Warm-up• 1. Given: y = x2 – 6x + 3

• Find: Vertex, AOS, y-intercept, and graph it

• 2. Given: y = -2(x – 3)2 + 4

• Find: Vertex, AOS, y-intercept, and graph it

• 3. Given: y = 2(x – 4)(x + 2)

• Find: Vertex, AOS, y-intercept, and graph it

Chapter 4

Section 4-9

Solving Quadratic Inequalities

Objectives

• I can graph quadratic inequalities with the assistance of a calculator

• I can solve quadratic inequalities with a calculator or with 3 test method

Review

• The solutions to any quadratic equation are the x-intercepts (where the graph crosses the x-axis)

• Now, the solutions to any quadratic inequality is where the shaded region crosses the x-axis. (It will usually be a range of values)

Review from Linear Inequalities• Remember when graphing linear inequalities we

used two types of boundaries:

• Solid Line: Used to include equality. Used when equations contained ( and )

• Dashed Line: Used when equations contained ( < and >), no equality

• We will be using same ideas with quadratic graphs.

Graphing Quadratic Inequalities

• To graph any quadratic inequality, you can use the following technique:

• 1. Graph the boundary (solid or dashed). Use the graphing calculator and data table to find key points.

• 2. Test a point. Point (0, 0) works best if that point is not on the boudary.

• 3. Shade the correct region based on the test results.

Example 1: y x2 – 6x + 2

-10 -8 -6 -4 -2 2 4 6 8 10

-10

-8

-6

-4

-2

2

4

6

8

10

1st: Draw the boundary. Shown here in blue. It is solid because of the .

Next test a point. I chose (0, 0).

0 (0)2 – 6(0) + 2

0 0 – 0 + 2

0 2

This is true!

Since (0,0) tested good, then shade the area outside the boundary which includes (0,0).

Solving by Graphingy > x2 – 9

-10 -8 -6 -4 -2 2 4 6 8 10

-10

-8

-6

-4

-2

2

4

6

8

10 Graph boundary. Dashed because >

Test point. I chose (0,0).

0 > -9 (yes)

So shade inside boundary.So solutions are all x-values inside parabola.

Solution: (-3, 3)

Notice that –3 & 3 are not in the solution because the are on the dashed line.

2nd Method: Three Test Points

• A 2nd method to solve the inequalities is by finding the roots and testing 3 regions.

• Consider the inequality below:

• x2 – x – 12 > 0 (Find solutions using 2nd Trace in calculator)

• So x = 4 or x = -3 are boundary points

Solve: x2 - x – 12 > 0• x = 4 or x = -3

-3 4

Test (-5)

(-5)2 - (-5) – 12 > 0

25 + 5 – 12 > 0

30 – 12 > 0

18 > 0

YES

Test (0)

02 - (0) – 12 > 0

-12 > 0

NO

Test (5)

52 –(5) – 12 > 0

25 – 5 – 12 > 0

20 – 12 > 0

8 > 0

YES

Solutions: (-∞,-3) U (4, ∞)

Solve: x2 - 8x – 33 > 0

• x2 - 8x – 33 > 0

• Use calculator to find the solutions!

• x = 11 or x = -3

• Test 3 areas x < -3, -3 < x < 11, x > 11

• Solve on next slide.

Solve: x2 - 8x – 33 > 0• x = 11 or x = -3

-3 11

Test (-5)

(-5)2 - 8(-5) – 33 > 0

25 + 40 – 33 > 0

65 – 33 > 0

32 > 0

YES

Test (0)

02 - 8(0) – 33 > 0

-33 > 0

NO

Test (15)

152 –8(15) – 33 > 0

225 – 120 – 33 > 0

105 – 33 > 0

72 > 0

YES

Solutions: (-∞, -3) U (11, ∞)

Solve: 2x2 - 3x - 4 0• Find solutions w/Calc.

• -.85, 2.35

-.85 2.35

Test (-1)

2(-1)2 - 3(-1) – 4 0

2 + 3 – 4 0

5 – 4 0

1 0

NO

Test (0)

2(0)2 - 3(0) – 4 0

-4 0

YES

Test (3)

2(3)2 –3(3) – 4 0

18 – 9 – 4 0

9 – 4 0

5 0

NO

Solutions: [-.85, 2.35]

Homework

• WS 4-5

• Quiz Wednesday