1539 graphs linear equations and functions

1
Graphs, Linear Equations, and
Functions
• 3-1 The Rectangular Coordinate System
• 3-2 The Slope of a Line
• 3-3 Linear Equations in two variables
• 3-5 Introduction to Functions

2
3-1 The Rectangular Coordinate System
Plotting ordered pairs.
An “ordered pair” of numbers is a pair of
numbers written within parenthesis in which
the order of the numbers is important.
• Example 1: (3,1), (-5,6), (0,0) are ordered pairs.
Note: The parenthesis used to represent an ordered pair are also
used to represent an open interval. The context of the problem tells
whether the symbols are ordered pairs or an open interval.
Graphing an ordered pair requires the use of graph paper and the
use of two perpendicular number lines that intersect at their 0 points.
The common 0 point is called the “origin”. The horizontal number
line is referred to as the “x-axis” or “abscissa” and the vertical line
is referred to as the “y-axis” or “ordinate”. In an ordered pair, the
first number refers to the position of the point on the x-axis, and the
second number refers to the position of the point on the y-axis.

3
Plotting ordered pairs.
The x-axis and the y-axis
make up a “rectangular
or Cartesian” coordinate
system.
• Points are graphed by moving the
appropriate number of units in the x
direction, than moving the appropriate
number of units in the y direction.
(point A has coordinates (3,1), the point
was found by moving 3 units in the
positive x direction, then 1 in the
positive y direction)
• The four regions of the graph are called
quadrants. A point on the x-axis or y-
axis does not belong to any quadrant
(point E). The quadrants are numbered.

4
Finding ordered pairs that satisfy a given
equation.
To find ordered pairs that satisfy an equation, select any
number for one of the variables, substitute into the equation
that value, and solve for the other variable.
• Example 2: For 3x – 4y = 12, complete the table shown:
Solution
X Y
0
0
-12
-4
Given: Equation Solution
3x - 4y =12
x = 0 3(0) -4y = 12 y = -3
y = 0 3x - 4(0) =12 x = 4
y = -12 3x - 4(-12) = 12 x = -12
x = -4 3(-4) -4y = 12 y = -6
X Y
0 -3
4 0
-12 -12
-4 -6

5
Graphing lines:
Since an equation in two variables is satisfied with an
infinite number of ordered pairs, it is common
practice to “graph” an equation to give a picture of
the solution.
Note: An equation like 2x +3y = 6 is called a “first degree”
equation because it has no term with a variable to a power
greater than 1.
The graph of any first degree equation in two variable is a
straight line.
• A linear equation in two variable can be written in the
form: Ax + By = C, where A,B and C are real numbers
(A and B both not zero). This form is called the “Standard
Form”.

6
Graphing lines:
• Example 3: Draw the graph of 2x + 3y = 6
Step 1: Find a table of ordered pairs that satisfy the equation.
Step 2: Plot the points on a rectangular coordinate system.
Step 3: Draw the straight line that would pass through the
points.
Step 1 Step 2 Step 3

7
Finding Intercepts:
In the equation of a line, let y = 0 to find the “x-intercept” and
let x = 0 to find the “y-intercept”.
Note: A linear equation with both x and y variables will have
both x- and y-intercepts.
• Example 4: Find the intercepts and draw the graph of 2x –y = 4
x-intercept: Let y = 0 : 2x –0 = 4 2x = 4 x = 2
y-intercept: Let x = 0 : 2(0) – y = 4 -y = 4 y = -4
x-intercept is (2,0)
y-intercept is (0,-4)

8
Finding Intercepts:
In the equation of a line, let y = 0 to find the “x-intercept” and
let x = 0 to find the “y-intercept”.
Example 5: Find the intercepts and draw the graph of 4x –y = -3
x-intercept: Let y = 0 : 4x –0 = -3 4x = -3 x = -3/4
y-intercept: Let x = 0 : 4(0) – y = -3 -y = -3 y = 3
3
x-intercept is (- ,0)
4
y-intercept is (0,3)

