangle with an area of 100 cm2 and sides of unknown length. We call the sides x and y. We know that xy = 100. We express it like this: y = 100 (For a given area, the sides are inversely proportional). x • There are 6 L of air in a tank at atmospheric pressure. If we compress the air (increase the pressure), the volume decreases. It satisfies the relationship P · V = 6. We express it like this: 6 V = P (The pressure and volume of a given mass of gas are inversely proportional).

30 25 20 15 10 5 5 10 15 20 25 30 V (litres) 6

Inversely proportional relationships are very common in nature, physics, economics, etc. Now we are going to analyse them theoretically.

1 1

P (atm)

6

ãã Studying the function y = k/x This is the graph for the function y = 1 : x

x

y

x

y

1 2 3 4 5 …

1 0.5 !

0.3 0.25 0.20 …

1 0.5 !

0.3 0.25 0.20 …

1 2 3 4 5 …

1 y=— x

• It is not defined for x = 0.

• If x gets close to 0, y takes on

ever-larger values. This is why we say that the Y axis is an asymptote.

1 1

• If x takes ever-increasing values, y

gets closer to 0. This is why the X axis is another asymptote.

When the sign of x changes, the sign of y changes as well.

• This curve is called a hyperbola. Its

asymptotes are the coordinate axes.

The following curves are from the same family: 6 y = –– x

1 y = – –– x

4 y = – –– x

The functions y = k are called functions of inverse proportionality. x They are represented with hyperbolas whose asymptotes are the coordinate axes.

Their domain of definition is formed by the sections (–@, 0) and (0, +@).

It is expressed like this: Dom = (–@, 0) « (0, +@) 120

Unit

5

ãã Functions related to y = k/x

2 , which related 2–d the magnification, M, produced by a magnifying glass to the distance, d, of the object being magnified. The graph of this function is also a hyperbola. Its asymptotes are the axis of abscissas and the line d = 2.

• In the previous unit we looked at the function M =

• The functions k y= x –a, y= k , y= k +b a–x x–a are generally hyperbolas related to functions of inverse proportionality. Let’s look at some examples.

anayaeducacion.es Extension: translation of hyperbolas.

Problems solved 1 Graph

the following function and indicate its domain of definition: y=

6 x–4

The graph of this function will be like the one of the function y = 6/x, shifted 4 units to the right. Let’s check this using a table of values: x

3 … 5

6

7

8 10 …

x – 4 –6 –4 –3 –2 –1 … 1

2

3

4

3

2 1.5 1 …

y

–2

0

1

2

–1 –1.5 –2 –3 –6 … 6

6 …

Its asymptotes are x = 4, y = 0.

Its domain of definition is (–∞, 4) « (4, +∞).

2 Graph this function and

indicate its definition: y=

domain

of

6 +2 x–4

The graph will be like the previous one, but shifted 2 units upwards. In other words, it can be obtained from the previous graph by shifting the X axis down two units.

Its domain of definition is (–∞, 4) « (4, +∞).

Think and practise

anayaeducacion.es GeoGebra. Graph functions of inverse proportionality.

1 Match each of the following graphs with its corresponding function. Indicate the domain of

each one: a) y = 2 b) y = – x c) y = 2 d) y = x–3 x

2 x 2 + 5 –3

A

B

2 Graph each function and indicate its domain:

a) y = 8 b) y = – 8 c) y = 8 x x x–2

d) y =

C

8 e) y = 8 – 3 2–x x

D

f) y =

8 +3 x–2

121

7

EXPONENTIAL FUNCTIONS ãã Increasing exponential functions: y = a x, a > 1

15

The graph on the right shows the exponential function of base 2: y = 2x.

y = 2x

x ≥ 0: 10

0 1 2 3 4 …

–1 x x 2x

5

–4

x < 0: 1

2

4

8

16 …

2x

2–1

–2

= 0.5

2–2

= 0.25

–3 2–3

= 0.125

When x takes increasingly larger values, 2x tends to infinity. However, when x takes the values –4, –5, –10, …, 2x becomes really small, it tends to 0.

0

4

Exponential functions are those that take the form y = ax.

• They are continuous, their domain is all Á and they pass through (0, 1) and

(1, a).

• If the base is greater than 1 (a > 1), then they are increasing.

• The greater a is, the more quickly they increase.

