Chapter 10
Graphing
Equations and
Inequalities
Chapter Sections
10.1 – Reading Graphs and the Rectangular
Coordinate System
10.2 – Graphing Linear Equations
10.3 – Intercepts
10.4 – Slope and Rate of Change
10.5 – Equations of Lines
10.6 – Introduction to Functions
10.7 – Graphing Linear Inequalities in Two Variables
10.8 – Direct and Inverse Variation
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§ 10.1
Reading Graphs and the
Rectangular Coordinate
System
Vocabulary
Ordered pair – a sequence of 2 numbers where the order
of the numbers is important
Axis – horizontal or vertical number line
Origin – point of intersection of two axes
Quadrants – regions created by intersection of 2 axes
Location of a point residing in the rectangular coordinate
system created by a horizontal (x-) axis and vertical (y-)
axis can be described by an ordered pair. Each number
in the ordered pair is referred to as a coordinate
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Graphing an Ordered Pair
To graph the point corresponding to a particular
ordered pair (a,b), you must start at the origin and
move a units to the left of right (right if a is
positive, left if a is negative), then move b units
up or down (up if b is positive, down if b is
negative).
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Graphing an Ordered Pair
y-axis
Quadrant I
Quadrant II
(0, 5)
(5, 3)
(-4, 2)
3 units up
(0, 0)
5 units right
(-6, 0)
x-axis
origin
(2, -4)
Quadrant III
Note that the
order of the
coordinates is
very important,
since (-4, 2)
and (2, -4) are
located in
different
positions.
Quadrant IV
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Vocabulary
Paired data are data that can be represented
as an ordered pair
A scatter diagram is the graph of paired data
as points in the rectangular coordinate system
An order pair is a solution of an equation in
two variables if replacing the variables by the
appropriate values of the ordered pair results
in a true statement.
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Solutions of an Equation
Example
Determine whether (3, – 2) is a solution of 2x + 5y = – 4.
Let x = 3 and y = – 2 in the equation.
2x + 5y = – 4
2(3) + 5(–2) = – 4
6 + (–10) = – 4
–4=–4
(replace x with 3 and y with –2)
(compute the products)
(True)
So (3, –2) is a solution of 2x + 5y = –4
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Solutions of an Equation
Example
Determine whether (– 1, 6) is a solution of 3x – y = 5.
Let x = – 1 and y = 6 in the equation.
3x – y = 5
3(– 1) – 6 = 5
–3–6=5
–9=5
(replace x with – 1 and y with 6)
(compute the product)
(False)
So (– 1, 6) is not a solution of 3x – y = 5
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Solving an Equation
If you know one coordinate in an ordered pair that is
a solution for an equation, you can find the other
coordinate through substitution and solving the
resulting equation.
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Solving an Equation
Example
Complete the ordered pair (4, ) so that it is a solution of –2x + 4y = 4.
Let x = 4 in the equation and solve for y.
–2x + 4y = 4
–2(4) + 4y = 4
–8 + 4y = 4
–8 + 8 + 4y = 4 + 8
(replace x with 4)
(compute the product)
(add 8 to both sides)
4y = 12
(simplify both sides)
y=3
(divide both sides by 4)
Martin-Gay, Developmental Mathematics
So the completed
ordered pair is
(4, 3).
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Solving an Equation
Example
Complete the ordered pair (__, – 2) so that it is a solution of 4x – y = 4.
Let y = – 2 in the equation and solve for x.
4x – y = 4
4x – (– 2) = 4
4x + 2 = 4
4x + 2 – 2 = 4 – 2
4x = 2
x=½
(replace y with – 2)
(simplify left side)
(subtract 2 from both sides)
(simplify both sides)
(divide both sides by 4)
Martin-Gay, Developmental Mathematics
So the
completed
ordered pair
is (½, – 2).
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§ 10.2
Graphing Linear
Equations
Linear Equations
Linear Equation in Two Variables
Ax + By = C
A, B, C are real numbers, A and B not both 0
This is called “standard form”
Graphing Linear Equations
Find at least 2 points on the line
y = mx + b crosses the y-axis at b (called “slopeintercept form”)
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Graphing Linear Equations
Example
Graph the linear equation 2x – y = –4.
