IB Mathematics AI SL Functions models MAI Study Notes - New Syllabus

LINEAR MODEL

◆ CONNECTION WITH ARITHMETIC SEQUENCES
Consider the arithmetic sequence with first term \( u_1 = q \) and common difference \( d = 3 \):
$ q, 12, 15, 18, 21, 24, … $

Let us represent the terms of this sequence in the Cartesian plane.

We observe that the points lie on a line. This is not an accident. The \( n \)-th term of the sequence is given by:
$ u_n = q + (n – 1) \times 3 $
that is:
$ u_n = 3n + 6 $
which is a linear expression. In fact, the points lie on the line:
$ y = 3x + 6 $
The gradient \( m \) of the line is equal to the common difference \( d = 3 \).

◆ A POINT SERIES ON A LINE
Consider the following data:

Let us place the points \((x, y)\) on the Cartesian plane.

It seems that the points lie on a line. Indeed:
 The gradient of the line segment AB is:
$ m_{AB} = \frac{\Delta y}{\Delta x} = \frac{1.1}{1} = 1.1 $
 The gradient of the line segment BC is:
$ m_{BC} = \frac{\Delta y}{\Delta x} = \frac{3.3}{3} = 1.1 $
 Even for non-consecutive points, for example C and F:
$ m_{CF} = \frac{\Delta y}{\Delta x} = \frac{7.7}{7} = 1.1 $

In fact, any pair of points gives the same gradient, and this is the gradient of the line in question.

What is the equation of this line?

Two points are enough to find the line passing through them. Let us select the endpoints \( A(1, 6.1) \) and \( F(12, 18.2) \).

First approach:
The gradient of the line segment AF is:
$ m_{AF} = \frac{\Delta y}{\Delta x} = \frac{12.1}{11} = 1.1 $

Thus, the equation of the line (in point-gradient form) is:
$ y – y_1 = m(x – x_1) $
$ \Rightarrow y – 6.1 = 1.1(x – 1) $

We carry on finding the gradient-intercept form:
$ y – 6.1 = 1.1x – 1.1 $
$ \Rightarrow y = 1.1x + 5 $

Second approach:
The equation of the line has the form:
$ y = ax + b $

We need to find the coefficients \( a \) and \( b \):
 By substituting \( A(1, 6.1) \), we obtain: \( a + b = 6.1 \)
 By substituting \( F(12, 18.2) \), we obtain: \( 12a + b = 18.2 \)

The solution of the system is \( a = 1.1 \) and \( b = 5 \). Therefore, the line is:
$ y = 1.1x + 5 $

Notice: Any pair of points would give the same result (check!).

Confirm: If we pick another point among our data, say \( E(10, 16) \), it satisfies the equation: For \( x = 10 \),
$ 1.1x + 5 = 1.1 \times 10 + 5 = 11 + 5 = 16 $

EXAMPLE 1
The relation between Celsius degrees (°C) and Fahrenheit (°F) is linear. That is:
$ °F = a °C + b $

We are told that:
 \( 4°C \) corresponds to \( 39.2°F \)
 \( 10°C \) corresponds to \( 50°F \)

Thus:
$ 39.2 = 4a + b $
$ 50 = 10a + b $

The solution is \( a = 1.8 \) and \( b = 32 \). Therefore, the formula is:
$ °F = 1.8 °C + 32 $

In other words, to convert temperatures in degrees Celsius to Fahrenheit, multiply by 1.8 (or 9/5) and add 32.

REGRESSION – BEST FIT LINE BY USING GDC
Later, in Statistics (Topic 4), we will discuss in further detail the method of regression. At the moment, we simply present the best fit line that our GDC gives. For our data above:

MENU: Statistics: Enter
List 1: \( Q \)-values
List 2: \( P \)-values

CALC: REG: X [F1]: \( ax + b \) [F1]

The best fit line is:
$ P = 1.98Q – 5.22 $

QUADRATIC AND CUBIC MODELS

◆ POLYNOMIALS AND GRAPHS
Remember the following functions:
Linear function: \( y = ax + b \)
Quadratic function: \( y = ax^2 + bx + c \)

We also call them polynomials of degree 1 and 2, respectively. The degree is the largest exponent of \( x \) in this expression (all the exponents are positive integers; we may have a constant term as well).