15

( )

1 y = –– 2

x

10

ãã Decreasing exponential functions: 0 < a < 1 x

y = c 1 m is also exponential. Since its base (1/2) is less than 1, the function is 2 decreasing. Its graph is symmetrical to the graph y = 2x with respect to the Y axis.

5

–4

The functions of the type y = a kx = (a k)x are exponential functions with base a k.

0

4

Functions y = ax where 0 < a < 1 also pass through (0, 1) and (1, a). They are continuous and their domain is all Á, but they are decreasing. They decrease more quickly the closer a is to 0.

The functions y = ax and y = (1/a)x are symmetric with respect to the Y axis.

Problem solved

Graph this function: y = 10 0.2x Then, plot the following function on the same axes: y=c

x

1 m 1.6

100.2x = (100.2)x = 1.6x It passes through points (0, 1) and (1; 1.6).

Since 1.6 > 1, the function is increasing.

We find some more points around the Y axis. x 1 m is symmetrical to The function y = c 1.6 the first one with respect to the Y axis.

Think and practise 1 Graph the following functions in your notebook.

What is the domain of definition of each of them? a) y = 1.25x b) y = 0.8x

122

Y

( )

1 y= — 1.6

x

y = 1.6x

X

anayaeducacion.es GeoGebra. Graphing exponential functions.

2 Write these expressions in exponential form:

a) 20.4x

b) 100.01x

x/28

c1m c) 1.0112x d) 2

Unit

5

ãã Applications of exponential functions Exponential growth is very common in nature (microorganism cultures, animal and plant populations, etc). They are also used to describe economic phenomena, among other things.

Let’s look at some examples. anayaeducacion.es Semi-logarithmic graph paper is very useful for graphing and comparing functions involving rapid growth, such as exponential functions.

Interesting fact In the Maths Workshop (p. 132) we study the application of the exponential function to radioactive decay and carbon-14 dating.

➜➜ example 1. growth of a population

Amoebas, as you know, are unicellular beings that reproduce by splitting in two (binary fission). This can happen at different rates.

Suppose that the conditions of a culture mean that the number of amoebas approximately doubles every hour, and that there is one amoeba in the beginning.

The number of amoebas that there will be after t hours is N = 2t. Its graph is like the first graph on the previous page, but it is only valid for t ≥ 0. ➜➜ example 2. capital growth

There are €20 000 in a bank with a monthly interest rate of 0.5 %, which means that each month the amount grows by 0.5 %. Therefore, the money in the account at the start of each month is multiplied by 1.005.

The expression that gives the money in the account after T months is: C = 20 000 · 1.005T, T ≥ 0

Problems solved 1 We deposit €60 000 in an

account with 5 % annual interest. How much money will we have in 8 years and 3 months?

An annual increase of 5 % means multiplying by 1.05 each year. The money will grow according to the equation: C = 60 000 · 1.05T The money in the account after 8 years and 3 months (8 and a quarter years) is the value of C for T = 8.25: C (8.25) = 60 000 · 1.058.25 = €89 735.23

2 The value of a machine

decreases at a rate of 8 % per year. If it is currently worth €120 000, how much will it be worth in 11 years?

An annual devaluation of 8 % means multiplying by 0.92 at the end of each year (1 – 0.08 = 0.92). The value of the machine changes as follows: V = 120 000 · 0.92T For T = 11, we get V (11) = 120 000 · 0.9211 = €47 956.49.

Think and practise 3 Write the equation expressing the approximate num-

ber of amoebas that there would be after t hours in a culture like the one in example 1, if there were 200 amoebas at the beginning.

How many amoebas would there be after 8 hours?

anayaeducacion.es More on the applications of exponential functions.

4 €130 000 is deposited in a bank account with an

annual interest rate of 12 %. Express the value of the capital, C, as a function of the time, T, expressed in years, that it is in the bank. How much money will there be after 6 years and 9 months?

123

8

LOGARITHMIC FUNCTIONS The graph of the function y = 2x is plotted on these coordinate axes in blue, and its symmetrical function with respect to the line y = x, in red.

20

Logarithms Remember that log a P is the exponent to which the base, a, must be raised to get P. In other words: log a P = x ï a x = P For example: • 23 = 8 8 log2 8 = 3

y=x

16

I

12 8

II

y = log2 x

4 –4

These two curves are related analytically in the following way:

y = 2x

4

8

12

16

(1, 0), (2, 1), (4, 2), (8, 3), (16, 4)… are points of II. In general, if point (a, b) belongs to I, then (b, a) belongs to II. This is why we say the function described by II is the inverse of I.