We find two ordered pair solutions (and a third
solution as a check on our computations) by choosing
a value for one of the variables, x or y, then solving
for the other variable. We plot the solution points,
then draw the line containing the 3 points.
Continued.
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Graphing Linear Equations
Example continued
Graph the linear equation 2x – y = – 4.
Let x = 1.
Then 2x – y = – 4 becomes
2(1) – y = – 4
(replace x with 1)
2–y=–4
(simplify the left side)
–y=–4–2=–6
y=6
(subtract 2 from both sides)
(multiply both sides by – 1)
So one solution is (1, 6)
Martin-Gay, Developmental Mathematics
Continued.
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Graphing Linear Equations
Example continued
Graph the linear equation 2x – y = – 4.
For the second solution, let y = 4.
Then 2x – y = – 4 becomes
2x – 4 = – 4
(replace y with 4)
2x = – 4 + 4
2x = 0
x=0
(add 4 to both sides)
(simplify the right side)
(divide both sides by 2)
So the second solution is (0, 4)
Martin-Gay, Developmental Mathematics
Continued.
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Graphing Linear Equations
Example continued
Graph the linear equation 2x – y = – 4.
For the third solution, let x = – 3.
Then 2x – y = – 4 becomes
2(– 3) – y = – 4
–6–y=–4
–y=–4+6=2
y=–2
(replace x with – 3)
(simplify the left side)
(add 6 to both sides)
(multiply both sides by – 1)
So the third solution is (– 3, – 2)
Martin-Gay, Developmental Mathematics
Continued.
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Graphing Linear Equations
Example continued
y
(1, 6)
(0, 4)
Now we plot all three of the
solutions (1, 6), (0, 4) and
(– 3, – 2).
x
And then we draw the
line that contains the
three points.
(– 3, – 2)
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Graphing Linear Equations
Example
3
Graph the linear equation y = 4 x + 3.
Since the equation is solved for y, we should choose
values for x.
To avoid fractions, we should select values of x that
are multiples of 4 (the denominator of the fraction).
Continued.
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Graphing Linear Equations
Example continued
3
Graph the linear equation y = 4 x + 3.
Let x = 4.
Then y =
3
4
x + 3 becomes
y=
3
4
(4) + 3
y=3+3=6
(replace x with 4)
(simplify the right side)
So one solution is (4, 6)
Continued.
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Graphing Linear Equations
Example continued
3
4
Graph the linear equation y = x + 3.
For the second solution, let x = 0.
Then y =
3
4
x + 3 becomes
y=
3
4
(0) + 3
y=0+3=3
(replace x with 0)
(simplify the right side)
So a second solution is (0, 3)
Continued.
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Graphing Linear Equations
Example continued
3
Graph the linear equation y = 4 x + 3.
For the third solution, let x = – 4.
Then y =
y=
3
4
x + 3 becomes
3
4
(– 4) + 3
y=–3+3=0
(replace x with – 4)
(simplify the right side)
So the third solution is (– 4, 0)
Continued.
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Graphing Linear Equations
y
Example continued
Now we plot all three of
the solutions (4, 6), (0, 3)
and (– 4, 0).
And then we draw
the line that
contains the three
points.
(4, 6)
(0, 3)
(– 4, 0)
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§ 10.3
Intercepts
Intercepts
Intercepts of axes (where graph crosses the axes)
Since all points on the x-axis have a y-coordinate
of 0, to find x-intercept, let y = 0 and solve for x
Since all points on the y-axis have an x-coordinate
of 0, to find y-intercept, let x = 0 and solve for y
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Intercepts
Example
Find the y-intercept of 4 = x – 3y
Let x = 0.
Then 4 = x – 3y becomes
4 = 0 – 3y
(replace x with 0)
4 = – 3y
(simplify the right side)

4
3
=y
(divide both sides by -3)
So the y-intercept is (0,  43 )
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Intercepts
Example
Find the x-intercept of 4 = x – 3y
Let y = 0.
Then 4 = x – 3y becomes
4 = x – 3(0)
(replace y with 0)
4=x
(simplify the right side)
So the x-intercept is (4,0)
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Graph by Plotting Intercepts
Example
Graph the linear equation 4 = x – 3y by plotting
intercepts.