Along the same lines, we define the polynomial of degree 3, also called:
Cubic function: \( y = ax^3 + bx^2 + cx + d \)

The simplest polynomials up to degree 3 are the following:

For any linear function \( y = ax + b \) with \( a \neq 0 \):
There are 2 parameters \( a \) and \( b \); so we need two points to determine the values of \( a \) and \( b \) (by solving a system).
 The graph is a straight line:
 Increasing if \( a > 0 \)
Decreasing if \( a < 0 \)

For any quadratic function \( y = ax^2 + bx + c \) with \( a \neq 0 \):
 There are 3 parameters \( a, b, c \); so we need three points to determine the values of \( a, b, \) and \( c \).
The graph is a parabola:
 Concave up if \( a > 0 \)
Concave down if \( a < 0 \)

For any cubic function \( y = ax^3 + bx^2 + cx + d \) with \( a \neq 0 \):
There are 4 parameters \( a, b, c, d \); so we need four points to determine the values of \( a, b, c, \) and \( d \).
 The graph has an S-shape:
 It looks like \( \cup \) if \( a > 0 \)
It looks like \( \cap \) if \( a < 0 \)

For example, the graph of the cubic function:
$ y = x^3 – 6x^2 + 11x – 6 $
is:

Notice: As \( x \to +\infty \) (going to the right), \( y \) tends to \( +\infty \), since \( a > 0 \).

◆ QUADRATIC MODEL
If they give us three points on a parabola \( y = ax^2 + bx + c \), 
Points: \( (1, -3) \), \( (3, -1) \), and \( (4, 6) \), we can find the parameters \( a, b, \) and \( c \) by substituting the coordinates of the points:
\( (1, -3) \) gives: \( a + b + c = -3 \)
 \( (3, -1) \) gives: \( 9a + 3b + c = -1 \)
 \( (4, 6) \) gives: \( 16a + 4b + c = 6 \)

We solve the system by GDC: \( a = 2 \), \( b = -7 \), \( c = 2 \). Hence, the quadratic function is:
$ y = 2x^2 – 7x + 2 $

If we have a number of points which form a U-shape with certain properties, for example:
 There seems to exist a vertical axis of symmetry.
There is a vertex.
 The points increase or decrease in a “smooth” way which is not exponential.

We may assume that the U-shape is a parabola. This will be our model. Of course, we are not certain that the assumption is correct, but it is a practical approach if we cannot find an analytic solution.

EXAMPLE 1
The perimeter of a rectangular region is 100 m. What is the area? Is it always the same? If not, what is the maximum possible area?

A model approach:
Suppose that the length of one side is \( x \). Choose some values for \( x \):

Let us observe the points \( (x, A) \) on the Cartesian plane:

It seems that the points lie on a parabola with axis of symmetry \( x = 25 \). Let us pick 3 representative points:
 The first point \( (5, 225) \)
 The last point \( (37, 481) \)
 The point \( (25, 625) \), which seems to be the vertex.

If \( A = ax^2 + bx + c \), then:
 Point \( (5, 225) \) gives: \( 5^2a + 5b + c = 225 \)
 Point \( (37, 481) \) gives: \( 37^2a + 37b + c = 481 \)
Point \( (25, 625) \) gives: \( 25^2a + 25b + c = 625 \)

The GDC gives the solution \( a = -1 \), \( b = 50 \), \( c = 0 \). That is:
$ A = -x^2 + 50x $

Indeed, the remaining points satisfy the equation. For example, let us check the point \( (10, 400) \):
$ A = -10^2 + 50 \times 10 = 400 $

◆ A POINT SERIES ALMOST ON A PARABOLA – REGRESSION
If the points do not perfectly fit into a parabola, we can pick three representative points and find a quadratic model that describes our data quite well.

Again, a better solution could be found by using regression. For our example:

The points, of course, perfectly fit into a parabola, so the GDC will give the exact solution. But let us try it just for practice:

MENU: Statistics: Enter
List 1: \( x \)-values
List 2: \( A \)-values

CALC: REG: \( X^2 \) [F3].

The (best fit) quadratic model is:
$ A = -x^2 + 50x $

◆ CUBIC MODEL – REGRESSION
Similarly, if we observe that our data form an S-shape, we can assume that our model is a cubic function of the form:
$ y = ax^3 + bx^2 + cx + d $

If we know that the points fit exactly on a cubic function:
We pick 4 points and substitute them into the model to find \( a, b, c, d \) (by solving a system of 4 equations).