20

–4

• 2–1 = 1 8 log2 1 = –1 2 2

(0, 1), (1, 2), (2, 4), (3, 8), (4, 16)… are points of I.

The function described by the red graph is called logarithmic function of base 2, and is written: y = log2 x • The logarithmic function y = loga x with a > 1

is the inverse of the exponential function y = a x.

1 1

a

• It is defined by values greater than 0. In other

words, its domain of definition is (0, +@).

• It passes through points (1, 0) and (a, 1).

• It is increasing, but as x increases, the velocity at which it does so drops to very small values.

• It has a branch tending to infinity on the Y axis.

Problem solved

Graph the following function: y = log1.5 x

We know that it passes through (1, 0) and (1.5; 1).

Y y = 1.5x

In order to graph it, we use the inverse, the exponential function y = 1.5x to help us: y = 1.5x

y = log1.5 x

(2; 2.25)

(2.25; 2)

(3; 3.38)

(3.38; 3)

(4; 5.06)

(5.06; 4)

(5; 7.59)

(7.59; 5)

(6; 11.39)

(11.39; 6)

Think and practise

y = log1.5 x 1 1

X

anayaeducacion.es GeoGebra. Graphing logarithmic functions.

1 In your notebook, plot each pair of functions on the same coordinate axes:

a) y = ex; y = ln x b) y = 3x; y = log3 x c) y = 10x; y = log x d) y = 2.2x; y = log2.2 x Find the domain of definition of each of them.

124

e) y = 4x; y = log4 x

Unit 5

BLEMS O R P D N A S E IS C R E EX

SOLVED

1 Piecewise continuous function

2x – 5 if x < 2 The function looks like this: y = ) –2x + 3 if x ≥ 2

–5 2x

2 · 2 – 5 = 2m + 3 8 m = –2

X

+3

2

–2x

Graph it for the value of m you get.

Y

y=

2x – 5 if x < 2 y=* mx + 3 if x ≥ 2

The function is represented with two lines. In order for it to be continuous, we just need these to coincide at x = 2, where it changes from the first to the second. To do this, 2x – 5 has to coincide with mx + 3 at x = 2:

y=

Calculate the value of m so that this function is continuous:

Your turn Find the value of k so that this function is continuous: 4x + 2 if x < –1 y=) Graph it for the value of k you get. kx – 3 if x ≥ –1 2 Finding the equation of a parabola from its vertex and one of its points

Find the equation of the parabola whose vertex is (3, –1) if it passes through the point (2, –2).

Graph it on coordinate axes.

The equation for a parabola has the form y = ax 2 + bx + c.

The abscissa of the vertex is x = 3.

Therefore: –b/2a = 3 8 b = –6a (*) The parabola passes through (3, –1) and (2, –2): (*)

–1 = 9a + 3b + c 8 –9a + c = –1 (*)

–2 = 4a + 2b + c 8 –8a + c = –2 9a – c = 1 ) 8 a = –1; b = 6; c = –10 –8a + c = –2 The equation is y = –x2 + 6x – 10.

Your turn Find the equation of the parabola whose vertex is at the point (–2, –9) and that passes through (0, 1) 3 Straight lines and parabolas

Show this system analytically and graphically:

*

y = x2 – 2 y=x

Describe and graph the continuous function that is obtained from both equations and that takes the form lineparabola-line.

We find the points where the parabola and the straight line intersect. In order to do this, we isolate y in each equation and equalise: y = x2 – 2 4 8 x 2 – 2 = x 8 x 2 – x – 2 = 0 8 x = –1, x = 2 y=x They intersect at (–1, –1) and (2, 2).

In the statement, it says that the first is a piece of straight line, then there is a piece of parabola, and then there is another piece of straight line. This is the piecewise-defined function: Z if x < –1 ]]x 2 y = [x – 2 if –1 ≤ x ≤ 2 ]x if x > 2 \ It is represented in the graph on the right. y = 4 – x2 Your turn Now do the same with this system: * y = –x – 2

Y

X

125

BLEMS O R P D N A S E IS EXERC

SOLVED

4 The absolute value of a function

Express the following function without using the absolute value sign and graph it:

In order to express it as a piecewise-defined function, we first need to see where the function within the absolute value intersects with the X axis. In this way, we can work out in which intervals it is positive and in which it is negative.

y = 1 x2 – 5 x + 9 2 4 4

1 x 2 – 5 x + 9 = 0 8 x = 1, x = 9 2 4 4 As the branches of the parabola point upwards, since the coefficient of x 2 is positive, the function is only negative between the points of intersection with the X axis.