We previously found that the y-intercept is (0,
the x-intercept is (4, 0).

4
3
) and
Plot both of these points and then draw the line through
the 2 points.
Note: You should still find a 3rd solution to check your
Continued.
computations.
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Graph by Plotting Intercepts
Example continued
Graph the linear equation 4 = x – 3y.
Along with the intercepts, for the third solution, let y = 1.
Then 4 = x – 3y becomes
4 = x – 3(1)
4=x–3
4+3=x
7=x
(replace y with 1)
(simplify the right side)
(add 3 to both sides)
(simplify the left side)
So the third solution is (7, 1)
Martin-Gay, Developmental Mathematics
Continued.
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Graph by Plotting Intercepts
y
Example continued
Now we plot the two
intercepts (0,  43 ) and (4, 0)
along with the third solution
(7, 1).
(4, 0)
4
And then we
draw the line
that contains
the three
points.
(0,  3 )
Martin-Gay, Developmental Mathematics
(7, 1)
x
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Graph by Plotting Intercepts
Example
Graph 2x = y by plotting intercepts
To find the y-intercept, let x = 0
2(0) = y
0 = y, so the y-intercept is (0,0)
To find the x-intercept, let y = 0
2x = 0
x = 0, so the x-intercept is (0,0)
Oops! It’s the same point. What do we do?
Continued.
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Graph by Plotting Intercepts
Example continued
Since we need at least 2 points to graph a line, we
will have to find at least one more point
Let x = 3
2(3) = y
6 = y, so another point is (3, 6)
Let y = 4
2x = 4
x = 2, so another point is (2, 4)
Martin-Gay, Developmental Mathematics
Continued.
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Graph by Plotting Intercepts
Example continued
y
(3, 6)
Now we plot all three
of the solutions (0, 0),
(3, 6) and (2, 4).
(2, 4)
(0, 0)
x
And then we
draw the line
that contains the
three points.
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Graph by Plotting Intercepts
Example
Graph y = 3
Note that this line can be written as y = 0·x + 3
The y-intercept is (0, 3), but there is no x-intercept!
(Since an x-intercept would be found by letting y = 0,
and 0  0·x + 3, there is no x-intercept)
Every value we substitute for x gives a y-coordinate
of 3
The graph will be a horizontal line through the
point (0,3) on the y-axis
Continued.
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Graph by Plotting Intercepts
Example continued
y
(0, 3)
x
Martin-Gay, Developmental Mathematics
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Graph by Plotting Intercepts
Example
Graph x = – 3
This equation can be written x = 0·y – 3
When y = 0, x = – 3, so the x-intercept is (– 3,0), but
there is no y-intercept
Any value we substitute for y gives an x-coordinate
of –3
So the graph will be a vertical line through the point
(– 3,0) on the x-axis
Continued.
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Graph by Plotting Intercepts
y
Example continued
(-3, 0)
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Vertical and Horizontal Lines
Vertical lines
Appear in the form of x = c, where c is a real
number
x-intercept is at (c, 0), no y-intercept unless
c = 0 (y-axis)
Horizontal lines
Appear in the form of y = c, where c is a real
number
y-intercept is at (0, c), no x-intercept unless
c = 0 (x-axis)
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§ 10.4
Slope and Rate of Change
Slope
Slope of a Line
rise y change y2  y1
m


run x change x2  x1
x2  x1
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Slope
Example
Find the slope of the line through (4, -3) and (2, 2)
If we let (x1, y1) be (4, -3) and (x2, y2) be (2, 2), then
2  (3)
5
m

24
2
Note: If we let (x1, y1) be (2, 2) and (x2, y2) be (4, -3),
then we get the same result.
3 2 5
m

42
2
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Slope of a Horizontal Line
For any 2 points, the y values will be equal to the
same real number.
The numerator in the slope formula = 0 (the
difference of the y-coordinates), but the denominator
 0 (two different points would have two different xcoordinates).
So the slope = 0.
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Slope of a Vertical Line
For any 2 points, the x values will be equal to the same
real number.