9
Recognizing equations of vertical and
horizontal lines:
An equation with only the variable x will always intersect the x-
axis and thus will be vertical.
An equation with only the variable y will always intersect the y-
axis and thus will be horizontal.
• Example 6: A) Draw the graph of y = 3
B) Draw the graph of x + 2 = 0 x = -2
A) B)

10
Graphing a line that passes through the
origin:
Some lines have both the x- and y-intercepts at the origin.
Note: An equation of the form Ax + By = 0 will always pass
through the origin. Find a multiple of the coefficients of x and y
and use that value to find a second ordered pair that satisfies the
equation.
• Example 7:
A) Graph x + 2y = 0

11
3-2 The Slope of a Line
Finding the slope of a line given two
points on the line:
The slope of the line through two distinct points
(x1, y1) and (x2, y2) is:
Note: Be careful to subtract the y-values and the x-values in the
same order.
Correct Incorrect
1
2 1
2 1
2( x )
x
y yrise change in y
slope m
run change in x x
x
−
= =
−
≠= =
2 1 1 2
2 1 1 2
y y y y
or
x x x x
− −
− −
2 1 1 2
1 2 2 1
y y y y
or
x x x x
− −
− −

12
Finding the slope of a line given two points on
the line:
• Example 1) Find the slope of the line through the points
(2,-1) and (-5,3)
2 1
2 1
3 ( 1) 4 4
=
( 5) 2 7 7
y yrise
slope m
run x x
− − −
= = = = = −
− − − −

13
Finding the slope of a line given an equation of
the line: The slope can be found by solving the
equation such that y is solved for on the left
side of the equal sign. This is called the slope-
intercept form of a line. The slope is the
coefficient of x and the other term is the y-
intercept. The slope-intercept form is
y = mx + b
• Example 2) Find the slope of the line given 3x – 4y = 12
3 4 12
4 3 12
3
3
4
x y
y x
y x
− =
− = − +
= −
3
The slope is
4
∴

14
Finding the slope of a line given an equation of
the line:
• Example 3) Find the slope of the line given y + 3 = 0
y = 0x - 3 ∴The slope is 0
• Example 4) Find the slope of the line given x + 6 = 0
Since it is not possible to solve for y, the slope is “Undefined”
Note: Being undefined should not be described as “no slope”
• Example 5) Find the slope of the line given 3x + 4y = 9
3 4 9
4 3 9
3 9
4 4
x y
y x
y x
+ =
= − +
= − +
3
The slope is -
4
∴

15
Graph a line given its slope and a point on the
line: Locate the first point, then use the slope
to find a second point.
Note: Graphing a line requires a minimum of two points. From the
first point, move a positive or negative change in y as indicated by the
value of the slope, then move a positive value of x.
• Example 6) Graph the line given
slope = passing through (-1,4)
Note: change in y is +2
2
3

16
Graph a line given its slope and a point
on the line: Locate the first point, then use the
slope to find a second point.
• Example 7) Graph the line given
slope = -4 passing through (3,1)
Note:
A positive slope indicates the line moves up from L to R
A negative slope indicates the line moves down from L to R

17
Using slope to determine whether two lines are
parallel, perpendicular, or neither:
Two non-vertical lines having the same slope are parallel.
Two non-vertical lines whose slopes are negative reciprocals
are perpendicular.
• Example 8) Is the line through (-1,2) and (3,5) parallel to the line
through (4,7) and (8,10)?
1 2
For line 1: For line 2:
5 2 3 10 7 3
Y
3 (
ES
1) 4 8 4 4
m m
− −
=
−
∴
= = =
− −

18
Using slope to determine whether two lines are
parallel, perpendicular, or neither:
Two non-vertical lines having the same slope are parallel.
Two non-vertical lines whose slopes are negative reciprocals are perpendicular.
• Example 9) Are the lines 3x + 5y = 6 and 5x - 3y = 2 parallel,
perpendicular, or neither?
For line 1: For line 2:
3 5 6 5 - 3 2
5 3 6 3 5 2
3 6 5 2
3 5
is the negative reciprocal of
5 3
5 5 3 3
x y x y
y x y x
y x
Perpendicula
y x
r
+ = =
= − + − = − +
= −
∴
+
−
= −

