Transcript Graphing an Inequality

Graphing Inequalities in Two
Variables
Objective
• Graph inequalities in the coordinate
plane.
Graphing an Inequality
1.
2.
3.
4.
5.
<, >, ≤, or ≥
Solve the equation for y (if necessary).
Graph the equation as if it contained an = sign.
Draw the line solid if the inequality is ≤, or ≥
Draw the line dashed if the inequality is < or >
Pick a point not on the line to use as a test point.
The point (0,0) is a good test point if it is not on
the line.
6. If the point makes the inequality true, shade that
side of the line. If the point does not make the
inequality true, shade the opposite side of the line.
Graph the inequality x < 3
4
AB: x = 3.00
2
5
-2
-4
10
15
Graph y > x + 1
4
f x  = x+1
2
5
-2
-4
x
-1
0
1
y
-6
0
1
2
-8
-10
The boundary line for this graph is
the line y = x + 1. since the
boundary is not part of the graph, it
is shown as a dashed line on the
graph. To determine which halfplane is the graph of y > x + 1, test a
point NOT on the boundary. For
example, you can test my favorite
point (0, 0) the origin. Since 0 > 0 +
1 is false, (0, 0) is not a solution of y
> x + 1. Thus, the graph is all points
in the half-plane that does NOT
contain (0, 0). This graph is called
an open half-plane since the
boundary is not part of the graph.
10
15
Graph y ≤ x + 1
4
f x  = x+1
2
5
-2
-4
x
-1
0
1
y
0
1
2
-6
-8
-10
The boundary line for this graph is
also the line y = x + 1. Since the
inequality y ≤ x + 1 means y < x + 1
or y = x + 1, this boundary is part of
the graph. Therefore, the boundary
is shown as a solid line on the
graph. The origin (0, 0) is part of
the graph of y ≤ x + 1 since 0 ≤ 0 +
1 is true. Thus, the graph is all
points in the half plane that contains
the origin and the line y = x + 1.
This graph is called a closed half
plane.
10
15
Graph y > -4x – 3
8
6
4
fx = -4x-3
2
-5
-2
x
-1
0
1
y
1
-3
-7
-4
-6
-8
Graph the equation y = -4x – 3.
Draw it as a dashed line since this
boundary is NOT part of the graph.
The origin, (0, 0), is part of the
graph since 0 >-4(0) – 3 is true.
Thus, the graph is all points in the
half 5plane that contains
the origin.
10
CHECK: Test a point on the other
side of the boundary, say (-2, -2).
Since -2 > -4(-2) – 3 or -2 > 5 is
false, (-2, -2) is NOT part of the
graph.
15
Ex. 2: Graph 15x + 20y ≤ 240 to answer the application at the
beginning of the lesson. How many of each ticket can Mr.
Harris purchase?
12
• First solve for y in terms of x.
15x + 20y ≤ 240
20y ≤ 240 – 15x
3
y ≤ 12 - x
4
Then graph the equation as if it
were equal to y as a solid line
since the boundary is part of
the graph. The origin (0, 0) is
part of the graph since 15(0) +
20(0) ≤ 240 is true. Thus, the
graph is all points in the halfplane that contains the origin.
10
8
6
4
2

3
fx = 12x
4
5
-2
10
15
Mr. Harris cannot buy fractional or
negative numbers of tickets.
• So, any point in the shaded region
whose x- and y-coordinates are whole
numbers is a possible solution.
• For example, (5, 8) is a solution. This
corresponds to buying five $15 tickets
and eight $20 tickets for a total cost of
15(5) + 20(8) or $235.
Is the boundary included in the
graphs of each inequality?
2x + y ≥ 3
3x – 2y ≤ 1
5x - 2 > 3y
Included or not?
• Signs with < or > are NOT included
• Signs with ≤ or ≥ are included
• If the graph is < or >, then it is said to be an
open half plane since the boundary is not
part of the graph.
• If a graph is ≤ or ≥, then it is a closed halfplane because the boundary is part of the
graph.
• Greater than → shade above
• Less than → shade below
Determine which (-2,2), (4,-1), or (3,1)
are solutions to the inequality x + 2y ≥ 3
x + 2y ≥ 3
-2 + 2(2) ≥ 3
-2 + 4 ≥ 3
2≥3
x + 2y ≥ 3
3 + 2(1) ≥ 3
3+2≥3
5≥3
x + 2y ≥ 3
4 + 2(-1) ≥ 3
4–2≥3
2≥3
Graph each inequality g(x)>0 and
f(x)≥(⅓)x
4
2
gx = 0
-5
fx =

1
3
5
10
x
-2
-4
-6
-8
-10
The overlap is where
these two meet which is
where your answer to both
of these lies.