The denominator in the slope formula = 0 (the
difference of the x-coordinates), but the numerator  0
(two different points would have two different ycoordinates),
So the slope is undefined (since you can’t divide by 0).
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Summary of Slope of Lines
If a line moves up as it moves from left to right, the
slope is positive.
If a line moves down as it moves from left to right, the
slope is negative.
Horizontal lines have a slope of 0.
Vertical lines have undefined slope (or no slope).
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Parallel Lines
Two lines that never intersect are called
parallel lines.
Parallel lines have the same slope
• unless they are vertical lines, which have no
slope.
Vertical lines are also parallel.
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Parallel Lines
Example
Find the slope of a line parallel to the line
passing through (0,3) and (6,0)
0  3  3 1
m


60 6
2
So the slope of any parallel line is also –½
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Perpendicular Lines
Two lines that intersect at right angles are called
perpendicular lines
Two nonvertical perpendicular lines have slopes that
are negative reciprocals of each other
The product of their slopes will be –1
Horizontal and vertical lines are perpendicular to
each other
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Perpendicular Lines
Example
Find the slope of a line perpendicular to the
line passing through (1,3) and (2,-8)
 8  3  11
m

2  (1)
3
So the slope of any perpendicular line is
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3
11
49
Parallel and Perpendicular Lines
Example
Determine whether the following lines are parallel,
perpendicular, or neither.
5x + y = 6 and x + 5y = 5
First, we need to solve both equations for y.
In the first equation,
y = 5x – 6
(add 5x to both sides)
In the second equation,
5y = x + 5
(subtract x from both sides)
1
y = 5x + 1
(divide both sides by 5)
The first equation has a slope of 5 and the second equation has a
slope of  1 , so the lines are perpendicular.
5
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§ 10.5
Equations of Lines
Slope-Intercept Form
Slope-Intercept Form of a line
• y = mx + b has a slope of m and has a y-intercept
of (0, b).
• This form is useful for graphing, since you have a
point and the slope readily visible.
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Slope-Intercept Form
Example
Find the slope and y-intercept of the line –3x + y = –5.
First, we need to solve the linear equation for y.
By adding 3x to both sides, y = 3x – 5.
Once we have the equation in the form of y = mx + b,
we can read the slope and y-intercept.
slope is 3
y-intercept is (0, – 5)
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Slope-Intercept Form
Example
Find the slope and y-intercept of the line 2x – 6y = 12.
First, we need to solve the linear equation for y.
– 6y = – 2x + 12
(subtract 2x from both sides)
y=
1
3
x–2
(divide both sides by –6)
Since the equation is now in the form of y = mx + b,
slope is 13
y-intercept is (0, –2)
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Point-Slope Form
The slope-intercept form uses, specifically, the yintercept in the equation.
The point-slope form allows you to use ANY point,
together with the slope, to form the equation of the
line.
y  y1  m( x  x1 )
m is the slope
(x1, y1) is a point on the line
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Point-Slope Form
Example
Find an equation of a line with slope –2, through the
point (–11, –12). Write the equation in standard form.
First we substitute the slope and point into the pointslope form of an equation.
y – (– 12) = – 2(x – (– 11))
y + 12 = – 2x – 22 (use distributive property)
2x + y + 12 = – 22 (add 2x to both sides)
2x + y = – 34 (subtract 12 from both sides)
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Point-Slope Form
Example
Find the equation of the line through (–4,0) and (6, –1).
Write the equation in standard form.
First find the slope.
y2  y1
1 0
1
m


x2  x1 6  (4) 10
Continued.
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Point-Slope Form
Example continued
Now substitute the slope and one of the points into the
point-slope form of an equation.
1
y  0   ( x  (4))
10
10 y  1( x  4)
10 y   x  4
x  10 y  4
(clear fractions by
multiplying both sides by 10)
(use distributive property)
(add x to both sides)
Martin-Gay, Developmental Mathematics
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Point-Slope Form
Example
Find the equation of the line passing through points
(2, 5) and (–4, 3). Write the equation using function
notation.
y2  y1
53
2 1
m

 
x2  x1 2  (4) 6 3
Continued.
