# graph of the given inequality if you rewrite it so that the absolute value expression

Document Sample

```					Advanced Mathematical Concepts                                                                Chapter 3

Lesson 3-3

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

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

y    x2 3                                        y    x2 3
4  3 2 3         x, y  =  3,4                2  11  2  3        x, y  = 11,2 
44                true                             26                   false

y    x2 3                                        y    x2 3
5 62 3             x, y  = 6,5               1  18  2  3        x, y  = 18,-1
55                   true                          1  7                 false

Of these ordered pairs, (3, 4) and (6, 5) and are solutions for y ≥ x - 2+ 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,
boundary equation to determine how the
boundary relates to the parent graph.
y   x  2  2
2

     
move 2 units right        move 2 units up

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.

y   x  2  2
2

0   0  2  2       Replace  x, y  with 0,0  .
2

02                    True

Since (0, 0) satisfies the inequality, the correct region

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                                                 Case 2

(x + 3)    <0                                        (x + 3)    >0
|x + 3| - 4   <2                                     |x + 3| - 4   <2
-(x + 3) - 4    <2     |x + 3| = -(x + 3)               x+3-4        <2   |x + 3| = (x + 3)
-x - 3 - 4   <2                                           x-1     <2
-x   <9                                               x   <3
x   > -9

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 h. 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 h ≥ 150 for h ≥ 0.

Let y1 = 1.2 x 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.

```
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
 views: 0 posted: 8/7/2012 language: Unknown pages: 5
How are you planning on using Docstoc?