warm up write as an inequality and interval notation

Warm UpWrite as an inequality and interval notation. 5 4x

5 or 6x x

5 or 0x x

7 4x

Graphs of Quadratic Inequalities

I can graph and find the solution set of quadratic

inequalities.

Remember having to graph

24

3y x

Remember having to graph

13

2y x

Steps for Graphing Steps for Graphing (quickly)(quickly)

1. Graph the quadratic (use your calculator for points)

2. For <> use DASHED for ≤≥ use SOLID line

3. Locate the roots/solutions of the quadratic

4. Shade the appropriate region using a test point

Graph: y ≤ x2 + 6x – 4* Vertex: (-3,-13)

* Solid Line

* Roots: X = {-6.6, 0.6}

* Interval Notation:

Graph: y > -x2 + 4x – 3

Vertex:

Solid or Dotted:

Roots:

Test Point:

Interval Notation:

Graph: y ≥ x2 – 8x + 12

Vertex:

Solid or Dotted:

Roots:

Test Point:

Interval Notation:

Graph: y > -x2 + 4x + 5Vertex:

Solid or Dotted:

Roots:

Test Point:

Interval Notation:

Warm UpPass HW forward

• Factor and solve:1. x2 – 5x = – 42. -x2 + 7x = 12 (hint: use rooftop)

• Sketch the graph-(x – 1)2 – 3

Solving a Solving a Quadratic Quadratic InequalityInequality

Steps for solvingSteps for solving

1. Write the original inequality as an equation

2. Set equal to 0, factor, and solve.3. Plot the points on a number line

and test points in each interval back into the original inequality.

4. Write the answer (the TRUE part) as an inequality

Solve: x2 – 5x ≤ – 4x2 – 5x = -4

Answer: 1 ≤ x ≤ 4

Interval Notation:

x2 – 5x + 4 = 0(x – 4) (x – 1) = 0

x = 1, 4

False True False

Solve: -x2 + 7x < 12-x2 + 7x = 12

Answer: x < 3 or x > 4

Interval:

-x2 + 7x – 12 = 0

(x – 4) (x – 3) = 0

x = 3, 4

True Fa

lse

True

What if it What if it can’t can’t

factor?factor?Graph it!Graph it!

Solve: -(x – 1)2 – 3 < 0-(x – 1)2 – 3 < y

y > -(x – 1)2 – 3

Answer: all real numbers

Solve: x2 + 4 ≤ 0x2 + 4 ≤ y

y ≥ x2 + 4

Answer: no solution