19
Solving Problems involving average rate
of change: The slope gives the average rate of change in y
per unit change in x, where the value of y depends on x.
• Example 10) The graph shown approximates the percent of
US households owing multiple pc’s in the years 1997-2001.
Find the average rate of change between years 2000 and 1997.
Use the ordered pairs:
(1997,10) and (2000,20.8)
20.8 10 10.8
200
%
0 1997 3
3.6m
yr
−
= = =
−

20
Solving Problems involving average rate
of change: The slope gives the average rate of change in y
per unit change in x, where the value of y depends on x.
• Example 11) In 1997, 36.4 % of high school students smoked.
In 2001, 28.5 % smoked.
Find the average rate of change in percent per year.
Use the ordered pairs:
(1997,36.4) and (2001,28.5)
28.5 36.4 7.9
2001
%
1.975
1997 4
m
yr
− −
= = =
−
−

21
3-3 Linear Equations in Two Variables
Writing an equation of a line given its slope
and y-intercept: The slope can be found by
solving the equation such that y is solved for
on the left side of the equal sign. This is called
the slope-intercept form of a line. The slope is
the coefficient of x and the other term is the y-
intercept.
The slope-intercept form is y = mx + b, where
m is the slope and b is the y-intercept.
• Example 1: Find an equation of the line with slope 2 and y-
intercept (0,-3)
Since m = 2 and b = -3, y = 2x - 3

22
Graphing a line using its slope and y-
intercept:
• Example 2: Graph the line using the slope and y-
intercept: y = 3x - 6
Since b = -6, one point on the line is (0,-6).
Locate the point and use the slope (m = ) to locate a
second point.
(0+1,-6+3)=
(1,-3)
3
1

23
Writing an equation of a line given its
slope and a point on the line: The “Point-
Slope” form of the equation of a line with slope m and passing
through the point (x1,y1) is:
y - y1 = m(x - x1)
where m is the given slope and x1 and y1 are the respective values of the given point.
• Example 3: Find an equation of a line with slope and a
given point (3,-4)
2
51 1( )
2
( 4) ( 3)
5
2 6
4
5 5
2 6 20 2 26
5 5 5 5 5
5 2 26 or in standard form 2 5 26y x
y y m x x
y x
y x
y x x
x y
− = −
− − = −
+ = −
= − − =
= − − =
−

24
Writing an equation of a line given two
points on the line: The standard form for a line was
defined as Ax + By = C.
• Example 4: Find an equation of a line with passing through
the points (-2,6) and (1,4). Write the answer in standard form.
Step 1: Find the slope:
Step 2: Use the point-slope
method:
1 1
4 6 2 2
1 ( 2) 3 3
( )
2
6 ( ( 2))
3
2 4
6
3 3
2 4 18 2 14
3 3
2
3 3 3
3 2 14 or 3 14
m
y y m x x
y x
y x
y x x
x x yy
− −
= = = −
− −
− = −
− = − − −
− = − −
= − − + =
+ += − =
− +

25
Finding equations of Parallel or
Perpendicular lines:
If parallel lines are required, the slopes are identical.
If perpendicular lines are required, use slopes that are
negative reciprocals of each other.
• Example 5: Find an equation of a line passing through the
point (-8,3) and parallel to 2x - 3y = 10.
Step 1: Find the slope Step 2: Use the point-slope method
of the given line
2 3 10
3 2 10
2 10
33
2
3
x y
y x
y mx
− =
− = − +
= − ∴ =
1 1
3 2 25 or
( )
2
3 ( ( 8))
3
2 16
3
3 3
-2 3 25
2 16 9 2 25
3 3 3 3 3
y y m x x
y x
y x
y x
y x x y
x
− = −
− = − −
−
= + +
+
= + + =
=
=
+