Therefore, we have three sections: (– ∞, 1] and [9, +∞) where the function is positive or zero, so there is no need to change the sign, and the interval (1, 9), in which the function is negative. In this Y section, we change the sign of the function.

Let’s see what it looks like:

Your turn Express as a piecewisedefined function and graph: y = |x 2 – 2x – 8|

Z ] 1 x2 – 5 x + 9 2 4 ]4 ] 1 2 5 y = [– x + x – 9 2 4 ] 4 5 9 1 2 ]] x – x + 2 4 \4

if x ≤ 1 if 1 < x < 9

1

9

X

if x ≥ 9

5 Functions with absolute values

Express each of the following functions as a piecewise-defined function: a) y = |x 2 – 1| + 1 b) y = |2x – 4| + |x|

Your turn Express each function as a piecewise-defined function: a) y = |x 2 – 4| – 2 b) y = |x + 3| + |x|

a) T he parabola of y = x 2 – 1 intersects with the X axis at x = –1 and x = 1.

Since its branches point upwards, the piecewise-defined function is: Z 2 Z 2 if x < –1 ]]x – 1 + 1 if x < –1 ]]x 2 2 y = [–x + 1 + 1 if –1 ≤ x ≤ 1 8 y = [–x + 2 if –1 ≤ x ≤ 1 ]x 2 – 1 + 1 if x > 1 ]x 2 if x > 1 \ \ b) W e make a table to study how the expression varies within each absolute value depending on the section. The endpoints of each section are those that make each of the expressions equal to 0: x<0

0≤x≤2

x>2

|2x – 4|

–2x + 4

–2x + 4

2x – 4

|x|

–x

x

x

|2x – 4| + |x|

–3x + 4

–x + 4

3x – 4

Therefore, the function is: Z ]]–3x + 4 if x < 0 y = [–x + 4 if 0 ≤ x ≤ 2 ]3x – 4 if x > 2 \

6 Functions related to the function of inverse proportionality

Express the following function k in the form y = + b. x–a Then, graph it. y = 2x + 3 x +1 Your turn Graph this function: y = 3x – 1 x+2

126

We start by dividing: 2x + 3 x + 1 This function can be expressed as: – 2x – 2 2 2x + 3 = 2 + 1 1 x +1 x +1 It is the function y = 1 shifted 1 unit to the left and x 2 units up.

m this resources fro r to choose Remembe portfolio.

unit for your

Unit 5

ROBLEMS

P EXERCISES AND

Practise

Quadratic functions

7

Linear functions

1

2

Graph the following linear functions: a) y = 2x – 3 b) y = 4 x 7 – 3 x + 10 c) y = d) y = 2.5 5

Match each expression to its graph: a) y = x 2 b) y = (x – 3)2 c) y = x 2 – 3

b) P (2, –1), m = –2

2

C

4

5

Find the equation of the straight line that passes through points A and B in each case: a) A (3, 0), B (5, 0)

b) A (–2, – 4), B (2, –3)

c) A (0, –3), B (3, 0)

d) A (0, –5), B (–3, 1)

Find the equation in each case and graph it:

b) Proportionality function that passes through (–4, 2). c) Constant function that passes through (18; –1.5).

Find the value of the unknown parameters so that the straight lines and points comply with these conditions. Graph them.

c) The lines of the equations y = 3x + c and y = cx + 3 intersect at ordinate 2. Which is the corresponding abscissa? d) The points (d, –2) and (4, e) belong to the line of the equation y = 1 x – 3 .

2

4

6

X

Graph the following functions by making a table of values like the one below for each of them: x

– 4

–3

–2

–1

0

1

2

3

4

y

…

…

…

…

…

…

…

…

…

a) y = x 2 + 1

b) y = –x 2 + 4

c) y = –3x 2 d) y = 0.4x 2 9

Graph the following parabolas by finding their vertex, some points close to it and the points where they intersect with the axes:

a) y = (x + 2)2 b) y = x 2 – 4x c) y = 1 x 2 + 2x + 1 d) y = x 2 – 9 2 10 Give the point (abscissa and ordinate) where the vertex of the following parabolas is found. In each case, state whether it is a maximum or a minimum.