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Point-Slope Form
Example continued
1
y  3  ( x  (4))
3
1
4
y 3  x 
3
3
1
4
1
13
y  x 3 x
3
3
3
3
1
13
f ( x)  x 
3
3
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§ 10.6
Introduction to
Functions
Vocabulary
• An equation in 2 variables defines a relation
between the two variables.
• A set of ordered pairs is also called a
relation.
• The domain is the set of x-coordinates of the
ordered pairs
• The range is the set of y-coordinates of the
ordered pairs
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Domain and Range
Example
Find the domain and range of the relation {(4,9), (–4,9),
(2,3), (10, –5)}
• Domain is the set of all x-values, {4, –4, 2, 10}
• Range is the set of all y-values, {9, 3, –5}
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Domain and Range
y
Example
Domain
Find the domain and
range of the function
graphed to the right.
Use interval notation.
Domain is [–3, 4]
x
Range
Range is [–4, 2]
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Domain and Range
y
Example
Find the domain and
range of the function
graphed to the right.
Use interval notation.
Range
x
Domain is (– , )
Range is [– 2, )
Domain
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Functions
Some relations are also functions.
A function is a set of order pairs that assigns
to each x-value exactly one y-value.
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Functions
Example
Given the relation {(4,9), (–4,9), (2,3), (10, –5)}, is it a
function?
Since each element of the domain is paired with only
one element of the range, it is a function.
Note: It’s okay for a y-value to be assigned to more
than one x-value, but an x-value cannot be assigned to
more than one y-value (has to be assigned to ONLY
one y-value).
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Vertical Line Test
• Relations and functions can also be described by
graphing their ordered pairs.
• Graphs can be used to determine if a relation is a
function.
• If an x-coordinate is paired with more than one ycoordinate, a vertical line can be drawn that will
intersect the graph at more than one point.
• If no vertical line can be drawn so that it intersects
a graph more than once, the graph is the graph of a
function. This is the vertical line test.
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Vertical Line Test
y
Example
Use the vertical line
test to determine
whether the graph to
the right is the graph of
a function.
x
Since no vertical line
will intersect this
graph more than once,
it is the graph of a
function.
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Vertical Line Test
y
Example
Use the vertical line
test to determine
whether the graph to
the right is the graph of
a function.
Since no vertical line
will intersect this
graph more than once,
it is the graph of a
function.
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x
70
Vertical Line Test
y
Example
Use the vertical line test
to determine whether the
graph to the right is the
graph of a function.
x
Since vertical lines can
be drawn that intersect
the graph in two points,
it is NOT the graph of a
function.
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Vertical Line Test
Since the graph of a linear equation is a line,
all linear equations are functions, except those
whose graph is a vertical line
Note: An equation of the form y = c is a
horizontal line and IS a function.
An equation of the form x = c is a vertical line
and IS NOT a function.
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Function Notation
• Specialized notation is often used when we know a
relation is a function and it has been solved for y.
• For example, the graph of the linear equation
y = –3x + 2 passes the vertical line test, so it
represents a function.
• We often use letters such as f, g, and h to name
functions.
• We can use the function notation f(x) (read “f of x”)
and write the equation as f(x) = –3x + 2.
Note: The symbol f(x) is a specialized notation that
does NOT mean f • x (f times x).
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Function Notation
When we want to evaluate a function at a particular
value of x, we substitute the x-value into the notation.
For example, f(2) means to evaluate the function f when
x = 2. So we replace x with 2 in the equation.
For our previous example when f(x) = –3x + 2,
f(2) = –3(2) + 2 = –6 + 2 = –4.
When x = 2, then f(x) = –4, giving us the order pair
(2, –4).
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Function Notation
Example
Given that g(x) = x2 – 2x, find g(– 3). Then write
down the corresponding ordered pair.
• g(– 3) = (– 3)2 – 2(– 3) = 9 – (– 6) = 15.
• The ordered pair is (– 3, 15).
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§ 10.7
Graphing Linear
Inequalities in Two
Variables
Linear Equations in Two Variables
Linear inequality in two variables
• Written in the form Ax + By < C
• A, B, C are real numbers, A and B are not both 0
• Could use (>, , ) in place of <
An ordered pair is a solution of the linear
inequality if it makes the inequality a true
statement.