26
Finding equations of Parallel or
Perpendicular lines:
• Example 6: Find an equation of a line passing through the
point (-8,3) and perpendicular to 2x - 3y = 10.
Step 1: Find the slope Step 3: Use the point-slope method
of the given line
Step 2: Take the negative reciprocal
of the slope found
2 3 10
3 2 10
2 10
33
2
3
x y
y x
y mx
− =
− = − +
= − ∴ =
1 1( )
3 ( ( 8))
3
3 1
3
2
2 3 18 or 3 18
2
2
9
2
2
3
y y m x x
y x
y x
y x
y x x y
− = −
− = − −
− = − −
= − −
−
= − − + = −1
2
3
3
2
mm ∴ = −=

27
Forms of Linear Equations
Equation Description When to Use
Y = mx + b
Slope-Intercept Form
slope is m
y-intercept is (0,b)
Given an equation, the slope
and y-intercept can be easily
identified and used to graph
y - y1 = m(x-x1)
Point-Slope Form
slope is m
line passes through (x1,y1)
This form is ideal to use when
given the slope of a line and
one point on the line or given
two points on the line.
Ax + By = C
Standard Form
(A,B, and C are integers, A 0)
Slope is -(A/B)
x-intercept is (C/A,0)
y-intercept is (0,C/B)
X- and y-intercepts can be found
quickly
y = b
Horizontal line
slope is 0
y-intercept is (0,b)
Graph intersects only the y
axis, is parallel to the x-axis
x = a
Vertical line
slope is undefined
x-intercept is (a,0)
Graph intersects only the x
axis, is parallel to the y-axis
≥

28
Writing an equation of a line that
models real data: If the data changes at a fairly
constant rate, the rate of change is the slope. An initial
condition would be the y-intercept.
• Example 7: Suppose there is a flat rate of $.20 plus a
charge of $.10/minute to make a phone call. Write an
equation that gives the cost y for a call of x minutes.
Note: The initial condition is the flat rate of $.20 and
the rate of change is $.10/minute.
Solution: y = .10x + .20

29
Writing an equation of a line that
models real data: If the data changes at a fairly
constant rate, the rate of change is the slope. An initial
condition would be the y-intercept.
• Example 8: The percentage of mothers of children
under 1 year old who participated in the US labor force
is shown in the table. Find an equation that models the
data.
Using (1980,38) and (1998,50)
Year Percent
1980 38
1984 47
1988 51
1992 54
1998 59
59 38 21
1998 1980
1.1
18
1.1
67
67
38
m
m
y x
−
= =
−
= +
=
∴

30
3-5 Introduction to Functions
Defining and Identifying Relations
and Functions: If the value of the variable y
depends upon the value of the variable x, then y is the
dependent variable and x is the independent variable.
• Example 1: The amount of a paycheck depends upon
the number of hours worked. Then an ordered pair
(5,40) would indicate that if you worked 5 hours, you
would be paid $40. Then (x,y) would show x as the
independent variable and y as the dependent variable.
A relation is a “set of ordered pairs”.
{(5,40), (10,80), (20,160), (40,320)} is a relation.

31
Defining and Identifying Relations and Functions:
A function is a relation such that for each value of the
independent variable, there is one and only one value of the
dependent variable.
Note: In a function, no two ordered pairs can have the same
1st component and different 2nd components.
• Example 2: Determine whether the relation is a function
{(-4,1), (-2,1), (-2,0)}
Solution: Not a function, since the independent variable
has more than one dependent value.