Then, graph them. a) y = 8 – x 2 b) y = 4 + (3 – x)2 c) y = –x 2 – 2x + 4 d) y = 3x – 1 x 2 + 1 2 e) y = 15 – 1 x 2 + 1 x f ) y = 1 x 2 + 2x + 3 2 3 4 4

a) The line that passes through points (4, 0) and (–2, a) has a slope of –1. b) The line y = bx + 2 passes through point (–3, 4).

2 –2

D

a) Line that passes through (2, –3) and is parallel to the one passing through (1, –2) and (– 4, 3).

6

–2

8

B

D

4

Calculate the equation of these linear functions: A

C

6

A

c) A (–2, 1), m = 1 d) A (1, 3), m = – 5 3 2 3

Y

B

Using the given slope and point, calculate the equation of each line: a) P (0, 0), m = 1

d) y = x 2 – 6x + 6

11

Calculate the vertex, the axis of symmetry and the points of intersection with the axes (if any) of each of these parabolas: a) y = 2x2 b) y = 2(x – 5)2 c) y = 2(x – 5)2 + 2

d) y = –x2 + 1

e) y = –(x + 1)2 + 1

f ) y = –3x + 2x2 127

EXERCISES AND PROBLEMS Piecewise-defined functions

Absolute value of a function

12

16

Graph these functions. Which are continuous?

Z ]] 2x if x ≤ –1 –3 if x < 0 a) y = [ –2 if –1 < x ≤ 3 b) y = ) 2x + 1 if x ≥ 0 ] x – 5 if x > 3 \ Z ]] –x + 3 if x < 1 if 1 ≤ x < 2 c) y = [ 2 ]x if x ≥ 2 \ Z if x < –2 ]] 0 d) y = [ x + 2 if –2 ≤ x ≤ 0 ] 3x – 2 if x > 0 \ 13 Graph the following functions: Z ]] –1 – x if x < –1 a) y = [ 1 – x 2 if –1 ≤ x ≤ 1 ] x – 1 if x > 1 \ x 2 if x < 0 b) y = * 2

–x if x ≥ 0

a) y = |2x – 2|

15

Write the equation of the piecewise function that corresponds to this graph.

A

17

a) y = |x 2 – 4|

b) y = |–x 2 – 10x – 22.75|

c) y = |x 2 – 6x + 5|

d) y = |–x 2 – 4x – 3|

A

C

D

Y 4 2 –6 –4 –2

2

4

6

8 X

Y

B

18

19 X

C

X

D

Y

X

Express each function as a piecewise-defined function. Remember that to define the intervals you need the points of intersection with the X axis. a) y = 4 – 1 x b) y = |2x + 2| 3

Y

X

128

Match each function to its graph:

B

Write the analytical expressions of each function:

Y

D

C

B

We know that the equations of the parabolas that appear in the graphs are: y = x 2; y = –x 2 – 4x; y = 4 – x 2; y = x 2 – 6x + 5.

A

b) y = |4 – x |

c) y = 1 x + 2 d) y = 1– 1 x 2 2

x 2 + 4x if x < 0 c) y = * 2 –x + 4x if x ≥ 0 14

Match each function to its corresponding graph:

c) y = |x 2 – 2x – 3|

d) y = |–x 2 + 2x – 1|

e) y = 1 x 2 – 4x + 7 2 2

f ) y = |9 – x 2|

Write the piecewise-defined function for each of the functions in the graph.

Express them as absolute value functions.

a

b

Unit 5

Other functions

20

Match these functions to their corresponding graph and give the domain of definition of each one: I) y = 1 II) y=3– 1 x–3 2–x III) y = 2 + 2 IV) y = – 1 x x +3 Y Y a) b) 4

4

2

2

–4 –2

2

c)

4 X

d)

Y 2

6 X

4

Y 2

X

X

a) y = – 1 b) y= 2 x x c) y = 1 – 3 d) y = 3 + 2 x x 24 Give the domain of definition and asymptotes of the following functions. Graph them. a) y = 1 b) y = – 3 x +1 x +3 c) y = 1 + 2 d) y = 1 + 2 x –1 1– x 25

–2

–4

–4

a) y = x + 2

2

4

b)

Y 2 –4 –2 2 Y

c)

