6.5 Slope intercept form
for Inequalities:
Linear Inequality: is a linear equation with
an inequality sign (< , ≤, >, ≥)
Solution of an Inequality: is an ordered
pair (x, y) that makes the inequality true.
GOAL:
Whenever we are given a graph we must be able
to provide the equation of the function.
Slope-Intercept Form: The linear equation of a
nonvertical line with an inequality sign:
y (<, ≤, >, ≥) m x + b
Slope =
𝒚𝟐−𝒚𝟏
𝒙𝟐−𝒙𝟏
=
𝑹𝒊𝒔𝒆 ⍙𝒚
=
𝑹𝒖𝒏 ⍙𝒙
y-intercept
y crossing
http://mathgraph.idwvogt.com/examples.html
Whenever we are given a graph we must be able
to provide the equation of the function.
y<mx+b
dash line
shade left or down
Whenever we are given a graph we must be able
to provide the equation of the function.
y>mx+b
dash line
shade right or up
Whenever we are given a graph we must be able
to provide the equation of the function.
y≤mx+b
Solid line
shade left or down
Whenever we are given a graph we must be able
to provide the equation of the function.
y≥mx+b
Solid line
shade Right or up
EX:
Provide the equation of the inequality.
Solution: Since line is dashed and shaded at
the bottom we use <. Also, the inequality
must be in
slope-intercept form:
Y < mx + b
1. Find the y-intercept
In this graph b = +1.
2. Find another point
to get the slope.
A(0,5)
A(0,1)
B(3,-2)
Use the equation of slope to find the slope:
𝒚𝟐−𝒚𝟏
𝒙𝟐−𝒙𝟏
=
−𝟐−𝟏
𝟑−𝟎
=
−𝟑
𝟑
= -1
The slope-intercept
form inequality is:
A(0,1)
B(3,-2)
y < -1x + 1
Remember:
This means that if you start a 1 and move down
one and over to the right one, and continue this
pattern. We shade the bottom since it is <.
When work does not need to be shown:
(EOC Test)
look at the triangle
made by the two
points.
Count the number
of square going up
or down and to the
right.
In this case 1 down
and 1 right. Thus slope is -1/1 = -1
YOU TRY IT
Provide the equation of the
inequality.
YOU TRY IT: (Solution)
The inequality is solid and shaded below:
Y ≤ mx + b
1. Find the y-intercept
In this graph b = + 4.
2. Find another point
to get the slope.
A(0,4) B(1,0)
A(0,4)
B(1,0)
Use the equation of slope to find the slope:
𝒚𝟐−𝒚𝟏
𝒙𝟐−𝒙𝟏
=
𝟎−𝟒
𝟏−𝟎
=
−𝟒
𝟏
=-4
The slope-intercept
form equation is:
A(0,4)
B(1,0)
y ≤ -4x + 4
Remember:
This means that if you start a 4 and move down
four and one over to the right. Solid line and
shaded down means we must use ≤.
When no work is required, you can use the
rise/run of a right triangle between the two
points:
Look at the triangle,
down 4 (-4) over to the
right 1 (+1)
slope = -4/+1 = -4
A(0,4)
B(1,0)
Remember:
You MUST KNOW BOTH procedures, the
slope formula and the triangle.
Given Two Points: We can also create an
inequality in the slope-intercept form from
any two points and the words: less than (<),
less than or equal to (≤), greater than(>),
greater than and equal to(≥) accordingly.