If the points almost fit on a cubic function:
 A less accurate model is found by picking 4 random points and working as above.
A better model can be obtained by using regression.

MENU: Statistics: Enter
List 1: \( x \)-values
List 2: \( y \)-values

CALC: REG: \( X^3 \) [F4].

POWER FUNCTION MODEL – DIRECT/INVERSE VARIATION
Let us concentrate on the functions of the simple form \( y = x^n \).

We have already seen the graphs of the functions \( y = x^2 \), \( y = x^3 \), but let us investigate more powers (i.e., values of \( n \)). Use your GDC to verify all the results below:

When \( n \) is positive:
\( n = 1 \): \( y = x \)
 \( n = 2 \): \( y = x^2 \)
\( n = 3 \): \( y = x^3 \)
\( n = 4 \): \( y = x^4 \)

When \( n \) is negative:
 \( n = -1 \): \( y = \frac{1}{x} \)
 \( n = -2 \): \( y = \frac{1}{x^2} \)
 \( n = -3 \): \( y = \frac{1}{x^3} \)
\( n = -4 \): \( y = \frac{1}{x^4} \)

The \( y \)-axis is a vertical asymptote for negative exponents.

We observe that:
When \( n \) is even (i.e., \( \pm 2, \pm 4, \pm 6, \ldots \)), the graph is symmetric about the \( y \)-axis.
When \( n \) is odd (i.e., \( \pm 1, \pm 3, \pm 5, \ldots \)), the graph is symmetric about the origin \( (0, 0) \).

◆ POWER MODELS
In real-life applications, we meet models of the form:
$ y = kx^n $
or
$ y = \frac{k}{x^n} $
where \( x \) takes only positive values.

For positive exponents, the models look like as follows:
\( y = kx^2 \)
\( y = kx^3 \)
\( y = kx^4 \)
 \( y = kx^5 \)

For negative exponents, the models look as follows:
 \( y = \frac{k}{x^2} \)
 \( y = \frac{k}{x^3} \)
 \( y = \frac{k}{x^4} \)
 \( y = \frac{k}{x^5} \)

◆ DIRECT AND INVERSE VARIATION
The relation:
$ y = kx^n $
shows a direct variation between (a power of) \( x \) and \( y \): when \( x \) increases, then \( y \) increases as well.

Notice:
 If \( y = kx \), we say \( y \) is (directly) proportional to \( x \).
 If \( y = kx^2 \), we say \( y \) is (directly) proportional to the square of \( x \).

The relation:
$ y = \frac{k}{x^n} $
shows an inverse variation between (a power of) \( x \) and \( y \): when \( x \) increases, then \( y \) decreases.

Notice:
 If \( y = \frac{k}{x} \), we say \( y \) is inversely proportional to \( x \).
 If \( y = \frac{k}{x^2} \), we say \( y \) is inversely proportional to the square of \( x \).

EXAMPLE 1
Consider the relation:
$ P = 5Q^2 $

The quantity \( P \) is proportional to the square of \( Q \).

EXAMPLE 2
Consider the relation:
$ P = \frac{5}{Q^2} $

The quantity \( P \) is inversely proportional to the square of \( Q \).

EXAMPLE 3
(a) \( P \) is proportional to the cube of \( Q \) (i.e., \( P = kQ^3 \)). Find the relation if we know that \( P = 5 \) when \( Q = 2 \).

(b) Suppose that \( P = kQ^n \). Find the relation if we know that:
 \( P = 6.4 \) when \( Q = 2 \),
 \( P = 48.6 \) when \( Q = 3 \).

Solution:
(a) \( 2^3k = 5 \Rightarrow 8k = 5 \Rightarrow k = \frac{5}{8} = 0.625 \). Hence:
$ P = 0.625Q^3 $

(b) \( 2^nk = 6.4 \) and \( 3^nk = 48.6 \). If we divide the two relations:
$ \frac{3^nk}{2^nk} = \frac{48.6}{6.4} \Rightarrow \left( \frac{3}{2} \right)^n = 7.59375 $
Hence, \( n = 5 \).

