Lesson 3-3

Example 1
Determine whether (3, 4), (11, 2), (6, 5) and (18, -1) are solutions for the
inequality y ≥ + 3.

Substitute the x-value and y-value from each ordered pair into the inequality.

		

Of these ordered pairs, (3, 4) and (6, 5) and are solutions for y ≥ + 3.

Example 2
Graph y ≤ (x - 2)2 + 2.

The boundary of the inequality is the graph of	
y = (x + 2)2 - 2. To graph the boundary curve,
start with the parent graph y = x2. Analyze the
boundary equation to determine how the
boundary relates to the parent graph.

Since the boundary is included in the inequality,	
the graph is drawn as a solid curve.

The inequality states that the y-values of the
solution are less than the y-values on the graph
of y = (x - 2)2 + 2. For a particular value of x,
all of the points in the plane that lie below the
curve have y-values less than y = (x - 2)2 + 2.
So this portion of the graph should be shaded.

To verify numerically, you can test a point not	
on the boundary. It is common to test (0, 0)
whenever it is not on the boundary.

Since (0, 0) satisfies the inequality, the correct region
is shaded.

Example 3
Graph y < -2 - |x - 1|.

Begin with the parent graph y = |x|.	

It is easier to sketch the graph of the given inequality
if you rewrite it so that the absolute value expression
comes first.

y < -2 - |x - 1|		y < -|x - 1| - 2

This more familiar form tells us the parent graph is	
reflected over the x-axis and moved one unit to the
right and two units down. The boundary is not
included, so draw it as a dashed line.

The y-values of the solution are less than the y-values	
on the graph of y = -2 - |x - 1|, so shade below the graph
of y = -2 - |x - 1|.

Verify by substituting (0, 0) in the inequality to obtain
0 < -3. Since this statement is false, the part of the graph
containing (0, 0) should not be shaded. Thus, the graph
is correct.

Example 4
Solve |x + 3| - 4 < 2.

There are two cases that must be solved. In one case, x + 3 is negative, and in the other, x + 3 is positive.

Case 1

	(x + 3)	< 0
	|x + 3| - 4	< 2
	-(x + 3) - 4	< 2	|x + 3| = -(x + 3)
	-x - 3 - 4	< 2
	-x	< 9
	x	> -9	Case 2

	(x + 3)	> 0
	|x + 3| - 4	< 2
	x + 3 - 4	< 2	|x + 3| = (x + 3)
	x - 1	< 2
	x	< 3	

The solution set is {x | -9 < x < 3}. {x | -9 < x < 3} is read as “the set of all numbers x such that x is between -9 and 3.

Verify this solution by graphing.

First, graph y = |x + 3| - 4. Since we are solving	
|x + 3| - 4 < 2 and |x + 3 | - 4 = y, we are looking
for a region in which y = 2. Therefore, graph
y = 2 and graph it on the same set of axes.

Identify the points of intersection of the two
graphs. By inspecting the graph, we can see
that they intersect at (-9, 2) and (3, 2).

Now shade the region where the graphs of	
the inequalities y < |x + 3| - 4 and y < 2
intersect. This occurs in the region of the
graph where -9 < x < 3. Thus, the solution
to |x + 3| - 4< 2 is the set of x-values such
that -9 < x < 3.

	

Example 5
A person’s height h, in feet, above the ground influences how far V, in miles, the person can see to the horizon. This relationship is modeled by the equation V = 1.2. How high must a person be in order to see at least 150 miles to the horizon?

Since we need to know how high a person should be in order to see at least 150 miles to the horizon, we can write the inequality 1.2 ≥ 150 for h ≥ 0.

Let y1 = 1.2 and y2 = 150. Graph both equations on the same set of axes using a graphing calculator.

Calculating the points of intersection, we
find that the two equations intersect at
(15,625, 150). Therefore, when x ≥ 15,625,
a person can see 150 miles or more to the
horizon. That is, a person must be at least
15,625 feet above the ground in order to
see a minimum of 150 miles to the horizon.