Slide 1 / 31
Slide 2 / 31
Algebra I
New Jersey Center for Teaching and Learning
Progressive Mathematics Initiative
This material is made freely available at www.njctl.org
and is intended for the non-commercial use of
students and teachers. These materials may not be
used for any commercial purpose without the written
permission of the owners. NJCTL maintains its
website for the convenience of teachers who wish to
make their work available to other teachers,
participate in a virtual professional learning
community, and/or provide access to course
materials to parents, students and others.
Graphing Linear
Inequalities
October 29, 2012
www.njctl.org
Click to go to website:
www.njctl.org
Slide 3 / 31
Slide 4 / 31
Graphing Linear Inequalities
Click on the topic to go to that section
· Graphing Linear Inequalities in Slope-Intercept Form
· Graphing Linear Inequalities NOT in Slope-Intercept Form
Graphing Linear Inequalities
in
Slope-Intercept Form
(y = mx +b)
Return to
Table of Contents
Slide 5 / 31
Graphing
Graphs of inequalities are similar to linear equations
because they both have points on a coordinate plane and a
line connecting the points. However, a linear equation is
ONLY the line but an inequality extends beyond that line.
Slide 6 / 31
The following are graphs of linear inequalities:
y > mx + b
y < mx + b
y > mx + b
y < mx + b
How do the
graphs at left
compare with
the graph below
for
y = mx + b?
Linear Equation: y = 2x + 1
Inequality: y < 2x-1
Next slide for
observations
Slide 7 / 31
Slide 8 / 31
How to Graph a Linear Inequality
The following are graphs of linear inequalities:
y < mx + b
y > mx + b
Shading is below a dotted line.
This means the answers are
below the line but NOT on it.
Shading is above a dotted line.
This means the answers are
above the line but NOT on it.
y > mx + b
How do the
graphs at left
compare with
the graph below
for
y = mx + b?
y < mx + b
Shading is above a solid line.
This means the answers are
above the line AND on it.
Shading is below a solid line.
This means the answers are
below the line AND on it.
1) Decide where the boundary goes:
Solve inequality for y, for example y > 2x - 1
2) Decide whether boundary should be:
solid (< or >: points on the boundary make the inequality true) or
dashed (< or >: points on the boundary make the inequality false)
3) Graph the boundary (the line)
4) Decide where to shade:
y > or y >: shade above (referring to y-axis) the boundary
y < or y <: shade below (referring to y-axis) the boundary
(Or, you can test a point, which will be explained later)
Slide 9 / 31
EXAMPLE 1
Graph y < -2x + 1
Think
y = mx + b
to graph the boundary
Slide 10 / 31
Graph 2x - y < 4
Example 2
Step 2: The line should be dashed because the inequality is <
Step 1: Solve fory
-y < -2x + 4
y > 2x - 4
Step 3: Graph boundary
Step 2: The line should be solid because the inequality is>
Step 1: Solve fory: Think y = -2x + 1, m = -2 and b = 1
Step 3: Graph boundary
Step 4: Shade below the boundary line because y <
Step 4: Shade above the boundary line because y >
Slide 11 / 31
Graph
Example 3
Step 1: Solve fory
Step 2: The line should be dashed because the inequality is >
Slide 12 / 31
1 Why are there dashed boundaries on some graphs of
inequalities?
A
B
C
D
Points on the line make the inequality false.
Points on the line make the inequality true.
The slope of the line depends on the line type.
The y-intercept depends on the line type.
answer
Step 3: Graph boundary
Step 4: Shade above the boundary line because y >
Slide 13 / 31
Slide 14 / 31
2 For which of these equations would the graph
have a solid boundary and be shaded above?
3 For which of these equations would the graph
have a dashed boundary and be shaded above?
A y < 3x-2
B y < 3x-2
B y < 3x-2
C y > 3x-2
C y > 3x-2
D y > 3x-2
D y > 3x-2
answer
answer
A y < 3x-2
Slide 15 / 31
Slide 16 / 31
4 Which inequality is
graphed?
5
A y < 3x-2
Which inequality is
graphed?