The first relation gives: \( 2^5k = 6.4 \). Hence, \( k = 0.2 \). Therefore:
$ P = 0.2Q^5 $

Notice also that the relation:
$ A = k \frac{BC^2}{D^3} $
between the quantities \( A, B, C, D \) (\( k \) constant) implies that:
\( A \) is proportional to \( B \).
\( A \) is proportional to the square of \( C \).
 \( A \) is inversely proportional to the cube of \( D \) (when the remaining quantities stay constant).

◆ AREA vs LENGTH
Look at the area \( A \) for several shapes below:

In all these relations, we observe that the area \( A \) is proportional to the square of a length (side, radius, etc.).

For a rectangle of sides \( x \) and \( y \), since \( A = xy \):
\( A \) is proportional to \( x \) (if \( y \) remains constant).
\( A \) is proportional to \( y \) (if \( x \) remains constant).
 \( y \) is inversely proportional to \( x \) (if \( A \) remains constant).

EXAMPLE 4
Consider all rectangles of area 50 with side lengths \( x \) and \( y \).

(a) Describe the relation between \( x \) and \( y \).
(b) Express the perimeter of the rectangle in terms of \( x \).

Solution:
(a) \( xy = 50 \Rightarrow y = \frac{50}{x} \). Thus, \( y \) is inversely proportional to \( x \).

(b) \( P = 2x + 2y = 2x + \frac{100}{x} \).

◆ VOLUME vs LENGTH
Look at the volume \( V \) for several shapes below:

In all these relations, we observe that the volume \( V \) is proportional to the cube of a length (side, radius, etc.).

For a cone of radius \( r \) and height \( h \), since \( V = \frac{1}{3}\pi r^2h \):
\( V \) is proportional to \( h \) (if \( r \) remains constant).
 \( V \) is proportional to the square of \( r \) (if \( h \) remains constant).

When \( V \) remains constant, say \( V = 100 \), then:
$ \frac{1}{3}\pi r^2h = 100 \Rightarrow \pi r^2h = 300 \Rightarrow h = \frac{300}{\pi r^2} $
Hence:
\( h \) is inversely proportional to the square of radius \( r \).

◆ A POINT SERIES ALMOST ON A POWER MODEL
Consider:

Given that the points are well described by a power function model of the form:
$ y = ax^n $
the best power model can be obtained by using regression:

MENU: Statistics: Enter
– List 1: \( x \)-values
– List 2: \( y \)-values

CALC: REG: [F6]: Power [F3].

The result is:
$ y = 1.98x^{2.95} $

2.16 EXPONENTIAL MODEL

◆ CONNECTION WITH GEOMETRIC SEQUENCES
Consider the geometric sequence with first term \( u_1 = 30 \) and common ratio \( r = 3 \):
$ 30, 90, 270, 810, … $

Let us represent the terms of this sequence in the Cartesian plane.

The \( n \)-th term of the sequence is given by:
$ u_n = 30 \times 3^{n-1} = 30 \times \frac{3^n}{3} $
that is:
$ u_n = 10 \times 3^n $
which is an exponential expression. In fact, the points lie on the line:
$ y = 10 \times 3^x $

◆ THE FORM OF AN EXPONENTIAL MODEL
A simple exponential model may have the form:
$ y = A \cdot b^x $ if \( y \) is increasing,
$ y = A \cdot b^x $ if \( y \) is decreasing (assuming that \( b > 1 \)).

We have to determine two parameters, \( A \) and \( b \).

Very often, the model is given in the form:
$ y = Ae^{kx} $

According to \( k \):
 \( y \) increases if \( k > 0 \),
\( y \) decreases if \( k < 0 \).

Still, we have to determine two parameters, \( A \) and \( k \).

EXAMPLE 1
Suppose that the graph of an exponential function:
$ y = A \cdot b^x $
passes through the points \( (1, 30) \) and \( (4, 810) \).

Then:
Point \( (1, 30) \) gives: \( A \cdot b^1 = 30 \) (1)
Point \( (4, 810) \) gives: \( A \cdot b^4 = 810 \) (2)

If we divide (2) by (1), we obtain:
$ \frac{A \cdot b^4}{A \cdot b^1} = \frac{810}{30} \Rightarrow b^3 = 27 \Rightarrow b = 3 $

Then (1) gives:
$ A \cdot 3^1 = 30 \Rightarrow A = 10 $

Thus, the function is:
$ y = 10 \times 3^x $

Things are even better if we know the \( y \)-intercept of the function (corresponding to \( x = 0 \)).