32
Defining and Identifying Relations and Functions:
Since relations and functions are sets of ordered pairs, they
can be represented as tables or graphs. It is common to
describe the relation or function using a rule that explains the
relationship between the independent and dependent variable.
Note: The rule may be given in words or given as an equation.
y = 2x + 4 where x is the independent variable
and y is the dependent variable
x y
0 4
2 8
4 12
6 16

33
Domain and Range:
In a relation:
A) the set of all values of the independent variable (x) is the domain.
B) the set of all values of the dependent variable (y) is the range.
• Example 3: Give the domain and range of each relation. Is the
relation a function?
{(3,-1), (4,-2),(4,5), (6,8)}
Domain: {3,4,6}
Range: {-1,-2,5,8}
Not a function
Domain: {0,2,4,6}
Range: {4,8,12,16}
x y
0 4
2 8
4 12
6 16

34
Domain and Range:
In a relation:
A) the set of all values of the independent variable (x) is the domain.
Domain:
{1994,1995,1996,1997,1998,1999}
Range:
{24134,33786,44043,55312,69209,86047}
This is a function
Year
Subscribers
(thousands)
1994 24,134
1995 33,786
1996 44,043
1997 55,312
1998 69,209
1999 86,047
Cell Phone Users

35
Domain and Range:
In a relation:
A) the set of all values of the independent variable (x) is the
domain.
• Example 5: Give the domain and range of each relation.
Domain: (-∞, ∞ ) Domain: [-4, 4]
Range: (-∞, 4] Range: [-6, 6]

36
Agreement on Domain:
Unless specified otherwise, the domain of a relation is assumed
to be all real numbers that produce real numbers when
substituted for the independent variable.
• The function
has all real numbers except x = 0
Note: In general, the domain of a function defined by an
algebraic expression is all real numbers except those numbers
that lead to division by zero or an even root of a negative
number.
• The function is not defined for values <
1
y
x
=
3 2y x= −
2
3

37
Identifying functions defined by graphs
and equations:
Vertical Line Test:
If every vertical line intersects the graph of a relation
in no more than one point, the relation represents a
function .

38
Identifying functions defined by graphs and
equations:
Vertical Line Test:
If every vertical line intersects the graph of a relation in no
more than one point, the relation represents a function.
Function Not a Function

39
Identifying functions defined by graphs and
equations:
• Example 5: Decide whether the equations shown define a function and give
the domain
Function Function Not a Function
Domain: Domain: Domain:
3 2y x= − 5
1
y
x
=
−
2
y x=
2
,
3
 
∞÷ 
( ) ( ),1 1,−∞ ∞U [ )0,∞

40
Using Function Notation:
When a rule or equation is defined such that y is dependent on
x, the “Function Notation” y = f(x) is used and is read as “y = f
of x” where the letter f stands for function.
Note: The symbol f(x) does not indicate that f is multiplied by
x, but represents the y-value for the indicated x-value.
If y = 9x -5, then f(x) = 9x -5 and f(2) = 9(2) -5 = 13
and f(0) = 9(0) -5 = -5
• Example 6:
3 5
( )
2
: ( 3)
3( 3) 5 9 5 14
( 3) 7
2 2 2
3 5
( )
2
( )
x
f x
Find f and f
f
r
f
r
r
− − + +
− = = = =
−
=
+
−
+
−
=

41
Using Function Notation:
When a rule or equation is defined such that y is dependent on
x, the “Function Notation” y = f(x) is used and is read as “y = f
of x” where the letter f stands for function.
• Example 7: f(x) = 5x -1 find: f(m + 2)
f(m + 2) = 5(m + 2) -1 = 5m + 10 - 1
f(m + 2) = 5m + 9
• Example 8: Rewrite the equation given and find f(1) and f(a)
2
2
2
2
4 3
: (1) ( )
4 3
3
4
3
( )
4
x y
Find f and f a
y x
x
y
x
f x
− =
− = − +
−
=
−
=
2
2
(1) 3 2 1
(
: (
1)
4 4 2
3
1)
4
( )
( )
Find f and f a
f
a
f a
− −
= = = −
−
=

42
Identifying Linear Functions:
A function that can be defined by f(x) = mx + b for real
numbers m and b is a “Linear Function”
Note: The domain of a linear function is (-∞, ∞).
The range is (-∞, ∞).
Note: Remember that m represents the slope of a line and (0,b)
is the y-intercept.
A function that can be defined by f(x) = b is called a “Constant
Function” which has a graph that is a horizontal line.
Note: The range of a constant function is {b}.

1539 graphs linear equations and functions

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