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Functions
• Mathematical models that link two variables are called
functions.
• A function is a relationship between two sets of
numbers: a first set and a second set. Each element ‘x’
of the first set is related to one and only one element
‘y’ of the second set.
• We will be exploring primarily
three types of functions:
– Linear
– Quadratic
Domain and Range
• Recall that a function is a relationship between two
sets: a first set and a second set. Thus,
– The first set is called the domain of the function. The
elements of the domain, often thought of as ‘x values’,
are the independent variables.
– For each value of x’ (input) there is one and only one
output. This value is sometimes called the image of ‘x.’
The set of all the images (all the outputs) is called the
range of the function. The elements of the range, often
though of as ‘y values’, are the dependent variables.
Domain and Range
INPUT
x
DOMAIN
OUTPUT
equation
y
RANGE
Domain and range values are written as sets inside curly brackets.
Domain = {Inputs} and Range = {Outputs}
The domain and range tell us all the possible values for x and y.
Practice
• Consider the function y = 3x + 2 .
a) Find the images (y) values of (i) x = 3 (ii) x = -4
a) y = 11 and y = 10
• Consider the function y = x2 .
b) Find the images (y) values of (i) x = 1 (ii) x = -2
b) y = 1 and y = 4
• Consider the function y = 2x2 – x + 1 .
c) Find the images (y) values of (i) x = 2 (ii) x = -3
c) y = 7 and y = 22
Function Notation
Sometimes you may see f (x) instead of y
y = f (x) means that the image of x through the function f is y.
x is the independent variable and y is the dependent variable.
For example, if f (x) = 2x – 5
• f (3) represents the image of x = 3. To find the value of f (3)
substitute x = 3. Thus f (3) = 2(3) – 5 = 1
• f (-1) represents the image of x = -1. To find the value of f (-1)
substitute x = -1. Thus f (-1) = 2(-1) – 5 = -7
Practice
• Consider the function f(x) = 4x – 6
a) Find: (i) f (4) (ii) f (-2)
a) f(4) = 10 and f(-2) = -14
• Consider the function f(x) = x2 + 3x - 7
b) Find: (i) f (2) (ii) f (-1)
b) f(x) = 3 and y = -9
Practice
Consider the function f (x) = -x2 + 3x
a) Find the image of x = -2.
b) Find f (1)
c) Show that the point (4, -4) lies on the graph of f
a) -10
b) 2
c) –(4)2 + 3(4) = -4
Graphing Functions
• The graph of a function f is the set of points (x, y)
on the Cartesian plane where y is the image of x
through the function f.
• Different letters can be used to name functions,
for example: f, g, h …
Graphing Functions
• Drawing Graphs:
–
–
–
–
Draw a table of values to find some points on the graph.
Draw and label the axes with suitable scales.
Plot the points.
Join the points with a straight line or a smooth curve.
Practice
Graph the following function y = 2x – 3
x
-2
-1
0
1
2
3
y
-7
-5
-3
-1
1
3
Linear Functions
• y = 2x – 3 creates a straight line. Equations that form
straight lines are called linear functions
• Linear functions can be written in the form:
y = mx + c
• m = slope
• c = the y-intercept (the point where the line crosses
the y-axis
• y = mx +c is called the gradient-intercept form
Practice
1 mile is equivalent to 1.6 km
a) Draw a conversion graph of miles to km.
b) Find the gradient of the line.
c) Hence, write down a model for k (x), where k (x)
is the distance in km and x is the distance in miles.
Practice - Solution
m=
x
-1
y
-1.6
0
1
0
1.6
𝒚𝟐 −𝒚𝟏
𝒙𝟐 −𝒙𝟏
=
c=0
k(x) = 1.6x
𝟏.𝟔 −𝟎
𝟏−𝟎
= 1.6
Practice
Sketch the graph of the function y = 3x – 1
x
y
-1
0
1
-4
-1
2
Linear Functions
Consider the following function:
• The slope is ½ .
• It crosses the y-axis at 1.
• Using y = mx +c, then:
1
y  x 1
2
Finding the Equation of Linear Function
1
Find the equation of the line that has a gradient of 2 and
passes through the point (2, 3). Give you answer in
gradient – intercept form .
y y
m
2
1
x2  x1
1 y 3

2 x2
2(y – 3) = 1(x – 2)
2y – 6 = x – 2
y =½x+2
Practice
1) Find, in gradient-intercept form, the equation of the
line that passes through (10, 0) with a gradient of 2/5.
y = 2/5x - 4
2) The line L has gradient 1/3 and passes through A (2, -1)
a) Find the equation of line L in gradient-intercept form.
b) Write down the point of intersection of line L with the
y-axis.
c) Find the point of intersection of L with the x-axis.
d) Draw the line L, clearly showing the information from
(b) and (c).
y = 1/3x - 5/3
(0, -5/3)
(5, 0)
Practice
Find the equations of the line with graphs.
y = 2/3x - 6
y = -1/6x - 2
Practice
a) Draw the graph of the function y = -x + 1.
b) Write down the coordinates of the point where the graph of this
function intercepts the x-axis and the y-axis.
c) Decide whether the point A (200, -199) lies on the graph of this
function.
d) The point B (6, y) lies on the graph of this function. Find y.
b) x-intercept (1,0) y-intercept (0,1)
c) yes
d) y = -5
Linear Functions
Linear models in the form:
f (x) = m x + c
• What is the equation of this line?
y = 3/5x + 2
Practice
• In a chemistry experiment, a liquid is heated and the
temperature at different times is recorded. The table of
results for one student is shown.
Time (x minutes)
2
4
6
9
Temperature (y ∘C)
30
40
50
65
a) Draw a graph for this data.
b) Find the model of T (x), the temperature with respect
to time, for this data.
c) Use the model to predict:
i. The temperature of the liquid after 8 minutes.
ii. The time taken for the liquid to reach 57 C.
Vertical and Horizontal Lines
All vertical lines have
equations of the form x = k
where k is a constant.
All horizontal lines have
equations of the form y = k
where k is a constant.
Practice
1) Find the equation of the horizontal line that goes through
the point (-2, 5)
2) Find the equation of the vertical line that goes through
the point (1, -6)
y =5
x=1
Using Technology
You can use your calculator to graph functions.
1. Press
Y=
2. Enter in the function using the
domain variable
3. Hit
X,T,Ɵ,n key for the
GRAPH
4. You can adjust the window by hitting WINDOW and
changing the Xmin, Xmax, Ymin and Ymax settings
5. You can also hit ZOOM ,scroll down and hit ZOOMfit
Practice
Use your GDC to graph the following functions
1.
2.
3.
4.
y=½x+2
y = 4x + 5
y = -x – 1
y = -2x - 4