4

2 X 6 X

e) y = –2 – x f ) y = –2 – 2 –x g) y = 2 + –x h) y = 2 –x + 2 26

Y 2 2

4

d)

c) y = –x – 1 d) y=2+ x +3

X

6 Y

e) y = 1 + x – 1 f ) y = 2(x – 1)

2

g) y = – – (x + 2) h) y = 1 + 1 – (x + 1) X

–6 –4 –2

27

Match these functions to the graphs: I) y = 3x II) y = 1.5x IV) y = 0.7x III) y = 0.4x

4

2

2

–4 –2

2

4

8

–4 –2 d

4

4

2

2 4

28

–4 –2

d) y = 0.75–x

Plot each pair of functions on the same coordinate axes. What is the relationship between them? x

a) y = c 1 m ; y = 3x b) y = 0.25x; y = 4x 3

4

8 6

2

x

2

6

Graph the following functions by making a table of values for each of them: a) y = 2–x b) y = 3x + 1 c) y = c 2 m + 3 3

6

4

–4 –2 c

b

6

Work out the domain of definition of the following functions and graph them: a) y = 2 – x b) y = 7 – 2x + 4

–2

a

b) y = 2 – x

c) y = 2 –x d) y = – –x

Match the functions to their corresponding graph and give their domains of definition: I) y = x – 3 II) y= x –3 a)

Graph the following functions:

–2

–2

III) y = 3 – –x IV) y = –3x

22

23

Graph the following functions and find the domain of definition of each one:

–4 –2

21

2

When completing the exercises below, we recommend using a calculator to create tables of values.

29

2

Say whether they are increasing or decreasing.

4

a) Graph the following functions: y = 3x and y = log3 x b) Check whether the following points belong to the graph of y = log3 x : (243, 5)

c 1 , –3m 27

` 3; 0.5j

(–3, –1)

129

EXERCISES AND PROBLEMS Problem solving 30

The boiling point of water is 100 ºC at 0 m above sea level. For every 300 m increase in altitude, the boiling point decreases by 1 ºC. In other words, at 3 000 m above sea level the boiling point of water is 90 ºC. a) Write the function that relates the boiling point of water, T, to the altitude, a. b) Graph the function. c) What temperature does water boil at where you live? What about at the top of Mount Everest?

41

An arrow is shot upwards at a velocity of 40 m/s. Its height, h, at each moment in time, t, is h = 40t – 5t 2. a) Graph the function. b) What it is its domain of definition? c) At what moment in time does it reach its maximum height? What is that height? d) At what moment in time does the arrow hit the ground? e) In what interval of time is the arrow above a height of 35 metres?

42

Andrea spent €100 on a birthday present for Carlos. The rest of the friends in the group decide to share the cost of the present.

Write a function that shows the amount of money each friend has to contribute depending on the number of friends in the group and graph it.

Does it make sense to join the points? Why?

43

Ana’s basic annual salary is €24 000. In her work contract, its says she will receive an 8 % raise per year. a) How much will she be earning in 10 years time? b) Write the function that relates the salary to time. c) For which values of the variable is it defined?

44

The intensity of the sound coming out of a speaker is inversely proportional to the square of the distance we are from it. For example: d, distance in metres I = 252 , d I, intensity of the sound Graph it taking 1 square = 1 m on the X axis, and 1 square = 5 u in the Y axis.

At what distance must a person with hearing disorders who can only hear sounds over 100 u be from the speaker?

Solve the following systems of equations analytically and graphically: a) *

31

40

y = 2x 2 – 5x – 6 y = –x 2 + 5x * b) 2 y = 3x + 4 y = x + 3x – (15/2)

a) Calculate b and c so that the vertex of the parabola y = x 2 + bx + c is at point (3, 1). b) What is its axis of symmetry? c) What are its points of intersection with the axes? ax 2

32

The parabola y = + bx + c passes through the origin. What would the value of c be?

If we also know that it passes through points (1, 3) and (4, 6), find a and b and graph the parabola.

33

Calculate a and b so that the function y = a passes through points (2, 2) and (–1, –1). x –b

34

The graph of an exponential function of the type y = ka x passes through points (0, 3) and (1; 3.6). a) Calculate k and a. b) Is it increasing or decreasing? c) Graph the function.

35

The exponential function y = ka x passes through points (0, 2) and (2; 1.28). Calculate k and a. Graph it.

36

a) Graph the exponential function y = 1.2x using a table of values to help you. b) What is the inverse function of y = 1.2x? Graph it on the same axes.