A y < 3x-2
B y < 3x-2
C y > 3x-2
D y > 3x-2
D y > 3x-2
Slide 17 / 31
answer
answer
B y < 3x-2
C y > 3x-2
Slide 18 / 31
6 Which inequality could match the given graph?
y>3
B
y<3
C
x<3
D
x>3
Graphing Linear Inequalities
NOT in Slope-Intercept Form
Return to
Table of Contents
answer
A
Slide 19 / 31
Slide 20 / 31
Inequalities can be graphedwithout converting to
slope-intercept form.
The steps to graph are the same but determining where to
shade involves a different step.
Test Points
The shaded region represents all of the points that make the
inequality true, so:
A) Select a point NOT on the boundary and substitute into
1) Decide where the boundary goes:
the inequality
2) Decide whether boundary should be:
solid (< or >: points on the boundary make the inequality true) or
dashed (< or >: points on the boundary make the inequality false)
(The point (0, 0) is an easy value to work with if it is not on the boundary)
B) If your selected test point makes the inequality TRUE,
shade the region containing your test point
3) Graph the boundary (the line)
4) Decide where to shade:
Choose a test point on the graph and shade accordingly
C) If your selected test point makes the inequality FALSE,
shade the region opposite your test point
Slide 21 / 31
Slide 22 / 31
Example 5
Example 4
Graph y > -4x
Graph y < 2x + 5
Step 1: Solve fory
y > -4x
Step 2: The line should be dashed because the inequality is>
Step 3: Graph boundary
Step 1: Solve fory
y < 2x + 5
Step 2: The line should be solid because the inequality is<
Step 3: Graph boundary
Step 4: Test a point
-Choose a point and substitute it into the inequality
(0,-5):
y > -4x
-5 > -4(0)
-5 > 0
Step 4: Test a point
-(0, 0) is an easy point to work with and is not on the
boundary
o
(0,-5)
Statement is FALSE so the opposite region will be shaded
Slide 23 / 31
Example 5 Continued
y < 2x + 5
Substitute (0, 0) into the inequality
0 < 2(0) + 5
0<5
o
(0,0)
This statement is TRUE so shade the region of the graph where
this point occurs
Slide 24 / 31
Example 6
Graph y - 2 < -2(x + 1)
Step 1: Solve fory
y - 2 < -2(x + 1)
y - 2 < -2x - 2
y < -2x
Step 2: The line should be dashed because the inequality is <
Step 3: Graph boundary
Step 4: Test a point
-(0, 0) is on the graph so it cannot be used but try to pick
a point with a 0 in it to make the substitution easier
Slide 25 / 31
Slide 26 / 31
Example 6 Continued
7 What point can be used as test points when graphing
ANY inequality?
Example test point:
A
B
C
D
o
(5,0)
Substitute the test point into the inequality:
Only the origin.
Any point with a zero in the ordered pair.
Any point.
Any point not on the boundary.
0 < -2(5)
answer
y < -2x
o
(5,0) FALSE
0 < -10
The test point produces a FALSE statement, so we
shade the opposite region.
Note: You always know with certainty which side to
shade when you use test points.
Slide 27 / 31
9 Given the inequality,
, which test
point should be used and where is the shading?
Given the inequality,
, which test point
should be used and where should the graph be
shaded?
B (0,0); shade below boundary
C (0,4): shade above boundary
A (0, 0), shade below boundary
(-4,2)
(0,4)
B (-4,2), shade below
(0,0)
C (0, 0), shade above boundary
boundary
(0,0)
D (-4,2), shade above
D (0,4); shade below boundary
boundary
answer
answer
A (0,0): shade above boundary
Slide 29 / 31
Slide 30 / 31
10 Given the inequality 6x + 10y > 30, which is the
easiest test point to use and where should the graph
be shaded?
11 Given the inequality,
of the solution region?
A
, which is the graph
B
A (5, 3), shaded below
B (0, 0), shaded below
C (5, 3), shaded above
D (0, 0), shaded above
answer
C
D
answer
8
Slide 28 / 31
Slide 31 / 31