EXAMPLE 2
Suppose that the exponential function:
$ y = Ae^{kx} $
has \( y \)-intercept \( (0, 25) \) and passes through the point \( (5, 100) \).

Then:
Point \( (0, 25) \) gives: \( Ae^0 = 25 \Rightarrow A = 25 \)
Point \( (5, 100) \) gives: \( Ae^{5k} = 100 \)
$ \Rightarrow 25e^{5k} = 100 $
$ \Rightarrow e^{5k} = 4 $

We can use the natural logarithm \( \ln \) (or directly GDC):
$ \Rightarrow 5k = \ln 4 $
$ \Rightarrow k = \frac{\ln 4}{5} \approx 0.277 $

Hence:
$ y = 25e^{0.277x} $

◆ EXPONENTIAL GROWTH OF A POPULATION
In many applications, a quantity increases or decreases exponentially according to time. For example:

The population \( P \) at time \( t \) (let’s say in years) is given by:
$ P = P_0 e^{kt} $
 If \( k > 0 \), the population increases.
 If \( k < 0 \), the population decreases.

For \( t = 0 \), then:
$ P = P_0 e^0 = P_0 $
So the coefficient \( P_0 \) is always the initial value of \( P \).

EXAMPLE 3
The number \( n \) of some particles at time \( t \) hours is given by:
$ n = 1000e^{0.2t} $

(a) What is the initial number of particles?
(b) What is the number of particles after 3 hours?
(c) The number of particles is 2500 after \( t \) hours. Find \( t \).
(d) The number of particles doubles after \( t \) hours. Find \( t \).

Solution:
(a) For \( t = 0 \), \( n = 1000 \).

(b) For \( t = 3 \), \( n = 1000e^{0.2 \times 3} = 1822 \).

(c) If \( n = 2500 \), then:
$ 2500 = 1000e^{0.2t} $
GDC gives \( t = 4.58 \) hours.

Otherwise:
$ 2500 = 1000e^{0.2t} \Leftrightarrow 2.5 = e^{0.2t} $
$ \Leftrightarrow \ln 2.5 = \ln e^{0.2t} $
$ \Leftrightarrow \ln 2.5 = 0.2t $
$ \Leftrightarrow t = \frac{\ln 2.5}{0.2} = 4.58 \, \text{hours} $

(d) It’s a similar question to (c). We let \( n = 2000 \) and find \( t \).
$ 2n_0 = n_0 e^{0.2t} \Leftrightarrow 2 = e^{0.2t} $
$ \Leftrightarrow \ln 2 = \ln e^{0.2t} $
$ \Leftrightarrow \ln 2 = 0.2t $
$ \Leftrightarrow t = \frac{\ln 2}{0.2} = 3.47 \, \text{hours} $

We can also use the graph mode of GDC to answer the questions.

EXAMPLE 4
The mass \( m \) of a radio-active substance at time \( t \) hours is given by:
$ m = 4e^{-kt} $

(a) The mass is 1 kg after 5 hours. Find \( k \).
(b) What is the mass after 3 hours?
(c) The mass reduces to a half after \( t \) hours. Find \( t \).

Solution:
(a) For \( t = 5 \), \( m = 1 \), thus:
$ 1 = 4e^{-5k} $
GDC-Solve N gives:
$ k = 0.277 $
$ m = 4e^{-0.277t} $

(b) For \( t = 3 \),
$ m = 4e^{-0.277 \times 3} = 1.74 $

(c) For \( m = 2 \),
$ 2 = 4e^{-0.277t} $

Either by GDC-Solve N: \( t = 2.50 \) hours or by using \( \ln \):
$ \Leftrightarrow e^{-0.28t} = 0.5 \Leftrightarrow \ln e^{-0.277t} = \ln 0.5 \Leftrightarrow -0.277t = \ln 0.5 $
$ \Leftrightarrow t = \frac{\ln 0.5}{-0.277} = 2.50 \, \text{hours} $

This time is known as half-life time.