EX:
Write the slope-intercept form of the line
that is greater than or equal to and
inequality that passes through the points
(0, -0.5) and(2, -5.5)
Use the given points and equation of slope:
A(0,-0.5) B(2,-5.5)
𝒚𝟐−𝒚𝟏
𝒙𝟐−𝒙𝟏
=
−𝟓.𝟓−−𝟎.𝟓
𝟐−𝟎
=
−𝟓
𝟐
= -
𝟓
𝟐
We now use the slope and a point to find the
y intercept (b).
y ≥ mx + b
-3 = Isolate b:
𝟓
𝟐
𝟓
-3 + = b
𝟐
𝟔
𝟓
b=- +
𝟐
𝟐
(𝟏) + b
=-
𝟏
𝟐
Going back to the equation:
y = mx + b
we replace what we have found:
𝟓
𝟐
m = - and b = - ½
To get the final slope-intercept form of the
line passing through (3, -2) and(1, -3)
y≥
𝟓
−
𝟐
x–½
We now proceed to graph the equation:
𝑹𝒊𝒔𝒆
𝑹𝒖𝒏
𝟓
𝟐
y≥- x-
𝟏
𝟐
Y-intercept
y crossing
YOU TRY IT:
Write the equation of the inequality.
Use the given points and equation of slope:
B(1,-3)
A(3,-2)
𝒚𝟐−𝒚𝟏
𝒙𝟐−𝒙𝟏
=
−𝟑−−𝟐
𝟏−𝟑
=
−𝟏
−𝟐
=
𝟏
𝟐
We now use the slope and a point to find the
y intercept (b).
y < mx + b
-3 =
Isolate b:
𝟏
𝟐
𝟏
𝟐
(𝟏) + b
-3 - = b
b=-
𝟔
𝟐
𝟏
𝟐
− =-
𝟕
𝟐
Going back to the equation:
y = mx + b
we replace what we have found:
m=
𝟏
𝟐
and b = -
𝟕
𝟐
To get the final slope-intercept form of the
line passing through (3, -2) and(1, -3)
y<
𝟏
𝟐
x -3.5
We now proceed to graph the equation:
y<
𝑹𝒊𝒔𝒆
𝑹𝒖𝒏
𝟏
𝟐
2
1
x-
𝟕
𝟐
Y-intercept
y crossing
Real-World:
A fish market charges $9 per
pound for cod and $12 per pound
per flounder. Let x = pounds of cod
and y = pounds of flounder. What
is the inequality that shows how
much of each type of fish the store
must sell per day to reach a daily
quota of at least $120?
Real-World(SOLUTION):
A fish market charges $9 per pound for cod
and $12 per pound per flounder. Let x = pounds of
cod and y = pounds of flounder. What is the
inequality that shows how much of each type of
fish the store must sell per day to reach a daily
quota of at least $120?
Cod
x
Flounder  y
At least $120  9x + 12y ≥ 120
SOLUTION:
y≥-
9x + 12y ≥ 120 
𝟑
x
𝟒
+ 10
10
9
8
10
8
4
4
8
10
12
Any point in the line or in the shaded
region is a solution.
YOU TRY IT:
A music store sells used CDs
for $5 and buys used CDs for $1.50.
You go to the store with $20 and
some CDs to sell. You want to have
at least $10 left when you leave
the store. Write and graph an
inequality to show how many CDs
you could buy and sell.
Real-World(SOLUTION):
A music store sells used CDs for $5 and
buys used CDs for $1.50. You go to the store
with $20 and some CDs to sell. You want to
have at least $10 left when you leave the
store. Write and graph an inequality to
show how many CDs you could buy and sell.
Bought CDs
Sold CDs
 -5x
 +1.5y
At least $10 left  -5x + 1.5y ≥ -10
NOTE: -10 since you spent this money.
SOLUTION:
-5x + 1.5y ≥ -10 
y≥
𝟓
x
𝟏.𝟓
– 6.6
10
8
6
4
2
1
2
3
4
Any point in the line or in the shaded
region is a solution.
VIDEOS:
Linear Inequalities
https://www.khanacademy.org/math/algebra/line
ar-equations-and-inequalitie/graphing-linearinequalities/v/graphing-inequalities
https://www.khanacademy.org/math/algebra/line
ar-equations-and-inequalitie/graphing-linearinequalities/v/solving-and-graphing-linearinequalities-in-two-variables-1
CLASSWORK:
Page 393-395
Problems: As many as needed
to master the
concept.