37

Graph the following functions: Z 2 ]]x if x < 1 x if x < 0 y = [1 a) y = ) b) x if x ≥ 0 ] x if x ≥ 1 \ 38 Calculate the value of the parameter k so that the following function is continuous: Z 2 ]]–x – kx – 5 if x ≤ –2 y = [1 if x > –2 ]2 x + 4 \ 39 Graph these functions: a) y = |x 2 – 1| – 2 b) y = 1 + |x | c) y = 1 – |x 2 – 6x + 5| 130

d) y = |x | – |4 – x |

Unit 5

Advanced problem solving

Remember the theory

45

51

Explain why these expressions cannot be piecewise-defined functions: Z ]]2x + 1 if x ≤ –2 x + 1 if x ≤ 1 a) y = [x 2 – 4 if –2 ≤ x ≤ 1 b) y = ) x – 1 if x > 0 ]2x + 1 if x ≥ 1 \

52

Graph and write the equation for each of the parabolas that meet these conditions: a) Its axis is x = 2, the coefficient of x 2 is –1 and intersects the X axis at a single point. b) It has the vertex at point (3, –2) and is of the type y = x 2. c) It has the vertex at the origin and passes through point (–3, –18).

53

Construct piecewise-defined functions that meet the following conditions and graph them:

Choose a suitable scale and graph the following: 2 a) y = x b) y = –75x 2 + 675 100 c) y = 0.002x 2 – 0.04x d) y = –10x 2 – 100x

46

Express the following functions in the form y = k + b and graph them: x–a a) y = 3x – 1 b) y = x + 1 c) y = 1 – x x –1 x +1 x+2 47 This year, Veronica has managed to harvest 240 kg of avocados that she can sell today at €1.20/kg. From now on, every day that goes by, 4 kg will go off, but the price will increase by €0.10/kg. When should she sell the avocados to get the most profit? What would that profit be?

48

49

50

The manufacturing cost per unit of a type of box decreases according to the number of units manufactured and is given by the function: y = 0.3x + 1000 x a) What values does x take ? b) Calculate the cost per unit and the total cost for manufacturing 10 and 100 000 boxes. c) How much do you think the cost per unit will get close to when the number of boxes gets very large?

What is the analytical expression of each of the parabolas graphed on the right?

a) It is continuous and made up of two pieces of straight lines. It passes through the origin and has a slope of –2 at x = 4. It has a maximum at (3, 7). b) It is continuous and is made up of a piece of parabola and a piece of straight line, in that order.

It has a relative minimum at (0, 0) and a relative maximum at (2, 4).

54

All the exponential functions of the type y = a x pass through the same point. State which point this is and explain why. For what values of a is the function decreasing?

55

True or false? a) The functions y = x and y = –x form a sideways parabola when plotted on the same axes. b) If the axis of a parabola is x = 2, it cannot pass through points (–1, 6) and (5, 8). c) The functions y = 4x and y = – 4x are symmetrical with respect to the Y axis. x d) The functions y = 4x and y = c 1 m are 4 symmetrical with respect to the Y axis. e) The function y = log3 x has two asymptotes, a vertical one and a horizontal one. f ) The functions y = log x and y = 10x are symmetrical with respect to the Y axis. g) The functions y = log x and y = 10 x are symmetrical with respect to the line y = x.

f

g

A pool has a diving board 8 m above the water.

Esther rolls a ball off the diving board and it falls 12 m from the vertical line of the diving board.

Write the equation for the trajectory of the ball from when 8m it leaves the diving board until it touches the water. What is its O 12 m domain of definition?

The trajectory is a parabola y = ax 2 + bx + c with its vertex at the point where it falls. Take O as the origin and bear in mind that the vertex is (0, 8).

131

OP

MATHS WORKSH LEARN Radioactive decay

Radioactive substances decay by emitting radiation and transforming into other substances. This process takes place over time and its rate varies greatly from one substance to another.

Uranium 92 radioactive substance 8 radiation + different substance 2 500 million The rate at which a radioactive substance decays is measured by its half-life, which is the time it takes for half of its original mass to decay. On the right, you can see the half-lives of some radioactive substances.

If we have an initial mass of 1 g, the amount of mass of this substance that will be left after a given time is:

U

238.029

89

Ac

227.028

t

M = c 1 m grams 2 where t is the time elapsed taking one half-life as a unit. If the substance was actinium and we wanted to express the time in years, the formula would be: t/28

M = c1m 2

where M is the amount of actinium remaining after t years.

