1. Warm-up: Fill in all degree and  radian measures of the unit circle
HW: page (2 – 8 Even, 17, 18, 20, 21 – 31 odd, 42, 44, 46)

2. Systems of Inequalities
7.4
Systems of Inequalities
Objective:
Graph an inequality
Graph systems of inequalities
Find points of intersection (vertices) of systems of inequalities.

3. Graphing an Inequality
y = x2
y  x2
y > x2
y < x2

4. Graphing an Inequality
The graph of an inequality, in general, consists of:
A region in the plane whose boundary is the graph of the inequality written as an equation that is either dotted (no equality) or solid (equality).
To determine which side of the graph gives the solution set of the inequality, we need only check test points.

5. Ex1) Graph x2 + y2 < 25
The graph of x2 + y2 = 25 is a circle of radius 5 centered at the origin.
The points on the circle do not satisfy the inequality because of < so we us a dashed circle.
(5, 0)
To determine whether the inside or the outside of the circle satisfies the inequality, we use the test points:
(0, 0) on the inside.
(6, 0) on the outside.

6. We substitute the coordinates of each point into the inequality to see if it satisfies the inequality.
Note that any point inside or outside the circle can serve as a test point.
We have chosen these points for simplicity.
For a circle < and ≤ shade inside, > and  outside

7. System of Inequalities

8. The solution of a system of inequalities is:
The set of all points in the coordinate plane that satisfy every inequality in the system.

9. Ex) Graph the solution of the system of inequalities.
We will graph only those points that simultaneously satisfy both inequalities.
The solution consists of the intersection of the graphs of the two equations.

10. Figure (a) both inequalities graphed on the same coordinate plane
Figure (b) the intersection of the inequalities (region that satisfies both inequalities).

11. The points (–3, 4) and (5, 0) in (b) are the vertices of the solution set.
They are obtained by solving the system of equations
We do so by substitution.

12. When y = 4, we have: x = 5 – 2(4) = –3
Solving for x gives x = 5 – 2y
Substituting this into the first equation gives:
(5 – 2y)2 + y2 = 25
(25 – 20y + 4y2) + y2 = 25
–20y + 5y2 = 0
–5y(4 – y) = 0
Thus, y = 0 or y = 4.
When y = 0, we have: x = 5 – 2(0) = 5
When y = 4, we have: x = 5 – 2(4) = –3
So, the points of intersection of these curves are (5, 0) and (–3, 4).

13. Note that, in this case, the vertices are not part of the solution set.
This is because they don’t satisfy the inequality x2 + y2 < 25.
So, they are graphed as open circles in the figure.
They simply show where the “corners” of the solution set lie.

14. Systems of Linear Inequalities
An inequality is linear if it can be put into one of these forms: ax + by ≥ c ax + by ≤ c ax + by > c ax + by < c

15. Ex) Graph the solution set of the system, and label its vertices.

16. The inequalities x ≥ 0 and y ≥ 0 say that x and y are nonnegative.
2) Find points of intersection and test a point such as (1, 1) to determine the shaded region.
1) Graph the lines given by the equations that correspond to each inequality.
The inequalities x ≥ 0 and y ≥ 0 say that x and y are nonnegative.
1 + 3(1) ≤ 12
1 + 1 ≤ 12

17. The coordinates of each vertex are obtained by simultaneously solving the equations of:
The lines that intersect at each vertex.
From the system
we get the vertex (6, 2).
The other vertices are the x- and y-intercepts of the corresponding lines, (8, 0) and (0, 4), and the origin (0, 0).
Here, all the vertices are part of the solution set.

18. E.g. 4—System of Linear Inequalities
Graph the solution set of the system.
We must graph the lines that correspond to these inequalities and then shade the appropriate regions—as in Example 3.

19. E.g. 4—System of Linear Inequalities
We will use a graphing calculator.
So, we must first isolate y on the left-hand side of each inequality.

20. E.g. 4—System of Linear Inequalities
Using the shading feature of the calculator, we obtain this graph.
The solution set is the triangular region that is shaded in all three patterns.
We then use TRACE or the Intersect command to find the vertices of the region.

21. E.g. 4—System of Linear Inequalities
The solution set is graphed here.

22. Bounded Regions
When a region in the plane can be covered by a (sufficiently large) circle, it is said to be bounded.

23. A region that is not bounded is called unbounded.
Unbounded Regions
A region that is not bounded is called unbounded.
It cannot be “fenced in”— it extends infinitely far in at least one direction.

24. Sneedlegrit: Graph the system. Find vertices of the solution.
HW: page (2 – 8 Even, 17, 18, 20, 21 – 31 odd, 42, 44, 46)
Find vertices for all graphs!