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Linear Equations in Two Variables
To Graph a Linear Inequality
1) Graph the related linear equality (forms the
boundary line).
•  and  are graphed as solid lines
• < and > are graphed as dashed lines
2) Choose a point not on the boundary line &
substitute into original inequality.
• If a true statement results, shade the halfplane containing the point.
• If a false statements results, shade the halfplane that does NOT contain the point.
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Linear Equations in Two Variables
y
Example
Graph 7x + y > –14
• Graph 7x + y = –14 as a
dashed line.
• Pick a point not on the
graph: (0,0)
(0, 0)
x
• Test it in the original
inequality.
7(0) + 0 > –14, 0 > –14
• True, so shade the side
containing (0,0).
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Linear Equations in Two Variables
y
Example
Graph 3x + 5y  –2
• Graph 3x + 5y = –2 as a
solid line.
• Pick a point not on the
graph: (0,0), but just
barely
• Test it in the original
inequality.
3(0) + 5(0) > –2, 0 > –2
• False, so shade the side
that does not contain (0,0).
Martin-Gay, Developmental Mathematics
(0, 0)
x
80
Linear Equations in Two Variables
y
Example
Graph 3x < 15
• Graph 3x = 15 as a
dashed line.
• Pick a point not on the
graph: (0,0)
(0, 0)
x
• Test it in the original
inequality.
3(0) < 15, 0 < 15
• True, so shade the side
containing (0,0).
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Linear Equations in Two Variables
Warning!
Note that although all of our examples allowed us to
select (0, 0) as our test point, that will not always be
true.
If the boundary line contains (0,0), you must select
another point that is not contained on the line as your
test point.
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§ 10.8
Direct and Inverse
Variation
Direct Variation
y varies directly as x, or y is directly proportional to
x, if there is a nonzero constant k such that y = kx.
The family of equations of the form y = kx are
referred to as direct variation equations.
The number k is called the constant of variation or
the constant of proportionality.
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Direct Variation
If y varies directly as x, find the constant of
variation k and the direct variation equation,
given that y = 5 when x = 30.
y = kx
5 = k·30
k = 1/6
1
So the direct variation equation is y =
x
6
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Direct Variation
Example
If y varies directly as x, and y = 48 when x = 6, then
find y when x = 15.
y = kx
48 = k·6
8=k
So the equation is y = 8x.
y = 8·15
y = 120
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Direct Variation
Example
At sea, the distance to the horizon is directly
proportional to the square root of the elevation
of the observer. If a person who is 36 feet
above water can see 7.4 miles, find how far a
person 64 feet above the water can see.
Round your answer to two decimal places.
Continued.
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Direct Variation
Example continued
We substitute our given value
for the elevation into the
equation.
d k e
7.4  k 36
7 .4
d
64
6
7.4  6k
7 .4
k
6
So our equation is
7.4
59.2
d
(8) 
 9.87 miles
6
6
7 .4
d
e
6
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Inverse Variation
y varies inversely as x, or y is inversely proportional
to x, if there is a nonzero constant k such that y = k/x.
The family of equations of the form y = k/x are
referred to as inverse variation equations.
The number k is still called the constant of variation
or the constant of proportionality.
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Inverse Variation
Example
If y varies inversely as x, find the constant of
variation k and the inverse variation equation,
given that y = 63 when x = 3.
y = k/x
63 = k/3
k = 63·3
k = 189
189
So the inverse variation equation is y =
x
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90
Powers of x
y can vary directly or inversely as powers of x,
as well.
y varies directly as a power of x if there is a
nonzero constant k and a natural number n
such that y = kxn
y varies inversely as a power of x if there is a
nonzero constant k and a natural number n
such that y  k
x
n
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Powers of x
Example
The maximum weight that a circular column
can hold is inversely proportional to the
square of its height.
If an 8-foot column can hold 2 tons, find how
much weight a 10-foot column can hold.
Continued.
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Powers of x
Example continued
k
w 2
h
k
k
2 2 
8
64
We substitute our given value
for the height of the column
into the equation.
128 128
w 2 
 1.28 tons
10
100
k  128
So our equation is
128
w 2
h
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