THE EXPONENTIAL MODELS \( y = Ab^x + c \), \( y = Ae^{kx} + c \)
If we add a constant number \( c \) to our simple exponential models they take the form
$ y = Ab^x + c $
if \( y \) increases
$ y = Ab^-x + c $
if \( y \) decreases
or
$ y = Ae^{kx} + c $
with \( k > 0 \) if \( y \) increases – \( k < 0 \) if \( y \) decreases

We simply shift the graph \( c \) units above.

If the function is increasing it looks like

In either case:
The horizontal line \( y = c \) is a horizontal asymptote to the graph.
For \( x = 0 \), the \( y \)-intercept is \( y = A + c \).

EXAMPLE 5
The population \( P \) after \( t \) years is given by
$ P = 1000e^{-0.2t} + 500 $

The graph of this function is

The initial population is
$ P = 1000e^0 + 500 = 1500 $

The population after 5 years is
$ P = 1000e^{-1} + 500 = 867.87… \equiv 868 $

The population drops below 600 when \( P = 600 \).

Either by GDC-GSolv-X Calc: \( t = 11.51 \)

Or by solving the equation
$ 1000e^{-0.2t} + 500 = 600 $
$ \iff 1000e^{-0.2t} = 100 $
$ \iff e^{-0.2t} = 0.1 $
$ \iff -0.2t = \ln 0.1 $
$ \iff t = \frac{\ln 0.1}{-0.2} = 11.51 $

As \( t \to +\infty \) the value of \( P \) approaches 500. It never drops below 500 (in mathematical terms the line \( P=500 \) is a horizontal asymptote).

EXAMPLE 6 (mainly for HL)
Consider the following table of data.

It is given that the points lie on the exponential function of the form
$ P = Ab^t + c $

There are three parameters to be determined, namely \( A \), \( b \), and \( c \).

Let us substitute the coordinates of the given points:
The point (1,20) gives: \( Ab + c = 20 \) (1)
The point (2,30) gives: \( Ab^2 + c = 30 \) (2)
 The point (3,50) gives: \( Ab^3 + c = 50 \) (3)

We firstly eliminate \( c \):
 (2) – (1) gives \( Ab^2 – Ab = 10 \) (4)
 (3) – (2) gives \( Ab^3 – Ab^2 = 20 \) (5)

Now we divide (5) by (4) to obtain
$ \frac{Ab^3 – Ab^2}{Ab^2 – Ab} = \frac{20}{10} $
$ \Leftrightarrow \frac{Ab^2(b-1)}{Ab(b-1)} = 2 \Leftrightarrow b = 2 $

Then (4) gives: \( 4A – 2A = 10 \Leftrightarrow 2A = 10 \Leftrightarrow A = 5 \)
Finally (1) gives: \( 10 + c = 20 \Leftrightarrow c = 10 \)

Therefore, the model is
$ P = 5 \times 2^t + 10 $

• A POINT SERIES ALMOST EXPONENTIAL – REGRESSION
Consider the example of the previous paragraph (where we found a power function model).

Mind the difference between:
Power model: \( y = ax^n \)
Exponential model: \( y = ab^x \) (or \( y = ae^{bx} \))

The power function model we found was \( y = 1.98x^{2.95} \).

Let us find an exponential model either in the form \( y = ae^{bx} \) or in the form \( y = ab^x \).

The GDC, regression gives:
MENU: Statistics: Enter
List 1: \( x \)-values
List 2: \( y \)-values
CALC: REG: [F6]: EXP [F2]: [F1] or [F2]

[F1] gives
$ y = 0.707e^{1.35x} $

[F2] gives
$ y = 0.707 \times 3.85^x $

2.17 SINUSOIDAL MODEL
This paragraph presupposes the study of the trigonometric numbers sine and cosine: see Topic 3 (Trigonometry).

For our purpose, we only present some basic notions of the functions \( f(x) = \sin x \), \( f(x) = \cos x \), and certain modifications of them.

• \( f(x) = \sin x^\circ \)
Look at the graph of the function \( f(x) = \sin x^\circ \) on your GDC by using the following settings:
Set up: Angle: degrees
Set V-Window: \( x \) from -360° to 720°
\( y \) from -2 to 2

We obtain:

The following table contains some values of the function which are also shown on the graph above.