88

years

226.025

81

Actinium 28 years 1

204.383

mass (g)

1 –– 2 1 –– 4

Carbon-14 dating There are three naturally occurring isotopes of carbon: C12, C13 and C14. The first two are stable but the third one is radioactive and has a half-life of 5 700 years. This means that over this period of time the amount of C14 reduces by half.

Just like the other isotopes of carbon, C14 is present in the atmosphere (CO2) and is absorbed by plants (photosynthesis) which incorporate it in a specific proportion. C14 is then incorporated in the same proportion, via these plants, into other living things. When a living thing dies and becomes fossilised, its C14 continues to decay according to the function on the right.

Every 5 700 years, the amount of C14 left reduces by half. Therefore, by finding out the proportion of C14 in a fossil and checking it against the initial proportion (that in a living plant), the equation above can be used to find the fossil’s geological age, in other words, the time when it was formed.

If we express the time, t, in centuries, the equation above can be expressed like this: P = 100 · c 1 m 2

= 100 · >c 1 m 2

1 57

t

H ≈ 100 · 0.988t 8 P = 100 · 0.988t, t in centuries

In other words, in 1 century there will be 100 · 0.988 = 98.8 % of the C14 left, which means that 1.2 % will have decayed.

• What percentage of C14 will a 33 000-year-old fossil have compared to a living plant?

• How old must a fossil be if it only has 10 % of the C14 of a living plant?

Substitute P in the formula for 10 and isolate t taking logarithms.

132

Thallium 3 minutes

Tl

1

t 57

Radon 1 620 years

Ra

2

3

4

time in half lives

P = 100 · c 1 m 2

t 5 700

Unit 5

PRACTICE MAKES PERFECT! • A candle lasts an hour. With the leftover wax of 10 candles you can make a new one. a) How many hours of light will 442 candles give you? b) How many candles are needed for 1 000 hours of light?

• A farmer went to the market to sell a basket of eggs. The first customer bought half her eggs plus half an egg. The second customer bought half of the remaining eggs plus half an egg, and the third one did the same. The seller then had no more eggs left.

How many eggs did she have at the beginning?

SELF-ASSESSMENT

anayaeducacion.es Answer key and interactive self-assessment.

1 Graph the piecewise function that has the following

equation:

Z ]] 2x + 6 if x < –2 y = [ x/2 + 3 if –2 ≤ x < 2 ] –x + 6 if x ≥ 2 \ Is it continuous? Come up with a function with the same sections that is not continuous.

2 Find the vertex of each of the following parabolas

and graph them:

2 a) y = x – 2 2 c) y = (5 – x)(x + 1)

b) y = x 2 + 4x – 5 d) y = –(x – 3)2 – 1

e) y = 2x 2 + 4x f ) y = 9 – (x – 1)2 g) y = 2(x – 1)(x + 3)

h) y = (x + 2)2 – 2x 2

3 Express as piecewise-defined functions and graph

them:

a) y = |2x + 1| c) y = |–x 2 + 4x – 3|

Commitment

• A farmer shared a flock of sheep among his children.

— He gave the eldest one sheep plus 1/7 of those remaining.

— He gave the second child two sheep plus 1/7 of those remaining.

— He gave the third child three sheep plus 1/7 of those remaining.

— And so on until he got to the youngest.

In this way, all of the children received the same inheritance and no sheep were left over.

How many children does the farmer have?

How many sheep were there in the flock?

b) y = 1 – x 4 d) y = |9 – (x – 2)2|

4 Graph the following functions and find their

domains of definition: a) y =

1 b) y = 3 – 2 x x +5

d) y = x + 2 e) y = 2 x – 1

c) y =

3 +1 x –1

f ) y = – x – 3

5 Graph these pairs of functions:

a) y = 1.2x; y = log1.2 x b) y = 2.5x; y = log2.5 x With respect to which line are the two functions of each pair symmetrical?

6 Using a 3-metre-long strip of wood, we want to

make a picture frame. a) If the base of the frame is 0.5 m long, how tall is it? What is its surface area?

b) Which expression gives us the surface area, S, for any base, b ? Graph it. c) For which value of the base do we obtain the maximum surface area? What is the value of that surface area?

Watch the video for target 6.a. Think of something you can do to contribute to achieve that goal. Make a commitment to put your idea into practice.

133

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