We have:
Domain: \( x \in \mathbb{R} \)
Range: \( y \in [-1,1] \) [since \( y_{\text{min}} = -1 \) and \( y_{\text{max}} = 1 \)]

We also observe that the curve is repeated every 360°.

Furthermore, we define:
Central axis: \( y = 0 \) (the average between \( y_{\text{max}} \) and \( y_{\text{min}} \))
Amplitude = 1 (distance between \( y_{\text{max}} \) and central line)
Period: \( T = 360^\circ \) (the length of a complete cycle)

• \( f(x) = \cos x^\circ \)
A similar curve is obtained for the function \( f(x) = \cos x^\circ \).

Again:
Domain: \( x \in \mathbb{R} \)
Range: \( y \in [-1,1] \) [\( y_{\text{min}} = -1 \) and \( y_{\text{max}} = 1 \)]
Central axis: \( y = 0 \)
Amplitude = 1
Period: \( T = 360^\circ \)

NOTICE:
For both functions \( y = \sin x \) and \( y = \cos x \), the horizontal distance between:
Two consecutive \( x_{\text{max}} = 360^\circ \) (one period)
Two consecutive \( x_{\text{min}} = 360^\circ \) (one period)
Consecutive \( x_{\text{max}} \) and \( x_{\text{min}} = 180^\circ \) (half a period)
The central axis is also known as the principal axis.

We distinguish four basic types of trigonometric functions:

Notice that the amplitude for all these functions is 1 (positive).

• SINGLE TRANSFORMATIONS OF \( \sin x \)
Compare the original function \( f(x) = \sin x \) (in black) with the following three transformations of \( \sin x \) (in red):

Notice that the amplitude of \( f(x) = -2 \sin x \) is still 2.

• THE MODELING FUNCTION \( f(x) = A \sin(Bx) + C \)
For the function
$ f(x) = A \sin(Bx) + C $
we have:

Notice:
\( f(x) \) ranges between the values \( C – |A| \) and \( C + |A| \).
Similar observations apply for \( f(x) = A \cos(Bx) + C \).

For example, for the function \( f(x) = 3 \sin(4x) + 5 \):

and the graph is:

EXAMPLE 1
The graph of \( f(x) = A \sin(Bx^\circ) + C \) is given below. Find \( A \), \( B \), \( C \).

The curve is of type \( +\sin x \) (since \( y \)-int central/up):
 Central axis at \( y = 1.5 \), so \( C = 1.5 \).
 Amplitude = 5, so \( A = 5 \).
 Period \( T = 8 \), hence \( B = \frac{360}{T} = \frac{360}{8} = 45 \).

Therefore, the equation of the function is:
$ f(x) = 5 \sin(45x^\circ) + 1.5 $

EXAMPLE 2
The graph of \( f(x) = A \cos(Bx) + C \) is given below. Find \( A \), \( B \), \( C \).

The curve is of type \( +\cos x \) (since \( y \)-int maximum):
 Central axis at \( \frac{y_{\text{max} + y_{\text{min}}}}{2} = 5 \),, so \( C = 5 \).
Amplitude = \( y_{\text{max}} – C = 1.5 \), so \( A = 1.5 \).
 Period \( T = 180 \), hence \( B = \frac{360}{180} = 2 \).

Therefore, the equation of the function is:
$ f(x) = 1.5 \cos(2x^\circ) + 5 $

Notice that the amplitude is always positive, but the coefficient \( A \) of \( \sin x \) or \( \cos x \) can be positive or negative.

EXAMPLE 3
Express the following graph as a trigonometric function.

The curve is of type \( -\sin x \) (since \( y \)-int central/going down):
Central axis: \( y = 5 \), hence \( C = 5 \).
 Amplitude = 15, hence \( A = -15 \).
Period \( T = 180 \), hence \( B = \frac{360}{180} = 2 \).

Therefore, the equation of the function is:
$ f(x) = -1.5 \sin(2x^\circ) + 5 $

• A POINT SERIES ALMOST SINUSOIDAL
If we are given a series of points with a periodical behavior, for example, we can estimate (approximately) a central axis, an amplitude, and a period, so we can find a sinusoidal model function for our data.

NOTICE: The GDC-Regression provides a sinusoidal model of the form \( f(x) = a \sin(bx + c) + d \), but only when \( x \) is measured in radians. So it is not appropriate for the SL syllabus.